Question

A, B, C, D cut a pack of 52 cards successively in the order given. If the person who cuts a spade first receives Rs.350, what are their respective expectations?
I. $A = Rs.128,B = Rs.90,C = Rs.78,D = Rs.54$
II. $A = Rs.128,B = Rs.96,C = Rs.72,D = Rs.54$
III.$A = Rs.120,B = Rs.96,C = Rs.80,D = Rs.54$
IV.$A = Rs.124,B = Rs.96,C = Rs.76,D = Rs.54$

Answer 1

Hint : The probability of choosing a card from the pack of 52 cards is equal to $= {}^{52}{C_1} = 52$
In this question find the probability of winning of each contestant starting from A to D and using the probability of each contestant find the expectation of each contestant.

Complete step-by-step answer :
Let E be the event of cutting a spade from the pack of 52 cards
Let S be the sample of cutting any card
Now since the pack contains 52 cards and a spade is to be cut from the pack, hence we can write
$n\left( E \right) = {}^{13}{C_1}$
$n\left( S \right) = {}^{52}{C_1}$
Now as we know probability of an event to occur is the ratio of the number of favorable outcome to the total numbers of outcome, hence we can write

$$P\left( E \right) = p = \dfrac{{n\left( E \right)}}{{n\left( S \right)}} = \dfrac{{{}^{13}{C_1}}}{{{}^{52}{C_1}}} = \dfrac{{13}}{{52}} = \dfrac{1}{4} \\\ P\left( E \right) = p = \dfrac{1}{4} \\\$$

Now let q be the probability of event if card drawn from the pack is not a spade, hence we can write

$$P\left( {\bar E} \right) = q = 1 - p = 1 - \dfrac{1}{4} = \dfrac{3}{4} \\\ P\left( {\bar E} \right) = q = \dfrac{3}{4} \\\$$

Now the probability of A winning the game when A starts the game will be $= p + \left( {qqqq} \right)p + {\left( {qqqq} \right)^2}p + ....\infty = p + {q^4}p + {q^8}p + ....\infty = \dfrac{p}{{1 - {q^4}}}$
By substituting the values we get $= \dfrac{p}{{1 - {q^4}}} = \dfrac{{\dfrac{1}{4}}}{{1 - {{\left( {\dfrac{3}{4}} \right)}^4}}} = \dfrac{{64}}{{175}}$
Therefore the expectation of A winning the game $= Rs.350 = 350 \times \dfrac{{64}}{{175}} = Rs.128$
Now if A starts the game and he fails so the probability of B winning the game if he goes next will be $= qp + qqqq\left( {qp} \right) + \left( {qqqq} \right){}^2\left( {qp} \right) + ....\infty = \dfrac{{qp}}{{1 - {q^2}}}$
Hence by substituting the values we get $= \dfrac{{qp}}{{1 - {q^2}}} = \dfrac{{\dfrac{3}{4} \times \dfrac{1}{4}}}{{1 - {{\left( {\dfrac{3}{4}} \right)}^4}}} = \dfrac{{48}}{{175}}$
Therefore the expectation of B winning the game $= Rs.350 = 350 \times \dfrac{{48}}{{175}} = Rs.96$
Now if both A and B fails to win so the probability of C winning the game if he goes next will be $= qqp + qqqq\left( {qqp} \right) + \left( {qqqq} \right){}^2\left( {qqp} \right) + ....\infty = \dfrac{{qqp}}{{1 - {q^2}}}$
Hence by substituting the values we get $= \dfrac{{qqp}}{{1 - {q^2}}} = \dfrac{{\dfrac{3}{4} \times \dfrac{3}{4} \times \dfrac{1}{4}}}{{1 - {{\left( {\dfrac{3}{4}} \right)}^4}}} = \dfrac{{36}}{{175}}$
Therefore the expectation of B winning the game $= Rs.350 = 350 \times \dfrac{{36}}{{175}} = Rs.72$
Now if A, B and C all fails so the probability of D winning the game $= qqqp + qqqq\left( {qqqp} \right) + \left( {qqqq} \right){}^2\left( {qqqp} \right) + ....\infty = \dfrac{{qqqp}}{{1 - {q^2}}}$
Hence by substituting the values we get $= \dfrac{{qqqp}}{{1 - {q^4}}} = \dfrac{{\dfrac{3}{4} \times \dfrac{3}{4} \times \dfrac{3}{4} \times \dfrac{1}{4}}}{{1 - {{\left( {\dfrac{3}{4}} \right)}^4}}} = \dfrac{{27}}{{175}}$
Therefore the expectation of D winning the game $= Rs.350 = 350 \times \dfrac{{27}}{{175}} = Rs.54$
So, the correct answer is “Option II”.

Note : A pack of cards contains 52 cards where each club, diamonds, spades and hearts have 13 cards, so the probability of choosing any one cards from club, diamonds, spades and hearts becomes same which is equal to $= {}^{13}{C_1} = 13$
These cards are also grouped by their colors, the pack contains 26 red and 26 black cards.