7
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I made this puzzle up in 1999 when I was a student.

Fort Alpha, on your border territories, was attacked by The Enemy. You don't know if it fell, or repulsed The Enemy's vicious onslaught. A totally trusted advisor, with a detailed knowledge of the military situation, gave counsel before the attack that the chance of a successful defence was 50%. Unfortunately, there are no rapid communication systems; news of the war relies completely on messengers.

Three of your messengers were given a true report of the battle and set out from Fort Alpha immediately. They travelled separately to forts Beta, Gamma and Delta, which are still under your command. Before collapsing dramatically with exhaustion, these three messengers then relayed the message to three more messengers each, so a total of nine fresh messengers then set out to convey the message to your headquarters.

Now the 12 messengers were hand-picked from the ultra-loyal elite of your troops, so it is very very unlikely, though not impossible, that any given one of them is loyal to The Enemy. Your totally trusted advisor says that the chance of a given messenger being a traitor is 1%. Any messenger who was a traitor would reverse the message relayed to them, and pass on the opposite message, so as to befuddle you into inaction.

None of the messengers spoke to one another except to relay the message. The Enemy's spies, if there are any, maintain total secrecy, so none of the messengers know whether any of the others is a traitor and they each assume each other to be loyal to you. Only traitors would lie.

The three messengers who arrived from Beta, one who arrived from Gamma, and one who arrived from Delta, recount that Fort Alpha fell tragically to The Enemy. The other four claim that it bravely withstood the attack and decimated The Enemy's marauding rabble.

Which story do you believe, and why?

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    $\begingroup$ To clarify, the traitors also don't know who the other traitors are, and so if a traitor passed their message to another traitor, the end result would be truth, correct? $\endgroup$ – StephenTG Feb 5 '16 at 16:20
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    $\begingroup$ Yes, the effect of two traitors in sequence would cancel out. $\endgroup$ – oggotron Feb 5 '16 at 16:22
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    $\begingroup$ I would be having serious words with the "totally trusted" adviser. The chances of any one messenger being a traitor are clearly higher than 1%, unless they are posted with a non-uniform distribution, but if that were the case he ought to be able to adjust the probabilities accordingly. $\endgroup$ – jhabbott Feb 5 '16 at 16:27
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    $\begingroup$ For the purposes of this puzzle, you just have to trust totally the totally trusted advisor's advice. $\endgroup$ – oggotron Feb 5 '16 at 16:29
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    $\begingroup$ I guess if he's totally trusted, we must just be really unlucky! $\endgroup$ – oggotron Feb 5 '16 at 16:32
4
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The answer is

Good new! Alpha bravely withstood the enemy attack.

Because:

In order for the story that Alpha fell to be true, a minimum of four traitors would be required. In order for the story that Alpha survived to be true, you would need only three traitors. So given those two scenarios, Alpha surviving is 99 times more likely than Alpha having fallen.

To clarify:

For the Alpha to have survived, it would mean that the messenger who went from Alpha to Beta was a traitor, one from Gamma to headquarters was a traitor and one from Delta to headquarters was a traitor. That's three. For Alpha to have fallen, there would be no traitors in the Beta Group, but for both the Delta and Gamma Groups, either the first messenger and one of the second messengers, or two of the second messengers would have to be traitors. That's four total.

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  • $\begingroup$ Why 4? What if there are two traitors coming from alpha? Heck, or 3? $\endgroup$ – Raystafarian Feb 5 '16 at 16:21
  • $\begingroup$ @Raystafarian But then you'd need a traitor from Gamma and Delta $\endgroup$ – StephenTG Feb 5 '16 at 16:22
  • $\begingroup$ The exact ratio of probabilities is $(99^4+1)(99^2+99^2)^2$ to $(99^3+99)^3$, so the probability of the more likely one is just over 96%. $\endgroup$ – f'' Feb 5 '16 at 19:25

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