I think that may be the difference - some of the full houses in that 2.6% will not use both hole cards. Maybe I'm not such a math dummy after all!

Maybe you should write math problems for a living.


(That's my MathMan Signal for Jim to defend his work.)

I had overlooked that condition but can't take the time right now to work on the solution.

From http://www.math.cornell.edu/~mec/200...holdemsol.htm:

To count the number of full house hands, we divide up the types of full houses by looking at the two cards that are not used as part of the final hand. These two cards can either be a pair (but of a different rank than the triple or the pair you are using for the full house, or else you would have four of a kind), one of the two cards could be of the same rank as your pair (giving you two triples and one card of some different rank), or the two cards could be of different ranks from each other, the triple, and the pair.

We first consider the case of the unused cards being a pair. We can choose the rank for the triple in 13 ways. Once a rank is chosen we can pick the three cards for the triple in C(4,3) = 4 ways. We can then choose the two ranks for the two pairs in C(12,2) = 66 ways. For each pair, once we have chosen the rank we can choose the cards for the pair in C(4,2) = 6 ways. So we have a total of 13 · 4 · 66 · 62 = 123,552 full house hands of this type.
Now we consider the case of two triples. We can choose the ranks for the triples in C(13,2) = 78 ways, and for each triple we can then choose the cards for the triple in C(4,3) = 4 ways. There are then 44 remaining cards from which to choose the last card of the hand, so we have a total of 78 · 42 · 44 = 54,912 hands of this type.
Finally we consider the case of two cards of different rank from each other, the triple, and the pair. As above, the cards for the triple can be chosen in 13 · C(4,3) = 52 ways and the cards for the pair can then be chosen in 12 · C(4,2) = 72 ways. We can choose the two ranks for the remaining two cards in C(11,2) = 55 ways, and for each rank we can choose any of the four cards of that rank. This gives a total of 52 · 72 · 55 · 42 = 3,294,720 hands of this type.
Therefore, we have a total of 3,473,184 full house hands. This gives a frequency of (3,473,184/133,784,560) = 0.02696.

This is very true. V's best strategy would be to try to see every flop, and call down any FH+ draw. Though you could counter by raising really large pre-flop.

Yes this is true, the calculations were done without the criteria of using both hole cards.

This is another interesting aspect of this bet.

Boat Chaser has an EV of roughly $1300 over 90 hands minus whatever EV he loses in the game itself by playing like a monkey. Obviously if the other players know about the bet, they would be smart to play wider ranges more aggressively against him.

OP's game is 5/10; the stakes are significant relative to the bet. So if the other players in the game adjust well, BC's overall EV can fall or even go negative. At a minimum, it's going to cause a big short-term increase in variance for a normally tight player.

It would be pretty tough to take a line where you try to see every flop. The other players just need to open to 5xBB, and pot every street, and the Boat Chaser will go negative trying to chase all his draws.

If I were to take the 60:1 bet, and I would, I'd just play a slightly looser and more passive limpy pre-flop game than normal and I still think it would be overall profitable to take that bet. But it would depend on how aggressive the other players adjust to the increased action created by the bet.

Yeah, if you take the bet, you don't have to go nuts to win. You are gambling $100 to win $6000. If it happens, score! If not, you still made a great bet.

I would certainly try to see every flop, but you are right, this would be a great bet for the entire table!

Other players in the game should love this.

Players could also open fold to screw with the bet. Or play slow.

not worth the risk
casinos offer these type of odds at their table games
and even those hit

the difference is the casino can absorb the loss can you?

No, casinos never offer these odds. That is the entire point of the posts in this thread. Casinos always make mathematically profitable bets, giving you the sucker end.

I'm still not sure what even money odds on this bet should be, but I'd say something closer to 5:1.

Anyone else have a guess?

not sure if it can be quantified since you can cut the number of hands at least in half just by tanking every decision, and there will be different equilibrium for different odds

he might be capable to adjust his game to benefit from this side bet, eg calling pf raises in pos and then floating all dry flops making it very tough for us to continue since we know he must have at least a pair when he calls a cbet

with all that in consideration I won't give anyone better than 3 to 1

I think it's more in the ballpark of 4:1, but I can't do the math right now. The ground work is laid out in this thread if you want to do it.

keep in mind, we only had half and half left which means 25-30 hands at the most... So shouldn't it be in the 30:1 range?