Probability puzzle

[WMDhamnekar]

TL;DR

Two players start 1 meter away from a target. They simultaneously begin moving towards the target at a same constant speed. Target hitting probability of both the players when they are X meters from the target are given. If both players misses or hits the target simultaneously, neither wins and game will be started over. The player whoever hit the target first will win.What is the probability that left player wins, if both player uses optimal strategies?

Two players start 1 meter away from a target. They simultaneously begin moving towards the target at a same constant speed. If the left player shoots when he is X meters from the target, his shot hit with a probability 1-X. If the right player shoots when he is X meters from the target, his shot hits with a probability [$1- X^2$]. Each player has exactly one bullet and may choose to shoot at any time during the walk. If exactly one player hit the target, that player wins. If both players shoot simultaneously and both hits , then neither player wins. Both are sent back to the starting positions and game starts over. Similarly, if both the players miss the shots, the game starts over from the beginning. The players don't have to shoot at the same time and they can see each other at all times. If they both hit the target, but at different times, whoever hits the target first will win. Assuming both players use optimal strategies, what is the probability that the left player wins? Note: Though, both players uses pistol, they are not criminals,violence supporter.

My answer: I don't under stand how to answer this question. I am still working on this question. If any physics forum's members knows the answer to this question may reply with correct answer.

 

[anuttarasammyak]

$$1-x^2=(1-x)(1+x)>1-x$$

I know that the right player has advantage. I do not think the left has a strategy to win with more than 50% probability.

 

[Hornbein]

At first both players are better off hoping for the other guy to fire first and miss. There comes a point where firing and waiting are of equal value. That's when x is about 0.618 m. After that both players are better off firing. Each is more likely to hit than the other guy is to miss so you'd better go for it even if the odds are against you. Unless you think the other guy won't figure it out and you can wait some more then fire first.

Lefty might hope that firing for a tie at 0.618 will help but it makes things worse. If righty can (unrealistically) fire at the same moment as lefty then that's a good plan until I'm not sure when.

 

[Dale]

I see the same. At any distance greater than D = 0.618 m each player's best strategy is to wait, regardless of the other player's strategy. At any distance less than D each player's best strategy is to fire just before the other player fires. Since each player knows that the other player is also playing optimally, they will both fire at D.

 

[anuttarasammyak]

$$p_1(r)+p_2(r)=1$$
gives optimal firing position r for both the players where $p_i(r)$ is the probability of success of i-th player at position r.

In this particular case
$$1-r+1-r^2=1$$
$$r=\frac{\sqrt{5}-1}{2}$$
$$p_1=\frac{3-\sqrt{5}}{2}$$
$$p_2=\frac{\sqrt{5}-1}{2}$$

Success probalities of fire (red ) and wait (blue ) for player 1

Success probalities of fire (red ) and wait (blue ) for player 2

PS　
Three players case,e.g.,
$$p_1(r)=1-r$$
$$p_2(r)=1-r^2$$
$$p_3(r)=1-r^3$$
seems interesting though I have not solved it yet.

 

[Hornbein]

I see the same. At any distance greater than D = 0.618 m each player's best strategy is to wait, regardless of the other player's strategy. At any distance less than D each player's best strategy is to fire just before the other player fires. Since each player knows that the other player is also playing optimally, they will both fire at D.

Say you are at D. If the other guy fires first then your best move is to not fire until at point blank. Firing first gives no advantage but firing second is always bad. It is not clear what the best plan is. Maybe to fake the other guy into firing first then hold your fire so you will never be second. Or fire slightly before D to reduce the chance of the other guy firing first. But then the other guy might do the same. So you try to anticipate what your opponent will do but this seems beyond mathematics.

This sort of thing is prevalent in expert-level real world games and sports.

 

[Dale]

Maybe to fake the other guy into firing first

We are given that both players play optimally. We were not given it, but I also assumed that both players know the other player is also optimal. That knowledge is essential for knowing that the optimal strategy is D.

If the opponent were random or had some other strategy then that would change the optimal strategy.

 

[Hornbein]

We are given that both players play optimally. We were not given it, but I also assumed that both players know the other player is also optimal. That knowledge is essential for knowing that the optimal strategy is D.

If the opponent were random or had some other strategy then that would change the optimal strategy.

But both firing at the same instant is not optimal for Lefty. It favors Righty. This is included as a possibility in the problem statement.

