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#11 Tiger Fan

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Posted 01 June 2009 - 08:39 PM

5 prisoners are on a death row. Let's call them A, B, C, D, and E. The warden gives them a chance of living. He gives them each a doctored gun and let them engage in a death match. Let's say the guns are modified so that their chances of hitting varies. A's chance of hitting and killing any other player is always 1/5. B's chance is always 2/5, C's is 3/5, D's is 4/5, and E's is 5/5. Assume that every single shot will either miss or kill. A player must shoot someone on his turn. Each player knows his gun's accuracy rate and the others' as well.

The players take turn shooting in the following order: A, B, C, D, and E. During his turn, a gun slinger can choose to shoot at anyone he wants. If a player is killed, then the order of shooting will continue in the same sequence but with the dead player skipped. Players take turn shooting until there is only 1 player remaining.

Assume that each player wants to maximize their own chances of living, and that each player knows that the others will do the same. Answer the following

1) What should A do on this first turn?

2) What should B do on the second turn?

3) What should C do on the third turn?

Super hard bonus: Suppose that the warden likes you, so the night before the game he allows you the chance to choose your gun. Essentially he allows you to choose your position as A, B, C, D, or E. Which position should you choose to maximize your chances of living, assuming that everyone plays optimally? Assume that the other prisoners don't know about this so they won't unduly target you out of spite.


Spoiler for My thoughts...

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#12 adiace

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Posted 01 June 2009 - 08:44 PM

The 1st question asks what should A do...


I have created a decision tree...
Now when A shoots some one, he praying for luck to be on his side so he is successful. So all future steps must be planed assuming he is successful.

In this tree:
If A kills B, then C will decide that it is best to kill A, because if C kills D, E will shoot at him, also if C kills E, D will shoot at him. So killing A gets D to kill E, and finally C gets to shoot D.
I.e At each step, the shooter decides what is best based the next shooters reaction to what he does.

Result of the decision tree... A should shoot at anyone but B.

Spoiler for ans for (2)

Attached Images

  • decisiontree.png

Edited by adiace, 01 June 2009 - 08:47 PM.

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#13 adiace

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Posted 01 June 2009 - 08:55 PM

Spoiler for My thoughts...


the probabilities do not combine that way, what you are saying is if B and C both shoot at E he will die for sure. But each shoot is a hit or miss, so they both could miss aswell. The probability that E dies after B and C shoot at him is (2/5 + 3/5*3/5 = 19/25), i.e. either B misses him or B misses then C kills him.

Edited by adiace, 01 June 2009 - 08:58 PM.

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#14 James8421

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Posted 01 June 2009 - 09:03 PM

Spoiler for I'll do..


What are the chances of that happening tho :lol:
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#15 final

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Posted 01 June 2009 - 09:20 PM

for a second there i thought you were saying worst case scenario is the worst so it must be the right answer and i had lost faith in you, but low and behold faith was restored
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#16 Dimederwin

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Posted 01 June 2009 - 09:24 PM

Spoiler for They should.

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#17 adiace

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Posted 01 June 2009 - 09:39 PM

Another tree:
From the previous tree, we know that A can shoot at anyone, if he succeeds then B or C will shoot at A.
THIS tree shows the decisions that will be made if B or C misses.

Spoiler for My attempt at the ans:

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  • dtree_2.png

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#18 final

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Posted 01 June 2009 - 10:15 PM

im confused y are u saying b shoot a. A only has a 1/5 chance of shooting B. so even if he thought A was going to shoot at him next round its more likely that someone else (say d or E) kill someone else then next round shoot at b and kill him then A kill b if he shoots at him twice in a row. A has to shoot between 3 and 4 times at someone to get a 50% chance at killing him. compared to c d and e who have a >50% chance at one shot... maybe i just dont understand what your trying to show with your trees tho
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#19 palmerc7

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Posted 01 June 2009 - 10:18 PM

im confused y are u saying b shoot a. A only has a 1/5 chance of shooting B. so even if he thought A was going to shoot at him next round its more likely that someone else (say d or E) kill someone else then next round shoot at b and kill him then A kill b if he shoots at him twice in a row. A has to shoot between 3 and 4 times at someone to get a 50% chance at killing him. compared to c d and e who have a >50% chance at one shot... maybe i just dont understand what your trying to show with your trees tho


I don't understand why everyone wouldn't just shoot at the one with the highest average. why would anyone waste their first shot at the one with the least chance of killing them.

A, B, C, D at E and E at D. one could figure out the chances of E surviving the first round easily, just combination of their averages, but I suck at statistics.
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#20 adiace

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Posted 01 June 2009 - 10:24 PM

im confused y are u saying b shoot a. A only has a 1/5 chance of shooting B. so even if he thought A was going to shoot at him next round its more likely that someone else (say d or E) kill someone else then next round shoot at b and kill him then A kill b if he shoots at him twice in a row. A has to shoot between 3 and 4 times at someone to get a 50% chance at killing him. compared to c d and e who have a >50% chance at one shot... maybe i just dont understand what your trying to show with your trees tho


If you look at the 1st table i posted...

If A succeeds at killing some one, then B must shoot A. Say I am right and A shoots D and kills him, now B can shoot A, C, or E. If B soots A and succeed, then C will have to shoot B, D or E. If C kills E, then D will kill C so that he can face the less successful B-bullet. If C kills D, then E will kill C (he too would rather have B shoot that him than C). So C must kill B. To prevent C from killing him, B must kill A.

Edited by adiace, 01 June 2009 - 10:26 PM.

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