If 11 cards are dealt from a 216 card deck (four 54 card decks, 2 jokers each), what is the probability of there being 2 of a kind (not including jokers)? Then, what is the probability of the next drawn card making 3 of a kind?
If 11 cards are dealt from a 216 card deck (four 54 card decks, 2 jokers each), what is the probability of there being 2 of a kind (not including jokers)? Then, what is the probability of the next drawn card making 3 of a kind?
Hello, c00ps!
I assume you mean exactly one 2-of-a-kind . . . and the other 9 cards do not match.If 11 cards are dealt from a 216 card deck (four 54 card decks, 2 jokers each),
what is the probability of there being 2 of a kind (not including jokers)?
Then, what is the probability of the next drawn card making 3 of a kind?
There are: .$\displaystyle {216\choose11}$ possible 11-card hands.
There are $\displaystyle {\color{blue}13}$ choices for the value of the pair.
Then there are: .$\displaystyle {\color{blue}{16\choose2}}$ ways to get that pair.
The other nine cards are 'singletons'.
There are: .$\displaystyle {12\choose9}$ choices for their values.
There are $\displaystyle 16$ choices for each value.
Hence, the nine cards can be selected in: .$\displaystyle {\color{blue}{12\choose9}\cdot(16)^9}$ ways.
Therefore: .$\displaystyle P(\text{one pair}) \;=\;\frac{13{16\choose2}\!\cdot\!{12\choose9}(16) ^9}{{216\choose11}} $
Hello, AGAIN, c00ps!
Here is the first part . . .
There are: .$\displaystyle {216\choose11}$ possible 11-card hands.More specifically, what's the probability there's AT LEAST one pair?
We will find the number of 11-card hands with no pairs.
There are two cases to consider.
(1) Eleven different values (no Joker).
There are: .$\displaystyle {13\choose11}$ choices for the 11 values.
There are: .16 choices for each value.
Hence, there are: .$\displaystyle 13\!\cdot\!{13\choose11}(16)^{11}$ ways.
(2) One Joker and ten different valuies.
There are 8 ways to select a Joker.
There are: .$\displaystyle {13\choose10}$ choices for the 10 values.
There are 16 choices for each value.
Hence, there are: .$\displaystyle 8\!\cdot\!{13\choose10}(16)^{10}$
Thus, there are: .$\displaystyle {13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} $ hands with no pairs.
So: .$\displaystyle P(\text{no pairs}) \:=\:\frac{{13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} }{{216\choose11}} $
Therefore: .$\displaystyle P(\text{at least one pair}) \;=\;1 - P(\text{no pairs}) $
$\displaystyle P(\text{no pairs}) \:=\:\frac{{13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} }{{216\choose11}} $
=$\displaystyle 5.03137*10^{15}$
$\displaystyle P(\text{at least one pair}) \;=\;1 - P(\text{no pairs}) $
=$\displaystyle -5.03137*10^{15}$
What am I getting wrong?