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> 4D Toto Magnum winning probability

MPIK
post Mar 1 2010, 11:10 AM


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Who here can enlighten me on the winning probability of 4D toto, magnum or 1+3D?

I understand if I buy 1 number only and Magnum only draws a single number. My probability is 1 out of 10,000.

But things get complicated when I buy 7 numbers and Magnum draws 23 numbers. So, how the calculation then? What is my probability of winning?

Any maths experts here?
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pkiensing
post Mar 1 2010, 11:21 AM


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not exactly know the calculation but normally i will only think of striking the first prize only, so still 1 over 10000 loh
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MPIK
post Mar 1 2010, 11:23 AM


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But I just want to know the probability of winning a prize regardless it is a 1st prize or conso prize.
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Omage007
post Mar 1 2010, 11:27 AM


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QUOTE(MPIK @ Mar 1 2010, 11:10 AM)
Who here can enlighten me on the winning probability of 4D toto, magnum or 1+3D?

I understand if I buy 1 number only and Magnum only draws a single number. My probability is 1 out of 10,000.

But things get complicated when I buy 7 numbers and Magnum draws 23 numbers. So, how the calculation then? What is my probability of winning?

Any maths experts here?
*
(23/10000) * 7 = ???


I think the toto 6/49, 6/52, 6/55 is more complicated. . .
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lin00b
post Mar 1 2010, 11:32 AM


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you have 23 winning numbers, so your chances of winning something is 23/10000 or 0.23%.

1st prize=2.5k, 2nd prize=1k, 3rd=500, special=200, consolation=60.

if you buy all the digits, you would have "won" rm6600 from a total output of 10k (or a return of 66%)

if you buy all digit big and small, you would have "won" 14100 from a total output of 20k (or a return of 70.5%)

either way, the company wins smile.gif

it is also interesting that if you buy mbox, you get slightly more winnings/rm due to rounding

This post has been edited by lin00b: Mar 1 2010, 11:34 AM
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MPIK
post Mar 1 2010, 11:38 AM


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QUOTE(lin00b @ Mar 1 2010, 11:32 AM)
you have 23 winning numbers, so your chances of winning something is 23/10000 or 0.23%.
*
Your calculation is correct only if I only buy a single number and Magnum draws 23 numbers

or if I buy 23 numbers and Magnum draws only 1 number. Either way, the probability is 23/10000.

how about if I buy 13 numbers and magnum draws 23 numbers? What's the probability then?
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ZeratoS
post Mar 1 2010, 12:01 PM


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How about you pull out a calculator and use it?
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cherroy
post Mar 1 2010, 05:26 PM


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QUOTE(MPIK @ Mar 1 2010, 11:38 AM)
Your calculation is correct only if I only buy a single number and Magnum draws 23 numbers

or if I buy 23 numbers and Magnum draws only 1 number. Either way, the probability is 23/10000.

how about if I buy 13 numbers and magnum draws 23 numbers? What's the probability then?
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Aiya, you have 13 No, means each draw you have 13/10000 winning chance.

Draw 23 times, then 13/10000 x 23 lor.
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alanyuppie
post Mar 1 2010, 05:28 PM


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This doesn't seem to be PHD material. maths Probability are taught in secondary school for ages.

Why not Mods move it to Education section?
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cyberskull
post Mar 1 2010, 05:30 PM


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Dont make them richer more
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teyren
post Mar 1 2010, 05:38 PM


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if wanna get complicated....then sometimes a same number open twice at the same draw.....then become 22/10000.
23/10000 only true if 2 same number cannot open at same draw...
wat if open 3 times.....then 21/10000... but harder than jackpot haha...
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MPIK
post Mar 1 2010, 05:51 PM


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QUOTE(cherroy @ Mar 1 2010, 05:26 PM)
Aiya, you have 13 No, means each draw you have 13/10000 winning chance.

Draw 23 times, then 13/10000 x 23 lor.
*
U mean my chance is 299 out of 10000? (13x23)

So much?

How about if I buy 500 numbers? = 500x23/10000 = 11500/10000? Confirm win??
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xzibit
post Mar 1 2010, 06:00 PM


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QUOTE(MPIK @ Mar 1 2010, 05:51 PM)
U mean my chance is 299 out of 10000? (13x23)

So much?

How about if I buy 500 numbers? = 500x23/10000 = 11500/10000? Confirm win??
*
If you have the budget to buy all the numbers then you'll sure win..for at least once. laugh.gif

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arthurlwf
post Mar 1 2010, 06:00 PM


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QUOTE(alanyuppie @ Mar 1 2010, 05:28 PM)
This doesn't seem to be PHD material. maths Probability are taught in secondary school for ages.

Why not Mods move it to Education section?
*
This thread should be in PHD because it requires a high level of statistics to compute magnum probability... icon_rolleyes.gif icon_rolleyes.gif
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lin00b
post Mar 1 2010, 09:14 PM


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QUOTE(MPIK @ Mar 1 2010, 05:51 PM)
U mean my chance is 299 out of 10000? (13x23)

So much?

How about if I buy 500 numbers? = 500x23/10000 = 11500/10000? Confirm win??
*
provided the above calculation is true (which i doubt, have to ponder a lil more, a little rusty in probability)

given that for rm1 bet, the total payout for 23 numbers is rm6600. that's an average of 286/winning number. if buying 500 numbers guarantee you a winning number(which i dont think it will), on average you would have "won" 286 after having to pay 500 thumbup.gif

i think that if you buy x number your chance of winning would be a binomial distribution with n=x, p=0.0023, r=1-0.0023

explanation: for each number you have an independent chance of hitting the winning number with a probability of 0.0023 and a chance of 1-0.0023 of not hitting it. so you have to account for hitting with 1 number, 2 numbers, 3 numbers and so forth

This post has been edited by lin00b: Mar 1 2010, 09:17 PM
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Jurlique
post Mar 2 2010, 12:07 AM


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QUOTE(lin00b @ Mar 1 2010, 09:14 PM)
provided the above calculation is true (which i doubt, have to ponder a lil more, a little rusty in probability)

given that for rm1 bet, the total payout for 23 numbers is rm6600. that's an average of 286/winning number. if buying 500 numbers guarantee you a winning number(which i dont think it will), on average you would have "won" 286 after having to pay 500  thumbup.gif

i think that if you buy x number your chance of winning would be a binomial distribution with n=x, p=0.0023, r=1-0.0023

explanation: for each number you have an independent chance of hitting the winning number with a probability of 0.0023 and a chance of 1-0.0023 of not hitting it. so you have to account for hitting with 1 number, 2 numbers, 3 numbers and so forth
*
But u haven answered MPIK's question:-

What is the probability or odds or winning if he buy 500 numbers and Magnum draws 23 numbers.
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lin00b
post Mar 2 2010, 01:37 AM


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QUOTE(Jurlique @ Mar 2 2010, 12:07 AM)
But u haven answered MPIK's question:-

What is the probability or odds or winning if he buy 500 numbers and Magnum draws 23 numbers.
*
i did. i just did not spoon feed you the number.

binomial distribution with n=x (in this case 500); p=0.0023; r=1-0.0023. input that into your calculator if you have one strong enough.

if you dont know how to solve a binomial distribution, perhaps google might be of help.

you might be able to solve using poisson approximation/normal approximation, but its been years since i touched it, and i cant be bothered to look it up.
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ZeratoS
post Mar 2 2010, 02:33 AM


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QUOTE(lin00b @ Mar 2 2010, 01:37 AM)
i did. i just did not spoon feed you the number.

binomial distribution with n=x (in this case 500); p=0.0023; r=1-0.0023. input that into your calculator if you have one strong enough.

if you dont know how to solve a binomial distribution, perhaps google might be of help.

you might be able to solve using poisson approximation/normal approximation, but its been years since i touched it, and i cant be bothered to look it up.
*
Yeah, and I think its bout time to stop spoonfeeding him any information. This section is DEFINITELY not the section for basic mathematical application. A simple google search would bring him the answers, but clearly he wants his homework done for him.


