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## Question

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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here goes

If E were to live through the first round he would most certainly shoo the person with the next best percentage (D if he is still alive, followed by C, then by B..) that gives E the best chance at continuing in subsequent rounds.

A should almost certainly shoot for E, A does not need to worry about someone coming after him in the first round, so he needs to take out the best shot if he can.

now assuming A misses

B should shoot for C. If he kills C, then D must go for E and then B would continue to the next round (either D kills E and we start at A again, or D misses and then E kills D).

if he misses C, well his chances remain the same as if he decides to shoot as someone else because ...

C should always shoot at B. This leaves D/E to shoot each other as described above.

now assuming A kills E.

B should shoot for A. he would then be guaranteed to move on as before (either C kills D, D kills C, or both C & D survive)

C of course tries to kill D

D of course tries to kill C.

my guess is C probably is the spot to be in if you had a choice.

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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

Edited by mollie13eth
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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.

shoot C. Then B and D will both shoot at E. If E survives he will shoot at B or D before shooting at A. Most likely D. This will all work to A's advantage. If B shoots at D and D does not survive, E will still have to shoot B before shooting A.

How many bullets do they have? Do they continue to get turns if they miss?

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shoot C. Then B and D will both shoot at E. If E survives he will shoot at B or D before shooting at A. Most likely D. This will all work to A's advantage. If B shoots at D and D does not survive, E will still have to shoot B before shooting A.

How many bullets do they have? Do they continue to get turns if they miss?

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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[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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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.

shoot C. Then B and D will both shoot at E. If E survives he will shoot at B or D before shooting at A. Most likely D. This will all work to A's advantage. If B shoots at D and D does not survive, E will still have to shoot B before shooting A.

B on the 2nd turn should shoot E,D , C in that order if they are alive, if E did not die on 1st turn, then D is certainly dead already. so B would shoot E and E would shoot C

C, if still alive on the 3rd turn should shoot E, D, B, A in that order if still alive...

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pick A. shoots at everybody, but not shot at until only one other is alive.

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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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i say it doesnt matter who a shoots at(shooting at E slightly increase his odds) either way at least one is gone first round (either e will die or e will kill d). Then next turn the same except not necessarily at least one (but most likely). But its all about increasing your odds. so no one will shoot at you till the end and that person is almost gaurenteed to be c or b. (d or e will die first round then have to survive 2/5,3/5,and then at least 2/5 or 3/5 again. So anyway a's chances of survival would be close to (for against b) 1/5+4/5*3/5(1/5+4/5*3/5(1/5+4/5*3/5(....

which equals a=12/25, 1/5+a/5+a^2/5=1/5/(13/25)=.385

and against c

1/5+4/5*2/5(1/5+4/5*2/5(1/5+4/5*2/5(.....

= same as above except a =8/25

so 1/5/(17/25)=.29

this doesnt take into acount the small possiblilty that d or e is left verse you or the probability difference of it being c or b against you but i think the maths right so A isnt too bad off. Also the enemy might get first chance which would change the answer just multiply them by the chance of A surviving the first shot of whoever. so i think a is best off.

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

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
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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.

A should shoot at B, assuming that B and C will both subsequently target E. B and C's combined prob (2/5 + 3/5) sounds a lot like a lock to kill E. D will then shoot C and with an 80% prob, should kill C. Thus, after the first round, A,B & D will be remaining. A&B have a 60% chance of killing D with there combined second round shots, thus D is dead. A then takes his second shot at B, resulting in 1/5 + 1/5 prob of killing him. B then shoots at A (assuming he has survived) and has a 40% chance of killing A. By A's third shot at B, he should finally kill him (60% likely) in the 4th round...

So, take position A if given the warden's choice...

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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.

From the tree it can be seen that the second shot (fired by either B or C) will be aimed at A (that is if A successfully kills someone)

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A should shoot at B, assuming that B and C will both subsequently target E. B and C's combined prob (2/5 + 3/5) sounds a lot like a lock to kill E. D will then shoot C and with an 80% prob, should kill C. Thus, after the first round, A,B & D will be remaining. A&B have a 60% chance of killing D with there combined second round shots, thus D is dead. A then takes his second shot at B, resulting in 1/5 + 1/5 prob of killing him. B then shoots at A (assuming he has survived) and has a 40% chance of killing A. By A's third shot at B, he should finally kill him (60% likely) in the 4th round...

So, take position A if given the warden's choice...

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.

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A worst case scenario for A,B,C,and D. If they all miss before hitting, then E will be the last one standing. A=4 misses before a hit, B=3,C=2, and D=1.

So A shoots at B(miss#1).

B shoot C(miss#1)

C shoot at D(miss#1)

D shoot at E(miss#1)

Now E wastes D.

round 2:

A shoot B(miss#2)

B shoot C(miss#2)

C shoot E(miss#2)

Now E takes care of C.

round 3:

A shoot B(miss#3)

B shoot E(miss#3)

Now E blows B away.

final round:

A shoot E(miss#4)

E blasts A

E is left alone.

What are the chances of that happening tho

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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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I would look at it a little differently:

There is no question E would always go for the next best shooter, so he would go for D first at least if E gets to shoot the first round.

