alphabeta1234 wrote:
What is the probability of a two pair in poker?
So we have 52 cards. 13 pairs.
We choose our first two pairs from the 13 pairs. (13C2)
From the first pair we choose our pair from the 4 cards (4C2)
From the second pair we choose our pair from the 4 cards (4C2)
Here is my question. Since we already have picked 4 cards from our deck. Can I just say from the remaining 48 cards, I will pick 1?
So (48C1)?
Is this wrong because of the remaining 48 cards are also members of the 2 pairs we picked from?
So we have to make sure we pick that last car from a group of cards that will definitly not include
cards from the two other pairs?
So we have 11 pairs remaining. From the 11 we choose 1. (11C1) and from the 4 cards we pick 1.
So answer: (13C2)(4C2)(4C2)(11C1)(4C1)/(52C4)
\(C^1_{48}\) is wrong because if you choose the value you already have in pairs you'd have full house.
What is the probability of a two pair in poker?Cards = 52.
Suits = 4.
Values = 13.
\(P(two pair)=\frac{C^2_{13}*C^2_4*C^2_4*C^1_{44}}{C^5_{52}}\)
\(C^2_{13}\) is the number of ways to choose which 2 values will be the pairs (for example, 2 kings, and 2 aces);
\(C^2_4\) is the number of ways to choose 2 suits for the first pair;
\(C^2_4\) is the number of ways to choose 2 suits for the second pair;
\(C^1_{44}\) is the number of ways to choose the remaining fifth card from 52-8=44 (exclude 4 cards we used for the first pair and 4 cards we used for the second pair). This can also be written as \(C^1_{11}*C^1_4\): choosing value for the fifth card and choosing suit for the fifth card.
For more check:
let-s-play-poker-84956.htmlHope it helps.
P.S. Please name topics properly. _________________