### The Good, the Bad and the Weird: Duels and the Gentleman's Draw

As I mentioned in the previous article "The Good, the Bad and the Weird: Duels, Truels and Utility Functions", a classic probability puzzle involves a 3-way duel (called a "truel").
A, B and C are to fight a three-cornered pistol duel. All know that A's chance of hitting his target is 0.3, C's is 0.5, and B never misses. They are to fire at their choice of target in succession in the order A, B, C, cyclically (but a hit man loses further turns and is no longer shot at) until only one man is left unit. What should A's strategy be?
There's a subtle issue involved in these types of problems in that we don't know how each participant values each outcome. If we allow duelists to deliberately miss there are $$2^3-1=7$$ possible outcomes; each person may or may not be shot and at least one person will not be shot. Even if deliberate missing isn't allowed, there are still 3 possible outcomes. A, for example, could conceivably value B winning more than C winning.

The classic solution concludes that with the given hit probabilities the optimal strategy for the first trueler is to deliberately miss. My contention is that this is an incomplete solution; for some sets of utility values this may not be the first trueler's optimal strategy. Let's examine duels using utility values as the first step towards addressing truels.

Let the two duelers have hit probabilities $$p_i$$ for $$i=1,2$$, and also let the values each assigns to a dueling win, loss or tie be $$W_i, L_i, T_i$$. I'm assuming these values are all known to both duelers and that $$W_i > T_i > L_i$$. Note that a strategy that optimizes expected utility is preserved under a positive affine transformation, so under the assumption that all values are finite, we may assume $$W_i=1$$ and $$L_i=0$$ for all $$i$$.

We'll declare the duel a "gentleman's draw" if both duelers deliberately miss in a single round. Let the expected utility value for dueler $$i$$ be $$U_i$$.

Trivially, if the optimal strategy for dueler 1 is trying for a hit, the optimal strategy for dueler 2 must be to also try for a hit. Conversely, the optimal strategy for dueler 1 can be to deliberately miss if and only if the optimal strategy for dueler 2 is also to deliberately miss. Assume the draw is taken if there's indifference (they are, after all, gentlemen).

If dueler 1 deliberately misses, dueler 2 has two choices. If he deliberately misses, it's a gentleman's draw and $$U_2 = T_2$$; if he tries for a hit, both duelers will subsequently try to hit each other and $$U_2 = p_2 + q_2 q_1 U_2$$. Solving, we get $U_2 = \frac{p_2}{1-q_1 q_2}.$ It's therefore optimal for dueler 2 to take the gentleman's draw if and only if $T_2 \geq \frac{p_2}{1-q_1 q_2}.$ As a consequence, dueler 1 will not deliberately miss if $T_2 < \frac{p_2}{1-q_1 q_2}.$ If dueler 1 tries for a hit, both will subsequently try to hit each other and we have $$U_1 = p_1 + q_1 q_2 U_1$$. Solving, we get $U_1 = \frac{p_1}{1-q_1 q_2}.$ Thus, the optimal strategy for dueler 1 is to deliberately miss if and only if $T_1 \geq \frac{p_1}{1-q_1 q_2}$ and $T_2 \geq \frac{p_2}{1-q_1 q_2}.$ These are, as expected, precisely the conditions under which dueler 2 will deliberately miss.

For example, if $$p_1=p_2=1/2$$, it'll be a gentleman's draw if and only if $$T_1,T_2 \geq 2/3$$. Paradoxically, both will fire under these hit probabilities if $$T_1 = 4/5$$ and $$T_2 = 1/2$$, even though this results in lower expected utility for both duelers than if they had agreed to a draw. This is a type of prisoner's dilemma.

### A Bayes' Solution to Monty Hall

For any problem involving conditional probabilities one of your greatest allies is Bayes' Theorem. Bayes' Theorem says that for two events A and B, the probability of A given B is related to the probability of B given A in a specific way.

Standard notation:

probability of A given B is written $$\Pr(A \mid B)$$
probability of B is written $$\Pr(B)$$

Bayes' Theorem:

Using the notation above, Bayes' Theorem can be written: $\Pr(A \mid B) = \frac{\Pr(B \mid A)\times \Pr(A)}{\Pr(B)}$Let's apply Bayes' Theorem to the Monty Hall problem. If you recall, we're told that behind three doors there are two goats and one car, all randomly placed. We initially choose a door, and then Monty, who knows what's behind the doors, always shows us a goat behind one of the remaining doors. He can always do this as there are two goats; if we chose the car initially, Monty picks one of the two doors with a goat behind it at random.

Assume we pick Door 1 and then Monty sho…

### What's the Value of a Win?

In a previous entry I demonstrated one simple way to estimate an exponent for the Pythagorean win expectation. Another nice consequence of a Pythagorean win expectation formula is that it also makes it simple to estimate the run value of a win in baseball, the point value of a win in basketball, the goal value of a win in hockey etc.

Let our Pythagorean win expectation formula be $w=\frac{P^e}{P^e+1},$ where $$w$$ is the win fraction expectation, $$P$$ is runs/allowed (or similar) and $$e$$ is the Pythagorean exponent. How do we get an estimate for the run value of a win? The expected number of games won in a season with $$g$$ games is $W = g\cdot w = g\cdot \frac{P^e}{P^e+1},$ so for one estimate we only need to compute the value of the partial derivative $$\frac{\partial W}{\partial P}$$ at $$P=1$$. Note that $W = g\left( 1-\frac{1}{P^e+1}\right),$ and so $\frac{\partial W}{\partial P} = g\frac{eP^{e-1}}{(P^e+1)^2}$ and it follows $\frac{\partial W}{\partial P}(P=1) = … ### Solving a Math Puzzle using Physics The following math problem, which appeared on a Scottish maths paper, has been making the internet rounds. The first two parts require students to interpret the meaning of the components of the formula $$T(x) = 5 \sqrt{36+x^2} + 4(20-x)$$, and the final "challenge" component involves finding the minimum of $$T(x)$$ over $$0 \leq x \leq 20$$. Usually this would require a differentiation, but if you know Snell's law you can write down the solution almost immediately. People normally think of Snell's law in the context of light and optics, but it's really a statement about least time across media permitting different velocities. One way to phrase Snell's law is that least travel time is achieved when \[ \frac{\sin{\theta_1}}{\sin{\theta_2}} = \frac{v_1}{v_2},$ where $$\theta_1, \theta_2$$ are the angles to the normal and $$v_1, v_2$$ are the travel velocities in the two media.

In our puzzle the crocodile has an implied travel velocity of 1/5 in the water …