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#1 bushindo

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Posted 01 June 2009 - 05:17 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.
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#2 preflop

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Posted 01 June 2009 - 05:53 PM

here goes

Spoiler for my initial thoughts

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Sent from my rotary dial phone.

#3 mollie13eth

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Posted 01 June 2009 - 06:58 PM

Round 1

I think A, B, and C all will shoot at E and E would be dead the first round based on 1/5 + 2/5 + 3/5 = 6/5 = a darn good probability that E is dead

Then D would shoot C, and with his 4/5 probability and there only being 4 people, C would probably be dead in the first round as well.


Round 2

Next A & B would shoot at D and 1/5 + 2/5 = 3/5 good chance that D is dead


Round 3

C wouldn't shoot anyone because he's probably dead.

A & B would continue shooting D until D is dead.

Overall best position:

B, because B and A are the last two standing, and B has a better chance of killing A.


- me :wub:

Edited by mollie13eth, 01 June 2009 - 06:59 PM.

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#4 chicory

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Posted 01 June 2009 - 06:59 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 A should

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#5 bushindo

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Posted 01 June 2009 - 07:04 PM

Spoiler for A should


Assume unlimited bullets. And players will continue to get turn on the next round if they miss. Players continue shooting until there's only 1 player left.
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#6 genmaximus2000

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Posted 01 June 2009 - 07:06 PM

[spoiler name='my guess']
Everyone will shoot for the guy with the best gun. So A will shoot for E, if he hits B will shoot for D, but if A misses B will shoot for E. Each person wants to take out the guy with the best gun to give his gun a better chance to win. If the player has the best gun, he will shoot for the second best. The end result will be that A will be one of the final contenders along with B, C, D, or E depending on who hits. It's very likely that it will be A and B or A and C (I don't have the percentages). So if I could choose my position I would choose A or B.
[\spoiler]
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#7 chicory

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Posted 01 June 2009 - 07:18 PM

Assume unlimited bullets. And players will continue to get turn on the next round if they miss. Players continue shooting until there's only 1 player left.

Spoiler for A should

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#8 chicory

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Posted 01 June 2009 - 07:20 PM

Spoiler for maybe

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#9 Minh08

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Posted 01 June 2009 - 07:38 PM

I would look at this situation backwards
E would go for the next best shooter
D would go for E because he knows that E will go for him
C would go for E because he doesn't want D to die since the next best shooter would be himself
B has no problem with E,D,C since they are the ones going to be killing each other so B would definitely go for A, because no one would be going for A
A would have to kill B since he knows that B will try and kill him.

So...

A would shoot B
B would go for A
and C would shoot E
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#10 final

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Posted 01 June 2009 - 08:10 PM

Spoiler for guesserino


havent really check to see if me thought process made too much sense just thought and wrote but thinking back on what i thought i think i was thinking thoughtfully

edit addition

to clarify i guesstimated that the chance of c being last to be 1/3 so his survival is close to .71*1/3=.23 and b's 2/3 (once again ignoring slight chance of d or e. ) so 2/3*.615=.41 and a's is 2/3*.385+1/3*.29= .351 so maybe b is best. b's chance and a's chance are actually lower then this and b's more overestimated then a's i think so i think there about equalish

Edited by final, 01 June 2009 - 08:15 PM.

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