Question 246889
If I want P1(4 cards same suit), I don't care what
the 1st card is, so I'll call P1(1st card) = {{{1}}},
which means I'm certain to get any one of the cards.
Now I have that card, and it is a certain suit, say
clubs. There are 12 clubs left, so my chances of getting
another club is 
{{{P2 = 12/51}}}
Now  there are 11 clubs left 
{{{P3 = 11/50}}}, and
{{{P4 = 10/49}}}
{{{P[ss] = P1*P2*P3*P4}}}
{{{P[ss] = 1*12/51*11/50*10/49}}}
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There are 13 groups of 4 cards which are the same rank
I don't care what the 1st card is
{{{P1 = 1}}}
I've got the 1st card. Say it is a 9
{{{P2 = 3/51}}} is P2(another 9)
{{{P3 = 2/50}}} is P3(another 9)
{{{P4 = 1/49}}} is P4(another 9)
{{{P[sr] = 1*3/51*2/50*1/49}}}
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given:
{{{P[ss] = k*P[sr]}}}
{{{k = P[ss]/P[sr]}}}
{{{k = (12/51*11/50*10/49)/(3/51*2/50*1/49)}}}
Multiply right side by {{{(51*50*49)/(51*50*49)}}}
{{{k = (12*11*10)/(3*2*1)}}}
{{{k = 220}}}