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AkamaiBrah wrote: Part 1: Gunman X and gunman Y are in a [#permalink]
23 Jul 2003, 03:45

Quote:

AkamaiBrah wrote: Part 1: Gunman X and gunman Y are in a duel. Let's say that gunman X will kill his target with one shot with probability P, and gunman Y will kill his target with one shot with probability Q. If gunman X shoots first, and they alternate shots until somebody dies, what is the probability that gunman X will survive?

Useful identity: if 0 < r < 1, then a*(1 + r + r^2 + r^3 + ... + r^infinity) converges to: a / (1 - r).

(A) P / (P + Q) (B) P / ((1 - P)(1 - Q)) (C) P / (1 - (1 - P)(1 - Q)) (D) P / (1 - P * Q) (E) P * (1 - Q)(1 - P)

C is the correct answer.

Part 2.

Gladiators A, B, and C are engaged in deadly game. AFter drawing straws, the order of play is A, then B, then C, then A again and so on. A is given a gun with one bullet and he must shoot at a target. After A shoots, B is given a gun with one bullet and he must shoot at target, and so on. A is not a very good shot and has a 50-50 chance of killing his target when he shoots to kill. B is a sharpshooter and is guaranteed to kill his target. C is a good shot, and will kill his target 80% of the time. There are three guards, one is a sharpshooter who never misses, and the other two kill their target 80% of the time. In addition, the guards have automatic weapons, an unlimited number of bullets, and will fire immediately and rapidly if either shot upon, or if a gladiator tries to escape. Assuming that B and C will always choose their best strategy, what strategy should you recommend for A in order to maximize his chances of survival? (Please justify your answer -- the results of part 1 may come in handy.)

(A) Try to kill B
(B) Try to kill C
(C) Try to escape
(D) Try to kill the sharpshooting guard, then try to escape
(E) Shoot himself in the foot.
_________________
Best,

AkamaiBrah
Newport Beach _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

A. kill B. --- A kills B, C misses, A kills C = 0.5*0.2*0.5=0.05
B. kill C. --- A kills C, B misses A = 0.5*0=0
C. escape --- The sniper guard kills A. Chances = 0
D. kill a sniper and escape. --- A kills sniper, the two other guard miss = 0.5*0.2*0.2=0.02
E. shoot his foot --- A shots his foot (does not affect the chances to survive), B kills C, A kills B = 1*0.5=0.5

Challenge Problem: Probability Part 2: [#permalink]
23 Jul 2003, 04:33

I think the answer is A

Reason: Lets look at A's probability of survival under each choice

E is trivial
D) A succeeds and the other two guards miss = 0.5*0.2*0.2=0.02
C) 0 since sharpshooter guard will kill him
B) Even if A succeeds, B will kill him. A will survive only if he fails, B chooses to shoot at C and then A succeeds in killing B = 0.125
A) A survives if he succeeds in killing B, then C fails and then A succeeds in killing C OR if A fails, B chooses to kill C and then A succeeds in killing B = 0.175

Therefore, the probability of A's survival is maximum under choice A

A. kill B. --- A kills B, C misses, A kills C = 0.5*0.2*0.5=0.05 B. kill C. --- A kills C, B misses A = 0.5*0=0 C. escape --- The sniper guard kills A. Chances = 0 D. kill a sniper and escape. --- A kills sniper, the two other guard miss = 0.5*0.2*0.2=0.02 E. shoot his foot --- A shots his foot (does not affect the chances to survive), B kills C, A kills B = 1*0.5=0.5

E seems to be the best one.

Hmmm.

Analysis is a little sloppy. What if A misses either B or C in first two scenarios? _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Re: Challenge Problem: Probability Part 2: [#permalink]
23 Jul 2003, 04:48

prashant wrote:

I think the answer is A

Reason: Lets look at A's probability of survival under each choice

E is trivial D) A succeeds and the other two guards miss = 0.5*0.2*0.2=0.02 C) 0 since sharpshooter guard will kill him B) Even if A succeeds, B will kill him. A will survive only if he fails, B chooses to shoot at C and then A succeeds in killing B = 0.125 A) A survives if he succeeds in killing B, then C fails and then A succeeds in killing C OR if A fails, B chooses to kill C and then A succeeds in killing B = 0.175

Therefore, the probability of A's survival is maximum under choice A

nice try. think it through a little more carefully.... _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

Re: Challenge Problem: Probability Part 2: [#permalink]
23 Jul 2003, 06:03

AkamaiBrah wrote:

prashant wrote:

I think the answer is A

Reason: Lets look at A's probability of survival under each choice

E is trivial D) A succeeds and the other two guards miss = 0.5*0.2*0.2=0.02 C) 0 since sharpshooter guard will kill him B) Even if A succeeds, B will kill him. A will survive only if he fails, B chooses to shoot at C and then A succeeds in killing B = 0.125 A) A survives if he succeeds in killing B, then C fails and then A succeeds in killing C OR if A fails, B chooses to kill C and then A succeeds in killing B = 0.175

Therefore, the probability of A's survival is maximum under choice A

nice try. think it through a little more carefully....

