## Qn on Probability

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• Hi all, I need help with a question. What is the probability of being dealt 2 pairs if we are dealt 5 cards from a pack of 52 cards? I calculated the
Message 1 of 2 , Feb 1, 2006
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Hi all,
I need help with a question.

What is the probability of being dealt 2 pairs if we are dealt 5 cards
from a pack of 52 cards?

I calculated the probability of being dealt AABBC, i.e. 2 pairs (AA
and BB) and an odd card (C) as 1 * (3/51) * (48/50) * (3/49) *
(44/48), and the number of ways to arrange the cards as 5!/(2!2!) = 30.

Therefore, I got the answer to be 30 * P(AABBC). However, my teacher
said that there is a "trap" in the question, and the number of ways to
arrange the cards should be 5!/(2!2!) * (1/2), due to overcounting. I
cannot see where the overcounting is, as I thought the 5!/(2!2!)
should have gotten rid of the overcounting problem.

Thank you.

Regards,
Rayne
• Rayne, Before you come to the arrangements of the cards you chose, you should be sure you have chosen without duplicates. When you calculate the probability
Message 2 of 2 , Feb 3, 2006
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Rayne,

Before you come to the arrangements of the cards you chose, you should be
sure you have chosen without duplicates. When you calculate the probability
P(AABBC), consider whether AABBC and BBAAC are counted as the same or
separate.

With regards,
Rino.

On 2/2/06 1:13 PM, "wu_weidong" <ghcgwgwg@...> wrote:

> Hi all,
> I need help with a question.
>
> What is the probability of being dealt 2 pairs if we are dealt 5 cards
> from a pack of 52 cards?
>
> I calculated the probability of being dealt AABBC, i.e. 2 pairs (AA
> and BB) and an odd card (C) as 1 * (3/51) * (48/50) * (3/49) *
> (44/48), and the number of ways to arrange the cards as 5!/(2!2!) = 30.
>
> Therefore, I got the answer to be 30 * P(AABBC). However, my teacher
> said that there is a "trap" in the question, and the number of ways to
> arrange the cards should be 5!/(2!2!) * (1/2), due to overcounting. I
> cannot see where the overcounting is, as I thought the 5!/(2!2!)
> should have gotten rid of the overcounting problem.
>