Question 1187199: Compute the probability that a five-card poker hand is dealt to you that contains 2 Aces.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
With poker hands, the order of the cards does not matter. That means we use the nCr combination formula
nCr = (n!)/(r!*(n-r)!)
First consider the aces. We have n = 4 aces to choose from and have r = 2 slots to fill for those aces. Order doesn't matter.
nCr = (n!)/(r!*(n-r)!)
4C2 = (4!)/(2!*(4-2)!)
4C2 = (4!)/(2!*2!)
4C2 = (4*3*2*1)/(2*1*2*1)
4C2 = 24/4
4C2 = 6
There are 6 ways to choose the two aces from a pool of four.
Since the list is short, we can list out all the possible combos- AC,AD
- AC,AH
- AC,AS
- AD,AH
- AD,AS
- AH,AS
A = ace
C = clubs
D = diamonds
H = hearts
S = spades
Something like "AC" means "ace of clubs"
Order doesn't matter. Something like AC,AD is the same as AD,AC.
Now we consider the three other cards that aren't aces. There are 52 cards total in a deck. Kicking out the aces leaves us with 52-4 = 48 left over.
We'll use the nCr formula again to figure out how many ways to pick the three non-ace cards.
nCr = (n!)/(r!*(n-r)!)
48C3 = (48!)/(3!*(48-3)!)
48C3 = (48*47*46*45!)/(3!*45!)
48C3 = (48*47*46)/(3!)
48C3 = (48*47*46)/(3*2*1)
48C3 = (103,776)/6
48C3 = 17,296
There are 17,296 ways to pick the three non-aces.
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To summarize, we found that there are- 6 ways to pick the two aces
- 17,296 ways to pick the three non-aces
Overall, we have 6*17,296 = 103,776 ways to form the five-card hand we're after.
This is out of 52C5 = 2,598,960 ways to form any five-card hand (without any restrictions). The steps to computing this are similar to the other set of steps above.
The probability we want to compute is therefore:
(103,776)/(2,598,960) = 0.0399298
Answer: Approximately 0.0399298
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