[Math] Rolling 7 with two dice

diceprobability

In certain board game, two fair dice are rolled. If a pair of the same number is obtained, the dice are rolled for the second time. The player continues rolling those two dice and he stops when the numbers turn up on the dice are different. Evaluate the probability that a player obtains a total of seven points when he rolls the dice as described.

I answered this twice and got it wrong. Can anyone please help me?

PS This lesson is about bayes theorem so that might help a little.

EDIT: It said the answer is $\frac{3997}{23328}$.

Best Answer

In certain board game, two fair dice are rolled. If a pair of the same number is obtained, the dice are rolled for the second time. The player continues rolling those two dice and he stops when the numbers turn up on the dice are different. Evaluate the probability that a player obtains a total of seven points when he rolls the dice as described.

There are apparently two interpretations of this statement, and I will discuss the probabilities associated with each. It is not made perfectly clear from the question how exactly the points are calculated.

  • Possibility 1: The points are just counted from the final roll and no other rolls matter.

E.g. The sequence of rolls $(1,1),(1,1),(1,2)$ would end in a failure since the "total" here would be $1+2=3$ whereas the sequence of rolls $(1,1),(3,4)$ would be considered a success since the final roll $(3,4)$ has $3+4=7$.

Under this interpretation, we may use a conditional probability argument to convince ourselves that we may look instead at the restricted sample space $\{(1,2),(1,3),\dots,(2,1),(2,3),\dots,(6,4),(6,5)\}$ which has only $30$ members, six of which correspond to success. This would have the probability be then $\frac{6}{30}=0.2$

If it is hard to convince yourself of this, then you may approach in a more tedious way: letting $p$ be the probability of rolling doubles on a specific roll and $q$ the probability of success on a specific roll, i.e. $p=q=1/6$, we would have the probability as $=q+pq+p^2q+p^3q+\dots+p^nq+\dots = \frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+\dots = \frac{1/6}{1-1/6}=0.2$, same answer as before

  • Possibility 2: The points are counted from all rolls combined

E.g. the sequence of rolls $(1,1),(1,1),(1,2)$ would end in a success this time since the "total" here would be $1+1+1+1+1+2=7$ whereas the sequence of rolls $(1,1),(3,4)$ would be considered a failure since the total here is $1+1+3+4=9$.

Under this interpretation, we use a great deal more casework. Fortunately, the desired total is small and we are guaranteed to have won or lost by the third roll.

The possibilities:

  • Win on the first roll
  • Win on the second roll after having rolled (1,1)
  • Win on the second roll after having rolled (2,2)
  • Win on the third roll after having rolled (1,1) twice in a row

Convince yourself that there are no other possibilities. We could not for example win after having rolled $(3,3)$ on the first turn because the total would be $3+3+\dots+a+b>7$.

So then, we continue by counting probabilities and combining for a final total probability of:

$$\frac{1}{6}+\frac{1}{36}\cdot\frac{4}{36} + \frac{1}{36}\cdot\frac{2}{36}+\frac{1}{36}\cdot\frac{1}{36}\cdot\frac{2}{36} = \frac{3997}{23328}$$

The terms in the above are calculated directly via multiplication principle and direct counting. For example, the final term corresponds to rolling a (1,1) followed by a (1,1) followed by something which adds to three (noting that no combination that adds to three will result in a pair).

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