# The Three-Cornered Duel

I am analyzing the following problem from the the book "Fifty Challenging Problems in Probability with Solution" by Frederick Mosteller.

It seems to me that the solution to the The Three-Cornered Duel Problem presented in the book is incomplete. Can anybody confirm if my calculation is ok?

The problem

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?

Mosteller's solution:

A is naturally not feeling cheery about this enterprise. Having the first shot he sees that, if he hits C, B will then surely hit him, and so he is not going to shoot at C. If he shoots at B and misses him, then B clearly shoots the more dangerous C first, and A gets one shot at B with probability 0.3 of succeeding. If he misses this time, the less said the better. On the other hand, suppose A hits B. Then C and A shoot alternately until one hits. A's chance of winning is $$(.5)(.3)+(.5)^2(.7)(.3)+(.5)^3(.7)^2(.3)+…$$ Each term corresponds to a sequence of misses by both C and A ending with a final hit by A. Summing the geometric series we get $$(.5)(.3)+\{1+(.5)(.7)+[(.5)(.7)]^2+… \}= \frac{(.5)(.3)}{1-(.5)(.7)}= \frac{0.15}{0.65}= \frac{3}{13} < \frac{3}{10}$$ Thus hitting B and finishing off with C has less probability of winning for A than just missing the first shot. So A fires his first shot into the ground and then tries to hit B with his next shot. C is out of luck.

My calculation
case 1 $$P(A survives \ in \ case \ A \ shots \ B )= \mathbf{0.3} [(.5)(.3)+(.5)^2(.7)(.3)+(.5)^3(.7)^2(.3)+…] = \mathbf{0.3} \frac{(.5)(.3)}{1-(.5)(.7)}= \frac{\mathbf{0.3}(0.15)}{0.65}=0.069$$ case 2 $$P(A \ survives \ in \ case \ A \ misses \ the \ shot \ at \ B )= (0.7) \ 1 \ (0.3) = 0.21$$

Probability of survival is higher in case 2 therefore A should miss the first shot at B.
Is my calculation correct?

You and Mosteller are calculating slightly different quantities.

Let's look at what you are calculating for case 1:

You write $P(\text{A survives in case A shots B})$ which means $P\text{(A survives given A kills B}) \equiv P(\text{A survives|A kills B})$

But in the right hand side of your equation you are calculating $P(\text{A survives}\ \cap\ \text{A kills B})$ which is equal to $P(\text{A survives|A kills B})\cdot P(\text{A kills B}) = \frac{3}{13}\cdot 0.3$

So you are calculating the intersection of two events (event 1 and event 2) instead of the conditional (event 1 given event 2). Note that if you use the right name/description for the probability you are calculating then your calculations are agreeing with Mosteller.

The same goes for case 2. You are calculating $P(\text{A survives}\ \cap \ \text{A shoots at B and misses})$

Now that we have named the probabilities more accurately, you can ask yourself: How do they help me in answering the question? Isn't it more useful if I know the conditional probabilities instead? Yes it is. This is what will help you decide on the optimal action for A.

So as Monteller says: If A has to pick a target, he has to go for B. If A misses B, then B kills C on the next round, and on the round after that A has a single chance at B. So A survives with probability $0.3$. If A kills B then we find that the probability of surviving a duel with C (where C goes first) is $\frac{3}{13}$. So we see it's better to miss. And indeed A can choose to deliberately miss, and this is what he should do.

Finally here's another way to find A's survival probability against a duel with C. Let's define two probabilities, $P_A$ and $P_C$: $$P_A \equiv P(\text{A shoots first and A survives at the end}) \\ P_C \equiv P(\text{C shoots first and C survives at the end})$$

Now note the relationship between the two, that creates a $2\times 2$ system that can be easily solved: $$\left. \begin{array}{} P_A = 0.3 + 0.7\cdot (1-P_C)\\ P_C = 0.5 + 0.5\cdot (1-P_A) \end{array} \right\} \iff \begin{array}{} P_A = \frac{6}{13}\\ P_C = \frac{10}{13} \end{array}$$

What we care about in our scenario is person A surviving given that person C goes first, which is equal to $1-P_C = \frac{3}{13}$

You can solve this problem more easily with the help of Probability-tree diagram. Through Right half of tree(i.e. A targets first to C)

Through left half of sub-tree

Prob. to Survive A= 0.7*0.3=0.21;
Prob. to Survive B= 0.7*0.7= 0.49;


Through right half of sub-tree

Prob. to Survive B=0.3;


So, with this Choice,

Total Prob. to Survive A= 0.21;

Total Prob. to Survive B=0.49+0.3= 0.79;


Through Left half of tree(i.e. A targets first to B)

Through left half of sub-tree

Prob. to Survive A= 0.7*0.3=0.21;
Prob. to Survive B= 0.7*0.7= 0.49;


Through right half of sub-tree

Prob. to Survive A=(.3*.5*.3)+(.3*.5*.7*.5*.3)+ --------------- = 0.0692;
Prob. to Survive C=(.3*.5)+(.3*.5*.7*.5)+ --------------------- = 0.2307;


So, with this Choice,

Total Prob. to Survive A= .21+.0692= 0.2792;
Total Prob. to Survive B= 0.49;
Total Prob. to Survive C= 0.2307;


From this we can refer that,

Prob. of survival when A targets B first(0.2792) > Prob. of survival when A targets C first(0.21)