There are 15 pigs are racing and they are wearing the number from 1-15 respectively. Here are 5 strong pigs are categorized as favorite, they are number 2, 6, 9, 10 and 12. The question is: calculate the probabilities to get an odd and even for the sum of number of champion, 1st runner-up and 2nd runner-up. Example, champion=number 6, 1st runner-up=9 and 2nd runner-up=1, then the sum=16, then 16 is even.

I think at least one additional information is required. You say there are favourites, so what is the additional probability for a favourite pig to win compared to an ordinary pig?

Those favoured are not necessarily to get into top 3. The top 3 may be all non-favoured, also can be all favoured, of course can be 2 favoured and 1 non-favoured, 1 favoured and 2 non-favoured.
The highest prob for a favourite pig to win also can be out of top 3 and at the same time other favourite pigs get into top 3.
So, is there the probability for favourite pig to win still needed?

Yes you need to, else it'll be a 15-variable problem which would be a pain in the arse (algebraically) to solve

I think the confusion here comes from the word favorite.

In this situation it could mean that the favored pigs are more likely to win than the regulars. This is why Thinkingfox is asking for more information regarding the probabilities of the favored pigs winning.

darkwall's definition of favorite could be that there is something else beside the winning probability that is making the pigs the favorites. Maybe the pigs have nice names or they all belong to a particular person. This make definition means that the favored pigs does not have a different winning probability compared to the normal pigs.

like tat ah...
then let say pig number 9 got 30% chance to get into top 3, number 2 and 10 20% respectively, number 6 and 12 15% respectively.
how can it gonna to be solved?
any idea to start the calculation?

This post has been edited by darkwall: Jul 15 2009, 01:23 PM

I'll provide the steps:

Firstly get all the listed probabilities of the pigs to win

Secondly, to simplify the calculation, we assume indepedence of each pig's performance in the race to all its contemporaries. Compute the probabilities of every possible combination that gives you either an even or odd number (which will be very very very long)

there are 455 combinations
do we nid the permutation? if yes, 2730 combinations

Yes permutations, because clearly pig no. 1 and pig no. 3 getting first and third respectively is different from third and first respectively.

But you've listed down the probabilities for a top 3 finish instead, which may help to simplify things further

how they know the pig is strong ?

If this is like most toy problems, it shouldn't be a 15-variable problem. Assuming that pigs within the same class have same probability of winning, then there's only one variable involved - the bias in favor of the strong pigs vs the weak pigs.

The problem can be simplified by classifying the pigs into {Strong,Weak} x {Odd,Even}. And the top three positions as {Even,Odd} x {Even,Odd} x {Even, Odd} - pick only the outcomes that result in an even sum.

Then the rest of the problem can be solved with a probability tree. Okay, that would be how i'd do it coz i'm stupid - but it will work!

actually i'm thinking to use kangaroo...

yes if u substitute the animal the problem becomes halal liao.

That's right, except that he's weighted each strong pig with different probabilities, which makes it more of a pain instead

Edit: I forgot an added simplification step:

Suppose pigs 1, 3, 5 get the top 3 positions. We don't really care about the arrangement of the other 12 pigs thereafter, so we can compute the number of possible permutations with pigs 1, 3, 5 being in the top 3 to be 12! * 3! (3! for the various arrangements of pigs 1, 3, 5)

This post has been edited by bgeh: Jul 15 2009, 02:00 PM

So, you're saying that a favourite pig has an equal chance to an ordinary pig? i.e. it's favourite not because of it's high chances of winning.

nope. its favourite because of it's higher chance to win if compare to others.

So what is the probability of a favourite pig compared to an ordinary pig?

hhmm. Any of you watch 21 - The Movie. They use to calculate the possibilities and they use it at casino. I dunno how it works coz i'm very bad in Maths but is quite interesting.

21 (The Movie)

only provide the prob is higher for fav pigs if compare to other pigs, no exact number.

Darkwall, hope you don't mind if I tumpang your thread.

A friend asked me earlier. Let's say you have these parents. The parents have two children. One of them is a girl. What is the probability that the other child is a boy?

Assume there's 1 girl to every 1 boy in the world (it's close). When calculating the the probability of the other child's gender, do we consider the population as a whole (6 billion, so the probability that it will be a boy is a tiny bit higher than 1/2), or do we consider this family in itself (so, 2/3 that it will be a boy because they already have a girl).