# Thread: Probability of getting 2 pairs

1. ## Probability of getting 2 pairs

Hello

You are dealt 4 cards (each card is numbered from 0 - 9). Each card is independed of the other card. What is the probability of getting 2 pairs?

P(2 pair) = ­4C2P(1st = 2nd)P(3rd ≠ 2nd = 1st) P(4th = 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.1)=0.054
Is this correct?

2. You've doubled it.

3. So it's supposed to be:

P(2 pair) = (0.5)(4 Choose 2)(0.1)(0.9)(0.1)= 0.027?

If so, why would that be?

Is it because I would be counting (for example) 1212 twice?

4. Yes, once with the ones chosen by (4 choose 2)(0.1) and the twos by (0.9)(0.1), and once again the other way.

5. Thank you for your help. Just one more thing, did I make a mistake in the following calculations:

P(4 different digits) = P(2nd ≠1st)P(3rd ≠1st ≠2nd) P(4th ≠ 2nd ≠ 3rd ≠ 1st) = (0.9)(0.8)(0.7) = 0.504

P(pair) = (4 Choose 2)P(1st = 2nd)P(3rd ≠ 2nd = 1st)P(4th ≠ 3rd ≠ 2nd = 1st) = (6)(0.1)(0.9)(0.8) = 0.432

P(Three-of-a-kind) =(4 Choose 3)P(1st = 2nd = 3rd ≠ 4th) = (4)(0.1)2(0.9) = 0.036

P(Four-of-a-kind) = P(1st = 2nd =3rd =4th ) = 0.001

6. No, those are correct.

7. Thank you for all your help.