One can pick up any probability book and that would explain the workings of 4D Magnum, don't be so lazy.
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cloudwan0
post Mar 2 2010, 12:25 PM


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dont think too much, the winning rate is 50:50
win or lose niah
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Jurlique
post Mar 2 2010, 10:32 PM


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Based on the binomial distribution, I found the answer for buying 20 numbers winning probability is equal to 3 out of 8433.
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lin00b
post Mar 2 2010, 11:27 PM


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the number is logically consistent (too lazy to check).

to those who understand probability, the odds that 4d offer is really much too biased to the banker.

compared to casino games such as roulette, you have a 1 in 37 chance of increasing your bet 36 times. common roulette has 37 numbers where if you get your number, you win 36x your bet. some casino have 38 numbers and even then there are comments about it being too biased towards the banker.
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Jurlique
post Mar 2 2010, 11:50 PM


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QUOTE(lin00b @ Mar 2 2010, 11:27 PM)
the number is logically consistent (too lazy to check).

to those who understand probability, the odds that 4d offer is really much too biased to the banker.

compared to casino games such as roulette, you have a 1 in 37 chance of increasing your bet 36 times. common roulette has 37 numbers where if you get your number, you win 36x your bet. some casino have 38 numbers and even then there are comments about it being too biased towards the banker.
*
All gamble games are favour to the dealer, banker or 4D operator. Otherwise, no one wanna be a casino owner.

For roulette with a single zero, there are 37 numbers including zero altogether. If you use probability. The dealer should pays you 37 to 1 if you bet the winning number. However, they would only pays u 35 to 1.

Similarily to 4D, there are 10,000 chances. But when u strike the 1st prize, they will only pay you 2,500 instead of 10,000 provided Toto only draws a single number just like roulette.
However, the calculation will become too complicated when it involved 2nd, 3rd, Special and Conso prize.

This is why I believe TS wanna get the answer to study the probability of 4D as this kind of university level of probability calculation is not being taught in Secondary School.
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lin00b
post Mar 3 2010, 02:32 AM


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roulette pays 36 to 1. the banker's advantage is a mere 1/37. compared to payment of magnum of 6600 for 10000. banker's advantage is 34/100.

and all the concept for this type of calculations is taught in secondary schools. introduction to binomial distribution and normal approximation is f5 add maths syllabus. more complex binomial distribution and approximation to normal and poisson distribution is f6 science stream maths syllabus.

you think uni would have course to calculate gambling odds?
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abubin
post Mar 3 2010, 11:57 AM


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if you pay more attention in school during the subject of probability then you would know out of the 23 numbers, each of the numbers are not related to one another. There are no proven pattern that works.

So with each numbers, you have 23 times of winning a number out of the odds of 1/10000.

But the thing that I believe is that for magnum, they do not publish their drawing of numbers. So they could have easily manipulate their system to select winning numbers based on buyers. That means the less buyers of a number will have higher chance or first prize. This way magnum will maximize their profit. But that does not take into consideration of those buying from third party 4d sellers.

So in theory, if you want to stand higher odds of winning, buy from third party 4d sellers. But these are illegals. Buy at your own risk.


Added on March 3, 2010, 12:03 pm
QUOTE(lin00b @ Mar 3 2010, 02:32 AM)
roulette pays 36 to 1. the banker's advantage is a mere 1/37. compared to payment of magnum of 6600 for 10000. banker's advantage is 34/100.

and all the concept for this type of calculations is taught in secondary schools. introduction to binomial distribution and normal approximation  is f5 add maths syllabus. more complex binomial distribution and approximation to normal and poisson distribution is f6 science stream maths syllabus.

you think uni would have course to calculate gambling odds?
*
roulettes now have number 0 and 00. In europe they even have 000. So the odds for let's say genting is 1/38. Eventhough for 1 person odds are pretty low, but when it comes to involving a big group of people, it will always become close to the odds. That means banker is always the winner.

that means if the total betting for the day is 1,000,000 then banker would have won 1000000 x 1/38 = 26,315. The scale will tip a little to + or - but in the end banker is always the big winner. Therefore, gambling is not the way to make money.

This post has been edited by abubin: Mar 3 2010, 12:03 PM
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MPIK
post Mar 3 2010, 01:49 PM


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QUOTE(lin00b @ Mar 3 2010, 02:32 AM)
roulette pays 36 to 1. the banker's advantage is a mere 1/37. compared to payment of magnum of 6600 for 10000. banker's advantage is 34/100.

and all the concept for this type of calculations is taught in secondary schools. introduction to binomial distribution and normal approximation  is f5 add maths syllabus. more complex binomial distribution and approximation to normal and poisson distribution is f6 science stream maths syllabus.

you think uni would have course to calculate gambling odds?
*
Have you been to the casino? Roulette pays only 35 to 1. For double bet (2 numbers), it pays only 17 to 1, corner bet (4 numbers), it pays only 8 to 1.

Unless you have been gambling online, certain online gambling may pays you 36 to 1.
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lin00b
post Mar 3 2010, 04:02 PM


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QUOTE(MPIK @ Mar 3 2010, 01:49 PM)
Have you been to the casino? Roulette pays only 35 to 1. For double bet (2 numbers), it pays only 17 to 1, corner bet (4 numbers), it pays only 8 to 1.

Unless you have been gambling online, certain online gambling may pays you 36 to 1.
*
house rules may vary from place to place, but the comparison between simple casino games such as roulette and big/small is much less biased towards the banker compared to 4d.

at the end of the day, banker will always get a profit. but you as a player may get profit some of the time if you beat the odds. the key to being a successful gambler is to know when to quit when you are ahead. for the longer you play, the higher chance for you to lose.

in your case, its best to bet corner then, cause you get 8*4/37(or 38/39 depending on wheel)=36/37; compared to 17*2/37 or 35*1/37

This post has been edited by lin00b: Mar 3 2010, 04:05 PM
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profdrahhen
post Mar 4 2010, 06:43 PM


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QUOTE(cloudwan0 @ Mar 2 2010, 12:25 PM)
dont think too much, the winning rate is 50:50
win or lose niah
*
rclxms.gif

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underworld
post Mar 4 2010, 07:27 PM


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4d toto kuda
i just blieve in luck
if u have the luck wateva number u buy it will strike
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lin00b
post Mar 4 2010, 08:43 PM


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scientific discussion kopitiam'd liao
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samteng
post Mar 5 2010, 12:41 AM


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The probability of winning Jackpot 1 is a gigantic factorial of 10000!

For those who don't understand the concept of a factorial, here is how it works.

For example, a factorial of five (5!) is expanded like this: 4*3*2*1=24. So there are possible combinations of 24 pairs should the grand prize require a paired number from 0-5.

Now, let's try to figure out 10000!

10000!=9999*9997*9996.....3*2*1

The multiplication of it gives you a total sum that is nearly as good as infinity which is why it is nearly impossible for one to strike Jackpot 1. Don't believe? Go google up factorial calculator of permutation and combination to compute it.

This post has been edited by samteng: Mar 5 2010, 12:44 AM
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anson lee
post Mar 5 2010, 04:51 AM


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Try this on Mbox

0123

3456

7890

5678

mostly each set of number will come twice a week...
just due to my reseach...

last 2 week came out 5687 n 6798..

so its about probelity...learn more bout it...lolx

just my 2 cents


Added on March 5, 2010, 4:53 am
QUOTE(samteng @ Mar 5 2010, 12:41 AM)
The probability of winning Jackpot 1 is a gigantic factorial of 10000!

