The player gets 8 cards of 32. What is the probability that it gets
a) all 4 aces
b) at least 1 ace

Correct result:

a =  0.1947 %
b =  70.4505 %


C4(4)=(44)=4!4!(44)!=11=1  C4(28)=(284)=28!4!(284)!=282726254321=20475  C8(32)=(328)=32!8!(328)!=323130292827262587654321=10518300  a=100 (44) (284)(328)=0.1947%
C8(28)=(288)=28!8!(288)!=282726252423222187654321=3108105 b=100 (1(3248)(328))=70.4505%

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