Alice, Bob and Carole are involved in a game of three way duel. They take turns to shoot in the order of Alice-Bob-Carole-Alice-... until only one surviver is left standing. The rule is very simple: in your turn, you must fire one shot at another opponent of your choice. If you hit, your target is immediately removed from the game.

Before the game, players must choose their guns. They can choose a gun with any hit probability from 1% to 100%, in integer percentage. Carole must choose his gun first, followed by Bob, and finally Alice. All players can see what guns the others choose. No communication is allowed.

Question 1: if players want to maximize their own surviving probabilities, what guns should they choose?

Question 2: if the worst gun has the option to pass his/her turn without shooting, how would the players' choices change?

Note: when two players choose the same gun, say for example Bob and Carole both choose 80% gun, the other player (Alice) will be indifferent between shooting either of them, if she must (Question 1) or decides (Question 2) to shoot. We stipulate that in such cases the shooter will choose her target randomly.


Answer to question 2: reasoning

If there is a player with the worst gun: Passing is the best strategy since that gives the first shot after the first casualty. So we have effectively two duels between the best two guns, and between the winner and the worst gun.
(note: I am using b,m,w for the best, middle and worst gun chosen respectively)
1 Passing will give at least w% chance of winning
2 Successfull shooting will give w/(w+m-mw)); worse than w/(w+w+ww)
3 This means that w can be at most 41% for this to be effective
If B or C chooses guns below 50%, A will choose the 100% gun over choosing the worst gun for more than 50% success (and it does not apply, since B or C dies immediately)
If B or C both choose 50+, It does not apply

if Alice fights the first duel, she should pick 100.
if Alice fights only the second duel, she should pick 1 below the worst of the guns chosen.

After Carole picks a gun, Bob has two possibly best choices:
Pick just low enough for Alice to kill Carole
Pick just low enough for Alice to let Bob and Carol fight it out first.


a bad gun is not good for Carole, Alice will pick 100% and Bob does not want certain death..
a good gun is not good for Carole, Bob will choose medium, so that Alice kills her and at least he will have the first shot in the second duel.

example calculation 1:

let's assume Carole picks 50
- assume Bob picks 100, Alice can:
-- kill Bob -> 50% victory
- let Bob kill Carole for 49% victory -> Worse than 50%; bad play of Bob
Since the percentages were close, let's drop Bobs choice a bit:
- Assume Bob picks 96, Alice can:
- kill Bob -> 50% victory
- let Bob and Carole fight:
-> Bob wins fight 1 96 out of 98; win% Alice against Bob 96/98 * 49/(49+5196%) (=48.9996)
-> Carole wins 2 out of 98; win% Alice against Carole 2/98 * 49/49+51
50% (= 1.3423)
So Alice wins 51.34% of the time, Carole wins almost 1% of the time; not often...

example calculation 2:

If Carole picks higher, Bob can also pick higher without being killed immediately by Alice: bad for Carole

lets assume Carole picks 40
- Assume Bob picks 40, Alice can:
- kill Bob/Carole -> 60% victory
- let Bob and Carole fight:
-> Bob wins fight 1 10 out of 16;
-> Carole winsfight 1 6 out of 16; Alice wins 39/39+61*40% (=61.51)
Since the percentages were close, Bob can drop his choice a bit and Alice will kill Carole:
Assume Bob picks 38, Alice can:
- kill Carole -> 62% victory
- let Bob and Carole fight
-> Alice will have less success than the previous 61.51%,so Alice will kill Carole and Bob scores 38%, better than him picking 40 There is also some room for Bob picking higher (and not getting killed), but not much, so Carole should probably pick higher

Brute force solution:

Carole should pick 46% for 5.01% success (rounded) at optimal play
Then Bob should pick 74% for 40.09% success (rounded) at optimal play Then Anna can not pick better than 45% for 54.90% success.
note: Carol can only pick 1,2,and 46-50 to avoid 0% success.

Fun fact: If Carol picks 2%, Bob can give - as a best choice together with 1% - Alice an arbitrary shooting choice by also picking 2%. Good for Carol if she believes 2 things: Alice likes her more. Bob thinks Alice likes him more. (Carole has a much better choice o.c.)

