A faultless die is thrown 12 times.
Evaluate the probability of each of the numbers (1 to 6) appearing twice.
(12! / 2^6) / 6^12
Numerically:
>> (factorial(12) / 2^6) / 6^12
ans =
0.00343828589391861
>>
for a 0.3438 % probability or about 1 in 291.
A simulation is consistent with this:
match='112233445566';
ct=0;
for trial=1:1000000
throw=match(randi(12,[1,12]));
if isequal(sort(throw),match)
ct=ct+1;
end
end
ct/trial
trial/ct
>> aPerfectAppearance
ans =
0.003495
ans =
286.123032904149
>>
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Posted by Charlie
on 2024-02-06 08:29:36 |
solution
