# Card Probability Challenge

• Aug 23rd 2007, 02:40 PM
c00ps
Card Probability Challenge
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?
• Aug 23rd 2007, 03:17 PM
Soroban
Hello, c00ps!

Quote:

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?

I assume you mean exactly one 2-of-a-kind . . . and the other 9 cards do not match.

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: .$\displaystyle P(\text{one pair}) \;=\;\frac{13{16\choose2}\!\cdot\!{12\choose9}(16) ^9}{{216\choose11}}$

• Aug 23rd 2007, 05:58 PM
c00ps

More specifically, what's the probability there's AT LEAST one pair, and what's the probability the next drawn card makes AT LEAST one of those pairs into 3 of a kind?
• Aug 23rd 2007, 07:04 PM
Soroban
Hello, AGAIN, c00ps!

Here is the first part . . .

Quote:

More specifically, what's the probability there's AT LEAST one pair?
There are: .$\displaystyle {216\choose11}$ possible 11-card hands.

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: .$\displaystyle P(\text{at least one pair}) \;=\;1 - P(\text{no pairs})$

• Aug 23rd 2007, 07:45 PM
c00ps
$\displaystyle P(\text{no pairs}) \:=\:\frac{{13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} }{{216\choose11}}$

=$\displaystyle 5.03137*10^{15}$

$\displaystyle P(\text{at least one pair}) \;=\;1 - P(\text{no pairs})$

=
$\displaystyle -5.03137*10^{15}$

What am I getting wrong?
• Aug 23rd 2007, 09:05 PM
CaptainBlack
Quote:

Originally Posted by c00ps
$\displaystyle P(\text{no pairs}) \:=\:\frac{{13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} }{{216\choose11}}$

=$\displaystyle 5.03137*10^{15}$

$\displaystyle P(\text{at least one pair}) \;=\;1 - P(\text{no pairs})$

=$\displaystyle -5.03137*10^{15}$

What am I getting wrong?

$\displaystyle P(\text{no pairs}) \:=\:\frac{{13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} }{{216\choose11}} \approx 0.00421015$

RonL
• Aug 23rd 2007, 10:27 PM
c00ps
oops :)
Thank you. I obviously miscalculated.

So, if I know I have one pair in the 11 cards dealt, then is the probability of drawing a third to make a set:

14 choices left / 205 in the deck

$\displaystyle 14/205 \approx 0.068$ ?
• Aug 23rd 2007, 10:34 PM
CaptainBlack
Quote:

Originally Posted by c00ps
Thank you. I obviously miscalculated.

So, if I know I have one pair in the 11 cards dealt, then is the probability of drawing a third to make a set:

14 choices left / 205 in the deck

$\displaystyle 14/205 \approx 0.068$ ?

Not quite because you have at least one pair, not exactly one pair. The above would be right if you had one pair.

RonL
• Aug 24th 2007, 11:07 AM
c00ps
Quote:

Originally Posted by Soroban
Hello, AGAIN, c00ps!

We will find the number of 11-card hands with no pairs.

There are two cases to consider.

(1) Eleven different values (no Joker).
.$\displaystyle 13\!\cdot\!{13\choose11}(16)^{11}$ ways.

(2) One Joker and ten different valuies.
.$\displaystyle 8\!\cdot\!{13\choose10}(16)^{10}$

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: .$\displaystyle P(\text{at least one pair}) \;=\;1 - P(\text{no pairs})$

Where did the 13 go from step (1) for no Jokers when you brought the two together:

$\displaystyle P(\text{no pairs}) \:=\:\frac{{13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10} }{{216\choose11}}$

Thank you for all your help, this is very interesting.