The game is texas hold em.

- There are 5 cards dealt
- You hold 1 club, 1 spade
- The board is 1 club, 4 hearts

What is the probability of one other person holding a heart? Is this correct:

Cards shown:
4 hearts
2 spades
1 club
0 diamonds

Cards left unshown:
9 hearts
11 spades
12 clubs
13 diamonds

26/33 * 25/32 = .61

1 - .61 = 38% chance of them holding one heart

Three non-clubs removed (you have two and board has one). 49 cards left, 13 clubs.

Four clubs on the board. 45 cards left, 9 clubs.

What is the probability that villain holds at least 1 club? Easiest to figure the probability that he does not hold any clubs:

36/45 * 35/44 = .636

There is the 63.6% chance of villain holding exactly 0 clubs. The probability of him holding > 0 clubs (i.e. one or two clubs) is then the compliment:

1 - .636 = .363

Thus, there is a 36.3% chance of villain holding at least one club by chance alone. The actual probability he holds a club probably varies based on the actual action in the hand.

Sherman

edit: Also, this is called Clarkmeister's theorem iirc.

FYP

Ok thanks. I hastily threw together my numbers and didn't really double check anything. Looking back over this I really don't know how I didn't add the total shown/unshown cards correctly or use the correct number of cards remaining.

Heh. Thanks though, this is helpful.