What are the odds of being dealt 2 aces in a game of poker?

Answer 1

Its 1:220

(4/52)(3/51) = 1/221 which is the probability
Thus the odds are 1/220 or 1:220

Answer 2

A poker hand is usually 5 cards, so are you asking for the odds of exactly 2 aces in 5 cards, or at least 2 aces in 5 cards, or 2 aces in the first 2 cards dealt?

Everyone seems to be answering the last question, but not getting that correct.

There are 4C2 ways (4!/2!(4-2)!) = 4 x 3 / 2 = 6 ways to pick the suits for two aces.
There are 52C2 ways (52!/2!(52-2)!) = 52 x 51 / 2 = 1326 ways to pick any two cards.

So the chance you pick two aces out of two cards is 6 / 1326 or 1 / 221.

Answer 3

P(exactly 2 aces in 5 cards) = P(AAXXX or AXAXX or AXXAX or ... or XXXAA) (These are the possible combinations of two aces and three other cards; you have to take order into account.)

= P(AAXXX) + P(AXAXX) + ... + P(XXXAA) (These different orders are independent.

P(AAXXX) = (4/52)(3/51)(48/50)(47/49)*
(46/48)

You will find that all combinations are equally likely; we just need to find how many combinations are possible and multiply it by P(AAXXX).

The number of combinations is 5C2 = 5!/2!3! = 10

So P(exactly 2 aces) = 10*P(AAXXX)
= 10*(4/52)(3/51)(48/50)(47/49)*
(46/48)
= 0.04 or 4%.

You can also use the hypergeometric distribution to get this probability.

P(exactly 2 aces) = (4C2)(48C3)/(52C5)
= 6*17296/2598960
= 0.04

So the odds of being dealt 2 aces is 0.04/(1-0.04) = 0.04167. Remember that odds is different from probability.

Answer 4

Poker Independent, The (London), December, 2004 by Nic Szeremeta A PLAYER can expect to be dealt a pair of aces face down once every 220 hands. This applies to both hold'em and seven-card-stud. In playing time, this represents about once every three-and-a- half hours in an online game, and double that in a real live casino card room.

Answer 5

As told to me my a friend:

"Seems like it would be 4/52 times 3/51 no matter what the game. That would be the odds of getting, in two cards, two cards of the same rank in a game with a normal, 52-card deck and no wild cards.

From the numerators of the fractions, subtract 1 for each ace you can see outside your hole cards from the numerators of the fractions."

Answer 6

In a 52-card deck, the probability is:

(4/52 * 3/51 * 48/50 * 47/49 * 46/48) * 10 for EXACTLY 2 aces, and (4/52 * 3/51) for AT LEAST 2 aces (or at least i think so...)

Answer 7

It is a combination function used in probability and stats. You have to find the probability of getting 1 card (the first ace) out of the entire 52. Then the probability of getting 1 card out of 51 (since the first ace is gone). That is your numerator. Then you divide that by the total possible number of poker hands which is 2,598,960 or COMBIN(52,1). That gives you .00102 probability or a .1%chance, or 1in 1000.

Answer 8

4 in 52

Answer 9

It depends on what type of poker you're playing.

The probability of being dealt two aces in a five-card hand is...
C(4,2) * C(48,3) / C(52,5)
...or approximately 3.993%.

The probability of being dealt two aces in a seven-card hand is...
C(4,2) * C(48,5) / C(52,7)
...or approximately 7.679%.

The C(4,2) is the combinations of 2 aces being dealt to you out of 4.
The C(48,3) and C(48,5) are the combinations of the remaining cards dealt to your hand.
The C(52,5) and C(52,7) are the combinations of the total possible ways the cards could be dealt.

C(n,r), the combination of n things taken r at a time is equal to:
n! / [r! * (n - r)!]

Answer 10

combinations of 4 taken by 2/combinations of 52 taken by 2
or
6/(26*51)=1/(13*17)=1/221