Incoming enemy! Scout, dispatch! [duplicate]

You are the leader in a group of 9 scouts.

There are rumors of an incoming enemy army and your unit was sent to investigate where they are coming from.

You are aware that there are only 4 paths to your base.
You must divide your members into groups to investigate which one it will be.
Two members of your unit(excluding you) are unpredictable jokers who cannot be trusted.
You don't have much time left so you must send everyone(including you) immediately.
Everyone must report back and then you must be able to tell for sure where the enemy will come from.

How will you proceed?

• unpredictable jokers who cannot be trusted So they don't always lie? – max8126 Mar 9 '17 at 22:02
• @max8126 that is correct. – stack reader Mar 9 '17 at 23:05
• I'm sure that I've seen this puzzle before but framed differently. Then I think it was something like archaeologists in an ancient temple and two of them got possessed, the temple was collapsing and you needed to find a way out quickly... Having trouble finding it with a search though. :( – Chris Mar 10 '17 at 11:28
• "You are the leader in a group of 9 scouts." so 10 total, or just 9? Also will the enemy only attack from one path? – schil227 Mar 10 '17 at 15:50
• @schil227 in a group a of 9. So yes 9 total. And only 1 path. – stack reader Mar 10 '17 at 16:03

Divide the scouts in this way:

You take one path because you trust yourself.
Split the other 8 into 3, 3, 2.

(Credit to the earlier answers for getting me that far)

Now the possibilities.

1. You found the enemy. This is trivial.
2. You did not find the enemy, and two of the other groups are reporting mixed results. Therefore, the jokers are split - trust the majority of each of the 3-groups, and use process of elimination to decide what the 2-group saw.
3. You did not find the enemy, and only one group is reporting mixed results. Trust the other groups, and use process of elimination to decide what the mixed group saw.
4. You did not find the enemy, and each of the other groups is in complete agreement. The only group that might not be trustworthy is the 2-group, so use what you and the 3-groups saw to decide what the 2-group saw.

• It looks like this answer restates content that is already thoroughly expressed in another answer. In the future, please be sure to read through other answers on the site before adding your own. Thanks! – Deusovi Mar 9 '17 at 22:26
• @Deusovi I made this answer because Marius' answer has an error, and yours... I haven't figured out whether yours has an error or not. It's really hard to understand. – Brilliand Mar 9 '17 at 22:34
• I found this answer to express the reasoning more clearly than the others. – supercat Mar 10 '17 at 0:00
• Good job, this answer is clear and to the point and also take into account that the unpredictable jokers can tell the truth too. – stack reader Mar 10 '17 at 1:07

I divide us like this:

Me on one path, 3 on the second, 3 on the third, and 2 on the fourth.

Here's why it works:

Assume that I don't see any enemy. (If I do, the answer is obvious.)

If there's a disagreement on path 4, you can trust majority opinion on the other two and use it to determine which is the right path. (If both are safe, then it's path 4.)

If there is no disagreement on path 4, then there are either disagreements on both remaining paths, on one of them, or neither.
- If both, then you can safely trust majority opinion, since the tricksters are divided.
- If one, everyone on one of paths 2 or 3 will say "safe". In that case, someone on the other path will say "enemies coming", so the enemy is on that path. Otherwise, there are people claiming "enemies coming" for both paths 2 and 3 - in that case, one of them will have all three people agreeing that enemies are coming, so you pick that path.
- If none, then the two tricksters must be on path 4 telling us it's safe, so it's path 4.

• I may be wrong (it took me a long time to process your answer), but doesn't your answer fail to address the case where [you don't see the enemy], [group 2 sees the enemy], [group 3 votes mixed] and [group 4 votes safe]? The solution is obvious in that case, but I'm pretty sure you failed to mention it. (You also missed the case where everyone is completely truthful, and group 4 saw the enemy.) – Brilliand Mar 9 '17 at 22:41
• @Brilliand: Whoops, you're right! I'll fix that. – Deusovi Mar 9 '17 at 23:18
• The "If one ..." line is rather confusing. You could simplify it by saying you can trust the path on which everyone agrees, whether they're saying it's safe or dangerous. – Rob Watts Mar 9 '17 at 23:45
• It's still not quite right. If there is no disagreement on path 4 and none disagreements on both remaining paths, two tricksters does not have to be on path 4. They can be anywhere, just agreeing with the rest. It's not a problem obviously, since then we know where the enemy is naturally, but the wording, is definitely wrong. – Andrew Savinykh Mar 10 '17 at 0:50
• Sorry for not picking yours or @Marius answer. You got the general idea, but the explanation is slightly confusing and doesn't seem to take into account that the jokers can also tell the truth. I am sorry if unpredictable jokers who cannot be trusted was not clear enough on that point. – stack reader Mar 10 '17 at 1:09

You take one path because you trust yourself.
Split the other 8 into 3, 3, 2.

Now the possibilities.

You took the path the enemy took, it's trivial.
You did not take the enemy path:
There are 4 options here:
1. Both groups of 3 agree with each other. This means the liars are in the group of 2. And it easy to see who to trust.
2. One group of 3 reports both yes/no results and the group of 2 reports a yes/no result. This means there is 1 liar in each of these groups. For the group of 3 believe the majority report (2 should say the same thing). Then you know if the enemy is coming or not from 3 paths (yours and the groups of 3.) You can determine if the enemy is coming from the group of 2.
3. One group of 3 reports reports yes/no results and the group of 2 reports the same result. This means that both liars are in the group of 3 that reported contradictory results. Take the least voted answer from this group.
4. Both 3 groups report yes/no results. This means there is a liar in each one of these. So take the most voted report from each group to get the truth.

• In case 3, you can't necessarily trust the minority opinion, because one liar might randomly decide to tell the truth. In that case, ignore the group with the liars and trust the other three groups (that's enough information). – Brilliand Mar 9 '17 at 19:44
• I guess you are right. I was working under the assumption that the liars just lie. – Marius Mar 9 '17 at 19:56
• While the conclusion is the same, the reasoning for 1 is wrong as well. A correct reasoning for 1 would be: if three people make a common statement, at least one is trustworthy, thus the statement is true. – Édouard Mar 10 '17 at 0:06

I would divide it so that both liars are with me, and every other path is 2 2 2

The possibilities of this:

1. You see the enemy, and if the two liars are saying a different answer than you, your vote wins since you're the leader.
2. If one of the other groups sees the enemy, we can trust them because they don't have any liars in the group.
• You don't know who the liars are. – Deusovi Mar 10 '17 at 15:50