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I'm having a problem understanding a solution based on probabilities in the following puzzle:

Puzzle: There is a "triangular" duel between the three shooters. Everyone shoots one by one, can shoot once and any person he wants. Smith always hits the target (100%) and shoots last. Brown hits the target 80% of the time and shoots second. Jones is the worst, hitting the target only 50% of the time and he shoots first. Who should Jones shoot in order to increase his chance of surviving?


Answer: Jones should shoot in the air. The worst shooter has the best chance of surviving in the triangular duel, which is Jones. After him goes Smith who never misses the target. Since Smith and Brown will be shooting each other when their turn comes, Jones must shoot no one until one of his enemies dies.

After that he shoots at his enemy, again having an advantage. First, it's easier to calculate the probability of surviving for Smith. In the duel with Brown he shoots first with a probability of 1/2. In this case he kills Brown. Brown also shoots first with a probability of 1/2.

Smith survives with a probability of 1/5. Thus, Smith will live over Brown with a probability of 1/2 + 1/2 * 1/5 = 3/5. With a probability of 1/2, Smith survives the duel with Jones. All in all: the probability to survive for Smith is 3/5 * 1/2 = 3/10.

The probability for Brown to survive the duel with Smith is 2/5.

Then Jones shoots Brown. In the first round, Brown has a probability of 1/2 * 4/5 = 4/10 to win. In the second round, he has a probability of 1/2 * 1/5 * 1/2 * 4/5 = 4/100.

Thus, Brown has a chance to survive Jones: 4/10 + 4/100 + 4/1000 + 4/10000 = 0.4444(4) = 4/9.

The probability of Brown to survive both of his opponents is equal to 2/5 (over Smith) * 4/9 (over Jones) = 8/45. The probability of Jones to survive = 1 - 3/10 - 8/45 = 47/90.


In bold are the calculations I'm confused about and don't know how to arrive at. Can someone please explain it? Thank you.

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  • $\begingroup$ I think it may be easier to do a backwards induction route, starting at what would smith would do if he still had two enemies, then with this knowledge derive what would Brown would do if he had two enemies, and we already know if there is only one enemy left that person will just shoot that enemy. With this you have all knowledge to calculate Jones survival of round for each possible strategy, shooting brown (missing or hitting), shooting smith (missing or hitting) and shooting no one $\endgroup$ – Kamster Aug 25 '14 at 19:45
  • $\begingroup$ You should end having that the optimal 2 person strategies for Smith and Bob are to shoot the other. Thus Jones has a probability of 1 to live next round $\endgroup$ – Kamster Aug 25 '14 at 19:49
  • $\begingroup$ and not only that if there is probability of 1 that Jones will be facing only one person next round. Thus with this strategy, Jone will always be in the "final duel" $\endgroup$ – Kamster Aug 25 '14 at 20:01
  • $\begingroup$ The order of shooting is best stated as the player who is the worst shot goes first. With Smith vs Brown you had them at 50/50 for the first shot. Brown goes first. And the Jones vs Brown can continue indefinitely. Summing that infinite series does in fact give you 4/9. Making those 2 changes should give you the final win probabilities: Smith = 0.10, Brown= 16/45, Jones=49/90. This assumes the given strategy. $\endgroup$ – Mr.Spot Aug 25 '14 at 20:29
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I found the explanation of the puzzle by its original author. The answer is in fact for Jones to shoot into the air until one of the opponents is dead.

The order is kept as per the question, but working with probabilities we need to consider all possible outcomes. Thus, we consider the case where Brown shoots Jones first instead of logically (as each will try to take away the strongest person) shooting Smith.

Smith: We start calculating the probability of surviving for Smith first as it is the easiest one to do. There are two ways Smith can win Brown in their duel: 1) Smith shoots first and kills Brown (since he is 100% accurate), giving him a chance of surviving of 1/2; 2) Brown shoots first and misses, then Smith kills him; this case gives Smith a chance of surviving of 1/2*1/5 (since Brown is 4/5 accurate). As a result, Smith can survive Brown with a probability of: 1/2 (case 1) + 1/2*1/5 (case 2) = 3/5. Next Smith faces Jones. Since Jones is only 50% accurate, Smith has a survival chance of 1/2 against Jones. Smith's overall chance of surviving in this triangular duel is 3/5 (case against Brown) * 1/2 (case against Jones) = 3/10.

Brown: Next comes Brown. He has a chance of surviving against Smith of 2/5 (simply deducting Smith's survival probability of 3/5 from total probability of 1). Here it gets complicated as we run into infinite series of possibilities with Brown against Jones. There is a chance of 1/2 that Jones will miss (he is 50% accurate), and Brown has a chance of 4/5 to kill Jones. At this point, Brown has a chance of surviving of 1/2*4/5 = 4/10. But! He can also miss with a probability of 1/5 (which is 1 - his accuracy of 4/5). That can go on forever, and will result in Brown's total survival chance against Jones yielding 4/10 + 4/100 + 4/1,000 + 4/10,000 + ... This can be written as 0.444444... which is the decimal expansion of 4/9. Brown's overall chance of surviving in this triangular duel is 2/5 (case against Smith) * 4/9 (case against Jones) = 8/45.

Jones: This leaves Jones with a probability of surviving of 1 (total probability) - 3/10 (Smith's chance) - 8/45 (Brown's chance) = 47/90.

Therefore, Jones has 52.22% of surviving, Smith - 30%, Brown - 17.78%; and it's true that it's better for Jones to fire into the air in order to increase his chance of surviving.

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