Maximum number of outs on the river in holdem.

October 11, 2011 12:27 PM

What is the maximum number of outs possible on the river in heads up Texas hold em? Assuming that out means a card which will take the player who is behind either level or ahead.
posted by therubettes to Sports, Hobbies, & Recreation (20 answers total)

I think that totally depends on the cards the player (who is behind) is holding.
posted by twinA at 12:30 PM on October 11, 2011

I'm doing this in my head, but 26? Assume your opponent has, say, pocket deuces ( 22, neither of which are your suit ), and you have 34 suited, and the board is 566 (8) with the 5 & 6 of your suit. Your outs are:

9 flush cards
6 straight cards (two of your straight outs are also flush outs)
3 threes
3 fours
3 fives
2 eights

Note that one of his deuce outs, above, actually makes you a straight flush and therefore isn't an out for him.
posted by mosk at 12:37 PM on October 11, 2011

as well as the cards on the flop and the turn...
posted by twinA at 12:38 PM on October 11, 2011

Oops - 27 outs; that should be 3 eights.
posted by mosk at 12:38 PM on October 11, 2011

twinA: Yes, that's exactly the question the OP is asking. Phrased more precisely: what set of cards (including your two cards, your opponent's two cards, and the four cards on the table) gives you the maximum number of outs on the river, and what is that number of outs?
posted by dfan at 12:45 PM on October 11, 2011

Best answer: if you have 3-4 and opponent has 4-5 and flop and turn come out 4-4-2-K, you are technically "behind" as of the turn but every card except for the three remaining 5's results in either a tie or you winning...
posted by lulz at 12:53 PM on October 11, 2011 [1 favorite]

Response by poster: dfan you phrased the question far better than I did thanks.
posted by therubettes at 12:54 PM on October 11, 2011

Best answer: change the K to a 6 and make your 3-4 suited and you can have one of the opponent's 5's result in a straight flush, reducing the number of non-outs to 2
posted by lulz at 12:57 PM on October 11, 2011 [2 favorites]

Response by poster: Bravo guys/gals.
posted by therubettes at 1:18 PM on October 11, 2011

Best answer: Well, it is possible to construct a hand where one guy is ahead, but drawing dead on the turn (he has no chance to win the hand outright whatsoever). For instance, current leader has 72, hero has 65, board is 4433. 2s and 7s make a straight, 6s and 5s make a better two pair and everything else is a chop.

In fact there are a number of these. One that isn't immediately obvious is 32 against 22 on an AAKK board. One out to the win, everything else is a chop. Or 43 against 42 on a KK42 board, two outs to win, everything else to chop.
posted by Lame_username at 2:26 PM on October 11, 2011

Also, if we had more poker questions, I'd be a MeFi superstar.
posted by Lame_username at 2:35 PM on October 11, 2011 [1 favorite]

Response by poster: Now that I have your attention lame! What are the most possible outs to come from behind and win outright. Thanks.
posted by therubettes at 2:42 PM on October 11, 2011

What are the most possible outs to come from behind and win outright. Thanks.

2 for 1? Very tricky of you. 25, I believe. For example 45dd vs 22 on a 6679dd board. I'm buzzed, so perhaps I'm missing something, but I think that's right. mosk's answer double counted a few outs.
posted by Lame_username at 4:32 PM on October 11, 2011

I'm doing this in my head, but 26? Assume your opponent has, say, pocket deuces ( 22, neither of which are your suit ), and you have 34 suited, and the board is 566 (8) with the 5 & 6 of your suit. Your outs are:

9 flush cards
6 straight cards (two of your straight outs are also flush outs)
3 threes
3 fours
3 fives
2 eights

Note that one of his deuce outs, above, actually makes you a straight flush and therefore isn't an out for him.

The number of straight cards isn't correct -- you're double counting.

1 deuce
3 treys
3 fours
3 fives (chop, meaning a tie)
3 eights (chop)
4 sevens
6 other flush cards (ace, king, queen, jack, ten, nine)

Total of 23, in this case, with 6 of those only giving you a tie.
posted by UrineSoakedRube at 4:35 PM on October 11, 2011

Whoops -- one of the eights gives a flush. So it's still 23, with 5 of those giving you a tie.
posted by UrineSoakedRube at 4:41 PM on October 11, 2011

Response by poster: Is there anyway of proving an answer to my second question about outs to win outright from behind?
posted by therubettes at 11:29 PM on October 11, 2011

Is there anyway of proving an answer to my second question about outs to win outright from behind?

I still like my answer of 25 above. You hold 54dd and opponent holds 22cs. Board is 6d6h7d9s

4 treys
3 fours
3 fives
3 sevens
4 eights
3 nines
5 remaining flush outs (T/J/Q/K/A diamonds)

That's 25 outs, making you slightly better than 50/50 to win despite being behind (51% to win)
posted by Lame_username at 1:38 AM on October 12, 2011

Ugh. My answer is flawed because I needed to give you a kicker higher than the board to avoid making some of those hands chop. Change my scenario to 98dd and make the board 4s6d6h7d and my count of 25 outs and enumerated list becomes correct. *sigh* It would also make you 57% to win when you have 25 clean outs.
posted by Lame_username at 1:42 AM on October 12, 2011

Clearly, I am not destined to be a MeFi superstar even in poker questions.
posted by Lame_username at 1:43 AM on October 12, 2011

Response by poster: Thanks a lot Lame and everyone who answered....
posted by therubettes at 1:47 AM on October 12, 2011