# A truel (part 1)

1. Mar 7, 2005

### msmith12

you have all heard of a duel... but what about a truel...

well... a truel is a duel with three people...

in this specific truel, there are three gentlemen players who each wait their turn before they shoot.

player 1 will hit the person he shoots at with probability x

player 2 will hit the person he shoots at with probability y

player 3 will hit the person he shoots at with probability z

x>y>z

a couple of other things about the game... each person knows how good of a shot the other is. (i.e. player 1,2,3 know x,y,z). also, there is no chance that player 2 will die if player 1 shoots at 3. that is, if player 1 misses his intended target, noone else will be hit. and finally, each player must try to hit someone each turn.

what is the probability that player 1 wins? player 2? player 3?

2. Mar 8, 2005

### K.J.Healey

Does everyone shoot at the same time, or do they take turns?

3. Mar 8, 2005

### msmith12

they take turns, starting with player 1, then 2, then 3...

4. Mar 8, 2005

### Bartholomew

I think this is an overly complicated problem that isn't a true brain teaser but more of a question for a mathematical journal. I've seen this before.

5. Mar 8, 2005

### Pseudopod

is it possible, for example, for both player 1 and player 2 to shoot at player 3?

6. Mar 8, 2005

### ToxicBug

I think that nobody will survive, because the last trueler standing will shoot himself in the head.

7. Mar 11, 2005

### RandallB

For player Three it would be something like:
xz
+(1-x)yz
+(1-x)(1-y)z(1-y)z
+(1-x)(1-y)z(1-y)(1-z)(1-y)z
+(1-x)(1-y)z(1-y)(1-z)(1-y)(1-z)(1-y)z
+(1-x)(1-y)z(1-y)(1-z)(1-y)(1-z)(1-y)(1-z)(1-y)z
etc.
+(1-x)(1-y)(1-z)xz
+(1-x)(1-y)(1-z)x(1-z)(1-y)z(1-y)z
+(1-x)(1-y)(1-z)x(1-z)(1-y)(1-z)(1-y)z
etc.
+(1-x)(1-y)(1-z)(1-x)yz
+(1-x)(1-y)(1-z)(1-x)y(1-z)(1-y)z
etc.

8. Mar 11, 2005

### AKG

Probability that player 1 wins...

... given 3 kills 2 (then eventually 1 kills 3)
$$\left( \sum _{i = 0} ^{\infty}(1 - x)(1 - z)\right )x = \alpha$$

... given 1 kills 2 (then eventually 1 kills 3)
$$(1 - z)\left( \sum _{i = 0} ^{\infty}(1 - x)(1 - z)\right )x = \beta$$

... given 2 kills 3 (then 1 kills 2)
$$\left( \sum _{i = 0} ^{\infty}(1 - x)(1 - y)\right )x = \gamma$$

... given 1 kills 3 (then 1 kills 2)
$$(1 - y)\left( \sum _{i = 0} ^{\infty}(1 - x)(1 - y)\right )x = \delta$$

Probability that 1 kills first:
$$\left (\sum _{i = 0} ^{\infty} (1 - x)(1 - y)(1 - z)\right )x = A$$

Naturally, 1 will want to kill 2, so P(first kill is 1K2) = A, P(first kill is 1K3) = 0.

Probability that 2 kills first
$$(1 - x)\left (\sum _{i = 0} ^{\infty} (1 - y)(1 - z)(1 - x)\right )y = B$$

Again, 2 will want to kill 1, so P(first kill is 2K1) = B, P(1st K is 2K3) = 0

Probability that 3 kills first
$$(1 - x)(1 - y)\left (\sum _{i = 0} ^{\infty} (1 - z)(1 - x)(1 - y)\right )z = C$$

So P(1st K is 3K1) = C, P(first K is 3K2) = 0.

