Andrew, I didn’t saw your comment while preparing my previous one. I’m happy you are full of work. I’m curious, when you got stranded at Denmark, the sky was its usual colour self, or did he had any odd colour?

Keep getting the good puzzles, while you can and when you can. They are awesome, and much much appreciated. But in the last one, you should have done fun on me instead of presenting the solution. In the end, I felt I was robbed of a good puzzle by my own stupidity… :-)

Andrew MW, part C is not that hard once you have the right strategy (which I hadn’t, and due to lack of time and motivation, never had the opportunity to rethink it again). You should have teased me… I would have tried to explain how I got there and you might have point my mistake to me.

Charlie strategy to win is to be the first to shoot in a “mano a mano”. He can always get there because the other players cannot avoid to shoot the others first if they want to have better chances of survival. Once he gets there, there is only two cases to consider:

Charlie x Andy: Charlie shoots first, he has 50% chances of survive and win (Andy has the rest because if he survives, he’ll kill Charlie in the spot).

Charlie x Bob: This is more difficult. If b and c are the accuracies of respectively Bob and Charlie, the probabilities of Charlie winning are the sum of the following terms:

c (Charlie win)
(1-c)(1-b)c (Charlie misses, Bob misses, Charlie win)
(1-c)(1-b)(1-c)(1-b)c
…
[(1-c)(1-b)]^n c
…
…and so on. In other words, the probability of Charlie win is equal to the sum of all the powers of r^n=[(1-c)(1-b)]^n times c. The first sum, (1+r+r^2+r^3+…) is the known and very useful geometric series and for r<1, it gives 1/(1-r). Since b=0.8 and c=0.5, then r=0.1, 1/(1-r)=1/0.9=10/9, and the probability of Charlie win against Bob became c/(1-r)=5/9 (more than half, so). The chances of Bob winning in that situation is of course, 1-c/(1-r) = 4/9.

Now we need to assert the probabilities to reach each of these situations. Because Charlie refuses to play the game until became alone with only one other player, it all resumes to discover the probabilities of each other player getting to that situation. Chances are, Bob will be first than Andy half the times, in which case, he'll have a probability b of shooting Andy. He will never have another chance so his chance of duelling with Charlie is b/2 or 2/5. His total chances are then b/2 * [1-c/(1-r)] = 2/5 * 4/9 = 8/45. Chances for Andy to reach the final duel is (1-b/2) (if Bob doesn't get there, Andy will), and so, is final probability of survival will be (1-b/2) * (1-c) = 3/5 * 1/2 = 3/10. And Charlie, of course will have the rest of the chances: 1- b/2 * [1-c/(1-r)] – (1-b/2) * (1-c) = 1 – 8/45 – 3/10 = 47/90.

As you see, part C is not hard, because Charlie strategy divides the game in two games of two players where he is the first one to play the second game. I would have loved to have thought that. Instead I had to calculate some twenty terms in excell, in the very brink of doing them wrongly.

Part D is a lot harder since it we need to solve two third degree equations for the down limit, if I'm not mistaken… too much work for the pay, so I went numerical :-)

Many apologies, chaps, I’ve been rather stressed the last few weeks with work hence the odd timings for the puzzles and solutions being posted. For the foreseeable future these puzzles won’t necessarily appear on Wednesday with solutions on Sunday; I will still try and get one out a week but you’ll have to keep popping back or subscribe to my feed. I really appreciate the comments, especially Joao, which I always read but don’t necessarily have time to respond to, but I will try and get better at that – just higher priorities right now.

Glad you all like the puzzles.

Hi,
I was also confused by Part A, but would have got it (I think) if you’d said “Charlie gets to go first” – although that seems to be the opposite of the original wording (“…when his turn comes around”)
Part B I got, intuitively, i.e. both B and A will aim at each other (until one dies) because he is a greater risk than C.
Part C was too hard for me.
I also figured Part D intuitively, but with no numbers!

I’ll not contest that shooting in the air is the best solution if indeed the final probabilities are the ones you said. I simply didn’t thought of other options beside shooting at someone. However, there’s something plainly wrong in the description of your solution. If by chance, Charlie is the last one to shoot, then surely you can not say to me that his chances are best if he shoots in the air…

(Being last, that means either Andy had the opportunity to shoot and shot already Bob, or that Bob had taken down Andy… in each case, Charlie will face one player alone in its turn… surely he is not best served shooting in the air)

I’m not saying the idea is wrong. Thinking carefully, if Charlie allows to be shoot before he shoots at, his probability of surviving is less, since any other is more accurate than him. If he is the first one to shoot, he’ll have at least 50% chances of surviving. Any other additional luck is a plus (and knowing that, is easy to disprove my strategy since I predicted less than 50% chances for Charlie). So, he has to warrant he’ll be the first one to shoot when the first looser is taken down, which will happen if he certifies it is the other surviving player that will do the killing… losing its turn to Charlie to shoot at him first. I should have though this more carefully. The math would have been even simpler… I know that because I redone the calculations easily with the new strategy, and indeed, obtained the probabilities presented by you.

Anyway, very nice puzzle. But you should have said something, Andrew… like “you are wrong!”. Others could have jumped in the opportunity :-)