Question

A and B throw with one die for a stake of Rs.11, which is to be won by the player who first throws 6. If A has the first throw, what are their respective expectations?
A.Expectations of A and B are Rs.6 and Rs.5, respectively.
B.Expectations of A and B are Rs.5 and Rs.6, respectively.
C.Expectations of A and B are Rs.7 and Rs.4, respectively.
D.Expectations of A and B are Rs.4 and Rs.7, respectively.

Answer 1

Hint : In this question, we need to determine the expectations (probabilities) for the success of A to the success of B. For this, we will use the general property of the probability, which states that the probability of an event is the ratio of the number of favorable outcomes to the number of total outcomes in the event.

Complete step-by-step answer :
For the success, the die should show 6, so the favorable number of outcomes is 1, while the total number of outcomes is 6.
The probability of success in the first throw of A is $P({A_1}) = \dfrac{1}{6}$ , and so, the probability of failure is the difference of 1 and the probability of success, which is given as $\left( {1 - \dfrac{1}{6}} \right) = \dfrac{5}{6}$.
The probability of the success of A in his second throw will be done when A and B in their first throw loses such that
$P\left( {{A_2}} \right) = P({\text{failure of A)}} \times P({\text{failure of B)}} \times P({\text{success of A)}} \\\ = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} \\\$
Similarly, the probability of the success of A in his third throw will be done when A and B in their first, as well as second throw, loses such that
$P\left( {{A_3}} \right) = P({\text{failure of A)}} \times P({\text{failure of B)}} \times P({\text{failure of A)}} \times P({\text{failure of B)}} \times P({\text{success of A)}} \\\ = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} \\\$
In general, we can write the probability of success of A as:
$P(A) = \dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + ... \\\ = \dfrac{1}{6}\left( {1 + {{\left( {\dfrac{5}{6}} \right)}^2} + {{\left( {\dfrac{5}{6}} \right)}^4} + ...} \right) - - - - (i) \\\$
Now, the probability of success of B in his first throw will be done when A in his first throw loses so,
$P({B_1}) = P(failure{\text{ }}of{\text{ }}A) + P(success{\text{ }}of{\text{ }}B) \\\ = \dfrac{5}{6} \times \dfrac{1}{6} \\\$
The probability of the success of B in his second throw will be done when A and B in their first throw loses and A also loses in his second throw such that
$P\left( {{B_2}} \right) = P({\text{failure of A)}} \times P({\text{failure of B)}} \times P({\text{failure of A)}} \times P({\text{success of B}}) \\\ = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} \\\$
Similarly, the probability of the success of B in his third throw will be done when A and B in their first, as well as second throw, loses and A also loses in his third throw such that
$P\left( {{B_3}} \right) = P({\text{failure of A)}} \times P({\text{failure of B)}} \times P({\text{failure of A)}} \times P({\text{failure of B)}} \times P({\text{failure of A)}} \times P({\text{success of B)}} \\\ = \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} \\\$ In general, we can write the probability of success of B as:
$P(B) = \dfrac{1}{6} \times \dfrac{5}{6} + \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + ... \\\ = \dfrac{1}{6} \times \dfrac{5}{6}\left( {1 + {{\left( {\dfrac{5}{6}} \right)}^2} + {{\left( {\dfrac{5}{6}} \right)}^4} + ...} \right) - - - - (ii) \\\$
Now, dividing equation (i) and (ii), we get
$\dfrac{{P(A)}}{{P(B)}} = \dfrac{{\dfrac{1}{6}\left( {1 + {{\left( {\dfrac{5}{6}} \right)}^2} + {{\left( {\dfrac{5}{6}} \right)}^4} + ...} \right)}}{{\dfrac{1}{6} \times \dfrac{5}{6}\left( {1 + {{\left( {\dfrac{5}{6}} \right)}^2} + {{\left( {\dfrac{5}{6}} \right)}^4} + ...} \right)}} \\\ = \dfrac{6}{5} \\\$
Hence, we can say that the expectations of A and B are Rs.6 and Rs.5, respectively.
So, the correct answer is “Option A”.

Note : It is interesting to note here that as A starts the throw of the dice so, the expectation of A is more than that of B, i.e., expectations of A and B are Rs.6 and Rs.5, respectively. However, if B starts with the throw of the dice, then the expectation of B will be more than A.