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The probability of being dealt 4 aces in a 5 card poker hand?
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%.
What is the probability that a random selected poker hand contains exactly 3 aces given that it contains at least 2 aces?
I will assume that you mean a five card poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408
A deck of ordinary cards is shuffled and 13 cards are dealt. What is the probability that the last card dealt is an ace?
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?
What is the probability of getting exactly three aces in a five-card poker hand dealt from an ordinary 52 card deck?
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%
What is the probability of not being dealt a queen in a 52 card deck?
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.
When you play poker cards game and you have straight Aces in your hand what that mean?
suited aces
In a poker hand consisting of 5 cards find the probability of holding 3 aces?
Approximately 2%
What wins in Texas holdem a pair of kings and a pair of queens or a pair of two and a pair aces?
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.
The probability of being dealt 4 aces in a 5 card poker hand?
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%.
What does Dead man's hand mean?
Aces and eights
What is the 'Dead Man's Hand' in poker?
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.)
Does two to of club and two aces beat a flush?
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.
What hand did Wild Bill Hickock hold when he died?
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
A poker hand five cards is drawn from an ordinary deck of 52 cards find the probability of the first four cards are the four aces?
4/52 x 3/51 x 2/50 x 1/49 About 0.00039%
What is the dead man's hand?
It is called the Deadman's Hand. Aces and 8s
Does five aces beat a straight in poker?
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.
What is the probability that a random selected poker hand contains exactly 3 aces given that it contains at least 2 aces?
I will assume that you mean a five card poker hand. We can label the cards C1, C2, C3, C4, and C5. We are basically told already that C1 and C2 are both aces. So we have to find the probability that exactly one of C3, C4, and C5 is an ace. Knowing that the first two cards in our hand are both aces means that there are only 50 cards left in the deck. The probability that C3 is an ace and that C4 and C5 are both not aces is (2/50)(48/49)(47/48)=0.03836734694. The same probability also applies to each of C4 and C5, considered independently of each other. Therefore, our final probability is 3* 0.03836734694=0.1151020408
