A hand consists of 5 cards (52 card deck). What is the probability of getting 2 aces and 2 kings?
There are [math]\binom{52}{5}[/math] total five card hands.There are [math]\binom42\binom42\binom{44}{1}[/math] five card hands with precisely two Aces, two Kings, and one other card.The probability is one divided by the other: [math]\dfrac{5!\cdot6^2\cdot44}{52\cdot51\cdot50\cdot49\cdot48}=\dfrac{33}{54145}\approx\dfrac1{1641}\approx0.06\%[/math]
A five-card poker hand is dealt at random from a standard 52-card deck.?
Note the total number of possible hands is C(52,5)=2,598,960. Find the probabilities of the following scenarios: (a) What is the probability that the hand contains exactly one ace? Answer= a/ C(52,5), where a= (b) What is the probability that the hand is a flush? (That is all the cards are of the same suit: hearts, clubs, spades or diamonds.) Answer= b/ C(52,5), where b= (c) What is the probability that the hand is a straight flush? Answer= c/ C(52,5), where c=
A poker hand of five cards is dealt from a standard pack of 52. Find the probability obtaining:?
There are ₅₂C₅ = 2,598,960 possible 5-card hands from a 52-card pack. a) Two cards have the same rank, and the remaining three cards have different ranks. Choose the rank of the pair in any of 13 ways; choose the ranks of the remaining 3 cards in ₁₂C₃ ways. There are ₄C₂ = 6 ways the pair may be suited; each of the remaining cards may be any of 4 suits. This makes 13 × ₁₂C₃ × ₄C₂ × 4³ = 1,098,240 ways to draw a 5-card pair, and the probability of doing so is: 1,098,240 / 2,598,960 = 352 / 833 ∎ ≈ 42.26% b) Two cards share one rank, another two share a different rank, and the fifth card has a third rank. Choose the ranks of the two pairs in ₁₃C₂ ways, and the rank of the singleton card in 11 ways. The suits of the pairs may be chosen in ₄C₂ ways (each), and the singleton card may be any of 4 suits: ₁₃C₂ × 11 × ₄C₂² × 4 = 123,552, and the probability is: 123,552 / 2,598,960 = 198 / 4,165 ≈ 4.754% c) Three cards share one rank; the remaining two cards have different ranks. Choose the three ranks in ₁₃C₃ ways. Choose one of these ranks to be that of the triple in 3 ways. The triple may be suited in ₄C₃ ways; the remaining two cards may have any of 4 suits. ₁₃C₃ × 3 × ₄C₃ × 4² = 54,912 Probability: 54,912 / 2,598,960 = 88 / 4,165 ∎ ≈ 2.113% d) All five cards share the same suit. Choose the ranks of the cards in ₁₃C₅ ways. Choose the suit of the flush in 4 ways. ₁₃C₅ × 4 = 5,148 Probability: 5,148 / 2,598,960 = 33 / 16,660 ∎ ≈ 0.1981% HOWEVER: This probability includes straight flushes (and royal flushes, which are a special case of the straight flush). These are considered different hands. So, we may wish to exclude flushes consisting of 5 consecutive ranks. Since the Ace can be either low or high, there are 10 ways of selecting 5 consecutive ranks and, again, 4 ways to choose a flush suit. This makes 10 × 4 = 40 straight/royal flushes, so the number of regular flushes goes down to: 5,148 - 30 = 5,108 Probability: 5,108 / 2,598,960 = 1,277 / 649,740 ∎ ≈ 0.1965%
What is the probability that a randomly drawn 5 card hand contains no diamonds or hearts?
a) All poker hands: C(52,5) = 2,598,960 hands Now you throw all the diamonds and hearts away, keep 26 cards in hands and draw 5 cards: C(26,5) = 65,780 hands. Probability: 65,780/2,598,960 ≈ 2.531% Your logic was flawed like this: You supposed that red and black cards play equally important roles, but the presence of just ONE red card will totally render the hand unacceptable, so even if 1 red card is present, there's a chance (although small) it will be dealt, ruining your hand's legitimacy. ADDITIONAL SOLUTION: First card being black: 26/52 chance 2nd card......................: 25/51 chance 3rd................................: 24/50 chance 4th.................................: 23/49 chance 5th.................................: 22/48 chance => 253/9996 ≈ 2.531% b) Since it has only black cards, there's only 2 ace in 26 cards: Total hands that can be dealt: 26C5 = 65,780 Now you discard 2 aces and deal a hand with 24 cards: 24C5 = 42,504 => Hands with 1 or 2 aces: 65,780 - 42,504 = 23,276 hands. Probability: 23,276/2,598,960 ≈ 0.0896%
If 5 cards dealt at random, what is the probability of exactly 3 spades?
