Answer to Puzzle #51: Three Way Duel or Truel

51. Alice, Bob, and Cheri, three smart people, arrange a three-way duel. Alice is a poor shot, hitting her target only 1/3 of the time on average. Bob is better, hitting his target 2/3 of the time. Cheri is a sure shot hitting 100% of the time. They take turns shooting, first Alice, then Bob, the Cheri, then back to Alice, and so on until one is left. What is Alice's best course of action? And her chances of survival?

I've been wrestling with this for a while because I couldn't find a way to explain the Maths:

Before reading the answer can I interest you in a clue?

Lets start with the answer. You may think the options available to Alice are to either aim at Bob or Cheri. Perhaps given that Cheri is the better shot Alice should aim at her. Alice is in fact better off deliberately missing until one of the other players dies. The other players of course, will target each other until one of them dies as the rational thing to do is to shoot the greatest threat, which for Cheri is Bob and Bob is Cheri.

Before we work the maths I have written what I believe to be the only three way duel or truel simulator. Have a play. If you want to know more about it click here.

 Probability of A hitting
 Probability of B hitting
 Probability of C hitting
 Time to Run milliseconds
 A's strategy is to...
Aim at B until they die
Aim at C until they die
Miss till someone dies
  

So we can see with reasonable confidence, if the simulator is accurate that Alice's best strategy is to miss both Bob and Cheri until one of them has dispatched the other, following this plan she will survive something like 39.7% of the time.

Essentially Alice, whatever she does, is going to find herself in a two person duel with Bob or Cheri. Even if she were to aim at someone and kill them, say she aims at Bob and hits, she is now in a two person duel with Cheri and it's Cheri's shot. Alice is dead. If however, she aims at nothing, after Bob or Cheri kill one or the other it is Alice's shot. And it is the first shot advantage that makes missing worth while. Lets move on to calculating this...

Solution to a Two Person Duel

Sometimes the probability of who survives a duel is easy to calculate say Bob and Cheri with Bob to shoot first. 2/3 chance Bob hits and wins, if not Cheri definitely hits and wins. This calculation is easier because Cheri is a sure shot and the game cannot go for another round. Otherwise though the generalised for looks like this:

Let p(X,Y) be the probability of X surviving a duel with Y where X has first shot-
p(X,Y) = Pr(X hits Y) + Pr(X misses Y) * Pr(Y misses X) * p(X,Y)
Which is essentially the probability of X killing Y in the first shot plus the probability of nobody being killed and the whole thing happening again.

p(X,Y) where Y has the first shot could reasonably be surmised to be 1 - p(Y,X) since only one person can survive a two person duel. or explicitly
p(X,Y) = Pr(Y misses X) * (Pr(X hits Y) + Pr(X misses Y) * p(X,Y))

For example p(Alice,Bob) the probability that Alice wins a two person duel against Bob where Alice has the first shot:
p(Alice,Bob) = Pr(Alice hits Bob) + Pr(Alice misses Bob) * Pr(Bob misses Alice) * p(Alice,Bob)
p(Alice,Bob) = 1/3 + 2/3 * 1/3 * p(Alice,Bob)
p(Alice,Bob) = 1/3 + 2/9 * p(Alice,Bob)
7/9 * p(Alice,Bob) = 1/3 
p(Alice,Bob) = 3/7 
All of the other calculations are similar. For convenience we will express the probabilities as fractions of 21, the lowest common denominator. If it's not clear the underline denotes first shot.
p(Alice,Bob)   = 3/7 = 9/21
p(Alice,Cheri) = 1/3 = 7/21 
p(Alice,Bob)   = 1/7 = 3/21
p(Alice,Cheri) = 0   = 0/21 

Remember Alice is going to be in a two person duel with one or the other of Bob and Cheri she has some control over who she fights, with the advantage being in fighting Bob rather than Cheri. But she has ultimate control over whether she has first shot. By not shooting anyone she can ensure that she has the first shot, and from the table it is clear that the first shot advantage is greater than the advantage of facing Bob over Cheri.

Back to the Truel

Now we have decided that Alice is going to miss we can calculate the odds of her survival. Bob and Cheri will be involved in a two person duel, survivor to face Alice with Alice to shoot first. We don't need to use the formula to work out the odds of Bob or Cheri surviving, it's trivial. There is a 2/3 chance of Bob killing Cheri, if he doesn't he dies with the return shot.
p(Alice survive) = 2/3 * p(Alice,Bob) + 1/3 * p(Alice,Cheri)
                 = 2/3 * 3/7 + 1/3 * 1/3
		 = 6/21 + 1/9
		 = 18/63 + 7/63
		 = 25/63 ≅ .3968

Remaining calculations

The remaining probabilities are worth a quick look. Easiest is Cheri. She has a 1/3 chance of surviving the duel with Bob, the a 2/3 chance of surviving her duel with Alice. Which is 2/9 or 14/63 ≅ 0.2222. Which leaves Bob with 24/63 ≅ 0.3810

Nice puzzle this once you break it down.

Assumptions

Assumptions here are things that we assume are omitted from the question for brevity or neatness. Such as there are no stray bullets. All hits are fatal etc. The phrase 'three smart people' or some versions say 'three logicians', is meant to indicate that the players will act rationally. That is to say Bob and Cheri will shoot at the person who poses the greatest threat. Again we assume that they know the relative skills of each player. Some versions will say something like 'Alice who is known to be a poor shot...'





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