To find the probability that the left player wins, we must analyze this as a sequential, continuous-time game of timing (often called a silent duel with asymmetric success probabilities), since players can see each other but only find out if the other has shot if they see the shot occur.
Let [imath]L[/imath] be the left player and [imath]R[/imath] be the right player. Both players start at a distance of [imath]1[/imath] meter and walk toward [imath]0[/imath] (the target). Their hit probabilities at distance [imath]X[/imath] are:
Left Player ([imath]L[/imath]):[imath]P_L(X) = 1 - X[/imath]
Right Player ([imath]R[/imath]):[imath]P_R(X) = 1 - X^2[/imath]
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Step 1: Understanding the Strategic Dynamics
Because the players can see each other at all times, neither player wants to shoot too early and miss, because a miss leaves them with [imath]0[/imath] bullets, allowing the other player to walk all the way to the target ([imath]X=0[/imath]) and hit with [imath]100\%[/imath] certainty.
Therefore, if one player shoots and misses at distance [imath]X[/imath], the other player automatically wins. This creates an equilibrium where both players will hold their fire until a critical threshold distance, [imath]X^*[/imath], is reached. At this critical distance, the danger of the opponent shooting and hitting balances the advantage of waiting to get closer.
Because they move simultaneously at the same speed, they will reach this critical point at the exact same time. Under optimal strategies, both players will fire simultaneously at this equilibrium distance [imath]X^*[/imath].
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Step 2: Finding the Critical Shooting Distance ([imath]X^*[/imath])
In a symmetric or optimally balanced game of timing, the equilibrium occurs when the sum of the probabilities of hitting the target equals [imath]1[/imath]. This is the point where a player becomes indifferent between shooting now or waiting an infinitesimal moment longer, given that the opponent might shoot.
[math]\quad P_L(X^*) + P_R(X^*) = 1 \quad[/math]
Substituting the given probability functions into the equation:
[math](1 - X^*) + (1 - (X^*)^2) = 1[/math]
Simplifying the equation:
[math]2 - X^* - (X^*)^2 = 1[/math]
[math](X^*)^2 + X^* - 1 = 0[/math]
Using the quadratic formula to solve for [imath]X^*[/imath] (keeping the positive root since distance [imath]X \in [0, 1][/imath]):
[math]X^* = \frac{-1 + \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{\sqrt{5} - 1}{2}[/math]
This is the reciprocal of the golden ratio, approximately [imath]X^* \approx 0.618[/imath] meters.
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Step 3: Calculating the Winning Probabilities at [imath]X^*[/imath]
At the optimal distance [imath]X^* = \frac{\sqrt{5}-1}{2}[/imath], both players fire simultaneously. Let's find their individual probabilities of hitting the target at this exact moment.
From our equilibrium equation, we know that [imath](X^*)^2 = 1 - X^*[/imath]. We can use this to simplify [imath]R[/imath]'s hit probability:
Probability that [imath]L[/imath] hits ([imath]p_L[/imath]):
[math]p_L = 1 - X^*[/math]
Probability that [imath]R[/imath] hits ([imath]p_R[/imath]):
[math]p_R = 1 - (X^*)^2 = 1 - (1 - X^*) = X^*[/math]
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Step 4: Determining the Game Outcomes
When both players fire simultaneously at [imath]X^*[/imath], there are four possible outcomes:
| Outcome | Probability | Winner |
| --- | --- | --- |
| [imath]L[/imath] hits, [imath]R[/imath] misses | [imath]p_L(1 - p_R)[/imath] | Left Player ([imath]L[/imath]) |
| [imath]R[/imath] hits, [imath]L[/imath] misses | [imath]p_R(1 - p_L)[/imath] | Right Player ([imath]R[/imath]) |
| Both hit | [imath]p_L \cdot p_R[/imath] | Draw (Game restarts) |
| Both miss | [imath](1 - p_L)(1 - p_R)[/imath] | Draw (Game restarts) |
The game only ends permanently if exactly one player hits. If a draw occurs (both hit or both miss), the game resets completely to the beginning, meaning the relative probability of [imath]L[/imath] winning a clean, decisive round remains constant across resets.
Using conditional probability, the final probability that the Left Player wins ([imath]W_L[/imath]) given that the game eventually ends is:
[math]W_L = \frac{\text{Prob}(L \text{ hits, } R \text{ misses})}{\text{Prob}(L \text{ hits, } R \text{ misses}) + \text{Prob}(R \text{ hits, } L \text{ misses})}[/math]
[math]W_L = \frac{p_L(1 - p_R)}{p_L(1 - p_R) + p_R(1 - p_L)}[/math]
Substitute the simplified values [imath]p_L = 1 - X^*[/imath] and [imath]p_R = X^*[/imath]:
[imath]1 - p_R = 1 - X^*[/imath]
[imath]1 - p_L = 1 - (1 - X^*) = X^*[/imath]
Now, substitute these into the winning fraction:
[math]W_L = \frac{(1 - X^*)(1 - X^*)}{(1 - X^*)(1 - X^*) + (X^*)(X^*)}[/math]
[math]W_L = \frac{(1 - X^*)^2}{(1 - X^*)^2 + (X^*)^2}[/math]
Recall that [imath]1 - X^* = (X^*)^2[/imath]. Substitute [imath](X^*)^2[/imath] in place of [imath](1 - X^*)[/imath]:
[math]W_L = \frac{((X^*)^2)^2}{((X^*)^2)^2 + (X^*)^2} = \frac{(X^*)^4}{(X^*)^4 + (X^*)^2}[/math]
Divide the numerator and the denominator by [imath](X^*)^2[/imath]:
[math]W_L = \frac{(X^*)^2}{(X^*)^2 + 1}[/math]
Finally, substitute [imath](X^*)^2 = 1 - X^*[/imath]:
[math]W_L = \frac{1 - X^*}{(1 - X^*) + 1} = \frac{1 - X^*}{2 - X^*}[/math]
Using the numerical value of [imath]X^* \approx 0.618034[/imath]:
[math]W_L = \frac{1 - 0.618034}{2 - 0.618034} = \frac{0.381966}{1.381966} \approx \frac{5 - \sqrt{5}}{10} \approx 0.2764[/math]
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Final Conclusion
Assuming both players play with optimal strategies, the probability that the left player wins is [imath]\frac{5 - \sqrt{5}}{10}[/imath], or approximately [imath]27.64\%[/imath].
