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I promised final and adiace a break after their exemplary effort on Prisoner on a death row 8. Well, break's over.

There are 5 prisoners on a death row, call them A, B, C, D, and E. The warden gives them a chance to live. 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 4/20. B's chance is always 7/20, C's is 10/20, D's is 13/20, and E's is 16/20. Assume that every single shot will either miss or kill.

The players shoot in this order, A, B, C, D, and E. Unlike Prisoners on a death row 8, prisoners don't have a choice of shooting anyone they want. Everyone must shoot someone right after him in the sequence. For example, A must shoot B, and if B is already dead, A must shoot C if C is alive, or go on to the next person if C is dead too. The person last in the sequence must shoot someone at the start of the sequence. So for instance, let's say it's E's turn, he must shoot A, or the next alive person at the start of the sequence if A's dead. The game continues until there's only 1 person left. The last person standing earns his freedom.

The warden likes you, so the night before the game he allows you to pick your position as A, B, C, D, or E.

1) What position should you pick?

Super hard bonus: What is the exact chance of survival for A, B, C, D, and E. Any method is allowed, but probabilistic method like simulation is not allowed since simulation provides an estimate of the rate but not the exact numbers. I'm not after the numbers, though. I'm looking for methods that would allow us us to compute the exact chance of survival. You don't have to implement the method, but you must describe it, and it must be doable within a reasonable amount of time.

Edited by bushindo

## 27 answers to this question

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ok i am challenging captian ed accepted answer on this one. i have only taken up to calculus III so i may be missing a few linear algebra lessens but here it goes

THE MATH

The chances of survival of any player is the Chances Of Survival (COS) of the eligible player(s) before you multiplied by the Killing Potential of that (those) player(s).

COD=Chances of Dying

1-COD=COS

Round 1

COD[a]=0

COD=COD[a]x KP[a]

COD[c]=CODx KP

and so on

Round 2

COD[a]=COD[e]x KP[e]

COD=1-Sumation of COD[a]x KP[a]

COD[c]=((1-SumCODx KP)+ COD)/2

and so on

Round 3

COD[d]= ((1-SumCOD[a]x KP[a])+(1-SumCODx KP)+ (1-SumCOD[c]x KP[c]))/3

/\//\/

After 3 rounds the chances of death decrease becouse almost all the players have the chance to shoot at you and there is a good chance you are dead so the CODs start to decrease and buffer.

So after three rounds the chances of survival are:

A-24%

B-53%

C-37%

D-19%

E-20%

So now the Killing Potential becomes the larger factor

So COS[abcde]xKP[abcde]

A=.048

B=.185283

C=.18447

D=.1235

E=.16

so E is not most likely to survive like Captian Ed said, it is B (or C)

if anyone would like to take it to Round 4 to get more accuratte between B or C be my guest.

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I have heard to get married you need to shut down alot of brain cells

and gk he changed his answer in his most recent post the answer he found was c

Edited by final

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