# Simple Probability Question [closed]

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?

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

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Usually you look at the probability, that no one gets the flush. –  Daniel Krenn Aug 10 '10 at 15:58
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

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

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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