# The most "outs" possible to win in Hold 'Em on the river **URL:** \ **Category:** Factual Questions\ **Created:** [November 24, 2005, 1:29pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581 "2005-11-24T13:29:29Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![John\_Stamos\_Left\_Ear](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/john_stamos_left_ear/32/3198_2.png) [@John\_Stamos\_Left\_Ear](https://boards.straightdope.com/u/John_Stamos_Left_Ear)\ **Post date:** [November 24, 2005, 1:29pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/1 "2005-11-24T13:29:29Z")
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](http://www.cardplayer.com/poker_odds/texas_holdem/index.php?stats=Tzo4OiJzdGRDbGFzcyI6OTp7czoxMToibnVtX3BsYXllcnMiO2k6MjtzOjk6InRoZV9ib2FyZCI7czoxMjoiN0QgVEQgVEggUUggIjtzOjM6ImNsaSI7czoyOToiOUQgOEQgN0MgMkMgIC0tIDdEIFREIFRIIFFIICAiO3M6MTQ6InBsYXllcl8xX2NhcmRzIjtzOjU6IjlEIDhEIjtzOjM6InBjdCI7YToyOntpOjE7czo1OiI1Mi4zJSI7aToyO3M6NToiNDcuNyUiO31zOjM6InRpZSI7YToyOntpOjE7czo0OiIwLjAwIjtpOjI7czo0OiIwLjAwIjt9czoxNDoicGxheWVyXzJfY2FyZHMiO3M6NToiN0MgMkMiO3M6OToibnVtYm9hcmRzIjtzOjI6IjQ0IjtzOjE1OiJwcm9jZXNzaW5nX3RpbWUiO3M6NToiMC4wMDQiO30=). 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? ---
**Author:** ![borschevsky](https://avatars.discourse-cdn.com/v4/letter/b/97f17d/32.png) [@borschevsky](https://boards.straightdope.com/u/borschevsky)\ **Post date:** [November 24, 2005, 3:13pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/2 "2005-11-24T15:13:33Z")
> [@John\_Stamos'\_Left\_Ear](#): > > How about if we include outs that will lead to avoiding a loss (i.e. a split pot), something I don’t consider above? Maybe something like this: Player A: 4,2 Player B: 3,2 Board: A,A,A,K Player B is behind, but the only cards he loses on are the three 4s left in the deck. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [November 24, 2005, 7:01pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/3 "2005-11-24T19:01:40Z")
> [@John\_Stamos'\_Left\_Ear](#): > > …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) But one remaining 6 is a diamond; ditto for the 8, 9, J and Q. So there are actually 18, not 23, cards that produce a win. ---
**Author:** ![Raygun99](https://avatars.discourse-cdn.com/v4/letter/r/6f9a4e/32.png) [@Raygun99](https://boards.straightdope.com/u/Raygun99)\ **Post date:** [November 24, 2005, 7:31pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/4 "2005-11-24T19:31:21Z")
> [@Xema](#): > > But one remaining 6 is a diamond; ditto for the 8, 9, J and Q. So there are actually 18, not 23, cards that produce a win. He’s excluded those that match in the count. There are four 6s still left, one is a diamond and already counted, so three more also help. ---
**Author:** ![Xema](https://avatars.discourse-cdn.com/v4/letter/x/9de053/32.png) [@Xema](https://boards.straightdope.com/u/Xema)\ **Post date:** [November 25, 2005, 3:56am UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/5 "2005-11-25T03:56:06Z")
> [@Raygun99](#): > > He’s excluded those that match in the count. Hmmm - good point. (How did I miss that?) ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [November 29, 2005, 12:40pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/6 "2005-11-29T12:40:01Z")
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. ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [November 29, 2005, 2:12pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/7 "2005-11-29T14:12:26Z")
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. ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [November 29, 2005, 3:16pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/8 "2005-11-29T15:16:12Z")
> [@Malacandra](#): > > This reminds me of the one time I held the only four-card hand in cribbage that no turn-up could possibly improve: Good one. I tried to puzzle it out but gave up quickly. Even then, if you cut a jack you still get the two. ---
