I'm trying to fight out what are the chances of double 1 or double 6 I know that for 2 six-faced dice (aka 2D6) the probability would be 2/36. What are the chances for 3D6? 4D6? 5D6? and etc?
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What does the notation $3D6$ mean? – Michael Albanese Jun 05 '14 at 06:53
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@MichaelAlbanese: I think it means 3 six-faced die; nD6 is n 6-faced die. – user99680 Jun 05 '14 at 07:14
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@user99680: correct – Eldwin Jun 05 '14 at 07:21
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Assume you have $nD6$ , i.e., that you throw n six-faced die . I am assuming also you want the probability of having exactly a double 1 or a double six. Then you have nP2 ways of having two die with the same outcome, each with probability $1/6$. The rest of the die must land in a face other than the repeated value, so all other $n-2$ outcomes will have probability $5/6$. Then, when you throw all $n$ die, you can have two equal outcomes in $nP2$ ways, EDIT the other $(n-2)$ throws can happen in $(6-1)^{n-2 }=5^{n-2}$ ways (the other outcomes can each be anything other than the repeated value) and the remaining $n-2$ outcomes each will have probability $5/6$ for a probability of $$nP2 (5^{n-2})(1/6)^2 (5/6)^{n-2} $$
user99680
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@André Nicolas: Edited; my bad, should have been a permutation, not combination. I was also doing a different problem with the distinguishable die. – user99680 Jun 05 '14 at 07:11
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It is not clear what OP is asking for, at least one double $1$ or double $6$? Exactly one? For neither of these will the above answer be in general right. – André Nicolas Jun 05 '14 at 07:16
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@user99680 hmm intersting but it doesn't seem to work or I've either misunderstood? In the first formula nC2=n!/2!(n−2)! is n = the number of dice or the number I want? Because if I plugged it in as double 1 or double 6 for 3D6 the result i get is 1.38% which is wrong? – Eldwin Jun 05 '14 at 07:18
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@Eldwin: Please ignore, I was thinking of something else, let me rewrite. Sorry. – user99680 Jun 05 '14 at 07:18
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@user99680: nP2 = (1/6)^2*(5/6)^n-2? sorry not how to do the power stuff. I've inputted that formula but doesn't spit out the right results...unless I've misunderstood? Sorry been a long time since I've down probability.. – Eldwin Jun 05 '14 at 07:39
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@Eldwin: No it is $nP2(1/6)^2(5/6)^{n-2}(5^{n-2})$. If this is not correct, I will just delete the post; sorry for taking so much of your time. I think there may be a factor $5^{n-2}$ there two that I did not put in (which I included now). – user99680 Jun 05 '14 at 07:41
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@user99680: What's the probability of getting 1 set of doubles for n six faced dices? – Eldwin Jun 05 '14 at 07:43
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@user99680: Thanks for the help! I think i got it. Just need to research some of the maths reference! Thanks! – Eldwin Jun 05 '14 at 07:48
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The way I would approach it is given $n$ six-sided dice, we need $2$ of them to show either both ones or both sixes, and $n-2$ of them to show anything but what is showing on the pair or the other double.
There are $n \choose 2$ ways to pick the ‘magic’ $2$ dice out of the $n$. These dice have a $\frac{1}{18}$ chance of showing either double 1's or double 6's: $\frac{1}{3}$ chance of picking a 1 or a 6 on the first throw, and a $\frac{1}{6}$ chance of the second matching the first.
In the remaining $n-2$ dice, we have two possibilities. One is that the remaining dice all show something other than 1 or 6. The second is that there is at most one 1 or one 6 is seen—not the one selected for the ‘magic’ pair.
Case 1 should have probability ${{n-2} \choose {n - 2}}\left(\frac{4}{6}\right)^{n - 2}$
Case 2 should have probability ${{n-2} \choose 1}\left(\frac{1}{6}\right) {{n-3} \choose {n - 3}}\left(\frac{4}{6}\right)^{n - 3}$
The sum of these two cases is: $$ \left(\frac{2}{3}\right)^{n - 2} + \left(\frac{2}{3}\right)^{n - 3} \frac{n - 2}{6}\\ =\left(\frac{2}{3}\right)^{n - 3}\left(\frac{2}{3} + \frac{n - 2}{6}\right)\\ =\left(\frac{2}{3}\right)^{n - 3}\left(\frac{n + 2}{6}\right) $$
If correct, the probability would be: $$ {n \choose 2}\frac{1}{18}\left(\frac{2}{3}\right)^{n - 3}\left(\frac{n + 2}{6}\right) $$
Avraham
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