Code used:

chance of B,C winning the first duel while fighting each other:
PB1(b,c) := b / (b+c-b*c/100.0);
PC1(b,c) := 1- PB1(b,c);
chance of A winning while fighting the first duel against c:
Afirst(b,c) $ (b < c):= 1-b/100.0;
chance of A winning while fighting the first duel against b:
Afirst(b,c) $ (b >= c):= 1-c/100.0;
chance of A winning second duel against b:
AB(b,c) $ (b >= c):= (c-1)/(c-1 + b*(101-c)/100);
AB(b,c) $ (b < c):= (b-1)/(b-1 + b*(101-b)/100);
chance of A winning second duel against c:
AC(b,c) $ (b >= c):= (c-1)/(c-1 + c*(101-c)/100);
AC(b,c) $ (b < c):= (b-1)/(b-1 + c*(101-b)/100);
chance of B winning while A chooses optimal:
PB2(b,c) := if Afirst(b,c) < PB1(b,c)*AB(b,c) + PC1(b,c)*AC(b,c) then PB1(b,c)*(1-AB(b,c)) else ((b/100.0) $ (b<c)) + ( (b/200.0) $ (b=c)) endif;
PB2(b,c) := if Afirst(b,c) < PB1(b,c)*AB(b,c) + PC1(b,c)*AC(b,c) then 1 else (b $ (b<c)) + ( (b/2) $ (b=c)) endif;
BestB(c) := max(b,PB2(b,c)); SuccessC(b,c) $ (BestB(c) = PB2(b,c) ) := if Afirst(b,c) < PB1(b,c)*AB(b,c) + PC1(b,c)*AC(b,c) then PC1(b,c)*(1-AC(b,c)) else (c $ (b>c)) + ( (c/2) $ (b=c)) endif;

| improve this answer | |
  • $\begingroup$ I'm not sure passing is always the best strategy for the worst gun. Let's say A, B and C have chosen 10%, 11% and 100% respectively. A's surviving probability without passing is about 17.5%. If A chooses to pass however, her surviving probability drops to about 14.4%. $\endgroup$ – Eric Sep 20 at 13:28
  • $\begingroup$ You are right, hopefully my result is still correct since not passing being optimal needs extreme choices.. $\endgroup$ – Retudin Sep 20 at 14:58
  • $\begingroup$ @Eric extended my reasoning to handle your observation $\endgroup$ – Retudin Sep 21 at 6:24
  • $\begingroup$ Won't C be better off choosing 100? Then B will choose 59, and A will choose 58. A waits to duel with the survivor of B and C. In this situation C will have 17.22% surviving probability, much better than 3.95% given by your strategy. $\endgroup$ – Eric Sep 22 at 5:14
  • $\begingroup$ No, in that case B is better of choosing 50%. Then A will choose 100%. 50% chance to win is better for B then he what he gets in your scenario. $\endgroup$ – Retudin Sep 22 at 15:25

Answer to Q2: all players choose 100%: C chooses 100%. Then B knows: A will never shoot B. Because, if A would hit, C would kill A. So B chooses 100% as well. Now A applies the same reasoning: no matter what A does on the first turn (shoot at C or pass), B will not shoot at A, because if B would hit, C would kill B. In order to maximize the survivability against the survivor of B and C player A chooses 100%.

Now all players hit 100%: the first to fire a shoot will be killed by the survivor. Since all players have equally `worse' probability to hit, they are all allowed to pass. So all players survive indefinitely.

| improve this answer | |
  • $\begingroup$ A could instead take a 99% win chance over a draw by selecting the 99% gun and passing. In most scoring systems this leads to a significant increase in expected value. $\endgroup$ – Sconibulus Sep 18 at 14:43
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    $\begingroup$ Well I guess you caught a bug in my description of the problem. When I say the worst gun is allowed to pass, I mean the worst in the strict sense. If not, there's no need for C to choose 100℅ gun. C choose any x℅ gun, then B will choose x gun too, knowing this will forcing A to choose x, making all three survive 100℅. Although the game never ends (yes, this IS a bug!) $\endgroup$ – Eric Sep 20 at 3:10
  • $\begingroup$ @Eric. No. Suppose C and B choose the 1% gun. A is not forced to choose the 1% gun; the 100% gun has a 99% chance of winning. $\endgroup$ – Mark Tilford Sep 20 at 12:58
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    $\begingroup$ @MarkTilford the problem is, 99% surviving probability is worse than 100%, if you forever passing counts as surviving. Players maximize their own surviving probabilities. The only way out is to allow only the strictly worst gun to pass turns. $\endgroup$ – Eric Sep 20 at 13:06

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