Duel Problem HINT: You will need to use geometric series formula: 1 + q+ q^2 + q

Question

Duel Problem

HINT: You will need to use geometric series formula: 1 + q+ q^2 + q^3 +... = 1/(1-q) Two gunmen A and B are fighting in a duel. The first one, A, has a probability of hitting 75% and the other one B has 45% chances of hitting a target. In each round they shoot simultaneously at each other. The game is played by rounds. At each round the players shoot at each other simultaneously, hence their speed doesn’t matter. The truel is over when ONLY ONE player is still alive or if both of them are dead. We assumed that each player has infinitely many bullets.

b) What is the expected number of rounds in a duel?

Part 2. Repeat part one but instead of given probabilities use pA for probability that A hits a target, pB for probability that B hits a target. Also use qA and qB for probability that A or B miss a target.

b) What is the expected number of rounds in a duel?

Explanation / Answer

Solution

Let X = number of rounds in the duel and p(x) = P(X = x). Then, E(X) = sum (over all possible values of x) of {x. p(x)}.

X = 1, if the duel ends in the first round itself i.e., either both of them kill each other or one kills the other, or put in a different way, at least one of them hits. So, P(duel ends in first round itself)

= P(at least one of them hits) = 1 – P(neither hits) = 1 – {(1- 0.75)(1 – 0.45)} = 1 – 0.1375 = 0.8625

X = 2, if the duel ends in the second round => neither hits in the first round and at least one hits in the second round

So, P(duel ends in second round) = P(neither hits in the first round)xP(at least one hits in the second round) = 0.1375x0.8625

Similarly, P(X = 3) = P(duel ends in third round) = P(neither hits in the first two rounds)xP(at least one hits in the third round) = 0.1375x0.1375x0.8625

Mathematically, the rounds may go on indefinitely. So,

E(X) = {1x0.8625} + {2x(0.1375x0.8625)} + {3x(0.1375)2x0.8625} + ….. to infinity …. (1)

E(X)x0.1375 =           {(0.1375x0.8625)} + {2x(0.1375)2x0.8625} +….. to infinity..…. …(2)

(1) - (2)

(0.8625)E(X) = {1x0.8625} + {(0.1375x0.8625)} + {0.1375)2x0.8625} + .. to infinity.…(3)

(3) is an infinite Geometric Series with first term, a = 0.8625 and common ratio, r = 0.1375 and hence its sum = a/(1 - r) = 0.8625/0.8625 = 1

So, E(X) = 1/0.8625 = 1.16 or rounding off to 2 ANSWER

To answer (b) part, we just need to substitute pA for 0.75, pB for 0.45, qA for 0.25 and qB for 0.55.

So,

E(X) = (1 - qA qB) + 2(qA qB)(1 - qA qB) + 3(qA qB)2(1 - qA qB) + …… to infinity …… (4)

Multiplying (4) by (qA qB) and then subtracting it from (4):

(1 - qA qB)xE(X) = (1 - qA qB) + (qA qB)(1 - qA qB) + (qA qB)2(1 - qA qB) + …… to infinity

= (1 - qA qB)/(1 - qA qB) = 1

So, E(X) = 1/(1 - qA qB) ANSWER

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