D would always go for E first, because D knows that he will be E's first victim if E gets to shoot.

C will not shoot E or D the first round, because if he does succeed in killing the one he goes for, he will be the one the other shoots at. For example if he shoots and kills E, D would then try to shoot C. So, unless A or B shoot and kill D or E, before C's first turn. C should go for B, because B has the next best chance of killing someone.

B will try to shoot C, unless A gets a lucky shot and kills someone (probably E). If A misses, then C will not want to chance killing E and highly likely getting shot by D, so because C will go for B, B must try to kill C before C gets a chance to kill B. If A gets a lucky shot and hits, B would go for A, because if B hits D, then C would have a better then average chance of eliminating B.

A will try for E, no one is going after A, because of the chance A has to affect anything, but A would want to kill the most dangerous, while he has the chance.

I would choose C as the best chance to survive.

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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.

From the frist tree... A will be shot at no matter who he shoots at. So he must survive a shot in the second round of shooting.

If he does survive the second shot then the decisions pan out as shown in the second tree...

From this it is evident that A should shoot D or E and that way he will face another bullet from B in the final round. He is still on the losing side, he will be shot at by B twice! But its better B than anyone else.

Further A should choose to shoot D over E so that C will shoot E at the next round.

Ans: (1) A will shoot D

(2) B will shoot A

(3).... need to draw more silly trees..

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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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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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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.

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My answer to the second question is wrong.. well incomplete. It is true only if A is successful

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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.

That makes no sense to me. If A kills someone, then his chances of killing next time are even lower, theoretically. His chances are much lower of killing 2 in a row. So he would be the absolute least threatening, besides the fact that A has already went and it makes more sense to try and take out someone yet to go.

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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 will have to face D-bullet rather than 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.

I don't get why B has to shoot A if A hits someone. I would think to myself, 'ok A got lucky and hit E or D, but his chances of hitting again are still 20%,' So I'd still go after D or E since they still have the higher chance of hitting.

If A,B,and C; between the three of them, could manage to eliminate D and E, Then C has the upper hand in the next rounds. But in reality A and B will probably miss, so C should shoot B, and let D and E shoot at each other. Then A and C try to eliminate E or D. C has a good chance at hitting. So that leaves A and C, with C being the likely victor. So I say C has best chance of surviving. Also thats what I would pick to be when the warden asks.

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I don't get why B has to shoot A if A hits someone. I would think to myself, 'ok A got lucky and hit E or D, but his chances of hitting again are still 20%,' So I'd still go after D or E since they still have the higher chance of hitting.

If A,B,and C; between the three of them, could manage to eliminate D and E, Then C has the upper hand in the next rounds. But in reality A and B will probably miss, so C should shoot B, and let D and E shoot at each other. Then A and C try to eliminate E or D. C has a good chance at hitting. So that leaves A and C, with C being the likely victor. So I say C has best chance of surviving. Also thats what I would pick to be when the warden asks.

That still doesn't make too much sense, or at least choosing C if you had the choice. D has a 4/5 chance of taking out E and if he doesn't, D is gone. So one of them is guaranteed to make it out of the first round, if you guess that ABC go after each other. second round, whichever is left is gunning for C now. Your logic assumes a kill, which in the case of A and B, very unlikely.

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That still doesn't make too much sense, or at least choosing C if you had the choice. D has a 4/5 chance of taking out E and if he doesn't, D is gone. So one of them is guaranteed to make it out of the first round, if you guess that ABC go after each other. second round, whichever is left is gunning for C now. Your logic assumes a kill, which in the case of A and B, very unlikely.

Anyway you test it out you have to assume someone other than E will get a kill. I'm just assuming C kills B in first round. and either D hits E or E hits D if D misses, either way the order starts over leaving A,C, and D or E. Now D or E will still be targeted because they have better chances of kill. So A tries, and fails. Now C goes and say he hits. That leaves A and C, with C having the best chance of surviving.

Basically the way I see it is the two guys with the worst guns (A,B) will try for the two guys with the best gun (D,E). The guys with the best gun will try to eliminate each other, and the order starts over, and C has yet to be targeted. That leaves at the most one guy with a good gun (DorE). So A B and C target him, before he gets a chance at shooting. Then that leaves A B C. C still has a better chance of surviving.

E will be targeted the most, so his chance of survival is slim.

D will be targeted the second most, once again slim chance.

C will be targeted 3rd most, but by the 2 worst guns, and if it came down to it I'd rather be C, and hope B misses, and blast B. And have A and C left in the final round where C has a better chance.

So I'll still stick with C.

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yah im pretty sure you wanna choose a,b,or c and im currently thinking b beats out A by a smidge. Logic in previous post but to reiterate a lil of it. but now c logic is simply in first round d or e are gone (most likely only one but not definely and very unlikely more then those two). Then you either have d or e shoot at c (c is prolly dead) or a and b shooting at c (decent chance at survival) but thats a decent chance of survival in a unlikely (not too unlikely) scenario. Where in my previous logic A will be in the last two (most likely with b or c) and a wins something like ... well logic is in previouos post but i say B is the winner closely followed by a. I used alot of close-ities in my math but im pretty sure that i applied them evenly enough across the board that the ratio of the percentages is pretty close to the ratios of the real ones

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