Probability that 1 wins is:
$$A(\beta + \delta) + B\gamma + C\alpha$$

$$K = (1 - x)(1 - y)(1 - z)$$

Then:

$$A = \frac{x}{1 - K}$$

$$B = \frac{y(1 - x)}{1 - K}$$

$$C = \frac{z(1 - x)(1 - y)}{1 - K}$$

$$\alpha = \frac{x}{1 - \frac{K}{1 - y}}$$

$$\beta = \frac{x(1 - z)}{1 - \frac{K}{1 - y}}$$

$$\gamma = \frac{x}{1 - \frac{K}{1 - z}}$$

$$\delta = \frac{x(1 - y)}{1 - \frac{K}{1 - z}}$$

So, P(1 wins):

$$A(\beta + \delta) + B\gamma + C\alpha$$

$$= \frac{x}{1 - K}\left (\frac{x(1 - z)}{1 - \frac{K}{1 - y}} + \frac{x(1 - y)}{1 - \frac{K}{1 - z}}\right ) + \frac{y(1 - x)}{1 - K}\frac{x}{1 - \frac{K}{1 - z}} + \frac{z(1 - x)(1 - y)}{1 - K}\frac{x}{1 - \frac{K}{1 - y}}$$

$$= \frac{x(x(1 - y) + y(1 - x))}{(1 - K)(1 - K/(1 - z))} + \frac{x(x(1 - z) + z(1 - x)(1 - y))}{(1 - K)(1 - K/(1 - y))}$$

$$= \frac{x}{1 - K}\left ( \frac{x(1 - y) + y(1 - x)}{1 - K/(1 - z)} + \frac{x(1 - z) + z(1 - x)(1 - y)}{1 - K/(1 - y)}\right )$$

$$= \frac{x}{1 - K}\left ( \frac{x(1 - y) + y(1 - x)}{1 - (1 - x)(1 - y)} + \frac{x(1 - z) + z(1 - x)(1 - y)}{1 - (1 - x)(1 - z)}\right )$$

$$= \frac{x}{1 - K}\left ( \frac{x(1 - y) + y(1 - x) - xK - y(1 - x)^2(1 - z) + x(1 - z) + z(1 - x)(1 - y) - xK - z(1 - x)^2(1 - y)^2}{\left (1 - (1 - x)(1 - y)}\right )\left (1 - (1 - x)(1 - z)\right )}\right )$$

... Ugh, this isn't going anywhere.

9. Mar 11, 2005

### honestrosewater

I've seen a puzzle very much like this except it asked for what you are missing- who does each player try to shoot? Player 1 can try to shoot either player 2 or player 3. Or is there a 50% chance Player 1 will try to shoot Player 2?

AKG,
the OP didn't say the players were rational or what their goal was; You are assuming they are rational and try to survive.

10. Mar 11, 2005

### AKG

I originally was thinking that there's a 50% chance that a player will shoot any other player, i.e. they shoot at random. But then I figured that the OP said:
for a reason. If not, then the probability that 1 wins is:

(A/2)(b + d) + (B/2)g + (C/2)a, which is just the probability given above, divided by 2

Actually, my original formula needs to be edited, it should be:

Ab + Bg + Ca, i.e., by the "rational" model, $\delta$ would never happen. so we would get what I have in the post above, but remove "x(1 - y) - xK" from the numerator. It will probably still be too ugly to simplify...

11. Mar 11, 2005

### BobG

If the players all know the odds and are smart enough to act on them, then:

Player 3 is the most likely to win followed by player 2. Player 1 is the least likely to win.

While player 3 is the least likley to hit his target, one of the other players has to be eliminated before anyone takes a shot at player 3. Player 3 has a better chance of surviving player 2's shots, so both players will shoot at player 1 as long as they're still alive in the game.

In other words, its essentially a duel of player 1 against the team of players 2 & 3 with 2 & 3 getting twice as many shots. Two-thirds of the shots are at player 1 and one-third of the shots are at player 2.

Player 1 is the only player that would eliminate player 2. Player 2, both being a better shot and shooting before player 3, is more likely to be the player on the two man team to eliminate player 1. That means player 3 gets the opening shot in the subsequent 2-man duel. Unless he's a lousy shot and the remaining player is a very good shot, having a first shot advantage gives player 3 the best chance of winning the 2-man duel.

But wait! That's only with the caveat that each player truly does try to hit another player on each shot (I guess however logical each player may be, honor out trumps logic in this case).

If players are high on logic, but low on honor and accidently miss on purpose, it's to their advantage not to be the first to actually hit an opponent. Whichever opponent they didn't shoot will then get the first shot in the two-man duel. Knowing the disadvantage associated with being the player to eliminate player 1, player 1 will miss with confidence. Player 2 will miss to avoid having player 3 take a shot at him. Player 3 who started out sitting in such a good position, now has to miss to avoid having player 1 or 2 take a shot at him. So in a duel of logical, but dishonorable players, the duel will go on endlessly with no one being shot.