For those who don't understand the concept of a factorial, here is how it works.

For example, a factorial of five (5!) is expanded like this: 4*3*2*1=24. So there are possible combinations of 24 pairs should the grand prize require a paired number from 0-5.

Now, let's try to figure out 10000!

10000!=9999*9997*9996.....3*2*1

The multiplication of it gives you a total sum that is nearly as good as infinity which is why it is nearly impossible for one to strike Jackpot 1. Don't believe? Go google up factorial calculator of permutation and combination to compute it.
*
Agree dude...

there is a very slim chance of it...

Wat to do,If easy give ppl win business,who 1 to do... hmm.gif hmm.gif


This post has been edited by anson lee: Mar 5 2010, 04:53 AM
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lin00b
post Mar 5 2010, 07:59 AM


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QUOTE(samteng @ Mar 5 2010, 12:41 AM)
The probability of winning Jackpot 1 is a gigantic factorial of 10000!

For those who don't understand the concept of a factorial, here is how it works.

For example, a factorial of five (5!) is expanded like this: 4*3*2*1=24. So there are possible combinations of 24 pairs should the grand prize require a paired number from 0-5.

Now, let's try to figure out 10000!

10000!=9999*9997*9996.....3*2*1

The multiplication of it gives you a total sum that is nearly as good as infinity which is why it is nearly impossible for one to strike Jackpot 1. Don't believe? Go google up factorial calculator of permutation and combination to compute it.
*
odds of jackpot 1 is more likely to be 1/10000 x 1/10000 rather than what you say. you need to get the first number right (1/10000) and then independently the second number right (1/10000)
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ZeratoS
post Mar 6 2010, 01:09 AM


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..and that's how they work. You cannot win by brute-forcing so you either have to be really really lucky or to consistently purchase numbers which ironically may not net you any winnings in the end.

All the same its a loser's game and I think only fools play it. Throwing money away man.
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C-Note
post Mar 11 2010, 12:27 AM


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statistics in this case is rubbish. say its 3/1000. what if all the 3 winning chances are accumulated at the 1st three times you buy toto? its 100% win aint it.
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nice.rider
post Mar 11 2010, 01:05 AM


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Mr. Sam went to a stadium to watch live football. Inside the stadium there were 10000 audiences (excludes Sam).

Halfway though the game, Sam realized that he has lost his mobile phone in the stadium. He sat at the far end corner and he remembered that his trips to purchase pop corn and also went to toilet made him going though one round of the entire stadium.

In the quest of getting his mobile phone back, he informed the organizer.

The following conditions apply by the organizer:
1) 1 out of the 10000 audiences found the phone
2) The organizer allows him to point/choose 23 audiences out of the 10000 audiences
3) If he manages to point out the audience that found his phone, he will get his phone back
4) Else, says bye-bye to the phone forever

Let's us wish him best of luck on getting his mobile phone back!

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lin00b
post Mar 11 2010, 01:47 AM


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QUOTE(C-Note @ Mar 11 2010, 12:27 AM)
statistics in this case is rubbish. say its 3/1000. what if all the 3 winning chances are accumulated at the 1st three times you buy toto? its 100% win  aint it.
*
the chance of that happening is approximately 27/1000000000 (in actuality 1/1000 * 1/999 * 1/998). while it can happen to you, i wouldnt bet on it

This post has been edited by lin00b: Mar 11 2010, 01:54 AM
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xkellyxemox
post Mar 11 2010, 01:59 AM


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QUOTE(lin00b @ Mar 1 2010, 09:14 PM)
provided the above calculation is true (which i doubt, have to ponder a lil more, a little rusty in probability)

given that for rm1 bet, the total payout for 23 numbers is rm6600. that's an average of 286/winning number. if buying 500 numbers guarantee you a winning number(which i dont think it will), on average you would have "won" 286 after having to pay 500  thumbup.gif

i think that if you buy x number your chance of winning would be a binomial distribution with n=x, p=0.0023, r=1-0.0023

explanation: for each number you have an independent chance of hitting the winning number with a probability of 0.0023 and a chance of 1-0.0023 of not hitting it. so you have to account for hitting with 1 number, 2 numbers, 3 numbers and so forth
*
i think this is by far the most logic explanation...

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FlayerZ
post Mar 11 2010, 03:43 AM


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dont buy lar ...how also u lose
unless u are lucky
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skon9
post Mar 11 2010, 10:34 AM


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Winning a lottery or contest is like one in a million.. Unless you're very lucky.

For example a contest. Lets say I submit 3k of form, it might be a lot for us, but its actually only a mere few percent of winning chance maybe less than 0.3%.. Coz we are not the only one who participate, there are thousand other of people outside who join the same contest..

same goes to lottery, there are so many sets of number, from 0001 to 9999... sweat.gif
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escay.h
post Mar 11 2010, 01:49 PM


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the Rules of Probability of winning in betting/gambling is always 50% by the punter or the person betting. It doesn't matter how the bet is being made or what is the condition of gambling rules adhere to (4d, 6d, i-box, m-box, etc.), it is always either you win or lose (50% chance). The simplest logical explanation is - If you bet 3000 4d nos with Magnum 4d, are you guaranteed a winning of 30% on that draw?? The answer is a very simple -NO!! Not even if you bet 5000 nos. or 7000 nos. that DOES NOT GUARANTEED A WINNING IN ANY PRIZE AT ALL!! Therefore the Rules of Probability of gambling is always 50% - 50% chance of winning (meaning, it is either you win or lose) simple.

The probability in raking in gambling bets is therefore the totally opposite of the above rules, as it is strictly for the betting operator/casino owner/bookies as they will determine on the statistic of punters or no. of people betting on their games. The more punters betting, the more income they will generate based on the odds calculation of the games they are providing. This rules is not applicable to the PUNTERS!! as you are not betting with everyone that placed a bets, but you're betting with the operator/bookies.


Added on March 11, 2010, 1:55 pmIf you bet, you get 50% chance of winning. But if you didn't bet you are 100% did not lost (therefore - didn't lose = wins!!).

Hahahaha

This post has been edited by escay.h: Mar 11 2010, 01:55 PM
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Bevan
post Mar 11 2010, 01:56 PM


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QUOTE(skon9 @ Mar 11 2010, 10:34 AM)
Winning a lottery or contest is like one in a million.. Unless you're very lucky.

For example a contest. Lets say I submit 3k of form, it might be a lot for us, but its actually only a mere few percent of winning chance maybe less than 0.3%.. Coz we are not the only one who participate, there are thousand other of people outside who join the same contest..

same goes to lottery, there are so many sets of number, from 0001 to 9999... sweat.gif
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actually its from 0000 to 9999 biggrin.gif
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escay.h
post Mar 11 2010, 02:08 PM


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Oh Btw, if you are thinking of winning Magnum4d Jackpot - then u should ask them if they allow you to buy Jackpot Pemutation of 0000-9999 and pay RM10,000 for it for a 100% SURE WIN GUARANTEED!!

HAHAHAHA.... If you do, pls sedekah me poor man 1% of your winning ok ah?
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amyhs99
post Mar 11 2010, 02:30 PM


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QUOTE(escay.h @ Mar 11 2010, 02:08 PM)
Oh Btw, if you are thinking of winning Magnum4d Jackpot - then u should ask them if they allow you to buy Jackpot Pemutation of 0000-9999 and pay RM10,000 for it for a 100% SURE WIN GUARANTEED!!

