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The number of times you have to roll a set of 3 die to see a 6 on each die appears to be around ~10.555 according to a implementation of this in c++ and c#. How would this be statistically proven as instinct would suggest it should be 6 ?

Each die has 6 sides that are evenly distributed. All three die are thrown at the same time counting as one roll. Once all three die have displayed 6 at least once you can stop.

What I am looking for is the statistical proof that you are required on average to roll the set 10.555 times to have seen 6 on each of the 3 die at least once.

Many Thanks

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Can you be more clear with your question ? Do you want to find out the probability or the number of times the set of dice is rolled ? –  lsp Apr 24 '13 at 12:17
    
Question updated :), hope that helps –  S_BatMan Apr 24 '13 at 12:19
1  
when you say, "a 6 on each die" you don't mean "three 6 on the dies", right? Die A may get a 6 on its third cast, die B a 6 on its first and fifth cast and die C a 6 only on its seventh cast? –  mau Apr 24 '13 at 12:29
    
I guess he means at least one time 6 appears on all dies. –  Seyhmus Güngören Apr 24 '13 at 12:31
    
every time you throw the 3 die together that is considered a roll, the requirement to stop rolling is that each die has at one point shown a 6, this dosn't have to be on the same roll. –  S_BatMan Apr 24 '13 at 12:34

2 Answers 2

up vote 6 down vote accepted

$$\mathbb{E}(T)=\sum_{n=0}^\infty \mathbb{P}(T>n) = \sum_{n=0}^\infty 1-\left(1-(5/6)^n\right)^3={10566\over 1001}=10.55544456. $$

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what is T here? –  Seyhmus Güngören Apr 26 '13 at 9:07
    
$T$ is the number of tosses until all three dice have shown a six. –  Byron Schmuland Apr 26 '13 at 17:43
    
of course)))) thx) –  Seyhmus Güngören Apr 26 '13 at 19:28

I can give you a hint why the expected value is not 6.

If you toss a single coin, the expected number of tosses $E_1$ before it shows a head is 2. Let's calculate the expected value $E_2$ that tossing two coins both of them have shown a head at least once. After the first toss, there are four cases, each with probability $\frac{1}{4}$: $HH, HT, TH, TT$. In the first case you are done, in the second and third case you need other $E_1$ tosses before the other coin shows a head, in the fourth case you are back to square 1. Therefore,

$$E_2 = \frac{1}{4}(1 + (1+E_1) + (1+E_1) + (1+E_2)) = \frac{1}{4}(8+E_2)$$

and therefore $E_2 = \frac{8}{3}$.

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