Followers 0

## 28 posts in this topic

Posted (edited) · Report post

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
0

##### Share on other sites

Posted · Report post

My gut feeling is wanting to be C, gut with some real quick mental run downs of scenarios, but will need some time for the bonus and actually figuring out an answer I am confident in for picking the position too.

0

##### Share on other sites

Posted · Report post

My gut feeling is wanting to be C, gut with some real quick mental run downs of scenarios, but will need some time for the bonus and actually figuring out an answer I am confident in for picking the position too.

I agree, but I won't be able to work the math till I get home later...

0

##### Share on other sites

Posted (edited) · Report post

B

0

##### Share on other sites

Posted · Report post

B

I mean A

0

##### Share on other sites

Posted (edited) · Report post

I mean A

B

0

##### Share on other sites

Posted · Report post

im going with c he has more of a buffer before someone gets a decent shot at him while haveing a decent enough shot to hopefully take out the best before they get to him.

the second part though will have to wait till after jeopardy and the Penguins hand Detroit there --- back to them in a handbasket

0

##### Share on other sites

Posted · Report post

I'll say B.

Since you said super hard, I'm sure I'm way off, but I'll still make an attempt.

Could you add the chances of not being hit (by the others) and divide that by 4. Basically taking an average of not being hit.

I get..

A=42.5% chance of surviving.

B=45%

C=50%

D=53.75%

E=57.5%

Damn that has to be wrong after reading it again....oh well

0

##### Share on other sites

Posted · Report post

1) A

2)

A = 57.7%

B = 2.2%

C= 12.0%

D= 28.4%

E= 41.7%

0

##### Share on other sites

Posted (edited) · Report post

1) A

2)

A = 57.7%

B = 2.2%

C= 12.0%

D= 28.4%

E= 41.7%

part one is right for sure.. the worst case is A kills B and C kills D. The chance of E killing A then is 8%. All other cases are less then 8% after the 1st round. So A has a better chance than any one else to survive. If E dies after the 1st round, A starts the next round without being shot at, A has the best advantage.

0

##### Share on other sites

Posted (edited) · Report post

1) A

2)

A = 57.7%

B = 2.2%

C= 12.0%

D= 28.4%

E= 41.7%

Shouldn't the survival probabilities for A-E add up to 1? The game requires that only 1 person survives, so survival probabilities are mutually exclusive.

part one is right for sure.. the worst case is A kills B and C kills D. The chance of E killing A then is 8%. All other cases are less then 8% after the 1st round. So A has a better chance than any one else to survive. If E dies after the 1st round, A starts the next round without being shot at, A has the best advantage.

I think you're missing a 0. E's chance of killing anyone is 16/20 = 80%. A's chance of killing anyone is only 4/20 = 20%. A can't be the best place to be because eventually the game comes down to a 2-people duel. Even if A is among the last two, his low chance of killing makes him lose way too often.

Edited by bushindo
0

##### Share on other sites

Posted · Report post

im thinking im not going to even try the bonus I just wrote a program and cant think of how to prove it even.

0

##### Share on other sites

Posted (edited) · Report post

1) D. ^^

2) Here's my explanation. Obviously, to survive this shoot out, you'd have positioned such that the person that shoots you will have a lower rate of success. so A is obviously out. E is out too, because, even though it has the highest chance of killing someone, D also has a high success rate. So D should be the best, since it has a 65% chance of sucess, and a 50% chance of survival, at least for round 1.

Let's test this out.

Round 1: (Prisoner - Chance of successfully killing someone - chance of survival)

A - 0.2 - 0.2

B - 0.35 - 0.8

C - 0.5 - 0.65

D - 0.65 - 0.5

E - 0.8 - 0.35

At the end of round 1, A is likely to die. Same goes for E. So that leaves the following possible scenario.

