Card Probability Challenge

**c00ps** · August 23rd 2007, 02:40 PM

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?

**Soroban** · August 23rd 2007, 03:17 PM

Hello, c00ps!

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: . ${216\choose11}$ possible 11-card hands.

There are ${\color{blue}13}$ choices for the value of the pair.
Then there are: . ${\color{blue}{16\choose2}}$ ways to get that pair.

The other nine cards are 'singletons'.
There are: . ${12\choose9}$ choices for their values.
There are $16$ choices for each value.
Hence, the nine cards can be selected in: . ${\color{blue}{12\choose9}\cdot(16)^9}$ ways.

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

**c00ps** · August 23rd 2007, 05:58 PM

Thank you for your response.

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?

**Soroban** · August 23rd 2007, 07:04 PM

Hello, AGAIN, c00ps!

Here is the first part . . .

More specifically, what's the probability there's AT LEAST one pair?

There are: . ${216\choose11}$ 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: . ${13\choose11}$ choices for the 11 values.
There are: .16 choices for each value.
Hence, there are: . $13\!\cdot\!{13\choose11}(16)^{11}$ ways.

(2) One Joker and ten different valuies.
There are 8 ways to select a Joker.
There are: . ${13\choose10}$ choices for the 10 values.
There are 16 choices for each value.
Hence, there are: . $8\!\cdot\!{13\choose10}(16)^{10}$

Thus, there are: . ${13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10}$ hands with no pairs.

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

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

**c00ps** · August 23rd 2007, 07:45 PM

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

= $5.03137*10^{15}$

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

= $-5.03137*10^{15}$

What am I getting wrong?

**CaptainBlack** · August 23rd 2007, 09:05 PM

Originally Posted by c00ps

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

= $5.03137*10^{15}$

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

= $-5.03137*10^{15}$

What am I getting wrong?

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

RonL

**c00ps** · August 23rd 2007, 10:27 PM

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

$14/205 \approx 0.068$ ?

**CaptainBlack** · August 23rd 2007, 10:34 PM

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

$14/205 \approx 0.068$ ?

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

RonL

**c00ps** · August 24th 2007, 11:07 AM

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).
. $13\!\cdot\!{13\choose11}(16)^{11}$ ways.

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

Thus, there are: . ${13\choose11}(16)^{11} + 8\!\cdot\!{13\choose10}(16)^{10}$ hands with no pairs.

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

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

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

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

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

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