Pistol Competition Probability pls help

Question

A, B, C and D are to fight a four-cornered pistol competition. All know that A's chance of hitting his target is 0.3, C's is 0.5, B never misses, and D is 0.7. The order of shooting is alphabetical, and it is assumed that the person will hit the other person with the higher hit rate. A wants to win, what is A's winning probability?

Note: There are two scenarios: A shoots B in the first shot, and A misses B in the first shot.

1. It involves Infinite Geometric Series, Tree Diagram can be used

Answer 1

Hi Rei,

My solution to this question doesn't involve a lot of difficult math, but it does require a lot of organization, so if it is confusing let me know and I will try to clear things up.

The strategy to understanding this problem is to break it up into smaller problems first, and build our way up. We're going to do something similar to how a computer program might do recursion.

The A + C Lemma

For a moment let's pretend that it is only a 2-player game and A and C are the only people playing. What is A's chance of winning? It depends on whether or not A goes first. For now suppose it is A's turn, so we'll call this probability AC1 since it involves A and C (and the "1" is to denote that A it is A's turn).

AC1 = (0.3)*∑(0.7*0.5)ⁿ = (0.3)*(1-(0.35))^-1

How did I get to this? Well for A to win, exactly 1 of A's shots needs to be a success so that where the 0.3 comes from. The (0.7)ⁿ(0.5)ⁿ come from that fact that A can also win if A and C continually miss one after another prior to A's successful shot. Note that A and C both need to have the same number of failures.

Ok so what is AC2 (A and C playing where C goes first)? Well as long as C misses their first shot, the situation turns into AC1 because it goes back to being A's turn!

AC2 = (0.5)*(AC1) ; The 0.5 factor here is to denote C missing their first shot.

These 2 example should illustrate how to solve the 2-player game for all relevant cases where A can win. We don't need to solve them all now, we just need to know that we CAN solve them, so that if it's relevant later we can solve for them and plug them in.

The 3-player Game

Ok so how do you solve the 3-player game... Don't be frightened, while it looks long each individual step is simple!

Suppose A, C, and D are playing. We know that the probability of A's victory depends on who does first, so we know there will be 3 probabilities involved: ACD1, ACD2, ACD3.

ACD1 (A goes first): If A hits D → AC2 is the probability of A winning because only A and C are left!

If A misses D → ACD2 is the probability of A winning now it's C turn and D is still playing!

Thus...

ACD1 = 0.3*(AC2) + 0.7*(ACD2)

Note that on its own, this is not solvable like the 2-player case since we still don't know ACD2. However if we create and expression of ACD1, ACD2, ACD3 we will have 3 equations and 3 unknowns which will allow us to solve for all the unknowns!

Hopefully that illustrates how to solve for any 3-player case that comes up. You don't need to solve for anything right now, but just know that it is possible and follow the steps what whatever 3 players are left.

Putting it all together

We are almost there! Now let's just play the 4 player game and see what comes up given the scenarios you provided!

Scenario 1: A hits B → ACD2 is probability of A winning since B is gone and it is now C's turn

Scenario 2: A misses B, but then B is will not miss D → ABC3 indicates D is gone and it is now C's turn

Thus the total chance of A winning is

(0.3)*(ACD2) + (0.7)*(ABC3) ; where the above steps tell you how to solve the 3-player game!

I know some of the notation and variable names are kind of confusing but hopefully that made sense, please let me know what parts didn't make sense to you (or if it just worked out completely!).

Thanks