Quiz: 989 to 1 draw to split
In the other 989 to 1 thread, I meant to talk about drawing to win, but MiamiKeith mentioned hands with certain odds to split. That made me think of the following question:
Suppose there is a player, say Player Z, and some other players in a pot. Suppose the following conditions hold:
1. All players' hole cards are exposed, including Player Z's.
and
2. An observer (other then Player Z) has enough information to correctly conclude both of the following with certainty:
- Player Z cannot win the pot outright.
- Player Z's odds of splitting the pot (i.e. winning any portion of the pot) are 989 to 1.
Question: What is the minimum number of players in the hand?
Now for the harder question. 8)
Replace "including" by "except for" in Part #1.
ScottyZ
Suppose there is a player, say Player Z, and some other players in a pot. Suppose the following conditions hold:
1. All players' hole cards are exposed, including Player Z's.
and
2. An observer (other then Player Z) has enough information to correctly conclude both of the following with certainty:
- Player Z cannot win the pot outright.
- Player Z's odds of splitting the pot (i.e. winning any portion of the pot) are 989 to 1.
Question: What is the minimum number of players in the hand?
Now for the harder question. 8)
Replace "including" by "except for" in Part #1.
ScottyZ
Comments
2. I gave examples of this in the other thread... eg Z has 23 (nice hand Z....), I have AA, and the flop is AKK. Player Z can't win, and only chop if he catches runner-runner K-K.
In fact, it must be EXACTLY two players. If there were 3, there would be 6 hole cards exposed, and 3 on the flop, and the longest possible odds against a live draw-out would be 902-1, thus falsifying your second second condition.
This is not possible. In order for your 2nd condition to be valid, there must be exactly 2 people whose hole cards are exposed, making it a 3-person pot. If we know (without looking at Z's holecards) that he has no chance of winning the pot, then one of his opponents must have at least 4 of a kind (otherwise Z might have a pocket pair that is not matched to any of the 7 exposed cards, and could go runner-runner for quads and the scoop). So there are two possibilites:
1. One of his opponents has quads. In this case the pot cannot be split (if you have quads, without all 4 of them being on the board, you cannot split the pot. You can only win it or lose it)
2. One of his opponents has a straight flush. The only way that you can chop with a straight flush is if a higher straight flush hits the board (eg you have 2h3h and the flop is 4h5h6h, and the tuen and river are 7h8h). But, if it is possible for these two cards to hit the board (ie they are not exposed in the 3rd players hand) then it is NOT possible to determine, without looking at Z's hole cards, that he does not have them in his hand himself, thus it would not be possible to determine that he has no chance of scooping)
Keith
Plus you're right that the number of exposed hands is going to change the perfect-perfect drawing odds slightly away from 989 to 1.
ScottyZ