Round 2:

B - 0.35 - 0.35

C - 0.5 - 0.65

D - 0.65 - 0.5

B is very likely to die. So, round 3 will be as follows:

Round 3:

C - 0.5 - 0.35

D - 0.65 - 0.5

Based on Round 2 & 3,

C's chance of survival: 0.65 * 0.35 = 0.2275

D's chance of survival: 0.5 * 0.5 = 0.25

So, D it is! ^^

Edited by Era
0

##### Share on other sites

Posted · Report post

im thinking im not going to even try the bonus I just wrote a program and cant think of how to prove it even.

1) Solve the case with 2-players. It is easy to compute the exact survival rates if you have two players with two known accuracy rate and with a given starting position (who has the first shot)

2) RECURSIVE functions

0

##### Share on other sites

Posted · Report post

1) D. ^^

2) Here's my explanation. Obviously, to survive this shoot out, you'd have positioned such that the person that shoots you will have a lower rate of success. so A is obviously out. E is out too, because, even though it has the highest chance of killing someone, D also has a high success rate. So D should be the best, since it has a 65% chance of sucess, and a 50% chance of survival, at least for round 1.

Let's test this out.

Round 1: (Prisoner - Chance of successfully killing someone - chance of survival)

A - 0.2 - 0.2

B - 0.35 - 0.8

C - 0.5 - 0.65

D - 0.65 - 0.5

E - 0.8 - 0.35

At the end of round 1, A is likely to die. Same goes for E. So that leaves the following possible scenario.

Round 2:

B - 0.35 - 0.35

C - 0.5 - 0.65

D - 0.65 - 0.5

B is very likely to die. So, round 3 will be as follows:

Round 3:

C - 0.5 - 0.35

D - 0.65 - 0.5

Based on Round 2 & 3,

C's chance of survival: 0.65 * 0.35 = 0.2275

D's chance of survival: 0.5 * 0.5 = 0.25

So, D it is! ^^

This is counter-balanced by the fact that C has a 1/2 chance of killing D on everyshot. Most of the time, C survives B's shot, so he gets to shoot D. If A kills B, that's even better for C. In this game, killing someone above you is not as important, or urgent, as having someone with low accuracy below you.

0

##### Share on other sites

Posted · Report post

so i found the probability that each player is the first to die (easy) and the probability of who wins each pair with each first and i thought i understood what you meant by recursive untill i thought about it and are u actually suggesting that for example for a shoot at b i look at b shot c,d,e there likelyhoods of reducing to ab and continue as such because that means each will have 3 forks(at most 3 i think or exactly three havent really thought that much and then for those each has at most two and then well ive done that), but anyway i can just work forward if thats the case right? find the probabilities that each person dies second within the first already died then third then fourth is done and multiply and add. anyway thoughts? advice?

0

##### Share on other sites

Posted · Report post

This is counter-balanced by the fact that C has a 1/2 chance of killing D on everyshot. Most of the time, C survives B's shot, so he gets to shoot D. If A kills B, that's even better for C. In this game, killing someone above you is not as important, or urgent, as having someone with low accuracy below you.

Additionally you said that likely A and E will die in the first round which is not possible. If E dies in the first round, no one can shoot at A because D has already fired his shot (killing E).

I figured this out for the first 3 rounds using good ole math but then my brain hurt. It kept getting worse with possibilities for each person getting shot at by someone else based on probabilities of each surviving all of the previous rounds.

After 3 B was looking the best though.

Prior to doing any math however my guess would have been C.

0

##### Share on other sites

Posted · Report post

Assuming I've done my math AND coding correctly,

E

Kx = probability of x killing in one shot (Ka = 4/20, Kb =7/20, etc.)

Exy (Expected survival of X shooting first against Y)

(X kills Y + (X misses and Y misses, and we wind up in the same state as before))

= Kx + (1-Kx)*(1-Ky)*Exy

Since Exy appears on both sides, we can solve for it in terms of basic kill probabilities

Exy *(1-(1-Kx)*(1-Ky)) = Kx

(1) Exy = Kx / (1-(1-Kx)*(1-Ky))