 

[Dale]

both firing at the same instant is not optimal for Lefty. It favors Righty

It does favor the right player, but it is still optimal for the left player also. In other words, the game is rigged. The optimal outcome for the left player still favors the right player.

 

[Hornbein]

It does favor the right player, but it is still optimal for the left player also. In other words, the game is rigged. The optimal outcome for the left player still favors the right player.

The difference comes when Righty can fire at the same time as Lefty. If if they both fire at exactly .618 it boosts Righty's chance of victory from 62% to about 71%. Lefty wants to avoid ties in firing time, which are allowed. If both fire at the same time and have the same result then there is a rematch. Righty wants to cause such situations.

 

[anuttarasammyak]

But both firing at the same instant is not optimal for Lefty. It favors Righty. This is included as a possibility in the problem statement.

I would estimate it using post #5
$$p_1=0.381966... , p_2= 0.618033...$$
where 1 is left and 2 is right.

Firing at the same time in that optimal case
Probabilities of
Player 1 wins : $p_1(1-p_2)=p_1^2$
Player 2 wins : $p_2(1-p_1)=p_2^2$
No contest : $1-p_1^2-p_2^2$
Relative probability of Player 1 wins : $$\frac{p_1^2}{p_1^2+p_2^2}=\frac{5-\sqrt{5}}{10}= 0.276393...$$
Relative probability of Player 2 wins : $$\frac{p_2^2}{p_1^2+p_2^2}=\frac{5+\sqrt{5}}{10}= 0.723606...$$

So co-firing at the optimal distance is advantageous for Player 2, right player.　The rich becomes richer and the poor becomes poorer.
Player 2 does not hesitate to fire at the optimal distance. Understanding it Player 1 fires at distance 0 in case Player 2 fails. The optimal probability $p_1,p_2$ take place.
The basic strategy behind : when you observe his firing (on purpose, in accident or in any manner) wait firing until you reach distance 0 if you have not fired yet. With this starategy one's failure probability equals the other's success probaility.

Player 2 thinks "He does not fire until distance 0. I will fire in closer distance for higher probability"
Player 1 thinks "He thinks so. So I should be a first shooter sometimes to break it"
and so on. Such a strategy struggle is a tough thing for mathematics and physics.

I think $p_1,p_2$ has meaning even in these tactics struggle. We may be able to disregard simultaneous shooting in chance because of its infinitesimal probability. I conjecture that such a chickin race would take players away from the optimal distance firing but not so much and as a kind of fluctuation. .

 

[Dale]

The difference comes when Righty can fire at the same time as Lefty. If if they both fire at exactly .618 it boosts Righty's chance of victory from 62% to about 71%. Lefty wants to avoid ties in firing time, which are allowed. If both fire at the same time and have the same result then there is a rematch. Righty wants to cause such situations.

Ah, I missed the tie -> rematch condition.

That makes the optimum a little difficult. I would guess that the left player would want to fire infinitesimally after D then. But I have not worked it out to confirm.

 

[Dale]

But both firing at the same instant is not optimal for Lefty. It favors Righty. This is included as a possibility in the problem statement.

Ok, so the issue isn’t that a tie favors the right player (the whole game does), but that for the left player a tie at D represents a finite “marginal” loss from the optimal D without a tie. Let's call that marginal loss L.

If the left player employs a strategy of firing at a non-D distance, then that strategy incurs a marginal loss, and there is some finite region around D where that marginal loss is less than L. If the left player chooses to fire at any distance in that region, then they are worse off than they would be at D without a tie, but better off than they would be at D with a tie. So they want to randomly choose a distance within that interval. The statistical distribution should have an expectation of D, be zero outside of the finite region (and maybe also at D), a minimal variance, and minimize the probability of a tie.

Any finite continuous distribution will have 0 probability of an exact tie. That doesn't mean that a tie is impossible, but its probability is so small that it won't affect the expectation. And there are plenty of distributions that are only non-zero in the region and with an expectation of D. So, the only problem is the minimal variance. There simply is no such thing. For any qualifying distribution with any finite variance there is another one with less variance.

So, I don't think that the left player has an optimal strategy, but there are strategies that are arbitrarily close to optimal. The right player would still stick with D since the expected value of trying to get the tie is 0.

I have to admit that I cannot prove any of this. I could be wrong. Also, it would be more interesting if the distances were discrete, then the probability of a tie could not be made 0, and there might be an actual optimal strategy for the left player.