HAHAHAHA....  If you do, pls sedekah me poor man 1% of your winning ok ah?
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If u buy 0000 - 9999 for jackpot, i dun think count like tat one..
Coz system bet counts like tat:
If you buy 0000 - 9999, means 10,000 numbers..
Den you'll have to pay:
10,000 * (10,000 - 1)
= 10,000 * 9999
=RM99,990,000

But max system bet number is 20? hmm.gif
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escay.h
post Mar 13 2010, 03:05 AM


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Wow RM99,990,000 to buy a guaranteed win, must make sure Jackpot Prize is more than RM99,990,001 to win RM1 GUARANTEED from Magnum4d then!!

But are u sure there is a limit on the system bet in the system?
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amyhs99
post Mar 13 2010, 03:10 AM


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Yes.. System bet max is 20 bet. But i haven't try 20 number with 24 permutations which makes 480 numbers.. But still far away form 10,000..

Magnum was my dad's business btw~ Haha~
I go to shop and check 1st~~

I try to screenshot of the jackpot betting screen.. biggrin.gif

Jackpot now only 30m.. rclxub.gif

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lin00b
post Mar 13 2010, 11:39 AM


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QUOTE(amyhs99 @ Mar 11 2010, 02:30 PM)
If u buy 0000 - 9999 for jackpot, i dun think count like tat one..
Coz system bet counts like tat:
If you buy 0000 - 9999, means 10,000 numbers..
Den you'll have to pay:
10,000 * (10,000 - 1)
= 10,000 * 9999
=RM99,990,000

But max system bet number is 20? hmm.gif
*
cant you have the same number for jackpot?

ie

1234 1234
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Winniekhoo89
post Mar 15 2010, 04:42 PM


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gosh, instead of spending time on these stuff
why not work hard ? -.-
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post Mar 19 2010, 01:08 AM


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QUOTE(Winniekhoo89 @ Mar 15 2010, 04:42 PM)
gosh, instead of spending time on these stuff
why not work hard ? -.-
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work hard cant get 30mil oso... whistling.gif
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lin00b
post Mar 19 2010, 10:56 PM


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mind you thins thread is not about encouraging people to gamble, but trying to work out the maths behind various gambling activities.

and showing proof that most of the risk is on the player and not the banker (not that i doubt any of you dont know that) and the only way to win, is not to play tongue.gif
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rafazafar
post Mar 21 2010, 03:45 AM


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use permutations and combinations of whatever number you want, and divide by the number of results that they give daily. then you should get the answer.

like :10000/23 - eg. 4 digit toto. since 23 results daily.
=1 in 435 will get something.

therefore each day, they get:
Rm20k.

to get a jackpot, its : (3/10000)^2
=1 in 150million. to get any jackpot.

to get special price its: (10/10000)x(3/10000)
=1 in 30million.

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Sifha238
post Mar 24 2010, 01:21 PM


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Spend Rm15 every week for probability to win thousand ringgit consider ok even the probability is too high

That is investment also. Smoker spent more than Rm15 every week and they can nothing
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post Mar 24 2010, 03:40 PM


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QUOTE(Sifha238 @ Mar 24 2010, 01:21 PM)
Spend Rm15 every week for probability to win thousand ringgit consider ok even the probability is too high

That is investment also. Smoker spent more than Rm15 every week and they can nothing
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They get lung cancer
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lin00b
post Mar 24 2010, 05:05 PM


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QUOTE(Sifha238 @ Mar 24 2010, 01:21 PM)
Spend Rm15 every week for probability to win thousand ringgit consider ok even the probability is too high

That is investment also. Smoker spent more than Rm15 every week and they can nothing
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you need to know the difference between investment and gambling. many investor dont, according to kiyosaki (among others)
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post Mar 24 2010, 11:31 PM


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QUOTE(lin00b @ Mar 24 2010, 06:05 PM)
you need to know the difference between investment and gambling. many investor dont, according to kiyosaki (among others)
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Gambling is haram laugh.gif
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99chan
post Mar 26 2010, 07:42 PM


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so we can discuss about gambling now? if that is so i would like to talk about card counting, implied odds, pot odds and so much more. may i go on?
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lin00b
post Mar 27 2010, 10:11 AM


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please continue
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faceless
post Mar 31 2010, 02:52 PM


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Ahhh finally a more acceptable math answer. Can you show the calculation Juelique.
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thus
post Apr 5 2010, 10:02 PM


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i dont think all these number gambling is truly help people....let say you own this toto..
would you like to give people RM15k for spending buying RM5 ticket of number?? of course not....
if me the owner.... i use statiscally math...to calculate or to find the number that nobody buy using computer program.... its sooo easyyy.....
and give the small2 prize to some people for getting trusted the buyers....
i will get richer n richer....

there is no fast money ...fast money, fast spending oso.... biggrin.gif
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post Apr 5 2010, 10:19 PM


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QUOTE(escay.h @ Mar 11 2010, 01:49 PM)
the Rules of Probability of winning in betting/gambling is always 50% by the punter or the person betting. It doesn't matter how the bet is being made or what is the condition of gambling rules adhere to (4d, 6d, i-box, m-box, etc.), it is always either you win or lose (50% chance). The simplest logical explanation is - If you bet 3000 4d nos with Magnum 4d, are you guaranteed a winning of 30% on that draw?? The answer is a very simple -NO!! Not even if you bet 5000 nos. or 7000 nos. that DOES NOT GUARANTEED A WINNING IN ANY PRIZE AT ALL!! Therefore the Rules of Probability of gambling is always 50% - 50% chance of winning (meaning, it is either you win or lose) simple.

The probability in raking in gambling bets is therefore the totally opposite of the above rules, as it is strictly for the betting operator/casino owner/bookies as they will determine on the statistic of punters or no. of people betting on their games. The more punters betting, the more income they will generate based on the odds calculation of the games they are providing. This rules is not applicable to the PUNTERS!! as you are not betting with everyone that placed a bets, but you're betting with the operator/bookies.


Added on March 11, 2010, 1:55 pmIf you bet, you get 50% chance of winning. But if you didn't bet you are 100% did not lost (therefore - didn't lose = wins!!).

Hahahaha
*
this made more sense .. lol ...
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lin00b
post Apr 5 2010, 11:24 PM


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QUOTE(thus @ Apr 5 2010, 10:02 PM)
i dont think all these number gambling is truly help people....let say you own this toto..
would you like to give people RM15k for spending buying RM5 ticket of number?? of course not....
if me the owner.... i use statiscally math...to calculate or to find the number that nobody buy using computer program.... its sooo easyyy.....
and give the small2 prize to some people for getting trusted the buyers....
i will get richer n richer....

there is no fast money ...fast money, fast spending oso....  biggrin.gif
*
on one hand you talked about statistical maths, and ramble onto computer rigging numbers. rclxub.gif

didnt toto have some representative rolling a barrel of physical number balls and see what drops out? or is that another game?
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teongpeng
post Apr 5 2010, 11:24 PM


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QUOTE(99chan @ Mar 26 2010, 07:42 PM)
so we can discuss about gambling now? if that is so i would like to talk about card counting, implied odds, pot odds and so much more. may i go on?
*

i can help with a bit of basic poker maths if u want. smile.gif

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lin00b
post Apr 5 2010, 11:26 PM


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mahjong maths is more interesting ^^
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post Apr 7 2010, 02:49 PM


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If you have all the opened numbers & winning position for past 10 years maybe we can come up with a tabulation of a "ranking system" that shows from 0000-9999 what are the "probablilities" of each number opening. For example numbers that came out in higher prize will have it's probability diminished greatly, while those like "consolidation prize" may have less probability deducted.

I don't see how difficult it is from calculation & processing point of view as theres only 10000 numbers.