But wait! How could they have an endless supply of bullets? Eventually, each player will have one bullet left. Player 1 will again intentionally miss to give himself the best chances of survival. Now player 2 will shoot at the only player remaining who could kill him, namely, player 3. If player 2 misses, player 3 can shoot at either opponent with impunity. Player 1's chances of being selected are no better than 50/50 since there is no logical advantage in selecting either player. With no logical advantage, one could assume that player 3 might be tempted to choose the player who pissed him off most recently. So, since it's impossible to force the players to try to shoot each other, player 1 actually has the best chance of surviving, followed by player 2, and player 3 is the most likely to die.

Edit: Actually, on 3's second to last shot, he has nothing to lose by taking a shot at 1. That fact won't have any effect on 1's actions, but it means 1's chances are still the worst. 2's chances decrease some, but they're still better than 3's.

Last edited: Mar 11, 2005
12. Mar 11, 2005

### BobG

If you want a mathematical answer, it's best to break it up into two scenarios: the three person duel followed by the subsequent two person duel.

The odds of each player being killed by the end of a three-person phase:

Player 1) $$\frac{y+z-xy-yz+xyz}{x+y+z-xy-yz+xyz}$$

Player 2) $$\frac{x}{x+y+z-xy-yz+xyz}$$

Player 3) 0% - he's guaranteed to survive the three person phase.

The same odds apply to whatever chances all 3 have survived. In other words, you just keep adding some fraction in the above proportions in a recurring sequence.

If all 3 players were very close in ability, so 1 had barely above a 50% accuracy, 2 exactly 50% accuracy, and 3 barely below 50% accuracy, 1 would have about a 55.6% chance of dying during the 3-man phase, 2 a 44.4% chance, and 3 a 0 chance.

1 would have to be significantly better than 2 or 3 to have a better chance of surviving than 2. 3 always has the best chance. (For example, if P1 shot at 90% accuracy, P2 at 20%, and P3 at 10%, x would only have about a 12% chance of dying).

Edit: I finally got back to looking at the second part of the problem.

The odds for winning the in the two man round:

1st shooter (with w accuracy): $$\frac{w}{w+v-wv}$$
2nd shooter (with v accuracy): $$\frac{v-vw}{w+v-wv}$$

It's another recurring series where you're just adding fractions to 2-man first round in the same proportions.

You apply the odds of the second to the survivors of the 3-man phase. (Which means I missed a step in the 3-man phase. You need to know more than the chances of dying. You need to know the chances of being the shooter that killed another player.)

The two-man odds show why it's going to take an awful lot of honor to be the first shooter to kill someone.

Player 1 can only be the second shooter in the two-man phase. If he survives the first round, it's because he shot player 2.

So his odds of winning are:

$$\left( 1 - \frac{y+z-xy-yz+xyz}{x+y+z-xy-yz+xyz} \right) \left(\frac{x-xy}{x+y-xy} \right)$$

Last edited: Mar 11, 2005
13. Mar 12, 2005

### Bartholomew

Why are you working with odds instead of probabilities? Not sure whether you're right or wrong.

For each 3 person cycle when you have all 3 people, player 1 shoots at player 2. If he hits player 2 then his chance of surviving the 3 person cycle is (1-z) since z will then shoot at him. If player 1 misses player 2 then his chance of surviving is (1-y)(1-z) since both y and z must miss. He then has the same chance at surviving the total sequence of 3 player cycles that he had starting the round--call that chance a. So:

a = x(1-z) + (1-x)(1-y)(1-z)a
a - (1-x)(1-y)(1-z)a = x(1-z)
a(1- (1-x)(1-y)(1-z)) = x(1-z)
a = x(1-z)/(1-(1-x)(1-y)(1-z))

Now if player 1 survives the 3 man cycles, then player 2 is dead and it's player 1's shot. The probability of player 1 killing player 3 is x, and the probability of player 1 missing player 3 and still surviving the round is (1-z). Player 1 then has the same chance of surviving the 2 player cycles as he did before they started, so (calling that chance b)

b = x + (1-x)(1-z)b
b-(1-x)(1-z)b = x
b(1-(1-x)(1-z)) = x
b = x/(1-(1-x)(1-z))

So player 1's chance of winning is ab, or x(1-z)/(1-(1-x)(1-y)(1-z)) * x/(1-(1-x)(1-z)). I don't care to simplify that.