Okay, you've got me there ..... I can't think of an alternative approach...

Re: Challenge Problem: Probability Part 2: [#permalink]
23 Jul 2003, 10:49

AkamaiBrah wrote:

Quote:

AkamaiBrah wrote: Part 1: Gunman X and gunman Y are in a duel. Let's say that gunman X will kill his target with one shot with probability P, and gunman Y will kill his target with one shot with probability Q. If gunman X shoots first, and they alternate shots until somebody dies, what is the probability that gunman X will survive?

Useful identity: if 0 < r < 1, then a*(1 + r + r^2 + r^3 + ... + r^infinity) converges to: a / (1 - r).

(A) P / (P + Q) (B) P / ((1 - P)(1 - Q)) (C) P / (1 - (1 - P)(1 - Q)) (D) P / (1 - P * Q) (E) P * (1 - Q)(1 - P)

C is the correct answer.

Part 2.

Gladiators A, B, and C are engaged in deadly game. AFter drawing straws, the order of play is A, then B, then C, then A again and so on. A is given a gun with one bullet and he must shoot at a target. After A shoots, B is given a gun with one bullet and he must shoot at target, and so on. A is not a very good shot and has a 50-50 chance of killing his target when he shoots to kill. B is a sharpshooter and is guaranteed to kill his target. C is a good shot, and will kill his target 80% of the time. There are three guards, one is a sharpshooter who never misses, and the other two kill their target 80% of the time. In addition, the guards have automatic weapons, an unlimited number of bullets, and will fire immediately and rapidly if either shot upon, or if a gladiator tries to escape. Assuming that B and C will always choose their best strategy, what strategy should you recommend for A in order to maximize his chances of survival? (Please justify your answer -- the results of part 1 may come in handy.)

(A) Try to kill B (B) Try to kill C (C) Try to escape (D) Try to kill the sharpshooting guard, then try to escape (E) Shoot himself in the foot. _________________ Best,

AkamaiBrah Newport Beach

Solution.

(C) and (D) are definitely out. C has probability 0 (guard will kill with prob 1), and D has probability approaching 0 (remember, guards have machine guns and lots of ammo).

For (A). 1/2 of the time he miss B. B will then kill C (C is bigger threat than A is). Now A will have one last chance to kill B will 1/2 prob of success.
1/2 of the time he will kill B. Then C and A will get into a duel with C going first. Using the results of part 1, we can calculate the C will kill A with prob 8/9, i.e., A has 1/9 chance of survival. Hence, the probability of survival for A if he shoots B first is: (1/2 * 1/2) + (1/2 * 1/9) or slightly more than 30%.

Note: we don't need to know the formula derived in part 1 for the duel between C and A. C goes first with prob 4/5, so C will have AT LEAST an 80% chance of killing A. This means that A will have AT MOST a 20% chance of surviving. Hence, the upper bound of A's chance of survival is: (1/2*1/2) + (1/2*1/5) or 35% (i.e., A has AT MOST a 35% chance of survival).

For (B), 1/2 of the time he will miss C. In which case, B will kill C and A will have one chance to kill B will probability 1/2 of surviving. The other 1/2 of the time, he will kill C, in which case, B will simply shoot A so there is NO chance of survival there. Hence, by choosing option B, A has a (1/2 * 1/2) = 25% chance of survival.

For (E) -- and E is NOT a trivial option -- A will definitely survive his first shot (he might even miss himself and not get hurt). B will then kill C and A will have one opportunity to kill B with probability 1/2 of surviving. Hence, under option (E), A has a 50-50 chance of survival.

From the above analysis, we see that A's best chance of survival is to shot himself in the foot, let B kill C, then take his one shot at killing B in the second round. Hence, the best answer is (E).

Hope you found this problem interesting! _________________

Best,

AkamaiBrah Former Senior Instructor, Manhattan GMAT and VeritasPrep Vice President, Midtown NYC Investment Bank, Structured Finance IT MFE, Haas School of Business, UC Berkeley, Class of 2005 MBA, Anderson School of Management, UCLA, Class of 1993

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Re: Challenge Problem: Probability Part 2:
[#permalink]
23 Jul 2003, 10:49

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