5 players see a flop (Hold'em)...

[Kristian]

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.

[AaronBrown]

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.

[Kristian]

Great work, thanks a lot!