Exyz (expected survival of X shooting first against YZ)=

X kills Y And Z misses: we get to Exz +

X misses Y and Y kills Z: we get to Exy +

X misses, Y misses, Z misses: we get to same state

Exyz = Kx*(1-Ky)*Exz +

(1-Kx)*Ky*Exy +

(1-Kx)*(1-Ky)*(1-Kz)*Exyz

Exyz * (1-(1-Kx)*(1-Ky)*(1-Kz)) = Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy

(2) Exyz = (Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy) / (1-(1-Kx)*(1-Ky)*(1-Kz))

Ewxyz (expected survival of W shooting first agains XYZ)

W kills X and Y kills Z: we get to Ewy +

W kills X and Y misses and Z misses: we get to Ewyz +

W misses X and X kills Y and Z misses: we get to Ewxz +

W misses X and X misses and Y kills Z: we get to Ewxy +

W misses X and X misses and Y misses and Z misses: we get to Ewxyz

Ewxyz = Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy +

(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz)*Ewxyz

(3) Ewxyz =

(Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy ) /

(1-(1-Kw)*(1-Kx)*(1-Ky))

Evwxyz (expected survival of v shooting first against wxyz)

V kills w, x kills y, z misses, get to Evxz +

v kills w, x misses, y kills z, get to Evxy +

v kills w, x misses, y misses, z misses, get to Evxyz +

v misses, w kills x, y kills z, get to Evwy +

v misses, w kills x, y misses, z misses, get to Evwyz +

v misses, w misses, x kills y, z misses, get to Evwxz +

v misses, w misses, x misses, y kills z, get to Evwxy +

all miss, get to Evwxyz

(4) Evwxyz = (

Kv*Kx*(1-Kz)*Evxz +

Kv*(1-Kx)*Ky*Evxy +

Kv*(1-Kx)*(1-Ky)*(1-Kz)*Evxyz +

(1-Kv)*Kw*Ky*Evwy +

(1-Kv)*Kw*(1-Ky)*(1-Kz)*Evwyz +

(1-Kv)*(1-Kw)*Kx*(1-Kz)*Evwxz +

(1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*Evwxy ) /

(1 - (1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz))

-------------------

Survivals:

(a) Evwxyz(a,b,c,d,e)

0.076

(b) A misses, then B starts the game:

(1-Ka)*Evwxyz(b,c,d,e,a)

0.194

© A kills B, then C starts with 4 players PLUS

A and B miss, then C starts the game:

Ka*Ewxyz(c,d,e,a) + (1-Ka)*(1-Kb)*Evwxyz(c,d,e,a,b)

0.181

(d) A Kills B, C misses, and D starts with 4 players, PLUS

A misses, B kills, D starts with 4 players PLUS

A, B, and C all miss, D starts the game

Ka*(1-Kc)*Ewxyz(d,e,a,c) +

(1-Ka)*Kb*Ewxyz(d,e,a,b)+

(1-Ka)*(1-Kb)*(1-Kc)*Evwxyz(d,e,a,b,c)

0.229

(e) A and C kill, E starts with 3 players PLUS

A kills, C and D miss, E starts with 4 players PLUS

A misses, B kills C, D misses, E starts with 4 players PLUS

A and B miss, C kills D, E starts with 4 players PLUS

A,B,C,D miss, E starts with 5 players

Ka*Kc*Exyz(e,a,c) +

Ka*(1-Kc)*(1-Kd)*Ewxyz(e,a,c,d)+

(1-Ka)*Kb*(1-Kd)*Ewxyz(e,a,b,d)+

(1-Ka)*(1-Kb)*Kc*Ewxyz(e,a,b,c)+

(1-Ka)*(1-Kb)*(1-Kc)*(1-Kd)*Evwxyz(e,a,b,c,d)

0.288

0

##### Share on other sites

Posted (edited) · Report post

Assuming I've done my math AND coding correctly, (edit: bad assumption, typos corrected below)

C (change of plans, due to more accurate typing)

Kx = probability of x killing in one shot (Ka = 4/20, Kb =7/20, etc.)

