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N. F. Taussig
N. F. Taussig
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elcharlosmaster
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Question:

What is the probability that a random bridge hand contains cards of exactly two suits?

My Attempt At A Solution Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be $$\frac{{13\choose n}{13 \choose 13-n}}{{52 \choose 13}} $$$$\frac{{26\choose 13}-2}{{52 \choose 13}} $$

We want some number of cards, $n$ of ourAs user @Lord Shark the Unknown hinted: there are $13$$26\choose 13$ ways to bechoose hands of two suits but one suit and the rest $13-n$ to be of another suitthose suits is only say hearts, and this is out of the total probability being ${52\choose 13}$another only spades, so we must compensate for those.

Thank you for any corrections/hints.

Question:

What is the probability that a random bridge hand contains cards of exactly two suits?

My Attempt At A Solution Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be $$\frac{{13\choose n}{13 \choose 13-n}}{{52 \choose 13}} $$

We want some number of cards, $n$ of our $13$ to be of one suit and the rest $13-n$ to be of another suit, and this is out of the total probability being ${52\choose 13}$.

Thank you for any corrections/hints.

Question:

What is the probability that a random bridge hand contains cards of exactly two suits?

My Attempt At A Solution Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be $$\frac{{26\choose 13}-2}{{52 \choose 13}} $$

As user @Lord Shark the Unknown hinted: there are $26\choose 13$ ways to choose hands of two suits but one of those suits is only say hearts, and another only spades, so we must compensate for those.

Thank you for any corrections/hints.

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elcharlosmaster
elcharlosmaster
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Random Bridge Hand w Cards of exactly two suits

Question:

What is the probability that a random bridge hand contains cards of exactly two suits?

My Attempt At A Solution Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be $$\frac{{13\choose n}{13 \choose 13-n}}{{52 \choose 13}} $$

We want some number of cards, $n$ of our $13$ to be of one suit and the rest $13-n$ to be of another suit, and this is out of the total probability being ${52\choose 13}$.

Thank you for any corrections/hints.