The probability, if the cards are dealt often enough, is 1.
On a single deal, the prob is 3.69379*10^-6
The odds are 220:1 of being dealt pocket aces.
Counting Aces as a face card, the answer is 0.0241 If Aces are not considered face cards, then the answer is 0.0181
The probability is 4/52 for the first ace and 3/51 for the second. So the probability of 2 aces is 4/52 x 3/51 = 1/221
Probability = Chance of Success / Total Chances (Chance of Success + Chance of Failure) There are 4 aces in a 52 card deck and 48 cards that are not aces. Probability of being dealt an ace = 4 / (4 + 48) = 4/52 = .0769 or about 7.7 percent
The probability of getting 3 aces in the order AAABB is; P(AAABB) = (4/52)∙(3/51)∙(2/50)∙(48/49)∙(47/48) = 0.0001736... There are 5C3 = 5!/(3!∙(5-3)!) = 10 different ways in which the aces can come out. So the probability of getting exactly three aces in a five card poker hand dealt from a 52 card deck is, P(3A) ~ 10∙(0.0001736) ~ 0.001736 ~ 0.1736%
Since there are only four aces in a standard 52 card deck, the probability of being dealt five aces is zero.
The odds are 220:1 of being dealt pocket aces.
The probability of being dealt pocket aces in a game of poker is approximately 1 in 221 hands.
Counting Aces as a face card, the answer is 0.0241 If Aces are not considered face cards, then the answer is 0.0181
The probability is 4/52 for the first ace and 3/51 for the second. So the probability of 2 aces is 4/52 x 3/51 = 1/221
Probability = Chance of Success / Total Chances (Chance of Success + Chance of Failure) There are 4 aces in a 52 card deck and 48 cards that are not aces. Probability of being dealt an ace = 4 / (4 + 48) = 4/52 = .0769 or about 7.7 percent
Aces and 9s are disjoint events, so the probability of either is the sum of the probabilities of each. P(A or 9) = P(A) + P(9) = 1/13 + 1/13 = 2/13
To find the probability of being dealt exactly 4 aces in a 13-card hand from a standard 52-card deck, we can use the hypergeometric distribution. The total number of ways to choose 4 aces from 4 available is ( \binom{4}{4} = 1 ), and the number of ways to choose the remaining 9 cards from the 48 non-aces is ( \binom{48}{9} ). The total number of ways to choose any 13 cards from 52 is ( \binom{52}{13} ). Thus, the probability is given by ( \frac{1 \times \binom{48}{9}}{\binom{52}{13}} ).
The odds of being dealt pocket aces in a game of Texas Hold'em poker are approximately 1 in 221, or about 0.45.
If dealt from a randomly shuffled pack it is 0.0399, approx.If dealt from a randomly shuffled pack it is 0.0399, approx.If dealt from a randomly shuffled pack it is 0.0399, approx.If dealt from a randomly shuffled pack it is 0.0399, approx.
The probability of getting 3 aces in the order AAABB is; P(AAABB) = (4/52)∙(3/51)∙(2/50)∙(48/49)∙(47/48) = 0.0001736... There are 5C3 = 5!/(3!∙(5-3)!) = 10 different ways in which the aces can come out. So the probability of getting exactly three aces in a five card poker hand dealt from a 52 card deck is, P(3A) ~ 10∙(0.0001736) ~ 0.001736 ~ 0.1736%
The odds of being dealt pocket aces in Texas hold 'em is (4 in 52) times (3 in 51) or 12 in 2652 which, reduced, is 1 in 221.