Question

A and B throw a pair of dice. A wins if he throws 6 before B throws 7 and B wins if he throws 7 before A throws 6. If A begins, find his chance of winning.

Answer 1

Here, we need to find the chance that A wins. To solve this question, we need to use the concept of probability. Probability can be defined as the certainty of an event to occur. First, we will find the favourable outcomes of getting a sum of 6. Then we will find the favourable outcomes of getting a sum of 7. We will then find the probability of A winning and losing and also the probability of B winning and losing on different throws. This will make a geometric progression with infinite terms. Using the sum of a geometric progression with infinite terms, we can simplify the expression and find A’s chance of winning.

Formula Used:
Here, we will use the formula sum of infinite terms of a geometric progression, $\dfrac{a}{{1 - r}}$ where $a$ is the first term and $r$ is the common ratio.

Complete step by step solution:
We know that the total number of outcomes on throwing a pair of dice are ${6^2} = 36$.
First, we will find the probability that A wins or loses on the first throw.
A wins if he gets a sum of 6 on throwing the pair of dice.
Thus, there are 5 favourable outcomes, that is $\left( {1,5} \right),\left( {5,1} \right),\left( {2,4} \right),\left( {4,2} \right),\left( {3,3} \right)$, where the numbers represent the numbers appearing on the two die.
Therefore, we get
$P\left( {{\rm{A wins}}} \right) = \dfrac{5}{{36}}$
We can get the probability that A loses by subtracting the probability that A wins from 1 (because the event that A loses is the complement of the event that A wins).
Therefore, we get
$\begin{array}{l}P\left( {{\rm{A loses}}} \right) = 1 - P\left( {{\rm{A wins}}} \right)\\\ = 1 - \dfrac{5}{{36}}\\\ = \dfrac{{31}}{{36}}\end{array}$
Now, we will find the probability that B wins or loses on the first throw.
B wins if he gets a sum of 7 on throwing the pair of dice.
Thus, there are 6 favourable outcomes, that is $\left( {1,6} \right),\left( {6,1} \right),\left( {2,5} \right),\left( {5,2} \right),\left( {3,4} \right),\left( {4,3} \right)$, where the numbers represent the numbers appearing on the two die.
Therefore, we get
$P\left( {{\rm{B wins}}} \right) = \dfrac{6}{{36}} = \dfrac{1}{6}$
We can get the probability that B loses by subtracting the probability that B wins from 1 (because the event that B loses is the complement of the event that B wins).
Therefore, we get
$\begin{array}{l}P\left( {{\rm{B loses}}} \right) = 1 - P\left( {{\rm{B wins}}} \right)\\\ = 1 - \dfrac{1}{6}\\\ = \dfrac{5}{6}\end{array}$
Now, we need to find the chance that A wins.
A can win on the first throw, on the third throw, on the fifth throw, and so on.
This requires that B does not win whenever he throws the pair of dice.
Therefore, we get probability that A wins on the first throw as
$P\left( {{\rm{A wins}}} \right) = \dfrac{5}{{36}}$
A can win on the third throw only if the first throw by both A and B results in a loss.
Therefore, we get the probability that A wins on the third throw as
$P\left( {{\rm{A loses}}} \right)P\left( {{\rm{B loses}}} \right)P\left( {{\rm{A wins}}} \right) = \dfrac{{31}}{{36}} \times \dfrac{5}{6} \times \dfrac{5}{{36}}$
A can win on the fifth throw only if the first two throws by both A and B results in a loss.
Therefore, we get the probability that A wins on the fifth throw as
$\begin{array}{l}P\left( {{\rm{A loses}}} \right)P\left( {{\rm{B loses}}} \right)P\left( {{\rm{A loses}}} \right)P\left( {{\rm{B loses}}} \right)P\left( {{\rm{A wins}}} \right) = \dfrac{{31}}{{36}} \times \dfrac{5}{6} \times \dfrac{{31}}{{36}} \times \dfrac{5}{6} \times \dfrac{5}{{36}}\\\ = {\left( {\dfrac{{31}}{{36}} \times \dfrac{5}{6}} \right)^2} \times \dfrac{5}{{36}}\end{array}$
This pattern repeats infinitely as long as either A or B wins.
We will add all these probabilities to find the chance that A wins.
Thus, we get the chance that A wins as
$P\left( {\rm{A}} \right) = \dfrac{5}{{36}} + \dfrac{{31}}{{36}} \times \dfrac{5}{6} \times \dfrac{5}{{36}} + {\left( {\dfrac{{31}}{{36}} \times \dfrac{5}{6}} \right)^2} \times \dfrac{5}{{36}} + \ldots$
Now, we can observe that this forms an geometric progression with infinite terms where the first term is $a = \dfrac{5}{{36}}$ and the common ratio is $r = \dfrac{{31}}{{36}} \times \dfrac{5}{6}$.
We know that the sum of infinite terms of a geometric progression is $\dfrac{a}{{1 - r}}$ where $a$ is the first term and $r$ is the common ratio.
Therefore, we get the chance that A wins as
$\begin{array}{l}P\left( {\rm{A}} \right) = \dfrac{{\dfrac{5}{{36}}}}{{1 - \dfrac{{31}}{{36}} \times \dfrac{5}{6}}}\\\ = \dfrac{{\dfrac{5}{{36}}}}{{\dfrac{{216 - 155}}{{216}}}}\\\ = \dfrac{{\dfrac{5}{{36}}}}{{\dfrac{{61}}{{216}}}}\end{array}$
Simplifying the above expression, we get
$P\left( {\rm{A}} \right) = \dfrac{{30}}{{61}}$

$\therefore$ The chance of A winning is $\dfrac{{30}}{{61}}$.

Note:
We should observe that the problem does not state the number of trials. Therefore, it will go on until one of the two players win. A common mistake we can make is to assume one throw of the pair of dice, and leave the answer at $P\left( {{\rm{A wins}}} \right) = \dfrac{5}{{36}}$. Here we need to find the probability after every throw so that we get a geometric progression series. Also, in the question, it is given that a pair of a die is thrown which means that the total outcome is 36. It would be a mistake if we think only one die is thrown as it will give a total outcome as 6. Hence, we will get the wrong answer.