Player 2's chance of surviving the 3 player cycles can be broken into two pats, depending on which way 1 died. The probability that 2 kills 1 in any round of the 3 player cycles is (1-x)y = c and the probability that 3 kills 1 in any round of the 3 player cycles is (1-x)(1-y)z = d.

If player 2 kills player 1, then player 3 shoots next. Given that player 3's move is next, 2's probability of surviving is:

e = y(1-z) + (1-z)(1-y)e
e = y(1-z)/(1-(1-z)(1-y))

Given that player 2's move is next, 2's probability of surviving is:

f = y + (1-y)(1-z)f
f = y/(1-(1-y)(1-z))

e will be multiplied by c/(c+d) (the probability that given 1 died, 2 killed him)
f will be multiplied by d/(c+d) (the probability that given 1 died, 3 killed him)

So player 2's chance of surviving the whole duel is (1-a)(ce + df)/(c+d), which I'm not going to write all the way out. Player 3's chance of surviving is thus 1-(ab + (1-a)(ce + df)/(c+d))

Last edited: Mar 12, 2005
14. Mar 13, 2005

### msmith12

someone had said...

Player 3 is the most likely to win followed by player 2. Player 1 is the least likely to win.

this all has to do with the values of x,y, and z...

imagine that player one is a great shot, and hardly ever misses, and that player 2 and 3 cant hit the broad side of a barn... who is the most likely to win?

what about if all of them are pretty good shots, would player 2 really have better odds of winning then player 1, when player 2 will likely be killed on the first shot?

also, if a player could miss on purpose...

player one never wants to miss on purpose, because both players 2 and 3 will be shooting at him... player 2 doesnt want to miss, because player 1 will always be shooting at him... player 3 is the only person who gains from this strategy, because since no one shoots at him until only two people are left, if he never kills the first person, he will always have the first shot in the duel, which give him a huge advantage.

~matt

15. Mar 13, 2005

### RandallB

Good point Matt – I’d need to revise my post #7 on:
Player three to win =
xz
+(1-x)yz
for (1-x)(1-y) because 3 would miss on the above would be time “0” and start over at top.
Then only once he’s in a two man duel. Add
+(1-x)y(1-z)(1-x)z
..etc. for missing again
+x(1-z)(1-x)z
..etc. for missing again
That would change the #’s for the other two as well.

3 would be most likely winner unless the difference between x & z was large.

HOWEVER - if player 2 survives the first shot, 2 may try to take advantage of the same statagy hoping that 3 would go for 1 in error.
And with an miss on purpose by 3 --
player 1 may get the same idea and then all start shooting in the dirt!!

16. Mar 13, 2005

### Bartholomew

All the players would refrain from shooting if they had that option. If 1 kills 2 or 3 then the other of 2 or 3 would be obliged to shoot at 1. Similar reasoning goes for 2 and 3; assuming that each player's strategy is to maximize his own probability of surviving, each player would skip his turn.

17. Mar 14, 2005

### BobG

Player 2 can still have a better chance of winning the entire truel than player 1, even if player 2 has the best chance of being eliminated first. Unless there's a big disparity in skill, the odds of both player 2 and 3 missing player 1 is less than the odds of player 1 missing, even allowing for the chance that player 2 may never get a chance to shoot at all.

Granted, as the skill level of player one increases, the disparity between all of the players' skill levels has to decrease (in other words, P2 and P3 have to be very good, also) in order to compensate for P1's first shot ability. Knowing each player's skill level, you would think the distance would be set far enough apart that death for player 2, followed by death for player 1 would not be automatic.

That's an assumption not included in the original problem, but if the truelists were stupid enough to stand close enough to each other to make every shot a virtual certainty, they would most likely be too stupid to apply any kind of logic at all, meaning each player's choice becomes a random 50/50 chance (Player 2 and player 3 wind up with about a 50/50 chance of winning while player 1 will almost certainly die).

