What are the odds of a at least a certain amount of aces being dealt depending on the number of players?

I'm interested to know what the odds are of at least a certain amount of aces being dealt anywhere, depending on the amount of players.

``````             Zero Aces | At least 1 Ace | At least 2 Aces | At least 3 Aces | 4 Aces
-------------------------------------------------------------------------------------------
10 Players |           |                |                 |                 |
9 Players |           |                |                 |                 |
8 Players |           |                |                 |                 |
7 Players |           |                |                 |                 |
6 Players |           |                |                 |                 |
5 Players |           |                |                 |                 |
4 Players |           |                |                 |                 |
3 Players |           |                |                 |                 |
2 Players |           |                |                 |                 |
``````

I'd like to expand on why I'm interested in knowing this. I'm trying to explain to friends of mine who are starting to play poker that the likelihood of facing an Ace is something to take into account, but is dependent on how many players are on the table.

I was told that when playing ten handed, the likelihood of facing at least one Ace is quite high. (90% or so) So, I'm interested in the exact numbers so I have something to talk about.

• Interested to know why this is important to you? Single ace in two hands is way different than two aces in one hand. If you have a pair of kings and there is an ace on the board you are beat by any ace - does it matter if three hands can beat you. Commented Jan 20, 2016 at 17:40
• Curiosity. How often do we have to deal with others having an Ace. Say I have KK, when in a ten handed game, what are the odds of someone having an Ace? Commented Jan 20, 2016 at 17:42
• Then why are you asking for the chances of 2,3,4 aces being out? Commented Jan 20, 2016 at 17:44
• Same story more or less. When holding Ax, what are the odds I'm facing someone with Ax as well. Commented Jan 20, 2016 at 17:49
• There's 52 cards, 4 of them are aces. It's approximately a 8% chance that any specific card is an ace. What more do you really care about?
– mah
Commented Jan 20, 2016 at 19:51

These are the chances (assuming you have no ace):

These are valid only preflop assuming that there are 50 cards left in the deck (you are holding 2) and you are one of the players (2 players = 1 opponent with 2 cards; 3 players = 2 opponents with 4 cards and so on)

• I don't think this is quite right...I'd be interested in seeing how you did it. For the 2players-0aces, this should simply be (46/50)(45/49)(44/48)(43/47) = .709 which doesn't match your .828. Also, on the 10 players line, I see that you have the odds of 4 aces being dealt as higher than the odds of zero aces being dealt, but just intuitively it seems like that inequality should be go the opposite way since less than half the deck is being dealt out. Commented Jan 29, 2016 at 17:43
• I finally tried out doing this whole table for myself and mine matches your exactly in the 0-aces, at-least-1-ace, and 4-aces columns but does not for the other two columns, especially as the number of players goes up. The thing that looks fishy now is that for the 10-players row, it's implied that there's only a 9% chance of there being exactly one ace, but 58% chance of there being exactly two? FYI, I tried it in Excel with the HYPEGEOMDIST function. Commented Jan 29, 2016 at 19:07
• Third time's the charm :) Commented Jan 29, 2016 at 21:20
• Well done, sir. Commented Jan 29, 2016 at 22:08
• Can you show your calcs? Commented Jan 30, 2016 at 20:10

You can't go by strictly odds, because the odds are good that an ace was not folded preflop.

My assumption, on a full table (8-10) is there are 2 aces in the pockets; I'll confirm more exactly, & who has them once we start doing post-flop betting.

I read this on another site & found it valid/useful, for the raw odds for 2 of you having aces(relevant to assessing your kicker, if you're one of them)

Depends on how many people are at the table originally, I think one of the things that doesn't get communicated properly is the filtering effect.

In micro, alot of people play an A no matter what its paired with if you want to assume that is true in a 9 way then your really asking what is the probability that any of my other 8 opponents were dealt an ace preflop even if there is only one left post flop. This is easiest calculated as the probability that none of the 8 people hit an ace. To do this we just imagine dealing out 16 cards out of a 47 card deck with 2 aces left in it.

The probability of not dealing an ace on the first card is 45/47. The probability of not dealing an ace on the second card is 44/46. On down to 30/32.

