5 players see a flop (Hold\'em)...
 

5 players see a flop. The flop has 2 spades. What is the probability that at least one of the 5 players holds 2 spades?

Chance of 1 player holding 2 spades is 5,09% (easy), but already by the 2nd players the calculations becomes tough, since you have to account for a possible spade in previous players hand.

I can't think of a clever way to calculate this.

Re: 5 players see a flop (Hold\'em)...
 

There's no clever way to do this, it's tedious.

The easiest way is to first compute the probability distribution of the number of spades among the five hands. There are C(11,k)*C(38,10-k) ways to get k spades. If you add these up from k=0 to k=10 you get C(49,10).

If there are zero or one spade, there is no chance of anyone having a pair. If there are six or more someone is certain to have a pair.

If there are two spades, the first one can go anywhere, the second one has 8 chances in 9 to avoid making a pair. So the chance of a pair is 1-8/9 = 1/9. If there are three spades, it's 1-(8/9)*(6/8) = 1/3. Four spades, 1-(8/9)*(6/8)*(4/7) = 13/21. Five spades 1-(8/9)*(6/8)*(4/7)*(2/6) = 55/63.

The table below shows for 0 to 10 spades, the number of ways it can occur, and the chance of it resulting in a pair. If you multiply each pair together and add them up, you get 1,796,390,889. Divide that by C(49,10) = 8,217,822,536 to get 0.2186.



0 472,733,756 0
1 1,793,128,040 0
2 2,689,692,060 1/9
3 2,082,342,240 1/3
4 911,024,730 13/21
5 231,897,204 55/63
6 34,102,530 1
7 2,783,880 1
8 115,995 1
9 2,090 1
10 11 1

This assumes everyone stayed in. If some people folded, the probability will be higher since people are more likely to stay in the pot with a suited hand than an unsuited one.

Re: 5 players see a flop (Hold\'em)...
 

Great work, thanks a lot!