Exy (Expected survival of X shooting first against Y)

(X kills Y + (X misses and Y misses, and we wind up in the same state as before))

= Kx + (1-Kx)*(1-Ky)*Exy

Since Exy appears on both sides, we can solve for it in terms of basic kill probabilities

Exy *(1-(1-Kx)*(1-Ky)) = Kx

(1) Exy = Kx / (1-(1-Kx)*(1-Ky))

Exyz (expected survival of X shooting first against YZ)=

X kills Y And Z misses: we get to Exz +

X misses Y and Y kills Z: we get to Exy +

X misses, Y misses, Z misses: we get to same state

Exyz = Kx*(1-Kz)*Exz + '------ replacing Ky with Kz

(1-Kx)*Ky*Exy +

(1-Kx)*(1-Ky)*(1-Kz)*Exyz

Exyz * (1-(1-Kx)*(1-Ky)*(1-Kz)) = Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy

(2) Exyz = (Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy) / (1-(1-Kx)*(1-Ky)*(1-Kz))

Ewxyz (expected survival of W shooting first agains XYZ)

W kills X and Y kills Z: we get to Ewy +

W kills X and Y misses and Z misses: we get to Ewyz +

W misses X and X kills Y and Z misses: we get to Ewxz +

W misses X and X misses and Y kills Z: we get to Ewxy +

W misses X and X misses and Y misses and Z misses: we get to Ewxyz

Ewxyz = Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy +

(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz)*Ewxyz

(3) Ewxyz =

(Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy ) /

(1-(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz)) '------------failed to append (1-Kz)

Evwxyz (expected survival of v shooting first against wxyz)

V kills w, x kills y, z misses, get to Evxz +

v kills w, x misses, y kills z, get to Evxy +

v kills w, x misses, y misses, z misses, get to Evxyz +

v misses, w kills x, y kills z, get to Evwy +

v misses, w kills x, y misses, z misses, get to Evwyz +

v misses, w misses, x kills y, z misses, get to Evwxz +

v misses, w misses, x misses, y kills z, get to Evwxy +

all miss, get to Evwxyz

(4) Evwxyz = (

Kv*Kx*(1-Kz)*Evxz +

Kv*(1-Kx)*Ky*Evxy +

Kv*(1-Kx)*(1-Ky)*(1-Kz)*Evxyz +

(1-Kv)*Kw*Ky*Evwy +

(1-Kv)*Kw*(1-Ky)*(1-Kz)*Evwyz +

(1-Kv)*(1-Kw)*Kx*(1-Kz)*Evwxz +

(1-Kv)*(1-Kw)*(1-Kx)*Ky*Evwxy ) / '---------------Replace (1-Ky) with Ky; Y kills!

(1 - (1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz))

-------------------

Survivals:

0.075

0.219

0.255

0.222

0.230

(a) Evwxyz(a,b,c,d,e)

(b) A misses, then B starts the game:

(1-Ka)*Evwxyz(b,c,d,e,a)

© A kills B, then C starts with 4 players PLUS

A and B miss, then C starts the game:

Ka*Ewxyz(c,d,e,a) + (1-Ka)*(1-Kb)*Evwxyz(c,d,e,a,b)

(d) A Kills B, C misses, and D starts with 4 players, PLUS

A misses, B kills, D starts with 4 players PLUS

A, B, and C all miss, D starts the game

Ka*(1-Kc)*Ewxyz(d,e,a,c) +

(1-Ka)*Kb*Ewxyz(d,e,a,b)+

(1-Ka)*(1-Kb)*(1-Kc)*Evwxyz(d,e,a,b,c)

(e) A and C kill, E starts with 3 players PLUS

A kills, C and D miss, E starts with 4 players PLUS

A misses, B kills C, D misses, E starts with 4 players PLUS

A and B miss, C kills D, E starts with 4 players PLUS

A,B,C,D miss, E starts with 5 players

Ka*Kc*Exyz(e,a,c) +

Ka*(1-Kc)*(1-Kd)*Ewxyz(e,a,c,d)+

(1-Ka)*Kb*(1-Kd)*Ewxyz(e,a,b,d)+

(1-Ka)*(1-Kb)*Kc*Ewxyz(e,a,b,c)+

(1-Ka)*(1-Kb)*(1-Kc)*(1-Kd)*Evwxyz(e,a,b,c,d)

