I am trying to figure out the most outs possible for a losing hand to win on the river of a Hold 'Em game. This is one scenario I came up with:
PLAYER A: 9d 8d
PLAYER B: 7c 2c
FLOP: 7d Td Th
TURN: Qh
Player A is behind now with a pair of tens to Player B’s two pairs (Tens and Sevens). However, with one card remaining, Player A will win with:
Any Diamond (9 Cards)
Any 6 (3 Cards)
Any J (3 Cards)
Any 9 (3 Cards)
Any 8 (3 Cards)
Any Q (2 Cards)
That means that Player A will win with 23 cards or about 52.3% of the time. Is there any other scenario that can create more than 23 outs? How about if we include outs that will lead to avoiding a loss (i.e. a split pot), something I don’t consider above?
I’ve given it a shot and I can’t find better than 52.3% for a player who is behind at the turn. The only trivial refinement I can offer is that it’s not important if the turn is a queen, king or ace; they all offer the same chance for a winning hand.
As a minor additional note, in the OP’s example, Player A has 66.7% odds after the flop, while this similar deal:
PLAYER A: 6d 5d
PLAYER B: 4c 2c
FLOP: 4d 7d 7h
…gets Player A up to 67.6%, though he still at that moment has the weaker hand.
This reminds me of the one time I held the only four-card hand in cribbage that no turn-up could possibly improve:
Four Aces. No turn-up can possibly make another pair, run, flush or fifteen. Any other hand is improvable, by making a fifteen with any four other than Aces, or a pair with anything but four cards of the same rank.
Yes, but as an independent award (“two for his heels”), not as part of the hand - you get the two points when the card’s cut, not when the hand’s scored.
Fair enough. I suppose the best possible outcome would be to be the dealer, cut the jack, and be in a four-player game when the other three end up with face cards and tens, so the plays goes “Ten” “Twenty” “Thirty” “Thirty-One” four times.
Twelve for double pair royal, eight for thirty-ones, two for heels… rock on!
Even better, have the first three cards add to twenty-seven (though this might put opponents up three or four points) and play four aces in a row for 2 + 6 + 12 + 2 = 22, plus the hand’s 12 and 2 for heels. Skunk 'em good!
Hey since, we’re posting trivial card game stuff, here’s a little poker exercise:
By the time the river card is dealt in standard Texas Hold’em, you can determine what the nut high hand is. If there’s three suited cards with five ranks of each other, then the nuts is the two remaining cards to the straight flush. If there’s a pair on the board, then the nuts is four of a kind, and so on.
What is the worst possible nut high hand by the river?
No, I think JSexton meant what is the worst 5-card hand that could possibly be the nuts after the river card has been dealt. A three-of-a-kind at least is bound to be possible, so I tried to find the lowest five cards that could not make a straight with any two other cards.
Right, which is why my first example doesn’t work. There’s no two-card combination that you can add to 76432 that doesn’t result in a higher hand. Any card higher than 7 and you improve to a better high card, any card 7 or lower and you improve to either a pair or a straight.
In my second example, if you’re holding, say, T2os and the board comes 87543 rainbow, then you can have the nuts, the highest possible hand, of T8754 assuming your opponent holds exactly 92os.
Nope, you’re misinterpreting my question. There’s no assumption about your opponent’s hand, it’s unknown. You need to construct a board where you hold the true nuts, regardless of your opponent’s cards. And that resulting hand needs to as low as you can get it. It’s difficult to explain the question properly. It took a few tries with my poker group to communicate it right.
Usram is correct. Well done. I award you two JSexton points.