Quant job question interview: unfair game probability

• Sequence probabilities;
• Computing stopping-times;
• Use of martingales for fair bets.

Eh.. what´s up, docs?
One more (not so easy) problem for fun:
Consider a game in which two players A and B have probabilities $P_A=$40% and $P_B=$60% of making a point, respectively. The winner will be the player who first makes 3 points. There is no possibility of the players draw a tie. Consider the following questions:

• (A) What is the probability of A wins a match? (piece of cake)
• (B) In average, how many matches are necessary until observes one in which player A wins with 3 points in a row?
• (C) Suppose that a fair bookmaker pays $10 if you win a bet against A. What is the value you need to bet with the bookmaker?

Answer item (A):
In order to calculate this probability, we need to consider every possible scenario in which A wins. They are:
Scenario 1) A makes 3 points, and B zero points $\to $ Probability: $P(B:0)$;
Scenario 2) A makes 3 points, and B 1 point $\to $ Probability: $P(B:1)$;
Scenario 3) A makes 3 points, and B 2 points $\to $ Probability: $P(B:2)$;
In the end, the probability of A wins will simply be the sum of probabilities of each scenario. Let´s take a look at them individually:
Scenario 1:
$P(B:0) = {P_A}^3 = {0.4}^3 = 0.064$
Scenario 2:
$P(B:1) = N {P_A}^3 P_B$
where N is the number of possibilities of having B with one point. In this case: AABA, ABAA, BAAA, thus $N=3$:
$P(B:1)  = 3 * {0.4}^3 * 0.6 = 0.1152$
Scenario 3:
$P(B:2) = N {P_A}^3 {P_B}^2$
In this case, N=6, because the following combinations are possible AABBA, ABBAA, BBAAA, ABABA, BABAA, BAABA. Hence:
$P(B:2) = 6 *{0.4}^3 *{0.6}^2 = 0.13824$

So the probability of A wins a match is: $P(B:0) + P(B:1)+P(B:2)=0.31744$

Answer item (B):
Consider $M$ the average number of matches until one observes A makes 3 points in a row. In other words, $M$ is the average number of games until one observes one of the following matches: AAA, BAAA, or BBAAA. The joint probability of these three matches is: $p={P_A}^3 + {P_A}^3 {P_B}+{P_A}^3 {P_B}^2$, resulting in: $p=0.12544$. Any other configuration add 1 new match to M with probability $(1-p)=0.87456$. Hence, one can obtain $M$ by means of:
$M=p+(1-p)(M+1)$
Therefore, the average number of matches until one observes A making three points in a row is $M \approx 8$.

Answer item (C):
Considering the bookmaker is fair, she will create a bet system which obeys a Martingale process. Therefore, the fair value of the bet is:
V =E[bet payoff]={value if wins the bet }*prob_win + {value if loses the bet }*prob_lose
The probability of  A wins is 0.31744 (calculated in item A), so the probability of B wins is 0.68256. Hence:
$V=\$10 * 0.68256 + \$0 * 0.31744 \to V=\$ 6.83$

That´s all folks! May the force be with you!

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Diogo de Moura Pedroso

LinkedIn: www.linkedin.com/in/diogomourapedroso