A fair coin is tossed:

  • If heads: an unbiased die is thrown three times. The sum of the outcomes of the three rolls is recorded.
  • If tails: an unbiased dice is thrown once. The result is recorded.

What are the possible outcomes of each action and the probability of each outcome?

My solution:

  • if heads:

    • Dice 1: $\Pr(\frac{1}{6})$
    • Dice 2: $\Pr(\frac{1}{6})$
    • Dice 3: $\Pr(\frac{1}{6})$

    This outcome will yield $\Pr(\frac{1}{6}) * \Pr(\frac{1}{6}) * \Pr(\frac{1}{6}) = \frac{1}{36}$

  • if tails:

    • Dice: $\Pr(\frac{1}{6})$

    This outcome will yield $\Pr(\frac{1}{6}) = \frac{1}{6}$

Am I on the right track?


2 Answers 2


There are a lot of possible outcomes playing this game so I tried a solution with generating functions. I hope I haven't made a mistake but there was a lot happening so please forgive any oversight.

The generating function for the random variable $X$ representing rolling a fair six-sided die is $G_X(t)=\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \frac{t^4}{6} + \frac{t^5}{6} + \frac{t^6}{6}$

Define $S$ to be the random variable for the sum of three rolls of the die.

As the thee rolls are independent random variables we have:



$ G_S(t)=\frac{t^3}{216} + \frac{t^4}{72} + \frac{t^5}{36} + \frac{5t^6}{108} + \frac{5t^7}{72} + \frac{7t^8}{72}+\frac{25t^9}{216} + \frac{t^{10}}{8} + \frac{t^{11}}{8} + \frac{25t^{12}}{216} + \frac{7x^{13}}{72} + \frac{5t^{14}}{72} + \frac{5t^{15}}{108} + \frac{t^{16}}{36} + \frac{t^{17}}{72} + \frac{t^{18}}{216}$

The flipping of a coin is independent from the rolling of the die. In order to reflect this in the probability generating function we can multiply all the coefficients by $\frac{1}{2}$ for $G_S(t)$ and $G_X(t)$

$\frac{1}{2}G_S(t)= \frac{t^3}{432} + \frac{t^4}{144} + \frac{t^5}{72} + \frac{5t^6}{216} + \frac{5t^7}{144} + \frac{7t^8}{144}+\frac{25t^9}{432} + \frac{t^{10}}{16} + \frac{t^{11}}{16} + \frac{25t^{12}}{432} + \frac{7t^{13}}{144} + \frac{5t^{14}}{144} + \frac{5t^{15}}{216} + \frac{t^{16}}{72} + \frac{t^{17}}{144} + \frac{t^{18}}{432}$

$\frac{1}{2}G_X(t)=\frac{t}{12} + \frac{t^2}{12} + \frac{t^3}{12} + \frac{t^4}{12} + \frac{t^5}{12} + \frac{t^6}{12}$

The final probability generating function is $G(t)$ is found (assuming you are only interested in the final total) by summing the previous generating functions because they represent mutually exclusive events or $G(t)=\frac{1}{2}G_S(t)+\frac{1}{2}G_X(t)$.

The final generating function is therefore:

$G(t)= \frac{t}{12} + \frac{t^2}{12}+\frac{37t^3}{432} + \frac{13t^4}{144} + \frac{7t^5}{72} + \frac{23t^6}{216} + \frac{5t^7}{144} + \frac{7t^8}{144}+\frac{25t^9}{432} + \frac{t^{10}}{16} + \frac{t^{11}}{16} + \frac{25t^{12}}{432} + \frac{7t^{13}}{144} + \frac{5t^{14}}{144} + \frac{5t^{15}}{216} + \frac{t^{16}}{72} + \frac{t^{17}}{144} + \frac{t^{18}}{432}$

As an example the probability of the outcome $14$ is $\frac{5}{144}$.

Hope this helps.

  • $\begingroup$ I have up-voted your answer, thank you! $\endgroup$
    – lucidgold
    May 7, 2015 at 3:59
  • $\begingroup$ You're welcome, it was an interesting problem. $\endgroup$
    – Karl
    May 7, 2015 at 13:41

As I have understood the problem, I explain the problem as follows. But please note that there may be alternative solutions, and please get others response as well !!!

I may get following outcomes:

Having $H$ or $T$: $P(H)=P(T)=\frac{1}{2}$. Having any number $N\in\{1,2,3,4,5,6\}$: $P(N)=\frac{1}{6}$.

It is easy to start with having tail, then possible outcomes (as only one die is thrown): $\{T,1\},\{T,2\},\cdots,\{T,6\}$, and thus for each out come: $P(\{T,N\})=\frac{1}{2}\times\frac{1}{6}=\frac{1}{12}$.

When we have $H$, we throw die 3 times and get the sum, i.e., if we get $1,1,1$, the sum is 3. Like that we may get $6,6,6$, the sum is 18. Therefore we may have sum $S\in\{3,4,\cdots,18\}$. Thus outcomes may be: $\{H,3\},\{H,4\},\cdots,\{H,18\}$.

Now consider the probability of each outcome. For example, we get sum 3 only when we have $1,1,1$, and thus $$P(\{H,3\})=\frac{1}{2}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}=\frac{1}{432}$$

we get sum 4, when we have $1,1,2$ or $1,2,1$ or $2,1,1$, and thus $$P(\{H,4\})=\frac{1}{2}\times\frac{1}{6}\times\frac{1}{6}\times\frac{1}{6}\times3=\frac{3}{432}$$

I think that you can get all outcomes, as most simple way, by using a tree diagram.

  • $\begingroup$ you are welcome !!! $\endgroup$
    – Frey
    May 7, 2015 at 4:31

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