Actually, setting the distance will be a critical part of the pre-truel negotiations. Player 1 will have to weigh his own ability against player 3's. He hopes to be able to kill player 2 with 1 shot, while being far enough away to make it likely player 3 will miss. Player 2 has to set the distance far enough apart that the combined skill of player 3 and player 2 outweighs player 1's skill, while still tipping the balance against player 3 in a two-man duel. Player 3 wants all the distances as close as possible - he doesn't want any misses by anyone. If the players are logical and know the skill of each opponent, I could see the negotiations dragging on for years, unless there's one distance that would set all three players' chances equal to each other (now that would be an interesting problem, especially if player 3 is virtually guaranteed to advance through to the two-man phase - I think player 3 will have to be a very bad shot to make a solution possible).

Last edited: Mar 14, 2005
18. Mar 14, 2005

### msmith12

so... the problem that you are proposing is,

given probability functions based on the distance from one player to another, what is the ideal initial setup to allow for an equal chance of each player living?

19. Mar 14, 2005

### BicycleTree

(Repost)
BobG, why are you working with odds instead of probabilities? Not sure whether you're right or wrong.

For each 3 person cycle when you have all 3 people, player 1 shoots at player 2. If he hits player 2 then his chance of surviving the 3 person cycle is (1-z) since z will then shoot at him. If player 1 misses player 2 then his chance of surviving is (1-y)(1-z) since both y and z must miss. He then has the same chance at surviving the total sequence of 3 player cycles that he had starting the round--call that chance a. So:

a = x(1-z) + (1-x)(1-y)(1-z)a
a - (1-x)(1-y)(1-z)a = x(1-z)
a(1- (1-x)(1-y)(1-z)) = x(1-z)
a = x(1-z)/(1-(1-x)(1-y)(1-z))

Now if player 1 survives the 3 man cycles, then player 2 is dead and it's player 1's shot. The probability of player 1 killing player 3 is x, and the probability of player 1 missing player 3 and still surviving the round is (1-z). Player 1 then has the same chance of surviving the 2 player cycles as he did before they started, so (calling that chance b)

b = x + (1-x)(1-z)b
b-(1-x)(1-z)b = x
b(1-(1-x)(1-z)) = x
b = x/(1-(1-x)(1-z))

So player 1's chance of winning is ab, or x(1-z)/(1-(1-x)(1-y)(1-z)) * x/(1-(1-x)(1-z)). I don't care to simplify that.

Player 2's chance of surviving the 3 player cycles can be broken into two pats, depending on which way 1 died. The probability that 2 kills 1 in any round of the 3 player cycles is (1-x)y = c and the probability that 3 kills 1 in any round of the 3 player cycles is (1-x)(1-y)z = d.

If player 2 kills player 1, then player 3 shoots next. Given that player 3's move is next, 2's probability of surviving is:

e = y(1-z) + (1-z)(1-y)e
e = y(1-z)/(1-(1-z)(1-y))

Given that player 2's move is next, 2's probability of surviving is:

f = y + (1-y)(1-z)f
f = y/(1-(1-y)(1-z))

e will be multiplied by c/(c+d) (the probability that given 1 died, 2 killed him)
f will be multiplied by d/(c+d) (the probability that given 1 died, 3 killed him)

So player 2's chance of surviving the whole duel is (1-a)(ce + df)/(c+d), which I'm not going to write all the way out. Player 3's chance of surviving is thus 1-(ab + (1-a)(ce + df)/(c+d))

20. Mar 19, 2005

### kleinwolf

Yes, but you made the implicit hypothese that both 2 and 3 shoot at 1...whereas 2 could shoot at three....this is just to point out that nothing indicates how the shooters react to the others skill in order to choose their target.

I would say that this problem is ill-defined : suppose the truel starts...then 1 has to choose 2 or 3 to shoot at...what is the probability he chooses 2 or 3...we just know it has to be a growing function of y or z, but we don't know which one : generally p(1->2)=xg(x,y,z)...where g(x,y,z) is the probability player 1 chooses 2...by analogy : p(1->3)=x*h(x,y,z)...aso...

where f(x,y,z) is the probability the actual shooter will choose 1
g(x,y,z) for choosing 2, and h(x,y,z) to choose 3...

The things eventually goes harder as soon as a player is actually shot...

or am i complicating the problem ?

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