What is the probably that a 5-card poker hand contains exactly one Ace (no more and no less)?
Please show your work.
Thanks in advance!
- 1 decade agoBest Answer
First, find the probability that the first card is an Ace and the 2nd to 5th are not:
(4/52) * (48/51) * (47/50) * (46/49) * (45/48).
Next, find the probability that the 2nd card is an Ace and the first, 3rd, 4th and 5th are not:
(48/52) * (4/51) * (47/50) * (46/49) * (45/48).
Next, the 3rd card is an Ace; the others are not:
(48/52) * (47/51) * (4/50) * (46/49) * (45/48).
Next, 4th card is an Ace; not the others:
(48/52) * (47/51) * (46/50) * (4/49) * (45/48).
Lastly, the 5th card is an Ace; not the others:
(48/52) * (47/51) * (46/50) * (45/49) * (4/48).
Adding these, the answer is:
0.29947... which is just under 30%.
- 4 years ago
The chance of one ace being drawn is 4/52 on the first draw. If non drawn the next card is 4/51, then next 450 and 4/49 and 4/48.each time an ace is drawn the3 top number will reduce by one. So if the first card is an ace the next odds would be 3/51 and so on. To draw 2 aces in 5 cards is (5 x 2/4)/52 = 2.1/2 in 52. or just under 5%.
- zex20913Lv 51 decade ago
We want one ace, and four "not aces". Let's split these up into the 4 aces and the 48 not aces.
From the 4 aces, we want one card. That's four choose one, or four.
From the 48 not aces, we want four cards. That's 48 choose 4, or 194580. Multiplying these gets us 778320. That is the number of hands with exactly one ace. However! This method doesn't take the order into account. It thinks that getting the ace first and, say, the king of spades second is the same as getting the king of spades first and the ace second. It counts each hand 5 times. So, we divide by 5 to get 155664 unique hands with one ace and 4 others.
Now, there are 52 choose 5 possible hands! This is 2598960. 155664/2598960=.059894727... or about 6%
This is more reasonable than 12%, because picking one card from a deck, there's a 1/13 chance that it will be an ace. With five cards selected, it should only go down.Source(s): Ma+h +eacher
- justainchoeLv 51 decade ago
I do probability problems in my own way since I figured out my methods before I formally learned probability, so here is my method:
What's the probability that the first card you draw is an ace, and no other cards are aces?
(1/13) * (39/51) * (38/50) * (37/49) * (36/48)
The probability of this is the same for the 2nd, 3rd, 4th, and 5th card being an ace. So, just multiply the formula above by 5 since there are 5 possible ways to do this.
(1/13) * (39/51) * (38/50) * (37/49) * (36/48) * 5
(I'm going to reduce fractions)
= (1/13) * (13/17) * (19/25) * (37/49) * (3/4) * 5
(In the numerator, the 13 and 5 cancel out and other numbers in the numerator are prime)
= (19*37*3) / (17*5*49*4)
= 2109 / 16660 = 0.126590636254501800720881152461
Roughly 12.659 percent.
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- 1 decade ago
It will be close to 0.27924. I used binomial distribution to work it out. The reason i say its close is that the probabilities in this instance might be changing and binomail distriubuation relies on the fact that they dont change. Having said that, the probabilities are not going to be changing a lot, so 0.28 is roughly right.