Daily Poker Quiz: Archive

Number of flushes #2
2008-08-15

Let's continue with yesterday's question and bring it one step further.

There's $16 in the pot and you're four players to the K 5 4 flop.

Let's assume that all of you go all-in with the exact same stack sizes. All of your opponents have either sets or flush draws, and your stacks are all huge so the $16 in the pot don't matter. This is just a though experiment anyway.

There are, in other words, four possible scenarios:
1) All opponents have sets
2) Two opponents have sets, one has a flush draw
3) One opponent has a set, two have flush draws
4) All opponents have flush draws.

Now, let's add a turn card to make the board K 5 4 2. Given the four possible scenarios, in how many of these scenarios can you expect to make a profit in the long run?





In zero cases (correct)
This is worse, since you only have one card left and not two.

If all opponents have sets and we don't know who has which one, everybody has the same expected value from our point of view. Seven hearts (not nine since K and 2 will give somebody a flush or four of a kind) and three treys give you the nuts. We know that there are 40 cards left (52 - 4*2 [hole cards] - 4 [the board]) and 10 out of 40 equals 25 %, so exactly one fourth. The other 75 % are split between the sets, and obviously the player with the highest set has the best EV, but since we don't know who that is, to us, it looks like everybody has a 25 % chance of winning. If we disregard the $16 in the pot and the rake, you'll break even if all opponents have unknown sets.

In the same way, you can construct a situation where you have exactly 25 % EV against two sets and one flush draw - that is, if the flush draw is K2 of hearts. But you can't get +EV.

Against one set and two flush draws, you're hopelessly behind. Heads-up against a set, you stand to win 22.78 % of the time, and against a third opponent having K2 of hearts, you have 23.81 %. That's as good as it gets, since every additional heart belonging to your opponents gives you fewer outs.

If all opponents have flush draws, you can create situations in which you're waaaaay behind and ones where you have a massive lead, depending on whether or not your opponents have pairs.

So the answer is a trick question again, and I'll give out correct answers to anyone who said 0 or 1.

In one case (correct)
This is worse, since you only have one card left and not two.

If all opponents have sets and we don't know who has which one, everybody has the same expected value from our point of view. Seven hearts (not nine since K and 2 will give somebody a flush or four of a kind) and three treys give you the nuts. We know that there are 40 cards left (52 - 4*2 [hole cards] - 4 [the board]) and 10 out of 40 equals 25 %, so exactly one fourth. The other 75 % are split between the sets, and obviously the player with the highest set has the best EV, but since we don't know who that is, to us, it looks like everybody has a 25 % chance of winning. If we disregard the $16 in the pot and the rake, you'll break even if all opponents have unknown sets.

In the same way, you can construct a situation where you have exactly 25 % EV against two sets and one flush draw - that is, if the flush draw is K2 of hearts. But you can't get +EV.

Against one set and two flush draws, you're hopelessly behind. Heads-up against a set, you stand to win 22.78 % of the time, and against a third opponent having K2 of hearts, you have 23.81 %. That's as good as it gets, since every additional heart belonging to your opponents gives you fewer outs.

If all opponents have flush draws, you can create situations in which you're waaaaay behind and ones where you have a massive lead, depending on whether or not your opponents have pairs.

So the answer is a trick question again, and I'll give out correct answers to anyone who said 0 or 1.

In two cases
This is worse, since you only have one card left and not two.

If all opponents have sets and we don't know who has which one, everybody has the same expected value from our point of view. Seven hearts (not nine since K and 2 will give somebody a flush or four of a kind) and three treys give you the nuts. We know that there are 40 cards left (52 - 4*2 [hole cards] - 4 [the board]) and 10 out of 40 equals 25 %, so exactly one fourth. The other 75 % are split between the sets, and obviously the player with the highest set has the best EV, but since we don't know who that is, to us, it looks like everybody has a 25 % chance of winning. If we disregard the $16 in the pot and the rake, you'll break even if all opponents have unknown sets.

In the same way, you can construct a situation where you have exactly 25 % EV against two sets and one flush draw - that is, if the flush draw is K2 of hearts. But you can't get +EV.

Against one set and two flush draws, you're hopelessly behind. Heads-up against a set, you stand to win 22.78 % of the time, and against a third opponent having K2 of hearts, you have 23.81 %. That's as good as it gets, since every additional heart belonging to your opponents gives you fewer outs.

If all opponents have flush draws, you can create situations in which you're waaaaay behind and ones where you have a massive lead, depending on whether or not your opponents have pairs.

So the answer is a trick question again, and I'll give out correct answers to anyone who said 0 or 1.

In three cases
This is worse, since you only have one card left and not two.

If all opponents have sets and we don't know who has which one, everybody has the same expected value from our point of view. Seven hearts (not nine since K and 2 will give somebody a flush or four of a kind) and three treys give you the nuts. We know that there are 40 cards left (52 - 4*2 [hole cards] - 4 [the board]) and 10 out of 40 equals 25 %, so exactly one fourth. The other 75 % are split between the sets, and obviously the player with the highest set has the best EV, but since we don't know who that is, to us, it looks like everybody has a 25 % chance of winning. If we disregard the $16 in the pot and the rake, you'll break even if all opponents have unknown sets.

In the same way, you can construct a situation where you have exactly 25 % EV against two sets and one flush draw - that is, if the flush draw is K2 of hearts. But you can't get +EV.

Against one set and two flush draws, you're hopelessly behind. Heads-up against a set, you stand to win 22.78 % of the time, and against a third opponent having K2 of hearts, you have 23.81 %. That's as good as it gets, since every additional heart belonging to your opponents gives you fewer outs.

If all opponents have flush draws, you can create situations in which you're waaaaay behind and ones where you have a massive lead, depending on whether or not your opponents have pairs.

So the answer is a trick question again, and I'll give out correct answers to anyone who said 0 or 1.

In all four cases
This is worse, since you only have one card left and not two.

If all opponents have sets and we don't know who has which one, everybody has the same expected value from our point of view. Seven hearts (not nine since K and 2 will give somebody a flush or four of a kind) and three treys give you the nuts. We know that there are 40 cards left (52 - 4*2 [hole cards] - 4 [the board]) and 10 out of 40 equals 25 %, so exactly one fourth. The other 75 % are split between the sets, and obviously the player with the highest set has the best EV, but since we don't know who that is, to us, it looks like everybody has a 25 % chance of winning. If we disregard the $16 in the pot and the rake, you'll break even if all opponents have unknown sets.

In the same way, you can construct a situation where you have exactly 25 % EV against two sets and one flush draw - that is, if the flush draw is K2 of hearts. But you can't get +EV.

Against one set and two flush draws, you're hopelessly behind. Heads-up against a set, you stand to win 22.78 % of the time, and against a third opponent having K2 of hearts, you have 23.81 %. That's as good as it gets, since every additional heart belonging to your opponents gives you fewer outs.

If all opponents have flush draws, you can create situations in which you're waaaaay behind and ones where you have a massive lead, depending on whether or not your opponents have pairs.

So the answer is a trick question again, and I'll give out correct answers to anyone who said 0 or 1.



Result:

In zero cases 7%
In one case 37%
In two cases 34%
In three cases 13%
In all four cases 9%

Totally 865 vote(s)