## Variation on A4 |
rahulchandrupatla |
---|---|

What if this problem asked us the probability that all 3 cards are different assuming that there is are at least two different card values in your hand? I got P(3 different)/ P(atleast two different card values are in your hand) and the bottom is 2 cases : 3 different, or 2 same and 1 different. So putting it all together is $\frac{(10choose3)(3choose1)^3}{(10choose3)(3choose1)^3+(10choose2)(3choose2)(27)}$. Can someone please help me and confirm or deny this? |

weisbart: March 20, 2015, 1:44 a.m. |
---|

This looks good. The 27 shows up because you remove all three cards with the same value, so you have 27 left. Good job. |