Edited by CaptainEd
0

##### Share on other sites

Posted · Report post

so i found the probability that each player is the first to die (easy) and the probability of who wins each pair with each first and i thought i understood what you meant by recursive untill i thought about it and are u actually suggesting that for example for a shoot at b i look at b shot c,d,e there likelyhoods of reducing to ab and continue as such because that means each will have 3 forks(at most 3 i think or exactly three havent really thought that much and then for those each has at most two and then well ive done that), but anyway i can just work forward if thats the case right? find the probabilities that each person dies second within the first already died then third then fourth is done and multiply and add. anyway thoughts? advice?

I seems to me that you thinking of going top down. That is, you start with 5 players, and consider pairing them to see how to reduce the number of players.

I'm suggesting that we work instead from the bottom up. Meaning that we assume that the game has only 2 players (lets say A and B). Then we can figure out the exact probability that A and B would survive. Once we have that,we work upwards. We start the game with 3 players (A, B, and C). Let's say that player A has the first shot. Either A is successful, in which case the game reduce down to a 2-player game for (A,C), or the 3-person game continues. For more hints, see the spoiler for CaptainEd below.

Assuming I've done my math AND coding correctly,

E

Kx = probability of x killing in one shot (Ka = 4/20, Kb =7/20, etc.)

Exy (Expected survival of X shooting first against Y)

(X kills Y + (X misses and Y misses, and we wind up in the same state as before))

= Kx + (1-Kx)*(1-Ky)*Exy

Since Exy appears on both sides, we can solve for it in terms of basic kill probabilities

Exy *(1-(1-Kx)*(1-Ky)) = Kx

(1) Exy = Kx / (1-(1-Kx)*(1-Ky))

Exyz (expected survival of X shooting first against YZ)=

X kills Y And Z misses: we get to Exz +

X misses Y and Y kills Z: we get to Exy +

X misses, Y misses, Z misses: we get to same state

Exyz = Kx*(1-Ky)*Exz +

(1-Kx)*Ky*Exy +

(1-Kx)*(1-Ky)*(1-Kz)*Exyz

Exyz * (1-(1-Kx)*(1-Ky)*(1-Kz)) = Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy

(2) Exyz = (Kx*(1-Ky)*Exz + (1-Kx)*Ky*Exy) / (1-(1-Kx)*(1-Ky)*(1-Kz))

Ewxyz (expected survival of W shooting first agains XYZ)

W kills X and Y kills Z: we get to Ewy +

W kills X and Y misses and Z misses: we get to Ewyz +

W misses X and X kills Y and Z misses: we get to Ewxz +

W misses X and X misses and Y kills Z: we get to Ewxy +

W misses X and X misses and Y misses and Z misses: we get to Ewxyz

Ewxyz = Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy +

(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz)*Ewxyz

(3) Ewxyz =

(Kw*Ky*Ewy +

Kw*(1-Ky)*(1-Kz)*Ewyz +

(1-Kw)*Kx*(1-Kz)*Ewxz +

(1-Kw)*(1-Kx)*Ky*Ewxy ) /

(1-(1-Kw)*(1-Kx)*(1-Ky))

Evwxyz (expected survival of v shooting first against wxyz)

V kills w, x kills y, z misses, get to Evxz +

v kills w, x misses, y kills z, get to Evxy +

v kills w, x misses, y misses, z misses, get to Evxyz +

v misses, w kills x, y kills z, get to Evwy +

v misses, w kills x, y misses, z misses, get to Evwyz +

v misses, w misses, x kills y, z misses, get to Evwxz +

v misses, w misses, x misses, y kills z, get to Evwxy +

all miss, get to Evwxyz

(4) Evwxyz = (

Kv*Kx*(1-Kz)*Evxz +

Kv*(1-Kx)*Ky*Evxy +

Kv*(1-Kx)*(1-Ky)*(1-Kz)*Evxyz +

(1-Kv)*Kw*Ky*Evwy +

(1-Kv)*Kw*(1-Ky)*(1-Kz)*Evwyz +

(1-Kv)*(1-Kw)*Kx*(1-Kz)*Evwxz +

(1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*Evwxy ) /

(1 - (1-Kv)*(1-Kw)*(1-Kx)*(1-Ky)*(1-Kz))

