Probability > 1? (Martin Gardner problem)

Question

I'm doing a version of this Martin Gardner puzzle

In this version, the order in which they shoot is by drawing lots, and a shooter always shoots the higher rated of his two opponents. Also, the accuracies are 100%, 75% and 50% for the 3 shooters. Let's call them M(50%), S(75%) and Y(100%).

The puzzle is to come up with the probability of each person winning.

There are 6 equally possible shooting orders. I chose the order MSY. However, when i calculate the probability of M, S & Y being the last man standing, i get a total > 1. Is this possible? or am i mistaken?

(I made a probability tree and added up all the branches where S wins (= 0.5625), M (=0.4375+epsilon) and Y wins(0.125). Total = 1.125+) Probability tree diagram

(In fact, there is a possibility of M&S infinitely continuing the game, with both of them missing all the time, right? - and that has a %age too - however small. So i ought to have a total P < 1. What am i missing?)

Is it that probability trees should only be counted to the same depth? Is a probability tree a correct representation where the events end at different depths?

Answer 1

HINT

Probability tree is probably not the way to go (unless you plan to do infinite summations). Instead I would suggest:

First, solve all the $2$-player cases. I.e. if players $A, B$ are left and it's $A$'s turn to shoot, what is probability of each winning? Do you know how to do this without using a probability tree, but instead using (in the most basic version) simultaneous equations / recursion?

Let $h_A$ be accuracy of $A$, and $p_{AB}$ be prob of $A$ winning when $\{A,B\}$ are left and $A$ shoots first. Then: $p_{AB} = h_A + (1-h_A) p_{BA}$ and $p_{BA} = h_B + (1 - h_B) p_{AB}$. These are $2$ equations in $2$ unknowns which you can solve. Note that $p_{AB} + p_{BA} \neq 1$ because in one case $A$ shoots first and in the other case $B$ shoots first, so they are not complementary events.

Then, solve the 3-player cases, again using simultaneous equations / recursion.