What Is the Probability Robin Hood Wins Without Exposing His Identity?

[uasaki]

Homework Statement

Although Robin Hood gets a bullseye with probability 0.9, he finds himself facing stiff competition in the tournament. To win he must get at least four bullseyes with his next five arrows. However, if he gets five bullseyes, he risks exposing his identity to the sheriff. Assume that if he wishes to, he can miss the bullseye with probability 1. What is the probability that Robin wins the tournament without risking exposing his identity?

Homework Equations

The Attempt at a Solution

From the problem description, it seems like Robin is free to miss anyone of the next five shots. However, I assumed that he would try to hit the first four and purposefully miss the last (I'm not sure if this assumption is valid or if I need to consider other cases). The chance of Robin hitting the first four and missing the last one is exactly 0.9 * 0.9 * 0.9 * 0.9. However, Robin could also accidentally miss one of the first four, which would force him to hit the rest. The probability that Robin misses a shot and hits the rest is 0.1 * 0.9 * 0.9 * 0.9 * 0.9. There are exactly 4 ways that this situation could occur (i.e., Robin misses the first shot, Robin misses the second shot, etc), so the total probability would be 4 * 0.1 * 0.9 * 0.9 * 0.9. Then, we can sum up the above two probabilities (they're disjoint since either Robin makes the first four or he doesn't).

 

[haruspex]

That all looks right to me.

 

[uasaki]

Thanks for the reply. Is there any reason why I don't need to consider the case(s) in which Robin tries to purposefully miss the first shot and make the rest, or purposefully miss the second shot and make the rest, etc?

 

[Ray Vickson]

Thanks for the reply. Is there any reason why I don't need to consider the case(s) in which Robin tries to purposefully miss the first shot and make the rest, or purposefully miss the second shot and make the rest, etc?

He should not deliberately miss any of the first four; if he does so, he must make all his remaining shots in order to meet his goal. However, if he waits until his last shot (if needed) he maximizes his opportunity to make 4 out of 5 successfully. In other words, leaving a deliberate miss until the end gives him the greatest chance of making 4 out of 5.

RGV

 

[uasaki]

That makes sense to me intuitively, but I'm not sure how I would go about proving that the strategy is optimal. Do I just need to compare the probability that Robin Hood will miss again if he intentionally misses one of the first four shots vs. the probability that Robin Hood will miss again if he accidentally misses one of the first four shots?

 

[Ray Vickson]

That makes sense to me intuitively, but I'm not sure how I would go about proving that the strategy is optimal. Do I just need to compare the probability that Robin Hood will miss again if he intentionally misses one of the first four shots vs. the probability that Robin Hood will miss again if he accidentally misses one of the first four shots?

You can formulate the problem as a 5-stage Markov decision problem and derive an optimal policy from the dynamic programming functional equations. That proves optimality of the policy I stated---but that seems like "overkill", applying fancy tools to a simple situation.

RGV