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Here is my guess: the probability of summing $7$ on two rolls is $\frac 16$. This means if I repeat the experiment many times I'll roll $7$ one sixth of them (approximately). Hence,

$$N \cdot \bigg(\cfrac 16\bigg) \cdot 7 = 7$$

where $N$ is the total number of rolls. That gives me a total number of $6$ rolls on average to sum $7$.

I'm not quite sure so I'm all open to suggestions! Thanks in advance.

Rodrigo de Azevedo
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Juan123
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    The probability that a pair of dice add to seven is indeed $\frac{1}{6}$. The intuitive interpretation of the probability here is infact the correct one, that on average one in six rolls will give a sum of seven, and it so happens that indeed $\frac{1}{p}$ or in this case $\frac{1}{1/6}$, i.e. $6$ is the expected number of rolls until you get a sum of seven. – JMoravitz Feb 02 '18 at 20:10
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    Sounds like a negative binomial distribution stopping at the first success. – Antoni Parellada Feb 02 '18 at 20:13
  • If you want a more formal proof of this, consider reading my answer to a related question here. – JMoravitz Feb 02 '18 at 20:14
  • @AntoniParellada more easily described as a geometric distribution. – JMoravitz Feb 02 '18 at 20:18
  • @JMoravitz Agree. – Antoni Parellada Feb 02 '18 at 20:19
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    I don't know why you are multiplying by seven. $7$ really has no numeric purpose in this question. You could have similarly asked about how many times you roll until you get snake eyes or a total of eight. That event is also probability $\frac{1}{6},$ and the expected rolls until you get that event is the same. So your multiplication is mysterious. – Thomas Andrews Feb 02 '18 at 20:27
  • I mean, if p=1/6 then I'll expect to succeed N/6 times so I then multiply by 7 to work out the expected value. – Juan123 Feb 02 '18 at 20:51
  • @Juan123 If, as you say, you "expect to succeed $N/6$ times", then what does "the total of excepted value" mean and why do you find it multiplying by $7$? – JiK Feb 02 '18 at 21:07
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    In case anybody landing here wondered about geometric v. negative binomial, the result is exactly the same. For the geometric distribution (probability that the first occurrence of success requires k independent trials, each with success probability p.), the expectation $E(X)=1/p=6.$ The negative binomial counts the total number of trials before $r$ successes - $r=1$ in this case - with the probability of failure $q= 1-p=5/6,$ as $\frac{r}{(1-q)}=6.$ I agree that NB is unnecessarily complex for this. – Antoni Parellada Feb 02 '18 at 21:26
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    @ThomasAndrews Did you mean to imply that the probability of rolling snake eyes is 1/6? Because it is not. – ale10ander Feb 02 '18 at 22:19
  • Did you read what I wrote? "snake eyes or a total of eight." @ale10ander – Thomas Andrews Feb 02 '18 at 22:29
  • Yes, the probability of getting snake eyes is 1/36 and that of a sum eight is 5/36. Neither is 1/6. Your statement was ambiguous - snarking others for it doesn't help. @ThomasAndrews – Nij Feb 03 '18 at 00:34
  • I apologize; I initially interpreted your comment as two scenarios, both equal to 1/6: the first being snake eyes, the second being a total of 8. I now understand you to mean the one scenario of either snake eyes or a total of 8. – ale10ander Feb 03 '18 at 00:34
  • @Nij The probability of getting snake eyes or eight is $1/6$. – Thomas Andrews Feb 03 '18 at 00:59

3 Answers3

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If $X$ is the number of rolls to get $7$ then the expected (or average) value of $X$ satisfies:

$$E(X)=1+\frac{5}{6}E(X)$$

That is, we always start with one roll, and $5/6$ of the time, we just start all over again. So $E(X)=6.$

Technically, as Heinrich comments below, this only proves that either $E(X)=6$ or $E(X)=+\infty.$ You might actually need some trick to prove that the expected value must be finite.

Thomas Andrews
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    Maybe there's no need to use power series, but there could just be the joy of power series to motivate us to use them. ;-) – Todd Wilcox Feb 02 '18 at 21:19
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    Even better, if you know the probability is 1/6 then you just take the inverse to get 6 immediately. – Robin Saunders Feb 02 '18 at 22:01
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    Sure, but this explains why you can do that. @RobinSaunders – Thomas Andrews Feb 02 '18 at 22:02
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    This assumes that the expected value of $X$ exists and is finite. – heinrich5991 Feb 03 '18 at 03:10
  • Well, the expected value exists, because any positive random variable has an expected value. However, you are correct it might be infinite, @heinrich5991 – Thomas Andrews Feb 03 '18 at 21:55
  • Just a guess is all, but if the expected value was indeed infinite in this case, wouldn't they diverge at the same rate? That is, if we let $f(x) = x$ and $g(x) = 1+5x/6$, wouldn't we expect the limit of $f(x) / g(x)$ as $x$ goes to infinity to be equal to $1$? – WaveX Feb 04 '18 at 14:29
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The probability of doing it after one roll is $1/6$, in two is $5/6 \times 1/6$, in three $(5/6)^2 1/6$ and so on ... we get

\begin{eqnarray*} E(7)=1 \times \frac{1}{6} + 2 \times \frac{5}{6} \times\frac{1}{6} + 3 \times \left(\frac{5}{6}\right)^2 \times\frac{1}{6}+\cdots = \frac{1}{6} \sum_{i=1}^{\infty} i \left( \frac{5}{6} \right)^i \\ \end{eqnarray*} Now recall that \begin{eqnarray*} \sum_{i=1}^{\infty} i x^{i-1} =\frac{1}{(1-x)^2}. \end{eqnarray*} So \begin{eqnarray*} E(7)= \frac{1}{6} \sum_{i=1}^{\infty} i \left( \frac{5}{6} \right)^{i-1} =\frac{1}{6} \frac{1}{(1-\frac{5}{6})^2}=6 \end{eqnarray*} So the expected value is $\color{red}{6}$ as expected.

Donald Splutterwit
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5

You are correct that it will take on average $6$ rolls, IF we are considering the tossing of both dice together as one roll. If we let $X$ be the number of rolls until we reach a sum of $7$, then we can model this using a Geometric Distribution.

You have already calculated that the probability of rolling a sum of $7$ to be $1/6$. Therefore the probability that you don't roll a sum of $7$ is $5/6$.

The distribution for the probability that it will take $k$ rolls to reach a sum of $7$ will be $$P(X=k) = (5/6)^{k-1} * (1/6)$$

You can then find the mean of this distribution, which turns out to be $$\frac{1}{1/6} =6$$ I'll leave the derivation of this for you to look up :)

WaveX
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