Difficult task is getting all the historical data, as you know sites like Magnum require manual writing down the numbers as they use jpeg instead of text.
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post Apr 8 2010, 12:54 AM


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why would historical trends have anything to do with luck? that is the gambler's fallacy.

if i have a coin, and for the past 10 flips, i get 4 heads, 6 tails; what do you think my next flip would most likely be?
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advocado
post Apr 8 2010, 06:30 AM


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You never studied probability do you?
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post Apr 8 2010, 07:34 AM


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read up on dependent and independent events.
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post Apr 8 2010, 02:57 PM


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QUOTE(abubin @ Mar 3 2010, 11:57 AM)
But the thing that I believe is that for magnum, they do not publish their drawing of numbers. So they could have easily manipulate their system to select winning numbers based on buyers. That means the less buyers of a number will have higher chance or first prize. This way magnum will maximize their profit. But that does not take into consideration of those buying from third party 4d sellers.
*
Magnum's numbers are not drawn by computers but by physical drums and balls. I suggest you go watch a live draw before making comments like that. The most they can do to minimize risk is to place a lower sales limit on some popular numbers.

On the most pairs of numbers which you can possibly buy on a single Jackpot ticket, it is:
PM24 x PM24 = 576 pairs, where each pair is RM2 = RM1,152

System bet 20 only allows 190 pairs = RM380. No permutation possible for system bet yet.

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teongpeng
post Apr 8 2010, 10:44 PM


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Here's an interesting probability problem i read somewhere:

A contestant on a game show is presented 3 doors, one of which has a car behind it and two do not.

The contestant is asked to pick one.

After choosing a door, the host then MUST open one of the two doors not picked by the contestant. The door opened does NOT have a car behind it.

Now if you're the contestant, and you get to change your choice to the other unopened door, would u do it? does it make a difference? Why?



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teongpeng
post Apr 8 2010, 10:55 PM


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QUOTE(lin00b @ Apr 8 2010, 12:54 AM)
why would historical trends have anything to do with luck? that is the gambler's fallacy.

if i have a coin, and for the past 10 flips, i get 4 heads, 6 tails; what do you think my next flip would most likely be?
*

ive read a long argument on that in another forum. and im not so sure its that clearcut.

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Noyze
post Apr 15 2010, 09:26 AM


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it all boils down to simple math and probability. You cannot determine the order nor the next number drawn as the actual draw by single numbers is determined by having the 1st number drawn which is 1/10 which is a 10% chance that the 1st number you choose will be drawn. same goes for the next till the 4th number. this is of course if you do an even distribution. hence to calculate the probability of your numbers getting selected would be 1/10 * 1/10 *1/10 *1/10 = 0.0001% of your 4 digits getting chosen which still equals to 1/10k. no need to think so much.

it's all a gamble nia.


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lin00b
post Apr 15 2010, 08:06 PM


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QUOTE(teongpeng @ Apr 8 2010, 10:44 PM)
Here's an interesting probability problem i read somewhere:

A contestant on a game show is presented 3 doors, one of which has a car behind it and two do not.

The contestant is asked to pick one.

After choosing a door, the host then MUST open one of the two doors not picked by the contestant. The door opened does NOT have a car behind it.

Now if you're the contestant, and you get to change your choice to the other unopened door, would u do it? does it make a difference? Why?

*
yes, you should change.

W=win L=lose

1st pick : result
W -> change -> L
L -> change -> W
L -> change -> W

there is 2/3 chance is winning if change

and the coin answer is simple, you will have 50:50 of getting head:tail regardless of past results.


Added on April 15, 2010, 8:12 pm
QUOTE(Noyze @ Apr 15 2010, 09:26 AM)
it all boils down to simple math and probability. You cannot determine the order nor the next number drawn as the actual draw by single numbers is determined by having the 1st number drawn which is 1/10 which is a 10% chance that the 1st number you choose will be drawn. same goes for the next till the 4th number. this is of course if you do an even distribution. hence to calculate the probability of your numbers getting selected would be 1/10 * 1/10 *1/10 *1/10 = 0.0001% of your 4 digits getting chosen which still equals to 1/10k. no need to think so much.

it's all a gamble nia.
*
you also have to realize the psychological factors involved. people often treat 4d as aiming for the right number, that is why you get a lot of "aiyah, nearly kena!" when results are announced.

these nearly kena includes, missed by 1 number, or numbers are jumbled up.

which would mean for a single bet of unique number (no repeat), you will have around 40+24=64 "nearly kena"

in case of (x)box when they automatically jumble up the number for you, for a single unique number, you will have around 24*40=960 "nearly kena"

so people think that they are nearly there and continue to bet.

This post has been edited by lin00b: Apr 15 2010, 08:12 PM
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teongpeng
post Apr 16 2010, 12:42 AM


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QUOTE(lin00b @ Apr 15 2010, 08:06 PM)
and the coin answer is simple, you will have 50:50 of getting head:tail regardless of past results.
To simpletons that might be the obvious thing to say. But in a sequence, when a coin comes up heads for the last 10 times (or whatever times)...it is getting more and more likely that the next flip will be tail.

50/50 means wutever goes up must come down. When u draw a graph that shows heads N number of times....you are getting closer and closer to the peak (unknown) when the switch would occur.

Perhaps someone with good indepth mathematical knowledge can explain this further for you (with forumlas even). smile.gif







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lin00b
post Apr 16 2010, 01:36 AM


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QUOTE(teongpeng @ Apr 16 2010, 12:42 AM)
To simpletons that might be the obvious thing to say. But in a sequence, when a coin comes up heads for the last 10 times (or whatever times)...it is getting more and more likely that the next flip will be tail.

50/50 means wutever goes up must come down. When u draw a graph that shows heads N number of times....you are getting closer and closer to the peak (unknown) when the switch would occur.

Perhaps someone with good indepth mathematical knowledge can explain this further for you (with forumlas even).  smile.gif
*
no, thats gambler's fallacy. the odds of getting 10 heads in a row is 0.5^10; however, given that you have already achieved the improbable results of 9 heads in a row, the odds of getting the 10th head is still 0.5. results that have already occurred has no effect on future results IF the events are independent (which a simple coin toss, is independent)
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post Apr 16 2010, 01:47 AM


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QUOTE(lin00b @ Apr 16 2010, 01:36 AM)
no, thats gambler's fallacy. the odds of getting 10 heads in a row is 0.5^10; however, given that you have already achieved the improbable results of 9 heads in a row, the odds of getting the 10th head is still 0.5. results that have already occurred has no effect on future results IF the events are independent (which a simple coin toss, is independent)
*
im not going to argue with you further on this because i cant elaborate. but i know i'm right (as always).

to put money where my mouth is....if u can flip heads 10 times on a normal coin...i'll give u 10:9 odds that the 11th throw will be tails. up for it? the 10 heads throw must be in one sequence.

if you are adamant that its still 50/50...then 10:9 odds would be a steal for you am i right?

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lin00b
post Apr 16 2010, 01:59 AM


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from wikipedia:

The gambler's fallacy, also known as the Monte Carlo fallacy (due to its significance in a Monte Carlo casino in 1913)[1] or the fallacy of the maturity of chances, is the belief that if deviations from expected behaviour are observed in repeated independent trials of some random process then these deviations are likely to be evened out by opposite deviations in the future. For example, if a fair coin is tossed repeatedly and tails comes up a larger number of times than is expected, a gambler may incorrectly believe that this means that heads is more likely in future tosses.[2] Such an expectation could be mistakenly referred to as being due. This is an informal fallacy. It is also known colloquially as the law of averages.

or more specific to this discussion:

We can see from the above that, if one flips a fair coin 21 times, then the probability of 21 heads is 1 in 2,097,152. However, the probability of flipping a head after having already flipped 20 heads in a row is simply 1/2 . This is an application of Bayes' theorem.
This can also be seen without knowing that 20 heads have occurred for certain (without applying of Bayes' theorem). Consider the following two probabilities, assuming a fair coin:
probability of 20 heads, then 1 tail = 0.520 × 0.5 = 0.521
probability of 20 heads, then 1 head = 0.520 × 0.5 = 0.521
The probability of getting 20 heads then 1 tail, and the probability of getting 20 heads then another head are both 1 in 2,097,152. Therefore, it is equally likely to flip 21 heads as it is to flip 20 heads and then 1 tail when flipping a fair coin 21 times. Furthermore, these two probabilities are as equally likely as any other 21-flip combinations that can be obtained (there are 2,097,152 total); all 21-flip combinations will have probabilities equal to 0.521, or 1 in 2,097,152. From these observations, there is no reason to assume at any point that a change of luck is warranted based on prior trials (flips), because every outcome observed will always have been equally as likely as the other outcomes that were not observed for that particular trial, given a fair coin. Therefore, just as Bayes' theorem shows, the result of each trial comes down to the base probability of the fair coin: .