Multiplying all these together gives a bunch of fractiosn most of the terms can be canceled and you get 31*30 / (47*46) = .43.

So the probability one of the other 16 cards is an ace given what you flopped is .57. And you have to figure they would limp in wiht it at 1bb or if they are loose play it at 3bb at a low stakes table. So even if you are down to one person you have to put him on it 57% of the time.

Now your not supposed to play an ace with a weak kicker on a big table, and incidentally this is exactly why. 57% of the time some one else has it, but if your kicker is weak more than not they got you dominated.

Don't fall in love with AT or A9, because the odds are good that another retained ace hasa better kicker; now, if you flop top pair, though, you're in good shape for thefirst post-flop betting round, but you've got a less than 5% chance of doing that -- of course, it's called gambling for a reason -- I tend to play a 9 or higher kicker to my ace, especially if suited(only less than a 1% chance of flopping a flush, but, hey, if I hit, or even get a 4-flush on the flop, I'm happy to push the pot up for a later payoff(look up "implied odds" & "pot odds")

I don't know why no one has answered this for you yet. It's a reasonable thing to want to know, but as others may have mentioned, should be more for novelty than for using to make your in-game decisions.

Anyways, here's a breakdown I put together. A lot of figures are fractions of percents, so I simplified it by rounding everything up to the nearest whole percent.

The likelihood of an Ace is always the same, what changes is the number of cards in play based on the number of players. The more players, the more cards dealt, meaning the more likely there is someone has at least one Ace. I made two charts; the first one is for 2-10 players in general, and the second is an adjusted chart which assumes that you are one of the players, and you have two non-Ace cards. This means there are effectively only 50 cards in the deck, instead of 52. These figures are, of course, based on pre-flop distribution.

The likelihood changes based on how many Aces are in play. 1st Ace = 1/13 (4/52) 2nd Ace = 1/17 (3/51) 3rd Ace = 1/25 (2/50) 4th Ace = 1/49 (1/49)

Hopefully this is what you were looking for!

• I think your math is off. #1 you can't have more than 100% chance when it comes to probability. And #2 there is no way 8 out 100 hands 4 aces are dealt. It's more like 8 out of 100 hands in a hand with 2 players one of the 4 cards would be an ace. Commented Jan 27, 2016 at 15:55
• It's possible I have done this wrong. I agree about not going over 100%, I should have double-checked that. This table is based on an auto-formula I made in Excel, calculating cards dealt/single-card probability. So if the odds of a single Ace is 1/13 (4 out of 52 cards), then with 7 players (14 cards dealt), I put 14/13. It does seem strange to me as well though, that these percentages are so high. Hmm, I wonder what I did wrong, then. Commented Jan 27, 2016 at 17:07
• Good attempt, I like the format and all but your approach is off. If you're not familiar with combinatorics, I'd suggest learning about it as it's how I would go about this. For example, the case of 2 aces being dealt to 4 players would amount to considering the total number of 8 card possibilities that can be dealt from a deck [52 choose 8] and divide into that the product of [48 choose 6] --(i.e. number of ways that 6 non-aces can be chosen)--and [4 choose 2] --(i.e. number of ways that the 2 aces can be chosen from the 4 total aces). The answer should be 9.8% Commented Jan 27, 2016 at 17:09
• One thing I should have mentioned is that when I calculated the odds of subsequent Aces (2nd, 3rd, 4th), it is supposed to be the odds of that Ace coming assuming the other Ace(s) are already dealt. So, for the 2-Players 2nd Ace (24%), this % is the remaining Aces/the remaining cards (not including the first Ace). If there are 52 cards, 4 of which are Aces, the odds for the first Ace are 4/52, which becomes 1/13 or 7.69% (which I rounded up to 8%). After that first Ace has been dealt, the odds for a second now become 3/51, which becomes 1/17 or 5.88% (which I rounded up to 6%). And so on. Commented Jan 27, 2016 at 17:09
• Thanks for the feedback, guys. I wasn't 154% sure my math was correct, but now I'm getting at least some direction on where I went wrong. Thanks also, Dr. Drfbaglll, I am looking more into combinatorics. :) Commented Jan 27, 2016 at 18:06