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This is solved by inclusion/exclusion. The probability that an example of each denomination occurs before any face card starts out by adding the sum of all the individual probabilities of a given denomination coming before any face card; from this is subtracted all the pairwise probabilities of either denomination A or denomination B before all the face cards. It goes on like this for triples, etc., alternately adding and subtracting from the total probability.
Part 1.
group prob of how overall
size each many contribution
1 1/4 10 5/2
2 2/5 45 -18
3 1/2 120 60
4 4/7 210 -120
5 5/8 252 315/2
6 2/3 210 -140
7 7/10 120 84
8 8/11 45 -360/11
9 3/4 10 15/2
10 10/13 1 -10/13
total 1/286
Part 2.
group prob of how overall
size each many contribution
1 1/11 3 3/11
2 1/6 3 -1/2
3 3/13 1 3/13
total 1/286
The probabilities are the same.
Verified by simulation:
succ=0;
for trial=1:1000000
deck=randperm(52);
had=zeros(1,10); hadCt=0;
for i=1:52
v=ceil(deck(i)/4);
if v>10, break, end
if had(v)==0
had(v)=1;
hadCt=hadCt+1;
end
end
if hadCt==10
succ=succ+1;
end
end
disp([succ/1000000 1000000/succ])
succ=0;
for trial=1:1000000
deck=randperm(52);
had=zeros(1,3); hadCt=0;
for i=1:52
v=ceil(deck(i)/4);
if v<11, break, end
if had(v-10)==0
had(v-10)=1;
hadCt=hadCt+1;
end
end
if hadCt==3
succ=succ+1;
end
end
disp([succ/1000000 1000000/succ])
A recognition of the fact that the individual, pairwise, etc. probabilities can all be reduced by dividing the numerator and denominator by 4 indicates the results are the same as if there were only one card of each denomination. That realization makes the problem equivalent to that of all the face cards coming at the end (or beginning) of the shuffling of a 13-card deck with one of each denomination. That probability is 1/C(13,3) = 1/286.
For the Extra Challenge:
X1:
The numerators and denominators get too large to show as rational numbers, and the probabilities are different for the differing denominations and combinations of denominations. So only the total of the contributions of each level (size of combinations) is shown.
size of total of
each subset contributions
1 4.41562604062604
2 -28.1965609286393
3 86.646713303523
4 -163.633524574637
5 205.890246287792
6 -177.206336737997
7 103.658920727275
8 -39.5678734065306
9 8.91600955226612
10 -0.901639344262295
The total probability is 0.0215809194152671 or 1 in 46.3372287694363.
X2:
The fractions here are amenable to showing as rational numbers:
contribution
1 4/59 + 1/56 + 1/56 = 171/1652 171/1652
2 1/12 + 1/12 + 2/57 = 23/114 - 23/114
3 6/61 6/61
total 673/5744004 ~= 0.000117165656569877
or 1 in 8534.92421991085
The total is the same as the sum of Jer's terms in the first comment.
Simulations are via
deck2=[1 2 2 3 3 3 4 4 4 4 5 5 5 5 5 6 6 6 6 6 6 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9
10 10 10 10 10 10 10 10 10 10 11 11 11 11 12 13];
tot=0;
succ=0;
for trial=1:1000000
deck=randperm(61);
had=zeros(1,10); hadCt=0;
for i=1:61
v=deck2(deck(i));
if v>10, break, end
if had(v)==0
had(v)=1;
hadCt=hadCt+1;
end
end
if hadCt==10
succ=succ+1;
end
end
disp([succ/1000000 1000000/succ])
succ=0;
for trial=1:1000000
deck=randperm(61);
had=zeros(1,3); hadCt=0;
for i=1:61
v=deck2(deck(i));
if v<11, break, end
if had(v-10)==0
had(v-10)=1;
hadCt=hadCt+1;
end
end
if hadCt==3
succ=succ+1;
end
end
disp([succ/1000000 1000000/succ])
With respective results of:
0.021826 45.8169156052415
0.000112 8928.57142857143
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