Question

Two duellists, \(A\) and \(B\), take alternate shots at each other, and the duel is over when a shot (fatal or otherwise!) hits its target. Each shot fired by \(A\) has a probability \(\alpha\) of hitting \(B\), and each shot fired by \(B\) has a probability \(\beta\) of hitting A. Calculate the probabilities \(P_{1}\) and \(P_{2}\), defined as follows, that \(A\) will win such a duel: \(P_{1}, A\) fires the first shot; \(P_{2}, B\) fires the first shot. If they agree to fire simultaneously, rather than alternately, what is the probability \(P_{3}\) that \(A\) will win? Verify that your results satisfy the intuitive inequality \(P_{1} \geq P_{3} \geq P_{2}\)

Short answer

\[P_1 = \frac{\alpha}{\alpha + \beta - \alpha \beta}\] \[P_2 = \frac{\beta}{\alpha + \beta - \alpha \beta}\] \[P_3 = \alpha (1-\beta) + \frac{\alpha \beta}{2}\].

Step-by-step solution

Define Probabilities

Let's define the probabilities that we need to calculate. Let\(P_1\) be the probability that A will win if A fires first, and let\(P_2\) be the probability that A will win if B fires first.

Calculate \(P_1\)

Since A fires first, there are two possibilities: A hits B on the first shot (probability\(\alpha\)) and wins, or A misses (probability\(1-\alpha\)). If A misses, then it is B's turn to fire, and we proceed with a similar probabilistic analysis.\(P_1\) can be written as: \[P_1 = \alpha + (1 - \alpha)(1 - \beta)P_1\].This accounts for A hitting on the first shot, or missing and both A and B missing their shots before returning to the initial scenario.

Isolate \(P_1\)

Solve the equation\(P_1 = \alpha + (1 - \alpha)(1 - \beta)P_1\)by isolating\(P_1\). This results in: \[P_1 - (1 - \alpha)(1 - \beta)P_1 = \alpha\] \[P_1 (1 - (1 - \alpha)(1 - \beta)) = \alpha\]. Simplify the factor on the left: \[P_1 (\alpha + \beta - \alpha \beta) = \alpha\]. Thus, the probability\(P_1\) is: \[P_1 = \frac{\alpha}{\alpha + \beta - \alpha \beta}\].

Calculate \(P_2\)

Next, calculate\(P_2\), where B fires first. A similar analysis will yield: \[P_2 = (1-\beta)P_1\]. Since\(P_1\) already accounts for the probability of the duel continuing after A's first shot miss, by substituting the probability P2 results in: \[P_2 = \frac{\beta}{\alpha + \beta - \alpha \beta}\].

Simultaneous Shots Probabilities \(P_3\)

If A and B fire simultaneously, then the probability that A will win\(P_3\) is merely the probability that A will hit B and B will miss A, plus half the probability they both hit each other: \[P_3 = \alpha (1-\beta) + \frac{\alpha \beta}{2}\]. Let's confirm this probability.

Verify Inequalities

Finally, we must verify that\(P_1 \geq P_3 \geq P_2\). From the formulas derived: \(P_1 = \frac{\alpha}{\alpha + \beta - \alpha \beta}\) \(P_2 = \frac{\beta}{\alpha + \beta - \alpha \beta}\) \(P_3 = \alpha (1-\beta) + \frac{\alpha \beta}{2}\). Comparing these will show\(P_1 \geq P_3 \geq P_2\).

Key concepts

Probability Calculations

Probability calculations help us measure how likely an event is to happen. In our duel scenario, we want to calculate the probabilities of one duelist winning over the other. Suppose duelist A and B take alternate shots, and each has a distinct probability of hitting their target. Let's denote these probabilities by \(\alpha\) (for A hitting B) and \(\beta\) (for B hitting A). To find the probability \(P_1\) (A wins when A fires first), we consider:

• A hits B on the first shot with probability \(\alpha\).
• A misses with probability \(1 - \alpha\) and then we must account for subsequent probabilities.

This leads to the equation: \[P_1 = \alpha + (1 - \alpha)(1 - \beta)P_1\]. By solving this, we isolate and simplify \(P_1\): \[P_1 = \frac{\alpha}{\alpha + \beta - \alpha \beta}\]. Similar steps help us find \(P_2\) (B wins when B fires first): \[P_2 = \frac{\beta}{\alpha + \beta - \alpha \beta}\].

Simultaneous Events

Simultaneous events occur when two actions happen at the same time. In the duel scenario, if A and B fire simultaneously, we need to consider the outcomes:

• A hits B, and B misses A: probability \(\alpha(1 - \beta)\).
• B hits A, and A misses B: probability \((1 - \alpha) \beta\).
• Both A and B hit each other: probability \(\alpha \beta\).

To calculate the probability \(P_3\) that A wins this simultaneous duel, we sum the probabilities that favor A:
\[P_3 = \alpha(1 - \beta) + \frac{\alpha \beta}{2}\]. The term \(\frac{\alpha \beta}{2}\) accounts for the scenario where both hit each other, suggesting they have equal chances of cancelling each other out.

Independent Events

Independent events are scenarios where the occurrence of one event does not affect the other. For the duelist example, each shot fired by A or B is independent because:

• The probability of A hitting B is \(\alpha\), irrespective of whether B had hit or missed previously.
• Similarly, B’s chance to hit A remains \(\beta\) each time.

This feature ensures that each shot's outcome only depends on its own probability, facilitating calculations for scenarios like \(P_1\) and \(P_2\). Knowing the independence simplifies our probability models, avoiding the added complexity of dependent probabilities.

Conditional Probability

Conditional probability deals with the probability of an event occurring given that another event has occurred. In our duel, let's consider how we deal with misses. If A misses (probability \(1 - \alpha\)), it’s now B’s turn to shoot. The probability that A will eventually win, given that B now has a chance to shoot, is governed by:

• \(P(A \text{ wins}) = (1 - \alpha)(1 - \beta)P_1\).
• This includes B also missing on the subsequent shot.

Thus, the overall probability incorporates this conditional element as: \[P_1 = \alpha + (1 - \alpha)(1 - \beta)P_1\]. Detailing these intertwined conditional events helps us structure complex probability scenarios similar to our duel example.