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teongpeng
post Apr 16 2010, 02:34 AM


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» Click to show Spoiler - click again to hide... «

Yes i know that. its obvious. and i've been holding on to that same idea since i was small kid. But in the last coupla years or so i came across an argument on another forum and made me rethink the whole concept.

Consider this:

is it more likely to get
a)21 heads or;
b)20 heads and a tail (non-consequetively).

Is it more likely to get 21 heads consequetively, or is it more like to get (lets say) 17 heads consequetively?

Do you see where im going?
mellow.gif

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slimey
post Apr 16 2010, 03:54 AM



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QUOTE(teongpeng @ Apr 16 2010, 02:34 AM)
» Click to show Spoiler - click again to hide... «

Yes i know that. its obvious. and i've been holding on to that same idea since i was small kid. But in the last coupla years or so i came across an argument on another forum and made me rethink the whole concept.

Consider this:

is it more likely to get
a)21 heads or;
b)20 heads and a tail (non-consequetively).

Is it more likely to get 21 heads consequetively, or is it more like to get (lets say) 17 heads consequetively?

Do you see where im going?
  mellow.gif
*
ai ai ai......
but in the end each and every toss the chance is 50:50......
mathematically of cause the chance of getting consequetives diminishes as more tosses are made. but the last toss the chance to get heads or tails is still 50:50
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post Apr 16 2010, 07:43 AM


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QUOTE(teongpeng @ Apr 16 2010, 01:47 AM)
im not going to argue with you further on this because i cant elaborate. but i know i'm right (as always).

to put money where my mouth is....if u can flip heads 10 times on a normal coin...i'll give u 10:9 odds that the 11th throw will be tails. up for it? the 10 heads throw must be in one sequence.

if you are adamant that its still 50/50...then 10:9 odds would be a steal for you am i right?
*
"to put money where my mouth is"

why don't u go to Genting and bet ur life savings on BIG/SMALL after u see 20 times SMALL/BIG in one sequence. ur chance is still less than 50% bro, on the 21st game laugh.gif
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teongpeng
post Apr 16 2010, 07:57 AM


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QUOTE(slimey @ Apr 16 2010, 03:54 AM)
ai ai ai......
but in the end each and every toss the chance is 50:50......
mathematically of cause the chance of getting consequetives diminishes as more tosses are made. but the last toss the chance to get heads or tails is still 50:50
*

its 50/50 if the the last toss were to be made singularly.

but if u had known the past sequence,
bet on the more likely sequential percentage and gain the upper hand? hmm.gif


Added on April 16, 2010, 8:05 am
QUOTE(acougan @ Apr 16 2010, 07:43 AM)
"to put money where my mouth is"

why don't u go to Genting and bet ur life savings on BIG/SMALL after u see 20 times SMALL/BIG in one sequence. ur chance is still less than 50% bro, on the 21st game  laugh.gif
*

ok, but why not i bet against you since to be fair, we both need to put money where our mouth is
rolleyes.gif



This post has been edited by teongpeng: Apr 16 2010, 08:06 AM
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lin00b
post Apr 16 2010, 08:39 AM


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QUOTE(teongpeng @ Apr 16 2010, 02:34 AM)
» Click to show Spoiler - click again to hide... «

Yes i know that. its obvious. and i've been holding on to that same idea since i was small kid. But in the last coupla years or so i came across an argument on another forum and made me rethink the whole concept.

Consider this:

is it more likely to get
a)21 heads or;
b)20 heads and a tail (non-consequetively).

Is it more likely to get 21 heads consequetively, or is it more like to get (lets say) 17 heads consequetively?

Do you see where im going?
  mellow.gif
*
you just failed to understand the whole "past result does not affect future result". before you make a series of tosses, the probability of getting a predetermined series of result is 0.5^n, where n is the number of tosses. however, regardless of series length and where you are in that series, the probability of getting head in the next toss is always 0.5
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cheecken0
post Apr 16 2010, 09:08 AM


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QUOTE(lin00b @ Apr 16 2010, 08:39 AM)
you just failed to understand the whole "past result does not affect future result". before you make a series of tosses, the probability of getting a predetermined series of result is 0.5^n, where n is the number of tosses. however, regardless of series length and where you are in that series, the probability of getting head in the next toss is always 0.5
*
If you are talking about tossing a coin DISREGARDING the previous results then yes, the probability is 50:50

but if you are talking about tossing a coin many times and determine the probability of getting a predefined result such as 3 heads and 3 tails then it would not be 50:50 (to be exact, it is 5/16)


Example:

1) what are the odds of getting a coin to show heads in 1 flip? 50:50

2) what are the odds of getting a coin to flip heads after knowing that it has given head, tails, tails, on its previous 3 throws? it is STILL 50:50 (note that the events are independent)

3) what are the odds of getting a coin to flip Head, Tail, Tail, Head in correct sequence? 1/16 : 15/16

See where I am going?

Edit: wrong probability

This post has been edited by cheecken0: Apr 16 2010, 09:27 AM
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acougan
post Apr 16 2010, 09:20 AM


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QUOTE(teongpeng @ Apr 16 2010, 07:57 AM)
ok, but why not i bet against you since to be fair, we both need to put money where our mouth is
rolleyes.gif
*
act i alrdy lost my shirt once on the 19th game couple yrs back(19 games consecutively against me till i ran out)

i was naive like u once, js becos u think ur theory is "right", doesn't mean real-life odds will support it
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teongpeng
post Apr 16 2010, 05:58 PM


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QUOTE(acougan @ Apr 16 2010, 09:20 AM)
act i alrdy lost my shirt once on the 19th game couple yrs back(19 games consecutively against me till i ran out)

i was naive like u once, js becos u think ur theory is "right", doesn't mean real-life odds will support it
*
lol. poor you. but saying something is more likely to happen does not mean it definitely will happen. to get 20 heads in a row is just bad luck on your part.


Added on April 16, 2010, 6:00 pm
QUOTE(cheecken0 @ Apr 16 2010, 09:08 AM)
If you are talking about tossing a coin DISREGARDING the previous results then yes, the probability is 50:50

but if you are talking about tossing a coin many times and determine the probability of getting a predefined result such as 3 heads and 3 tails then it would not be 50:50 (to be exact, it is 5/16)
Example:

1) what are the odds of getting a coin to show heads in 1 flip? 50:50

2) what are the odds of getting a coin to flip heads after knowing that it has given head, tails, tails, on its previous 3 throws? it is STILL 50:50 (note that the events are independent)

3) what are the odds of getting a coin to flip Head, Tail, Tail, Head in correct sequence? 1/16 : 15/16

See where I am going?

Edit: wrong probability
*
thank you.


