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December 21

Texas Hold 'em question...

I was wondering if there is any good way to calculated the following in Texas Hold 'em. What percentage of the possible 5 community cards will have a pair of aces as hole cards beat *all* other possible hold cards (that aren't a pair of aces?). An example of that type of set of community cards would be AS JD 8C 6H 3S. No straight or flush can be made.Naraht (talk) 12:14, 21 December 2017 (UTC)[reply]

In Texas hold 'em, "Each player seeks the best five card poker hand from any combination of the seven cards of the five community cards and their own two hole cards."

I see you updated your example to include an ace in the community cards, which is good since your previous example allowed your aces to be beat by a three of a kind. Additionally, community cards which give you an ace-high flush also guarantee your win. Those giving you an ace high straight, however, could leave you tied with the holder of a single ace. -- ToE 13:36, 21 December 2017 (UTC)[reply]

I believe a complete rule would be that you are guaranteed a win if the community cards:

include no aces but give you an ace-high flush,
or include the other two aces but exclude the possibility of a straight flush,
or include exactly one ace, don't contain any pair, exclude the possibility of a flush (i.e., has no three cards of any one suit), and exclude the possibility of a straight (~~i.e., its cards span a range of at least 8, i.e., high - low >= 7~~).

Does that sound complete? -- ToE 14:38, 21 December 2017 (UTC)[reply]

It looks complete, but I think your last "i.e." is erroneous - the community cards A, 2, 3, 4, 10 span a range of at least 8, but allow for a straight if anyone holds a 5. Excluding a straight requires that any 3 cards are not within 5 of each other, the range of the totality of the community cards is not limiting. MChesterMC (talk) 15:07, 21 December 2017 (UTC)[reply]

Ah, yes! I must have been building a seven card straight. Oops! -- ToE 15:18, 21 December 2017 (UTC)[reply]

I missed the case where where the board shows an ace-high straight that you can convert to a straight flush. -- ToE 15:18, 21 December 2017 (UTC)[reply]

(ec) A few more mistakes. For the first bullet point, the hand must include no aces and no pairs and give you a flush, or must include no aces and give you an ace high straight flush, as otherwise your flush can be beaten by a full house (e.g. 2S 2H 3S 7S 8S will cause you to lose to someone with 2D 7H if you hold AS AH) or by an 6 high straight flush (where you have an 5-high straight flush you will lose to someone with a 6 in that suit).

For the third bullet point, hands which give you a flush will win, as that will be the only possible ace high flush. e.g. if you hold AS AH, then KH QH JH 10H AD is a winning set of community cards for you and you alone.

The second bullet point looks complete, as you can't get a straight flush yourself with the other two aces out. MChesterMC (talk) 15:32, 21 December 2017 (UTC)[reply]

Wait, no, for the third bullet that should be "hands which give you a flush which is not a 5-high straight flush will win", for the same reason as in the second bullet. Also noting that the original question excludes another person holding the other two aces, a set of cards with no aces which gives you an ace high straight and which forbids a flush would also win. MChesterMC (talk) 15:43, 21 December 2017 (UTC)[reply]

I fold! (Until late tonight, at least.) -- ToE 16:01, 21 December 2017 (UTC)[reply]

Original Poster clarification

For this purpose, the win should be equally true for *all* pairs of aces in the hole that are still available. So in the discussion of the third bullet point. AS AH may be the winning pairing if the community cards are KH QH JH TH AD, but not if the hold cards are AS AC. This completely knocks out bullet point 1 as well.Naraht (talk) 15:56, 21 December 2017 (UTC)[reply]

So, any AA must take the whole pot against any non-AA hand (such as AK, A2, 83...)? In that case, based on this article that used to exist, I'd guess around 4-5%. This is only possible if either three-of-a-kind or four-of-a-kind are nuts and there is an ace on the board. In the first case 1.8% of all boards satisfy the condition, and as the lowest such board is Q-high (any 5 cards jack or lower have a straight possible), probably around 1/3 are A-high (K-high probably makes more combinations as with ace you can only have one card 2-5). In the second case 45% of boards have a four-of-a-kind as nuts, which is the pair on the board. AA will then take 1/13 of single-paired boards, 1/6 of double-paired ones and 1/13 of full house boards (e.g. AA888), but as the latter two are less frequent, the number is probably a bit above 1/13 of 45%, which is ~3.5%. 78.1.153.199 (talk) 04:05, 22 December 2017 (UTC)[reply]

So the only hands that AA will always win (not sure what nuts means)

no pair boards with no straight possible (which has to be ace high since otherwise a three of a kind could be made with any of them) and no flush possible,
AAXYZ no flush possible,
AAXXY no flush possible,
AAXXX

Naraht (talk) 13:31, 22 December 2017 (UTC)[reply]

Yes, except that flushes or straight can be possible in 2nd and 3rd case, but no straight flushes should be possible - for example AcAsKcQd8c qualifies, but AcAsKcKdQc doesn't. The nuts are the best hand that can be formed with the community cards given, i.e. AA in the situations you're looking for. 78.1.153.199 (talk) 22:54, 22 December 2017 (UTC)[reply]

Naraht, also note that these are figures for a completely random draw of five cards for the board. When you're holding AA yourself, you reduce the chances greatly because there are only two aces left in the deck. You'll have 2x smaller chance of hitting a three-of-a-kind and a 6x smaller chance of hitting a four-of-a-kind, so the probability will only be around 1% (if my previous guess is correct). 78.1.153.199 (talk) 20:14, 23 December 2017 (UTC) — Preceding unsigned comment added by 78.1.151.149 (talk) [reply]

Thanx. Turns out to be a much more difficult question than I had first thought.Naraht (talk) 15:20, 25 December 2017 (UTC)[reply]

Probably the best way to get an answer is to simulate it on a computer for a large number of hands. Bubba73 ^{You talkin' to me?} 23:17, 25 December 2017 (UTC)[reply]