Probability Exercise/Challenge

[Ken R]

OK, here's the challenge I've been working on lately. With one card to come in Hold'em, what is the lowest possible probability of a split pot, given that the 2 players have an equal chance to win the hand outright.

I believe that a pure 50/50 solution with no possibility of a split is impossible, but don't know what the best solution is. I'm guessing somewhere in the 40/40/20 range but would love to hear any thoughts.

I'm not sure that this exercise has any real practical benefit, other than playing with complex out situations, so I thought this was the right forum for the posting.

Ideas?????

[LetYouDown]

Your post is very hard to make sense of. After about 9 reads, I think I get what you're asking for. Define "equal chance of winning". Exactly equal? That's a huge requirement. Examples here would be beneficial to understanding what you're looking for.

[danzasmack]

I'm pretty sure he is asking the following:

Event A = player A wins the pot
Event B = player B wins the pot
Event S = split pot

With one card left to be dealt:
given that P(A) = P(B), minimize P(S)

[danzasmack]

and here is your answer....

http://twodimes.net/poker/?g=h&b=3h+3d+Jh+9h&d=&h=Qc+Kh%0D%0A2h+2s

Board = 3h 3dc 9h Jh
Player A = Kc Qh
Player B = 2s 2h

Both players have 22 outs, 50/50 chance to win, 0 chance to split

[LetYouDown]

nh

[moomoocow]

Here's another one /images/graemlins/smile.gif

Player 1: QJ hearts
Player 2: 4 club 7 heart
Board: 10H 9H 10S 4D

same thing - 50/50

[Beavis68]

I had this one last night.

Me 8 /images/graemlins/diamond.gif 7 /images/graemlins/diamond.gif
Him 9 /images/graemlins/heart.gif 9/images/graemlins/club.gif

Board: K /images/graemlins/diamond.gif 8 /images/graemlins/heart.gif 7 /images/graemlins/heart.gif T /images/graemlins/heart.gif

Also 50/50 no chance of split.