To calculate the probability of a random selected poker hand containing exactly 3 aces given that it contains at least 2 aces, we first need to determine the total number of ways to choose a poker hand with at least 2 aces. This can be done by considering the different combinations of choosing 2, 3, or 4 aces from the 4 available in a standard deck of 52 cards. Once we have the total number of ways to choose at least 2 aces, we then calculate the number of ways to choose exactly 3 aces from the selected hand. Finally, we divide the number of ways to choose exactly 3 aces by the total number of ways to choose at least 2 aces to obtain the probability.
Poker hands are combinations of cards (when the order does not matter, but each object can be chosen only once.)The number 52C5 of combinations of 52 cards taken 5 at a time is (52x51x50x49x48) / (5x4x3x2x1) = 2,598,960.The number of hands which contain 4 aces is 48 (the fifth card can be any of 48 other cards.)So there is 1 chance in (2,598,960 / 48) = 54,145 of being dealt 4 aces in a 5 card hand.The odds are 54,144 to 1 against. The probabilityis 1/54145 = (approx.) 0.000018469 or 0.0018469%.
The odds of any card pulled from an ordinary deck of 52 cards being an Ace is 4 in 52 (4 aces in a deck of 52). This can be reduced to a 1 in 13 chance of any random card pulled from the deck being an Ace (or any other specific value, for that matter). That 13th last card dealt in a hand is no different than picking a random card out of the pack, regardless of what cards you deal before (face down or blindfolded or even face up, it doesn't matter). A more interesting question would be "what would the probability be of ANY of those 13 cards being an Ace?" Any takers?
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 answer will depend on the exact situation.If you are dealt a single card, the probability of that single card not being a queen is 12/13 - assuming you have no knowledge about the other cards.Here is another example. If you already hold three queens in your hand (and no other cards have been dealt), the probability of the next card being dealt being a queen is 1/49, so the probability of NOT getting a queen is 48/49 - higher than in the previous example.
suited aces
Approximately 2%
In Texas Hold'em, the hand with the higher pair wins. In this case, a pair of aces is higher than a pair of kings, queens, or twos. Therefore, the hand with a pair of aces and a pair of twos would win over a hand with a pair of kings and a pair of queens.
To calculate the probability of a random selected poker hand containing exactly 3 aces given that it contains at least 2 aces, we first need to determine the total number of ways to choose a poker hand with at least 2 aces. This can be done by considering the different combinations of choosing 2, 3, or 4 aces from the 4 available in a standard deck of 52 cards. Once we have the total number of ways to choose at least 2 aces, we then calculate the number of ways to choose exactly 3 aces from the selected hand. Finally, we divide the number of ways to choose exactly 3 aces by the total number of ways to choose at least 2 aces to obtain the probability.
Poker hands are combinations of cards (when the order does not matter, but each object can be chosen only once.)The number 52C5 of combinations of 52 cards taken 5 at a time is (52x51x50x49x48) / (5x4x3x2x1) = 2,598,960.The number of hands which contain 4 aces is 48 (the fifth card can be any of 48 other cards.)So there is 1 chance in (2,598,960 / 48) = 54,145 of being dealt 4 aces in a 5 card hand.The odds are 54,144 to 1 against. The probabilityis 1/54145 = (approx.) 0.000018469 or 0.0018469%.
Aces and eights
The hand Wild Bill Hickok was holding when he got shot; aces and eights of clubs and spades. The suit and value of the fifth card is uncertain. (Wikipedia has a picture of the hand with the fifth card face down.)
Suit (club) does not matter unless you have a total of 5 same suited cards between your hand and the table. If you only have 2 clubs, you do not have a flush. Therefore, a pair of Aces does not beat a flush.
Aces over Eights; Full Hosue Wild Bill was holding a Full House, Aces and Eights. As a point of trivia, this is now referred to as the Dead Man's Hand. the hand was two pair, aces and eights
4/52 x 3/51 x 2/50 x 1/49 About 0.00039%
It is called the Deadman's Hand. Aces and 8s
It depends on the type of poker game being played.In a game where aces are low (they are equivalent to the number 1). In that case yes, a pair of fives beats a pair of aces.In a game where aces are high, then no, the pair of aces definitely wins. A pair of aces is the highest single pair you can get in the game of poker, before getting two pair or higher.Both of these types of games are played in poker.