http://math.stackexchange.com/questions/529141/conditional-probability-sheldon-ross-example-2h
Just imagine that 4 aces are in different piles and then follow this approach:
answer
= $\frac{\binom{48}{12}*\binom{36}{12}*\binom{24}{12}*\binom{12}{12}*4!}{\binom{52}{13}*\binom{39}{13}*\binom{26}{13}*\binom{13}{13}}$
$\approx 0.105$
23240 Points
17038 Points
7096 Points
6008 Points
5430 Points
4928 Points
4762 Points
4278 Points
3954 Points
3744 Points
Gatecse
X->YZ , Y->XZ , ...