Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hi,

I hope this question isn't too easy for this forum. I've looked at all the titles for the other questions and they seem pretty hard.

Anyway, I'm trying to work out some probabilities for the following scenario:

I'm playing Texas Holdem poker. There are 5 community cards on the board of which 3 are spades. Everybody has been dealt two cards. Neither of my cards are spades. Since we need 5 cards to make a flush, what is the probability of one or more of my three other opponents having a flush (ie that one or more of them has two spades for their hole cards).

In order to answer this, I solved the problem for if I was facing 1 opponent:

There are 45 unseen cards, of which 10 are spades. So the probability of a single opponent having two spades is:

10/45 x 9/44 = 1/22 or 4.55%

Then I (hope I've) solved the problem for 2 opponents:

If p(A) is the probability of my first opponent having 2 spades and if p(B) is the probability of my second opponent having 2 spades, then I need to find p(A ∪ B).

p(A ∪ b) = p(A) + p(B) - p(A ∩ B)

so:

p(A ∩ B) = p(A) x p(B|A)

p(A ∩ B) = 1/22 x (8/43 x 7/42) = 2/1419

therefore:

p(A ∪ B) = 1/22 + 1/22 - 2/1419 = 127/1419 = 8.95%

Is this right so far?

Now I'm really getting lost, because for 3 players I presumably need to find out p(A ∪ B ∪ C).

Is this the same as p((A ∪ B) ∪ C)?

If so then since I had p(A union B), I could use the p(x ∪ y) = p(x) + p(y) - p(x ∩ y) to find out the answer couldn't I?

The problem is that this seems wierd to me because if I draw some Venn diagrams then it occurs to me that as you add more opponents, the area which you add on to the sample space decreases from 1/22 for each opponent. But instead the intersection gets less and less, so my calcs are making extra area of the sample space CLOSER to 1/22 instead. Something is wrong.

Should my equation for 3 opponents instead be:

p(A ∪ B ∪ C) = p(A) + p(B) + p(C) - 3(p(A ∩ B)) - p(A ∩ B ∩ C)

If so, is there are more generalised formula for adding extra events that I should be aware of?

share|improve this question

closed as too localized by Victor Protsak, Robin Chapman, Igor Pak, Noah Snyder, S. Carnahan Aug 11 '10 at 12:29

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

    
Usually you look at the probability, that no one gets the flush. –  Daniel Krenn Aug 10 '10 at 15:58
4  
Since this question is getting voted down and attracting votes to close, probably because it is not really at the right level for MO, may I suggest that you have a look at math.stackexchange.com if you have similar questions –  Yemon Choi Aug 10 '10 at 17:59

1 Answer 1

up vote 1 down vote accepted

Your intuition is almost right. The inclusion exclusion principle solves your question:http://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle#In_probability

share|improve this answer
    
Thanks very much for your response. Thats exactly what I'm looking for. I'd like to know why I've been marked down for the question by the way. I've done my best to answer the question by myself and written out a clear question. No pleasing some people. :) –  Hmmmmm... Aug 10 '10 at 17:25

Not the answer you're looking for? Browse other questions tagged or ask your own question.