You can use the hyper geometric distribution to find the solution you have K = number of items to be drawn = 5 cards are dealt N = total objects = 52 cards in the deck M = number of objects of a given type = 13 spades The probability mass function for the hypergeometric distribution is defined as: P(X = x | N, M, R) = ( M C x ) * ( (N - M) C (K - x) ) / ( N C K ) for x = {0, 1, 2, 3, ..., K} P(X = 0 | N, M, R) = 0 otherwise If you have n objects and chose r of them, the number of combinations is: n! / ( r! (n-r)! ) this can be written as nCr the N C K is the total number of possible combinations of K objects drawn from N objects. the M C x is the number of combinations of getting x objects of the given type the (N - M) C ( K - x) is the number of combinations of non typed objects to be drawn. Let X be the number of spaded dealt to you in a five card hand. P(X = 0 ) = 0.2215336 P(X = 1 ) = 0.4114196 P(X = 2 ) = 0.2742797 P(X = 3 ) = 0.08154262 P(X = 4 ) = 0.01072929 P(X = 5 ) = 0.0004951981 P(X ≥ 1) = 1 - P(X = 0) = 0.7784664
A 3-card hand is drawn from a standard deck of 52 cards. Find the probability that the hand contains the given?
if you have a 52 card deck there is 13 cards of each suit. so if you want the prob of drawing a hand with three of the same suit the prob is (13/52)x(12/51)x(11/50) this is because when you pick a card you do not replace it.
A poker hand consist of 5 cards chosen at a random from a standard deck of cards. How many poker hands containing all face cards are possible?
The number of 5-card poker hands from a standard deck is (52 | 5) (read: “52 pick 5”) as there are 52 cards and you’re picking 5 of them. [There’s an easier notation for the above, but I dunno how to type it in plain text.] Anyway, that is a combinatorial and calculates out to 2,598,960 unique hands.c = n!/k!/(n-k)! = 52!/5!/47! = 2,598,960Your question, though, is how many 5-card hands made up of only face cards, which is the same as saying “what if the deck only hand face cards in it?” Face cards: King, Queen, Jack, across 4 suits. that is a 3x4 = 12 card deck. So the combinatorial is (12 | 5) (read: “12 pick 5”)c = 12!/5!/7! = 792There are 792 unique poker 5-card hands made up of only face cards. That’s only about .03% of all possible hands in a standard deck
What is the probability the hand will contain 3 spades and 2 hearts if five cards are dealt from 52 playing cars?
5 choose 3 and 5 choose 2 are the same. They both equal 5(4)/2 = 10. That’s the number of ways to partition a 5-card hand into a set of 3 and a set of 2. Let’s pick one of them — the first 3 and the last 2. We can calculate the probability that the first 3 cards dealt are spades and the last 2 are hearts. Then we need to multiply that by 10 to cover all the ways of getting 3 spades and 2 hearts.The probability that the first card dealt is a spade is 1/4, because there are 4 suits which have the same number of cards each. To put it another way, it is 13/52. Given that, the probability that the second is a spade is 12/51. Similarly, the third is 11/50. Then the probabilities that the last 2 are hearts are 13/49 and 12/48. These need to be multiplied to get the probability that we are dealt 3 spades followed by 2 hearts. This is 13(12)11(13)12/52(51)…48 = 267,696/311,875,200. Finally, as we said, since there are 10 orders in which 3 spades and 2 hearts can be dealt, we multiply by 10 and get 267,696/31,187,520. That’s a little bit less than 1%. Frankly, I’m surprised it isn’t a lot lower. But that’s my answer, if I’m not mistaken.