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What is the probability that a hand of 13 cards contains no pairs?
Wiki User
∙ 14y ago
Well the short answer is that your chances are pretty slim. The long answer:
The total number of 13-card hands is 52-choose-13, which equals 52! / (13! * 39!). Any hand without any pairs must have exactly one of each value from 2 to ace. All there is to choose is the suit of each card, for 413 possible pairless 13-card hands. The probability of choosing one of these hands is the quotient of these two numbers, or 413 * 13! * 39! / 52!, which equals 4194304/39688347475, approximately 0.00010568 or 1 in 9462. Another explanation:
Say the first card in your hand is an ace. For the second card, you have to avoid the other three aces in the deck. There are 51 cards, and 48 of them are OK, so the probability that the second card doesn't pair the first card is 48/51. Suppose your second card was a king. For the third card, you now have to avoid all remaining aces and kings. There are 50 cards left in the deck and you have to Dodge six of them. So the probability that the third card doesn't pair either of the first two cards is 44/50. Each time you pick a card, there are three more cards you have to avoid, and one fewer card in the deck. So for the 4th card, you have to avoid 9 cards out of 49, giving a probability of 40/49. This continues until the 13th card, when the probability is 4/40. All these things have to happen one after the other, so you have to multiply the probabilities together:
48/51 x 44/50 x 40/49 x 36/48 x 32/47 x 28/46 x 24/45 x 20/44 x 16/43 x 12/42 x 8/41 x 4/40.
This gives the same probability as above (about 1 in 9462).
Wiki User
∙ 14y ago
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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
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.
What are chances of 4 aces being in a hand of 9 cards?
The probability of 4 aces being in a hand of 9 cards is: 9C4 ∙ (4/52)∙(3/51)∙(2/50)∙(1/49)∙(48/48)∙(47/47)∙∙∙(44/44) = 0.0004654... ≈ 0.0465% where 9C4 = 9!/[(9-3)!∙3!] = 126
What is the probability that a five-card poker hand contains the two of diamonds and the three of spades Enter the value in decimal format and report it to four decimal places?
How many solutions are there to the equation , where , i = 1, 2, 3, 4, 5, is a nonnegative integer such that
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 in a regular 52 deck of getting a hand with no diamonds given that the cards are red?
The answer depends on how many cards are in the hand.
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
In a poker hand consisting of 5 cards find the probability of holding 3 aces?
Approximately 2%
What is the Probability of getting a 6 of clubs and 7 of clubs?
If you are drawing two cards from a full deck of cards (without jokers) then the probability will depend upon whether the the first card is replaced before the second is drawn, but the probability will also be different to being dealt a hand whilst playing Bridge (or Whist), which will again be different to being dealt a hand at Canasta. Without the SPECIFIC context of the two cards being got, I cannot give you a more specific answer.
How many ways can a hand of five cards consisting of four cards from one suit and one card from another suit be drawn from a standard deck of cards I think combination?
The probability of five cards being four cards from one suit and one card from another suit is the same as the probability of drawing four cards from one suit multiplied by the probability of drawing one card from another suit, multiplied by 5 (for each of the possible positions this other card can be drawn in). The probability of drawing four cards from one suit is 12/51 x 11/50 x 10/49. The probability of drawing a fifth card from another suit is 39/48. All these numbers multiplied together (and multiplied by 5) come to 0.0429. So the probability of drawing a hand of five cards with four cards from one suit and one card from another is 5.29%
What is the probability of being dealt a five card hand from a standard deck of cards will included all five cards of the same suit?
The probability that five cards chosen at random from a standard deck will all be the same suit is 12/51 * 11/50 * 10/49 * 9/48 = about 0.2%.
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%
Does a Qcb Qsp Jcb Jht 5sp hand beat 10cb 10sp 9cb Kcb Aht in poker?
Yes. That is two pairs against one pair, basically, and the pairs are higher in the first hand. ... If the king or the Ace were paired in the second hand, then that hand would win. But, basically, the value and suits of the cards don't matter in this particular hand, because two pairs always beats one pair. Actually the suits do not matter at all in poker. It is the ranking of the cards that matter. For example, if there are two flushes (all same suit) it is the highest card that wins, not the suit.
What is an additional hand of cards?
a hand of cards
Which hand in bridge contains no card higher than nine?
A bridge hand with no high cards in it is called a Yarborough hand, named after the 2nd Earl of Yarborough.
A card game using a pack with one card removed?
Old maid or pairs. Take one out. Deal out the cards, then remove all the pairs from each players hand. Then select card from player with most cards discarding pairs until one player is left with a single card (the pair to the card you removed)
What are the rules of Badugi?
The rules of Badugi can be classified into three namely blinds, the deal and basic strategy. The strongest hand is that one which has four low cards and no pairs.
