3

What are the odds of another player at an 8 or 10 player table having an ace when you're dealt pocket aces?

  • Why downvote this question? It seems fine to me – TmKVU Feb 5 '16 at 10:34
3

Frisbee's answer is correct. But it only answers for eight players (although the other numbers are hidden in that table). Since the OP asked for 8-10, I will add the answers for nine and ten players.

The odds of an ace not occurring in the next x cards dealt (after you're dealt two aces) is ( 48!/(48-x)! )/( 50!/(50-x)! ). That's the probability that no one else has an ace, so the probability that someone else has an ace is the complement of that. We can easily check the probabilities using this lookup table.

The number of cards left to be dealt is 2*(p - 1) where p is the number of players, so for 8, 9, and 10 players, the number of cards left is 14, 16, and 18 respectively.

The lookup table gives us the probabilities 0.458714, 0.542041 and 0.595102 respectively. For completeness, the probability of at least one opponent having at least one ace, when the table has p players is:

 p | P( Another player has an ace | We have pocket aces)
-------------------------------------------------------
 2 |  7.91837%
 3 | 15.5102%
 4 | 22.7755%
 5 | 29.7143%
 6 | 36.3265%
 7 | 42.6122%
 8 | 48.5712%
 9 | 54.2041%
10 | 59.5102%
  • 10 is kind of hidden in my answer 31 / 33 * = 0.5951 – paparazzo Feb 7 '16 at 6:39
  • Help push for getting mathjax on this site meta.poker.stackexchange.com/questions/137/… – paparazzo Feb 7 '16 at 6:44
  • @Frisbee 9 players is in there too: 33 / 35 * = 0.5420. :) I upvoted your answer, and your post about mathjax. – Paul Feb 7 '16 at 7:17
  • And I up voted your answer – paparazzo Feb 7 '16 at 7:17
2

Pick is it 8 or 10

At 8 I think it is 0.4114
Or 1.4306 : 1
This is for exactly one ace out - (not two)

Using combination

(2/1) * (48/13) / (50/14)

(2 aces need 1) (48 non ace need 13) / (50 cards need 14)

Both other aces out would be 0.0743

Add them for 1 or 2 aces out = 0.4857

You can get that same number with

1 -  
48  /   50  *    =  0.0400
47  /   49  *    =  0.0792
46  /   48  *    =  0.1176
45  /   47  *    =  0.1551
44  /   46  *    =  0.1918
43  /   45  *    =  0.2278
42  /   44  *    =  0.2629
41  /   43  *    =  0.2971
40  /   42  *    =  0.3306
39  /   41  *    =  0.3633
38  /   40  *    =  0.3951
37  /   39  *    =  0.4261
36  /   38  *    =  0.4563
35  /   37  *    =  0.4857
34  /   36  *    =  0.5143
33  /   35  *    =  0.5420
32  /   34  *    =  0.5690
31  /   33  *    =  0.5951
30  /   32  *    =  0.6204
29  /   31  *    =  0.6449
28  /   30  *    =  0.6686
27  /   29  *    =  0.6914
26  /   28  *    =  0.7135
25  /   27  *    =  0.7347
24  /   26  *    =  0.7551
23  /   25  *    =  0.7747
22  /   24  *    =  0.7935
21  /   23  *    =  0.8114
20  /   22  *    =  0.8286
19  /   21  *    =  0.8449
18  /   20  *    =  0.8604
17  /   19  *    =  0.8751
16  /   18  *    =  0.8890
15  /   17  *    =  0.9020
14  /   16  *    =  0.9143
13  /   15  *    =  0.9257
12  /   14  *    =  0.9363
11  /   13  *    =  0.9461
10  /   12  *    =  0.9551
9   /   11  *    =  0.9633
8   /   10  *    =  0.9706
7   /   9   *    =  0.9771
6   /   8   *    =  0.9829
5   /   7   *    =  0.9878
4   /   6   *    =  0.9918
3   /   5   *    =  0.9951
2   /   4   *    =  0.9976
1   /   3   *    =  0.9992
0   /   2        =  1.0000

See this approaches one at 48 cards as it should

The answer from TmKVU of 1 - (48/50)^14 is wrong in my opinion
At 48 cards 1-(48/50)^48 does not approach zero - it is 85.9%

we need mathjax on this site

  • There's no way it's that high. I believe .56 or .54 are the odds of someone having a pocket pair, yes. But another ace? – Jack Bauer Feb 5 '16 at 2:12
  • 1
    Then don't believe it. And I will not bother you with any more answers. There are still 2 aces and 14 cards out from 50. What is 2*14 / 50 ashat? That is not how the calc is done but clearly it is close to that. – paparazzo Feb 5 '16 at 2:18
  • @JackBauer I did not guess. I used combinatmetircs. Your answer is wrong. – paparazzo Feb 5 '16 at 10:37
  • @Frisbee how did you get 0.56? – TmKVU Feb 5 '16 at 10:43
  • Ok as a suggestion you should include your Maths, however there is no need to downvote this at all. This seems fairly correct. You're not talking about the probability of heads-up, you're talking about 9 other players having an ace. It is extremely likely. Each player gets 2 cards, after everyone has been dealt their cards there are only 30 cards left. You are calculating the combined probability. Just because you feel it is 'to high' doesn't mean it's not true. – Grinch91 Feb 5 '16 at 10:55
-4

Jack is definitely right, Frisbee, thus you're very wrong. The odds of getting dealt pocket aces are 1/221 (4/52*3/51). Odds that another ace has been dealt out preflop is 2/50, so roughly 4%.

  • 2
    2/50 is the chance of the next card being and ace. This is an 8 or 10 player table. – paparazzo Feb 5 '16 at 10:08
  • @John, yes in headsup play you'd be absolutely correct. The question specifically mentioned 8 or 10 player table. You're not just doing 2/50 as a single instance. You need to factor in all the players. – Grinch91 Feb 5 '16 at 10:58
  • 1
    @Grinch91 It's not even correct for heads-up unless you're opponent only gets one card. – Paul Feb 8 '16 at 23:33

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