Hi Alan and guests
,
I can explain the Monte-Carlo simulation result
The probability of getting at least one of each number is equal to
1-P(not getting one of the numbers)
I am going to try and determine the probability of not getting one of the numbers.
A) Now the prob of not getting any individual number in 11 tosses would be \((\frac{5}{6})^{11}\)
so the prob of not getting a 1 or a 2 or a 3....or a 6 would be
\(6*(\frac{5}{6})^{11}\)
B) Trouble is, there is a lot of double counting here.
For instance, if there were no sixes and no 2s that would be counted twice instead of just once.
There are 6C2 = 15 ways that 2 numbers can be NOT thrown and for each the probability will be \((\frac{4}{6})^{11}\)
The probability that there are two numbers missing is
\(15*(\frac{4}{6})^{11}\)
This probability must be subtracted.
C) Now what about if there are 3 numbers missing Lets say 1,2,and 3 are all not thrown.
Well in A this combination got counted 3 times. 1+2+3 In be it got subtracted 3 times -[12]-[23]-[13]
So now it needs to be added back in again. There are 6C3=20 ways that three numbers can be selected and the probability of getting one combination any of them is (3/6)^6 So the probability of thrre numbers all missing will be
\(20*(\frac{3}{6})^{11}\)
D) Now what if there are 4 numbers missing. Lets say a 1,2,3, and 4 are all not thrown.
In A this combination was added 4 times.
In B it was subtracted 4C2 = 6 times
In C it was added 4C3 = 4 times
so altogether it has been added 4-6+4=2 times so we must subtract it once
There are 6C4 = 15 different combinations of 4 numbers. So we need to subtract
\(15*(\frac{2}{6})^{11}\)
E) What if there are 5 numbers missing Lets say 1,2,3,4,5, are all not thrown.
In A this combination was added 5 times.
In B it was Subtracted 5C2=10 times
In C it was added 5C3 = 10times
In D it was subtracted 5C4=5 times
5-10+10-5=0 So we need to add it 1 time.
There are 6 combinations of 5 numbers so we need to add
\(6*(\frac{1}{6})^{11}\)
So what do we have
\(\mbox{P(not all numbers are thrown)}=6*(\frac{5}{6})^{11} -15*(\frac{4}{6})^{11}+20*(\frac{3}{6})^{11}-15*(\frac{2}{6})^{11}+6*(\frac{1}{6})^{11} \)
= 0.643793581390032
P(all the numbers are thrown at least once) = 1-0.643793581390032 = 0.356206418609968 
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Now can someone please fill me in on these Monte Carlo simulations I should i just google it :// 
