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?

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- August 23rd 2007, 02:40 PMc00psCard 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?

- August 23rd 2007, 03:17 PMSoroban
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?

**exactly**one 2-of-a-kind . . . and the other 9 cards do not match.

There are: . possible 11-card hands.

There are choices for the value of the pair.

Then there are: . ways to get that pair.

The other nine cards are 'singletons'.

There are: . choices for their values.

There are choices for each value.

Hence, the nine cards can be selected in: . ways.

Therefore: .

- August 23rd 2007, 05:58 PMc00ps
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? - August 23rd 2007, 07:04 PMSoroban
Hello, AGAIN, c00ps!

Here is the first part . . .

Quote:

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: . choices for the 11 values.

There are: .16 choices for each value.

Hence, there are: . ways.

(2) One Joker and ten different valuies.

There are 8 ways to select a Joker.

There are: . choices for the 10 values.

There are 16 choices for each value.

Hence, there are: .

Thus, there are: . hands with no pairs.

So: .

Therefore: .

- August 23rd 2007, 07:45 PMc00ps

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What am I getting wrong? - August 23rd 2007, 09:05 PMCaptainBlack
- August 23rd 2007, 10:27 PMc00psoops :)
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

?

- August 23rd 2007, 10:34 PMCaptainBlack
- August 24th 2007, 11:07 AMc00ps