Question

A makes a bet with B of 5s. to 2s. that in a single throw with two dice he will throw seven before B throws four. Each has a pair of dice and they throw simultaneously until one of them wins: find B’s expectation.

Answer 1

In this particular question if the ratio of their bet is a:b then the share of first person is $\left( {\dfrac{a}{{a + b}}} \right)$ and the share of second person is $\left( {\dfrac{b}{{a + b}}} \right)$, then find all the possibilities to come 7 and 4 when a pair of dice is thrown, then use the concept that probability is the ratio of favorable number of outcomes to the total number of outcomes so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
A makes a bet with B of 5s. to 2s.
Therefore, A:B = 5:2
Therefore, S (A) = $\dfrac{5}{{2 + 5}} = \dfrac{5}{7}$
And, S (B) = $\dfrac{2}{{2 + 5}} = \dfrac{2}{7}$
Now when we throw a pair of dice the possibilities to come seven are
7 = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), and (6, 1) = 6 possibilities.
And possibilities to come 4 are
4 = (1, 3), (2, 2), and (3, 1) = 3 possibilities.
Now the total possibilities when we throw a pair of dice = $\left( {6 \times 6} \right) = 36$
So the probability to come 7, P (7) = $\dfrac{6}{{36}} = \dfrac{1}{6}$
And the probability to come 4, P (4) = $\dfrac{3}{{36}} = \dfrac{1}{{12}}$
So probability to not come 7, P’ (7) = $\left( {1 - \dfrac{1}{6}} \right) = \dfrac{5}{6}$
And the probability to not come 4, P’ (4) = $\left( {1 - \dfrac{1}{{12}}} \right) = \dfrac{{11}}{{12}}$
Now in a single throw with two dice A will throw seven before B throws four until one of then wins.
Now we have to find the B’s expectation.
So, B’s expectation = $\left[ {P\left( 4 \right) \times S\left( B \right)} \right]$ + $\left[ {P'\left( 4 \right) \times S\left( B \right) + P'\left( 7 \right) \times S\left( A \right)} \right]\left[ {P\left( 4 \right) \times S\left( B \right)} \right]$+ ${\left[ {P'\left( 4 \right) \times S\left( B \right) + P'\left( 7 \right) \times S\left( A \right)} \right]^2}\left[ {P\left( 4 \right) \times S\left( B \right)} \right]$+....................... up to infinite
Now substitute the values we have,
So, B’s expectation =$\left( {\dfrac{1}{{12}} \times \dfrac{2}{7}} \right) + \left( {\dfrac{{11}}{{12}} \times \dfrac{2}{7} + \dfrac{5}{6} \times \dfrac{5}{7}} \right)\left( {\dfrac{1}{{12}} \times \dfrac{2}{7}} \right) + {\left( {\dfrac{{11}}{{12}} \times \dfrac{2}{7} + \dfrac{5}{6} \times \dfrac{5}{7}} \right)^2}\left( {\dfrac{1}{{12}} \times \dfrac{2}{7}} \right) + ...........\infty$
Now simplify we have,
So, B’s expectation =$\left( {\dfrac{1}{{42}}} \right) + \left( {\dfrac{{11}}{{42}} + \dfrac{{25}}{{42}}} \right)\left( {\dfrac{1}{{42}}} \right) + {\left( {\dfrac{{11}}{{42}} + \dfrac{{25}}{{42}}} \right)^2}\left( {\dfrac{1}{{42}}} \right) + ...........\infty$
So, B’s expectation =$\dfrac{1}{{42}}\left[ {1 + \left( {\dfrac{{11}}{{42}} + \dfrac{{25}}{{42}}} \right) + {{\left( {\dfrac{{11}}{{42}} + \dfrac{{25}}{{42}}} \right)}^2} + ...........\infty } \right]$
So, B’s expectation =$\dfrac{1}{{42}}\left[ {1 + \left( {\dfrac{{36}}{{42}}} \right) + {{\left( {\dfrac{{36}}{{42}}} \right)}^2} + ...........\infty } \right]$
So, B’s expectation =$\dfrac{1}{{42}}\left[ {1 + \left( {\dfrac{6}{7}} \right) + {{\left( {\dfrac{6}{7}} \right)}^2} + ...........\infty } \right]$
Now the series represents the infinite G.P series, with first term, a = 1, common ratio (r) = $\left( {\dfrac{6}{7}} \right)$
Now as we know that the sum of an infinite series of G.P = $\dfrac{a}{{1 - r}}$
So, B’s expectation = $\dfrac{1}{{42}}\left[ {\dfrac{1}{{1 - \dfrac{6}{7}}}} \right]$
So, B’s expectation = $\dfrac{1}{{42}}\left[ {\dfrac{7}{{7 - 6}}} \right] = \dfrac{7}{{42}} = \dfrac{1}{6}$
So this is the required answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall that the B’s expectation is nothing but the B’s chance of winning which is given as, $\left[ {P\left( 4 \right) \times S\left( B \right)} \right]$ + $\left[ {P'\left( 4 \right) \times S\left( B \right) + P'\left( 7 \right) \times S\left( A \right)} \right]\left[ {P\left( 4 \right) \times S\left( B \right)} \right]$+ ${\left[ {P'\left( 4 \right) \times S\left( B \right) + P'\left( 7 \right) \times S\left( A \right)} \right]^2}\left[ {P\left( 4 \right) \times S\left( B \right)} \right]$+....................... up to infinite, so just calculate these values as above and substitute in this formula and simplify we will get the required answer.