Added on April 16, 2010, 6:02 pm
QUOTE(lin00b @ Apr 16 2010, 08:39 AM)
you just failed to understand the whole "past result does not affect future result". before you make a series of tosses, the probability of getting a predetermined series of result is 0.5^n, where n is the number of tosses. however, regardless of series length and where you are in that series, the probability of getting head in the next toss is always 0.5
*

its not that i fail to understand. i already told u i understood. all im asking is for ppl to explore an alternate view to the same problem. haih.....

Learn to think out of the box ppl!


This post has been edited by teongpeng: Apr 16 2010, 06:02 PM
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post Apr 16 2010, 06:17 PM


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QUOTE(Marcian @ Apr 8 2010, 02:57 PM)
Magnum's numbers are not drawn by computers but by physical drums and balls.  I suggest you go watch a live draw before making comments like that.  The most they can do to minimize risk is to place a lower sales limit on some popular numbers.

On the most pairs of numbers which you can possibly buy on a single Jackpot ticket, it is:
PM24 x PM24 = 576 pairs, where each pair is RM2 = RM1,152

System bet 20 only allows 190 pairs = RM380.  No permutation possible for system bet yet.
*
hmm.. meaning if I go to Magnum, I can only buy 190 pairs of numbers max?
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lin00b
post Apr 16 2010, 08:22 PM


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QUOTE(teongpeng @ Apr 16 2010, 05:58 PM)
lol. poor you. but saying something is more likely to happen does not mean it definitely will happen. to get 20 heads in a row is just bad luck on your part.


Added on April 16, 2010, 6:00 pm
thank you.


Added on April 16, 2010, 6:02 pmits not that i fail to understand. i already told u i understood. all im asking is for ppl to explore an alternate view to the same problem. haih.....

Learn to think out of the box ppl!
*
there is no "out of the box" to think of, this is not a creative question. this is how much you understand statistics and probability (as well, as how sometimes, whats "logical" is not always true)


Added on April 16, 2010, 8:30 pm
QUOTE(cheecken0 @ Apr 16 2010, 09:08 AM)
If you are talking about tossing a coin DISREGARDING the previous results then yes, the probability is 50:50

but if you are talking about tossing a coin many times and determine the probability of getting a predefined result such as 3 heads and 3 tails then it would not be 50:50 (to be exact, it is 5/16)
Example:

1) what are the odds of getting a coin to show heads in 1 flip? 50:50

2) what are the odds of getting a coin to flip heads after knowing that it has given head, tails, tails, on its previous 3 throws? it is STILL 50:50 (note that the events are independent)

3) what are the odds of getting a coin to flip Head, Tail, Tail, Head in correct sequence? 1/16 : 15/16

See where I am going?

Edit: wrong probability
*
i'm not so sure you got the 3 head/3 tail (in any sequence) right, but too lazy to check at the moment.

true, now answer my original question,

QUOTE
if i have a coin, and for the past 10 flips, i get 4 heads, 6 tails; what do you think my next flip would most likely be?


which some genius answered,
QUOTE
ive read a long argument on that in another forum. and im not so sure its that clearcut.


what do you think?



This post has been edited by lin00b: Apr 16 2010, 08:30 PM
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cheecken0
post Apr 17 2010, 01:50 AM


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QUOTE(lin00b @ Apr 16 2010, 08:22 PM)
i'm not so sure you got the 3 head/3 tail (in any sequence) right, but too lazy to check at the moment.
*
I have worked it out. I used the binomial distribution to find out the possible combinations of 3 heads and 3 tails. If there is any error, it should fall under incorrect substitution not misuse of formulas.



QUOTE(lin00b @ Apr 16 2010, 08:22 PM)
what do you think?
*
Your next coin flip would still give equal chances of both heads and tails. because it is the same as not knowing the results of the previous flips and flipping it as normal. so yeah.


Answered from an SPM student's point of view.

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skeleton202
post Apr 18 2010, 10:32 PM


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tis kind of thread oso got in phd section?
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cozeyzero
post Apr 19 2010, 06:28 AM


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QUOTE(MPIK @ Mar 1 2010, 11:10 AM)
Who here can enlighten me on the winning probability of 4D toto, magnum or 1+3D?

I understand if I buy 1 number only and Magnum only draws a single number. My probability is 1 out of 10,000.

But things get complicated when I buy 7 numbers and Magnum draws 23 numbers. So, how the calculation then? What is my probability of winning?

Any maths experts here?
*
go ask pakcik2 at kedai kopi..they all master lo doh.gif
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advocado
post Apr 21 2010, 02:50 PM


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I think it depends how you see the results & individual numbers as related or independant events. Say the older results are:

1st issue) 1234
2nd issue) 1234

If you regard each digit's as independant event, then the probability will be 1/10.

From the 1st issue result, the probability of result "1234" coming out would be 1/10*1/10*1/10*1/10 = 0.0001

If you view the 2nd result "1234" as another independant event, the probability would still be 0.0001

But if you take into the account of the probability of 2nd result "1234" coming out after the 1st result "1234", they the probability will become dependant of previous result, so the probability of "1234" to come out again after the 1st "1234" would be 0.0001*0.0001, which will become 0.00000001 (1^-8).

obviously the chances of the 2nd result coming to be "1234" right after the 1st result "1234" is pretty much impossible, and that other numbers stand higher chances than it.

So the crucial point is to determine whether the winning numbers (or even individual digits) are independant events or dependant to the previous results, which we have been discussing so far.

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f4tE
post Apr 21 2010, 03:34 PM


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this also can become phd topic?
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lin00b
post Apr 21 2010, 07:18 PM


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better a discussion in practical statistics and probability than another "do you believe in XXX"
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cheecken0
post Apr 21 2010, 08:47 PM


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QUOTE(skeleton202 @ Apr 18 2010, 10:32 PM)
tis kind of thread oso got in phd section?
*
QUOTE(f4tE @ Apr 21 2010, 03:34 PM)
this also can become phd topic?
*
If the topic title is named Practical applications of theoretical Statistics: gambling with 4D as an example, will you comment any differently?


Then again, 4D is more to... classical statistics anyway.

Just a question:


How do you know if the lottery is skewed to pick a number which was not bought?

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kenwei
post Apr 24 2010, 01:49 PM


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Didn't read through the whole thread so I don't know if the question is answered already. But anyway, to answer this question:

But things get complicated when I buy 7 numbers and Magnum draws 23 numbers. So, how the calculation then? What is my probability of winning?

This is an elementary probability question actually; since Magnum draws 23 numbers the chances of an individual win is 23/10000 or 0.0023. A simple way to compute the probability is by a calculation of Bernoulli trials, where you can find the probability of winning more than once,i.e. 1-(probability of losing all 7 times)

1-(0.9977^7)=0.0159893349, or approximately 1.6%

However, to calculate for expected value, you must take the average return; it is 100% mathematically correct, because in all probability calculations, we are calculating the effect in the long run anyway, so it's no point considering each prize individually (if we did, the calculations would be 10 times longer, but I guarantee that if you combined them you'll end up with the same final answer as me). Therefore, you'd still end up losing in the long run (as you probably know), if the cost of playing a single number is RM1 and the return from a win is the average of RM6600/23(from post #5 by lin00b)=RM287

Your average return on a RM7 bet, i.e. risk RM7 to win RM287 with probability 0.0159893349:

0.0159893349*(RM6600/23)=RM4.58824393

E.V. is RM4.58824393/RM7=0.655463419, meaning you get back about 65.5 sen for every RM1 you gamble in the long run

I should point out that the difference in probability of winning more than once in a single game compared to winning exactly once is only marginal. Consider the probability of winning exactly once:

(7C1)*(0.0023)*(0.9977^6)=0.0158790936, or approximately 1.6%(again!)

therefore 287*0.0158790936=RM4.5566093

Your E.V. is 0.650944186, meaning you get back about 65 sen for every RM1 you gamble. Just a 0.5% difference with that of winning more than once.

