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 [math]1- X^2[/math]. 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. Assuming both players use optimal strategies, what is the probability that the left player wins?

1 hour ago, Dhamnekar Win,odd said:

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−X2. 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. Assuming both players use optimal strategies, what is the probability that the left player wins?

What kind of homework is this?

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 [math]L[/math] be the left player and [math]R[/math] be the right player. Both players start at a distance of [b]1[/b] meter and walk toward [b]0[/b] (the target). Their hit probabilities at distance [b]X[/b] are:

• Left Player ([b]L[/b]): [math[]P_L(X) = 1 - X[/math]

• Right Player ([b]R[/b]): [math]P_R(X) = 1 - X^2[/math]

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 [b]0[/b] bullets, allowing the other player to walk all the way to the target ([b]X=0[/b]) and hit with [math]100\%[/math] certainty.

Therefore, if one player shoots and misses at distance [b]X[/b], the other player automatically wins. This creates an equilibrium where both players will hold their fire until a critical threshold distance, [math]X^*[/math], 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 [math]X^*[/math].

Step 2: Finding the Critical Shooting Distance ([math]X^*[/math])

In a symmetric or optimally balanced game of timing, the equilibrium occurs when the sum of the probabilities of hitting the target equals $1$. This is the point where a player becomes indifferent between shooting now or waiting an infinitesimal moment longer, given that the opponent might shoot.

$$\quad P_L(X^*) + P_R(X^*) = 1 \quad$$

Substituting the given probability functions into the equation:

$$(1 - X^*) + (1 - (X^*)^2) = 1$$

Simplifying the equation:

$$2 - X^* - (X^*)^2 = 1$$

$$(X^*)^2 + X^* - 1 = 0$$

Using the quadratic formula to solve for $X^*$ (keeping the positive root since distance [math]X \in [0, 1][/math]):

$$X^* = \frac{-1 + \sqrt{1^2 - 4(1)(-1)}}{2(1)} = \frac{\sqrt{5} - 1}{2}$$

This is the reciprocal of the golden ratio, approximately [math]X^ {\approx 0.618}[/math] meters*.

Step 3: Calculating the Winning Probabilities at [math]X^*[/math]

At the optimal distance [math]X^* = \frac{\sqrt{5}-1}{2}[/math], 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 [math](X^*)^2 = 1 - X^*[/math]. We can use this to simplify [b]R[/b]'s hit probability:

• Probability that $L$ hits ([math]p_L[/math]):

  $$p_L = 1 - X^*$$

• Probability that $R$ hits ([math]p_R[/math]):

  $$p_R = 1 - (X^*)^2 = 1 - (1 - X^*) = X^*$$

Step 4: Determining the Game Outcomes

When both players fire simultaneously at $X^*$, there are four possible outcomes:

Outcome	Probability	Winner
[b]L[/b] hits, [b]R[/b]misses	[math]p_L(1 - p_R)[/math]	Left Player ([b]L[/b])
[b]R[/b] hits, [b]L[/b] misses	[math]p_R(1 - p_L)[/math]	Right Player ([b]R[/b])
Both hit	[math]p_L \cdot p_R[/math]	Draw (Game restarts)
Both miss	[math](1 - p_L)(1 - p_R)[/math]	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 $L$ winning a clean, decisive round remains constant across resets.

Using conditional probability, the final probability that the Left Player wins ([math]W_L[/math]) given that the game eventually ends is:

$$W_L = \frac{\text{Prob}(L \text{ hits, } R \text{ misses})}{\text{Prob}(L \text{ hits, } R \text{ misses}) + \text{Prob}(R \text{ hits, } L \text{ misses})}$$

$$W_L = \frac{p_L(1 - p_R)}{p_L(1 - p_R) + p_R(1 - p_L)}$$

Substitute the simplified values [math]p_L = 1 - X^*$ and $p_R = X^*[/math]:

• [math]1 - p_R = 1 - X^*[/math]

• [math]1 - p_L = 1 - (1 - X^*) = X^*[/math]

Now, substitute these into the winning fraction:

$$W_L = \frac{(1 - X^*)(1 - X^*)}{(1 - X^*)(1 - X^*) + (X^*)(X^*)}$$

$$W_L = \frac{(1 - X^*)^2}{(1 - X^*)^2 + (X^*)^2}$$

Recall that [math]1 - X^* = (X^*)^2[/math]. Substitute [math](X^*)^2[/math] in place of [math](1 - X^*)[/math]:

$$W_L = \frac{((X^*)^2)^2}{((X^*)^2)^2 + (X^*)^2} = \frac{(X^*)^4}{(X^*)^4 + (X^*)^2}$$

Divide the numerator and the denominator by [math](X^*)^2[/math]:

$$W_L = \frac{(X^*)^2}{(X^*)^2 + 1}$$

Finally, substitute [math](X^*)^2 = 1 - X^*[/math]:

$$W_L = \frac{1 - X^*}{(1 - X^*) + 1} = \frac{1 - X^*}{2 - X^*}$$

Using the numerical value of [math]X^* \approx 0.618034[/math]:

$$W_L = \frac{1 - 0.618034}{2 - 0.618034} = \frac{0.381966}{1.381966} \approx \frac{5 - \sqrt{5}}{10} \approx 0.2764$$

Final Conclusion

Assuming both players play with optimal strategies, the probability that the left player wins is math]\frac{5 - \sqrt{5}}{10}[/math], or approximately [math]27.64\%[/math].

Edited Sunday at 06:31 AM5 days by Dhamnekar Win,odd