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