-------------------

Survivals:

(a) Evwxyz(a,b,c,d,e)

0.076

(b) A misses, then B starts the game:

(1-Ka)*Evwxyz(b,c,d,e,a)

0.194

© A kills B, then C starts with 4 players PLUS

A and B miss, then C starts the game:

Ka*Ewxyz(c,d,e,a) + (1-Ka)*(1-Kb)*Evwxyz(c,d,e,a,b)

0.181

(d) A Kills B, C misses, and D starts with 4 players, PLUS

A misses, B kills, D starts with 4 players PLUS

A, B, and C all miss, D starts the game

Ka*(1-Kc)*Ewxyz(d,e,a,c) +

(1-Ka)*Kb*Ewxyz(d,e,a,b)+

(1-Ka)*(1-Kb)*(1-Kc)*Evwxyz(d,e,a,b,c)

0.229

(e) A and C kill, E starts with 3 players PLUS

A kills, C and D miss, E starts with 4 players PLUS

A misses, B kills C, D misses, E starts with 4 players PLUS

A and B miss, C kills D, E starts with 4 players PLUS

A,B,C,D miss, E starts with 5 players

Ka*Kc*Exyz(e,a,c) +

Ka*(1-Kc)*(1-Kd)*Ewxyz(e,a,c,d)+

(1-Ka)*Kb*(1-Kd)*Ewxyz(e,a,b,d)+

(1-Ka)*(1-Kb)*Kc*Ewxyz(e,a,b,c)+

(1-Ka)*(1-Kb)*(1-Kc)*(1-Kd)*Evwxyz(e,a,b,c,d)

0.288

Good work, CaptainEd. I didn't actually expect you to actually implement the method, so your solution makes it all the more impressive. Well done.

0

##### Share on other sites

Posted (edited) · Report post

Good work, CaptainEd. I didn't actually expect you to actually implement the method, so your solution makes it all the more impressive. Well done.

I seems to me that you thinking of going top down. That is, you start with 5 players, and consider pairing them to see how to reduce the number of players.

I'm suggesting that we work instead from the bottom up. Meaning that we assume that the game has only 2 players (lets say A and B). Then we can figure out the exact probability that A and B would survive. Once we have that,we work upwards. We start the game with 3 players (A, B, and C). Let's say that player A has the first shot. Either A is successful, in which case the game reduce down to a 2-player game for (A,C), or the 3-person game continues. For more hints, see the spoiler for CaptainEd below.

Thank you, I certainly couldn't figure out the right answer without getting the math right (after all, my choice changed from post 18 to post 19). A couple of the others picked C, though. But I'm not that clear thinking.

Edited by CaptainEd
0

##### Share on other sites

Posted (edited) · Report post

yah i think i was doing what captain ed was doing i just got confused and frustrated half way through

I was working bottom to top i just had already done the probability of the first kill because if i understand this right you need that at the end and that part was easy

good work captain ed

p.s. im starting to hate this board Im used to being the smartest math person in the room.

Edited by final
0

##### Share on other sites

Posted · Report post

p.s. im starting to hate this board Im used to being the smartest math person in the room.

I feel the same way you do

0

##### Share on other sites

Posted · Report post

I feel the same way you do

I lost that status over 40 years ago. But my wife loves me...

0

##### Share on other sites

Posted · Report post

jeez guys i understand the math but taking a good shot at dying in the first round seems a bit much. the chances at getting shot being E decrease slowly over time. while the chance of getting shot being A decrease rapidly over time.

no math but C seems the best choice becouse whether it be a blood bath (lucky shots) or a dust up (unlucky shots). C never gets shot at by any one good. with a 13% chance of dying in the first round C is in line to kill the biggest and be left with the smallest.

0

## Create an account

Register a new account