**Author:** ![Malacandra](https://avatars.discourse-cdn.com/v4/letter/m/45deac/32.png) [@Malacandra](https://boards.straightdope.com/u/Malacandra)\ **Post date:** [November 29, 2005, 3:21pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/9 "2005-11-29T15:21:45Z")
> [@Bryan Ekers](#): > > Good one. I tried to puzzle it out but gave up quickly. > > Even then, if you cut a jack you still get the two. 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. ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [November 29, 2005, 3:40pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/10 "2005-11-29T15:40:57Z")
> [@Malacandra](#): > > 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! ---
**Author:** ![JSexton](https://avatars.discourse-cdn.com/v4/letter/j/96bed5/32.png) [@JSexton](https://boards.straightdope.com/u/JSexton)\ **Post date:** [November 29, 2005, 5:11pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/11 "2005-11-29T17:11:43Z")
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? (I hope I explained that right.) ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [November 29, 2005, 5:57pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/12 "2005-11-29T17:57:29Z")
> [@JSexton](#): > > What is the worst possible nut high hand by the river? Three Queens? (e.g. 2 3 7 8 Q showing with no flush possible) ---
**Author:** ![Otto](https://avatars.discourse-cdn.com/v4/letter/o/bbe5ce/32.png) [@Otto](https://boards.straightdope.com/u/Otto)\ **Post date:** [November 29, 2005, 5:59pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/13 "2005-11-29T17:59:40Z")
> [@JSexton](#): > > What is the worst possible nut high hand by the river? If I’m understanding the puzzle, I believe it would be 75432 for a 7-high. Or, if you actually have to beat someone and not split, 76432. ---
**Author:** ![Otto](https://avatars.discourse-cdn.com/v4/letter/o/bbe5ce/32.png) [@Otto](https://boards.straightdope.com/u/Otto)\ **Post date:** [November 29, 2005, 6:06pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/14 "2005-11-29T18:06:17Z")
Oh, never mind, I totally misunderstood the puzzle. Seven cards, not five. Pardon me while I smack myself in the head. If splits are possible, then 9-8-7-5-4-3-2 for a 9-high. If no splits then T-8-7-5-4-3-2 for a ten-high. And if that’s wrong I give up. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [November 29, 2005, 6:10pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/15 "2005-11-29T18:10:36Z")
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. ---
**Author:** ![Otto](https://avatars.discourse-cdn.com/v4/letter/o/bbe5ce/32.png) [@Otto](https://boards.straightdope.com/u/Otto)\ **Post date:** [November 29, 2005, 6:53pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/16 "2005-11-29T18:53:20Z")
> [@Usram](#): > > 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. 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. ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [November 29, 2005, 7:35pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/17 "2005-11-29T19:35:47Z")
> [@Usram](#): > > Three Queens? (e.g. 2 3 7 8 Q showing with no flush possible) I have to vote for this, too. ---
**Author:** ![JSexton](https://avatars.discourse-cdn.com/v4/letter/j/96bed5/32.png) [@JSexton](https://boards.straightdope.com/u/JSexton)\ **Post date:** [November 29, 2005, 7:50pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/18 "2005-11-29T19:50:53Z")
> [@Otto](#): > > 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. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [November 29, 2005, 7:54pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/19 "2005-11-29T19:54:08Z")
> [@Otto](#): > > 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. If the board is 87543 unsuited, the nuts is 56789. ---
**Author:** ![Bryan\_Ekers](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/bryan_ekers/32/183_2.png) [@Bryan\_Ekers](https://boards.straightdope.com/u/Bryan_Ekers)\ **Post date:** [November 29, 2005, 7:59pm UTC](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581/20 "2005-11-29T19:59:55Z")
As a quickie (and fairly obvious, I’m sure), the fastest nuts are; Holding AK suited and flopping the QJT. You’re bulletproof, baby. [Next page](https://boards.straightdope.com/t/the-most-outs-possible-to-win-in-hold-em-on-the-river/332581.md?page=2)