But anyway, to answer the question, the mathematical probability of winning a 23/10000 game with 7 trials at least once is 0.0159893349, or approximately 1.6%

I should also add that the game IS independent of previous results as long as the drawn numbers are genuinely random; the reason why we seldom see the same number included in the winners twice in a row is because the probability for such a coincidence to occur is 1 - (0.9977^23) = 5%, which does happen actually...once in 20 times. As for the same number being the first prize winner twice in a row, that's a simple 0.01% chance.

And this type of game is not rigged as far as I know, since for the drawn numbers to be biased(i.e. deliberately picked based on the frequency of the numbers people bought), Magnum (or any similar company) must have access to the numbers people have bought. But, if I'm not mistaken, that is not the case; to claim your winnings all you have to do is bring your number slip as proof of purchase, and there is no record of exactly which numbers were bought. Either that, or there are witnesses to watch the numbers being randomly drawn. Now everything I just mentioned in this paragraph is 100% my theory without reference, and there's a good chance that I'm downright wrong about the procedure that they take.

But again, mathematically, the numbers drawn can't be effectively rigged because statistically, every number combination was probably bought with an evenly distributed frequency! Maybe with exceptions of 1111,2222 etc but that only marginally affects the distribution, and besides, if that was the case, those would be the published winners. Given the scenario, the increase of their edge would be only a small amount, and it has to be weighed against the risk of getting caught, which is highly likely, and will result in bankruptcy (who would gamble with a cheating company if they were exposed?). Long story short, it's not profitable for these companies to cheat when probability says they get to keep RM1 for every RM3 that is 'invested' onto them.

This post has been edited by kenwei: Apr 24 2010, 02:05 PM
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faceless
post Apr 26 2010, 03:30 PM


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QUOTE(advocado @ Apr 21 2010, 02:50 PM)
I think it depends how you see the results & individual numbers as related or independant events. Say the older results are:

1st issue) 1234
2nd issue) 1234

If you regard each digit's as independant event, then the probability will be 1/10.

From the 1st issue result, the probability of result "1234" coming out would be 1/10*1/10*1/10*1/10 = 0.0001

If you view the 2nd result "1234" as another independant event, the probability would still be 0.0001

But if you take into the account of the probability of 2nd result "1234" coming out after the 1st result "1234", they the probability will become dependant of previous result, so the probability of "1234" to come out again after the 1st "1234" would be 0.0001*0.0001, which will become 0.00000001 (1^-8).

obviously the chances of the 2nd result coming to be "1234" right after the 1st result "1234" is pretty much impossible, and that other numbers stand higher chances than it.

So the crucial point is to determine whether the winning numbers (or even individual digits) are independant events or dependant to the previous results, which we have been discussing so far.
*
Ahhh, talking probability. I like. Have you any idea how Magnum conducts the 4D draw? I don't think so. If you did, you will not have so much question on independent events. Let me iron it out for you. There are 6 drums. Lets name them P, SC, d1, d2, d3, d4. SC contain 13 blue balls (marked A to M) and 10 white balls. The digit drums (d1-d4) each contained 10 balls maked 0-9. The SC drum and the digits drum will draw a ball for 23 times. If the white ball is drawn on SC, the the corresponding numbers balls that are drawn will be a consolation prize. The balls drawn on the digit durms will be put back into the drum to draw the next number. Balls on the SC drum will not put back into the SC drum. Once balls had been depleted from the SC drum, we had 13 special prizes and 10 consolation prizes. The consolation prizes were labeled A to J accordingly. The P drum now has 13 blue balls, Label A to J. It will be drawn 3 times. The 3 of the corresponding 13 special prizes with the same letter will be selected as 1st, 2nd, 3rd prize. The balance 10 will be the special prize.

I am not good in math and would appriciate you work out the probability.

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lin00b
post Apr 26 2010, 08:03 PM


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QUOTE(faceless @ Apr 26 2010, 03:30 PM)
Ahhh, talking probability. I like. Have you any idea how Magnum conducts the 4D draw? I don't think so. If you did, you will not have so much question on independent events. Let me iron it out for you. There are 6 drums. Lets name them P, SC, d1, d2, d3, d4. SC contain 13 blue balls (marked A to M) and 10 white balls. The digit drums (d1-d4) each contained 10 balls maked 0-9. The SC drum and the digits drum will draw a ball for 23 times. If the white ball is drawn on SC, the the corresponding numbers balls that are drawn will be a consolation prize. The balls drawn on the digit durms will be put back into the drum to draw the next number. Balls on the SC drum will not put back into the SC drum. Once balls had been depleted from the SC drum, we had 13 special prizes and 10 consolation prizes. The consolation prizes were labeled A to J accordingly. The P drum now has 13 blue balls, Label A to J. It will be drawn 3 times. The 3 of the corresponding 13 special prizes with the same letter will be selected as 1st, 2nd, 3rd prize. The balance 10 will be the special prize.

I am not good in math and would appriciate you work out the probability.
*
seems the be a rather roundabout ways of doing things, i'm not sure what the P drum adds to the results.

however, to the buyer, the odds of winning something should remain the say. i see nothing that affects the independence of the winning numbers.
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faceless
post Apr 27 2010, 11:01 AM


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QUOTE(lin00b @ Apr 26 2010, 08:03 PM)
seems the be a rather roundabout ways of doing things, i'm not sure what the P drum adds to the results.
*
Agree. The method I describle was used in the 1990s. They previously (1980's) used only 5 drums. Four digits drums and Prize drum. The prize drum had balls marked 1st, 2nd, 3rd, ten balls marked "S" for special and 10 unmarked balls for consolation. Their reason was it adds to the suspense. In the 80s you will have no idea when the first prize is drawn. The 90s system, leaves the last moments to draw the top 3 prizes. Their draw is open to public. I am not sure if they had change the system today as I no longer work there. The system is only applicable to Magnum. I had not seen how Toto or Kuda conduct their draw.

So, Avocado working is correct? I had given back all the math I learnt to the teacher.
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lin00b
post Apr 27 2010, 01:59 PM


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avocado is wrong, as previous result do not affect the next result. thats why for magnum you sometime see 1 number hitting 2 prize. toto discard the repeated number.

kenwei is more logical. or you can refer to my solution in earlier post
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faceless
post Apr 27 2010, 02:27 PM


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Hahahah, I can remember the first time that happens. The draw committee were looking at each other dumbstruck. They were silent for 1 minute. Finally the decision was a number that had been drawn should be deem as drawn, therefore it should qualify to win two prizes.
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CANONPIXMA
post Apr 28 2010, 12:27 AM


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instead of calculating probabilty.. it is better that u consider 4d is gambling and it is juz pure luck
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faceless
post Apr 28 2010, 10:45 AM


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Anything that had odds can be turned into gambling, like football. Calculation probability will keep the mind thinking.
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sakaic
post May 1 2010, 12:59 AM


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Funny that this discussion came up. After all it was gambling that gave birth to probability.

Regarding the phenomena where when something keeps increasing it will eventually fall, its called regression towards the mean. I means that no matter how much a value drifts it will eventually zero and swing so that it will average out towards the mean. Thats why families that have tall kids will have very short ones etc.

Having said that, given that nothing in this world is perfect, there will be a slight skew in the numbers. That is given. Even roulette tables can have slight tile or balance problems leading to certain numbers being favoured. There was a maths professor that tried that out and actually won some money from a casino.

But having said that.......It has been worked out on how to make sure that the eventual winner is the seller of the numbers. And when enough repetitions have been performed, the method or the balls will be changed to prevent the mean number due to imperfections will be shown.
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