# Probability of getting 2 pairs

• Mar 8th 2011, 01:47 PM
statmajor
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?
• Mar 9th 2011, 10:23 AM
tom@ballooncalculus
You've doubled it. (Wink)
• Mar 9th 2011, 05:35 PM
statmajor
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?
• Mar 9th 2011, 10:49 PM
tom@ballooncalculus
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.
• Mar 10th 2011, 09:29 AM
statmajor
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
• Mar 10th 2011, 12:08 PM
tom@ballooncalculus
No, those are correct.
• Mar 10th 2011, 12:26 PM
statmajor
Thank you for all your help.