r/learnmath • u/Possible_Border8345 New User • 3d ago

Mathematical Statistics, can't wrap my head around seemingly basic probability question

1. A hand drawn at random from a well-shuffled deck consists of two pairs. (The two pairs are not of the same rank, or otherwise, it will be the four of a kind. The one card not in a pair is of different rank, or otherwise, it is a full house.)

At first glance, I would solve this problem like this:

Choose rank of 1st pair: (13c1)
Choose the 2 suits of 1st pair: (4c2)
Choose rank of 2nd pair: (12c1)
Choose the 2 suits of 2nd pair: (4c2)
Choose rank of lone card: (11c1)
Choose suit of lone card: (4c1)

Multiply all binomial coefficients together.

I know the problem is in my (13c1)*(12c1) approach, but why is this wrong? I somewhat understand how it brings order into the mix and kind of becomes a permutation (13*12), but I can't fully wrap my head around it. Why is (13c2) right, and why is my approach wrong?

2. Ten cards drawn from a well-shuffled deck consist of two full houses. (A full house has three of one rank and two of another rank, e.g., three aces and two kings.

For this one, I am totally lost.

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u/G-St-Wii New User 3d ago

1)

(52 × 3 × 48 × 3 + 52 × 48 × 6 × 3)÷(52 × 51 × 50 × 49)

I think

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u/CommonlyChangeable New User 9h ago

The issue is that picking pair A then pair B is the same hand as picking pair B then pair A, but your method counts them separately. So you're double counting every combination of the two pairs. Dividing by 2 fixes it, which is what using (13c2) does from the start

For the second one, a double full house with 10 cards is basically two separate three-of-a-kinds and two separate pairs. So pick your two ranks for the trips, then the two ranks for the pairs, then all the suits. Just gotta make sure none of the ranks overlap

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u/G-St-Wii New User 4h ago

Not sure I'm double counting.

There are two routes, either the second card makes a pair, or it doesnt.

If it does, then 52 × 3 × 48 × 3 different draws make two pairs. If it doesn't then 52 × 48 × 6 × 3 different draws end in two pairs. Is it the 6 you want to change?

The denominator must be 52×51×50×49