If you have two cards which are a pair, how would you calculate the probability that you would get a full house with the next three cards drawn. (One card that matches your pair and another pair.)
Thanks in advance for any responses.
If you have two cards which are a pair, how would you calculate the probability that you would get a full house with the next three cards drawn. (One card that matches your pair and another pair.)
Thanks in advance for any responses.
This is a little tricky. You can either draw another card that matches your pair and then draw another pair or draw three cards that all match each other but don't match your pair.
So, to start off with, what is the probability of drawing three cards that match given you have already drawn a pair, keeping in mind that there are four of each number, but only twelve possible numbers (since none of the three can match your pair)?
Thanks.
This is a complete shot in the dark on the next one, but would it be:
P(M) = Probability of drawing card matching pair
P(P) = Probability of drawing another pair
P(M) * P(P)
P(M) = 2/50
P(P) = ((4 C 2) *12) / 49 C 2 = 72/1176
So the answer would be .0024.
Hello, bimmer54!
Is that the exact wording of the problem?
If you have two cards which are a Pair,
calculate the probability that you get a Full House with the next 3 cards.
(One card that matches your pair and another pair.)
This is not the only way to get a Full House.
There are: .$\displaystyle {50\choose3} = {\color{red}19,\!600}$ possible outcomes.
There are two ways to get a Full House:
. . [1] One card that matches your Pair and another Pair.
. . [2] You draw a Triple (three-of-a-kind)
Case 1: One card that matches your Pair, and another Pair.
You must get one of the 2 remaining cards that match your Pair: .$\displaystyle {2\choose1} = {\color{blue}2}$ ways.
Your second pair has a choice of 12 values.
Then you must get two of the 4 cards of that value: .$\displaystyle {4\choose2} = {\color{blue}6}$ ways.
. . So there are: .$\displaystyle 2\cdot 12\cdot 6 \:=\:{\color{blue}144}$ ways.
Case 2: Draw a Triple.
There are 12 choices for the value of the Triple.
You must get 3 of the 4 cards of that value: .$\displaystyle {4\choose3} = {\color{blue}4}$ ways.
. . So there are: .$\displaystyle 12\cdot4 \:=\:{\color{blue}48}$ ways.
Hence, there are: .$\displaystyle 144 + 48 \:=\:{\color{red}192}$ ways to get a Full house.
$\displaystyle P(\text{Full House}\,|\,\text{Pair}) \:=\:{\color{red}\frac{192}{19,\!600}} \;=\;\boxed{\frac{12}{1225}}$