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probability

Customer Question

Three safari hunters are captured by a sadistic tribe of natives and forced to participate in a duel to the death. Each is given a pistol and tied to a post the same distance from the other two. They must take turns shooting at each other, one shot per turn. The worst shot of the three hunters (1 in 3 accuracy) must fire first. The second turn goes to the hunter with 50-50 (1 in 2) accuracy. And (if he's still alive) the third turn goes to the crack shot (100% accuracy). The rotation continues until only one hunter remains, who is then rewarded with his freedom. Which hunter has the best chance of surviving, and why?
Submitted: 11 years ago.
Category: General
Expert:  NancyH replied 11 years ago.
If the first and second hunters miss their first shots, then the hunter with 100% accuracy would be the most likely survivor as he would be likely to take out, with his 100% accuracy, the next best shot, the one with 50% accuracy.
If the second hunter kills the third hunter with his first shot then he has the highest likelihood of survival as even though the first hunter will get to the second round of shots first his aim is still poor.
If both the first and second hunter choose to shoot at the third hunter then again the most likely survivor is the second hunter as the hunter with 100% accuracy could be killed in the first round of shots.
So I would say in 2 out of 3 scenarios hunter #2 has the best odds of survival.
Customer: replied 11 years ago.
Reply to NHolmes's Post: more explanation please or refund my money. Use finite state machine to explain your solution
Expert:  NancyH replied 11 years ago.
Please be aware I answered your question as given.
So just give me a complete and concise explanation of what "finite state machine" is, what that has to do with a general logic riddle, and how you wish the information to be formulated and I'll be happy to see what I can do or perhaps someone else here will.
If you were looking for an answer using a specific method then knowing that up front would have been helpful and that information would also have changed whether or not I chose to answer. As asked, there are many ways to come up with a solution to the question. I used one that integrated knowledge of human behavior under stress.
You may wish to re ask the question in the homework category giving a full explanation of what method you need used to get the required results.
You may be happy to know you have not as yet 'paid' anything for your answer you just made a deposit giving you answer access - only when you accept an answer is payment actually made.
Customer: replied 11 years ago.
Reply to NHolmes's Post: At this time i would like to withdraw my question. It was nice using your service and hope to use it again in future.
Thank you
Expert:  NancyH replied 11 years ago.
Just post a request about that in the tech support area and they should help you right out.
Sorry I couldn't help you! Someone over in the homework area might be able to though!
Expert:  Don replied 11 years ago.
I know that you wanted to withdraw your question, but I had already started working on this before you posted that. Hopefully this will be of a little more use to you.

First, label the shooters A, B and C, where C is the worst shooter, B is the second best, XXXXX XXXXX is the deadeye. Assume that each person shoots at the person who poses the greatest threat to them. That is, C and B shoot A before they shoot at each other, and A shoots at B before C. Now if someone hits on their turn, I'll write an H after the letter, an M if they miss.

Here are the possible scenarios:

C wins by
CM, BM, AH, CH   2/3 * ½ * 1* 1/3 = 1/9

A wins by
CM, BM, AH, CM, AH   2/3 * ½ * 1 * 2/3 * 1 = 2/9

B wins by
CH, BH   1/3 * ½ = 1/6

The other possibilities look like this:

CH, BM, CM, BM, CM, …1/3 * ½ * …

CM, BH, CM, BM, CM, …2/3 * ½ * …

Obviously once A is out of the way, this could go on for any number of turns. Because of this, it isn't a finite problem—-there are infinitely many different possible states. But it's still possible to compute the probabilities.

Note that these last two scenarios have a combined chance of ½. And really, they're both the same thing since A is gone after two turns, and they are both alive. The probability of B or C winning in either of these cases can be calculated by an infinite sum, and it turns out that they both have a 50% chance.

So the totals are:

A - 8/36( = 2/9)
B - 15/36 ( = (1/2)*(1/2) + 1/6)
C - 13/36 ( = (1/2)*(1/2) + 1/9)

Note that the probabilities sum to 1, like they should.
So B has a slightly greater chance of living, followed (surprisingly) by C.
Customer: replied 11 years ago.
This makes more sense now,by the way if you were to draw FSM, how could you go about?
Thanks alot sir
Expert:  Steve -- a.k.a. Oreport replied 11 years ago.
Customer



'if you were to draw FSM, how
could you go about?'
sounds like a new question -- and you
haven't yet accepted Don's apparently helpful answer?



What gives?



Thanks in advance.



Steve













Expert:  Don replied 11 years ago.
Since this isn't exactly a finite problem, I'm not quite sure how you would go about drawing the FSM. A site that might help is http://en.wikipedia.org/wiki/Finite_state_machine

Good luck!

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