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My daughter threw a die (D6) six times and got: 6,2,2,6,2,6.

This got us wondering whether having only two distinct numbers come up in six throws was very unlikely or would happen fairly often. How would we go about working out the answer to that question? (short of doing it lots of times and counting!)

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    $\begingroup$ Real dice are different that abstract dice. $\endgroup$ – Masacroso Oct 28 '14 at 9:20
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    $\begingroup$ Right, but presumably not in a way that I can quantify or determine, so the 'theoretical' answer is still the best I can do, isn't it? $\endgroup$ – user76445 Oct 28 '14 at 9:22
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    $\begingroup$ See here or here, maybe interesting. $\endgroup$ – Masacroso Oct 28 '14 at 9:32
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    $\begingroup$ @Masacroso I think it's fairly safe to say that real dice are more or less fair. You are overcomplicating this answer by quite a margin, I think. The question, as far as I see it, is simply "what is the probability of rolling only two numbers when rolling a 6 sided die six times?" and it has a simple solution that does not even include experimenting. $\endgroup$ – 5xum Oct 28 '14 at 9:37
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    $\begingroup$ @Masacroso Citation, please, for your implicit claim that real dice are not very well approximated by independent uniform random variables. $\endgroup$ – David Richerby Oct 28 '14 at 11:23
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There are a total of $6^6$ possible outcomes of rolling a die six times. How many of those contain exactly $2$ distinct numbers?

Well, there are ${6\choose 2} = 15$ different pairs of numbers from $1$ to $6$, and for each pair of numbers, there are $2^6$ rolls of dice that contain only those two numbers. However, two of those contain only one of the numbers, so $2^6- 2$ is the number of rolls of two numbers that contains both of those numbers.

All together, this means there are $15\cdot (2^6-2)$ good outcomes out of $6^6$, so the probability of this occuring is $$\frac{15(2^6-2)}{6^6} \approx 1.99331\%$$ so what happened is something that happens rougly every once in $50$ tries. Rare, but not much rarer than rolling a double six with two dice, for example.

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    $\begingroup$ +1 for the comparison with a familiar probability at the end. $\endgroup$ – David Richerby Oct 28 '14 at 11:21
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    $\begingroup$ Here's an embarrassing question - what is (6 2). Six over two in parentheses? You don't have to explain it. I can look it up if I know what it is. :-) $\endgroup$ – Mayo Oct 28 '14 at 21:10
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    $\begingroup$ @Mayo en.wikipedia.org/wiki/Binomial_coefficient $\endgroup$ – peaceoutside Oct 28 '14 at 21:15
  • $\begingroup$ thx peaceoutside. $\endgroup$ – Mayo Oct 28 '14 at 21:20
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Here is an alternative solution to the (very good) answer already given here:

The number of ways that you can choose $2$ out of $6$ values is $\dbinom{6}{2}=15$.

Given $2$ values A and B, the number of combinations that you can get in $6$ rolls is:

  • Value A appearing $1$ time and value B appearing $5$ times: $\dbinom{6}{1}=6$
  • Value A appearing $2$ times and value B appearing $4$ times: $\dbinom{6}{2}=15$
  • Value A appearing $3$ times and value B appearing $3$ times: $\dbinom{6}{3}=20$
  • Value A appearing $4$ times and value B appearing $2$ times: $\dbinom{6}{4}=15$
  • Value A appearing $5$ times and value B appearing $1$ time : $\dbinom{6}{5}=6$

So you can get $6+15+20+15+6=62$ combinations containing A and B.

And you can get $15\cdot62=930$ combinations containing any $2$ out of $6$ values.

The total number of combinations that you can get in $6$ rolls is simply $6^6=46656$.

Therefore, the probability of having exactly $2$ values in $6$ rolls is $\dfrac{930}{46656}\approx0.0199$.

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    $\begingroup$ Just a note: the $62$ calculated here through sumation is precisely the $2^6 - 2$ calculated in my answer. This is due to the fact that $\sum_{i=1}^6{6\choose i} = 2^6$. $\endgroup$ – 5xum Oct 28 '14 at 12:12
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    $\begingroup$ @5xum: I agree (except for the fact that you need to change i=1 to i=0 in the equation above). I just figured it might help if I showed OP another way for calculating that value... $\endgroup$ – barak manos Oct 28 '14 at 12:28
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    $\begingroup$ Of course, $i$ must be set to $0$. I fully agree it is useful to show the OP another way, I just added a clarification so he wouldn't think $62$ is somehow magical in that it appears where you need it:P $\endgroup$ – 5xum Oct 28 '14 at 12:48
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    $\begingroup$ @5xum Well, it is interesting that the answers keep on using $2^6$, 62, and other combination of 2 and 6 - the same numbers the OP kept on rolling. $\endgroup$ – Ypnypn Oct 29 '14 at 1:44
  • $\begingroup$ @Ypnypn: It's also worth noting that so far I have asked $62$ questions on this website. In addition, I have answered $216$ questions, and in my last ($216$th) answer on a probability question, the size of the sample-space was $216$ (math.stackexchange.com/a/995427/131263)... Nevertheless, it is generally just a matter of chance. OP could have just as well asked about $1$ and $5$ (for example). $\endgroup$ – barak manos Oct 29 '14 at 5:53
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This was merely a coincidence as by no principle it's common to happen. total possibilities can be easily extracted from binomial distribution as taking 1/3 a success for 6 trials which obviously will give you the truth.

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  • $\begingroup$ The binomial distribution isn't what is useful here. Binomial distributions are sums of random variables taking the values $0$ and $1$, here, you have $6$ variables. There is no easy and obvious way to implement binomial distributions in here. $\endgroup$ – 5xum Oct 28 '14 at 13:25
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    $\begingroup$ it's about success and failure of these two events including 2,6 $\endgroup$ – Ved Prakash vats Oct 28 '14 at 14:12
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    $\begingroup$ But the question is how often can you have a sequence that only contains any two numbers, not how you can have a sequence with only $2$s and $6$s $\endgroup$ – 5xum Oct 28 '14 at 14:14
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    $\begingroup$ "Easily" and "obviously" aren't very helpful on a Q&A site, unless they're accompanied by answers. $\endgroup$ – Dancrumb Oct 28 '14 at 22:16

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