Puzzle time- Numbers on foreheads

[jon-nyc]

The idea is that the median is only slightly affected by your number, less so if your number is close to it.

Imagine if there were 1,000 people doing that. Clearly the differences in median that each would calculate would be a tiny rounding error, thus they would all converge on the same number, including at least one person who had that number.

[jon-nyc]

You can also improve over the random probability by doing something with modes like Horace suggested. Even with just 10.

I still haven't been bothered to work out probabilities. I'm hoping the hints reveal a more clever intuitive approach.

[Horace]

A lot would hinge on whether the median was a single-occupancy number or multi-occupancy. In the case of a single-occupancy median, that strategy is a guaranteed loss. It's a guaranteed win for multi-occupancy median when it's still the median after removing one of them. It begins to be similar to the mode idea from that perspective.

[Klaus]

I think I got a solution that works with probability 1. Good old pigeonhole principle.

click to show

The prisoners number themselves 0 to 9.

Let s the the sum of all numbers on all heads and let d be its rightmost digit.

Now consider prisoner number p. He computes the sum t of all numbers he sees on the other's heads.
He guesses that d = p. Hence he guesses (p - t) modulo 10 as his number.

This will yield a correct guess for prisoner d.

Proof. Consider prisoner d.

His guess will be (d-t) modulo 10, which is the same as s - t, which is the number on his forehead.

[jon-nyc]

That's clever.

[Klaus]

@jon-nyc said in Puzzle time- Numbers on foreheads:

The idea is that the median is only slightly affected by your number, less so if your number is close to it.

Imagine if there were 1,000 people doing that. Clearly the differences in median that each would calculate would be a tiny rounding error, thus they would all converge on the same number, including at least one person who had that number.

I don't think so.

Let's consider a uniform distribution of the numbers. Then the median of the 1000 would very likely be 4 or 5. But that number is not more likely to show up than any other number.

[jon-nyc]

You choose the visible number closest to the median.

[jon-nyc]

I think your solution isn't right. They each (potentially) have a different value of t.

[Klaus]

@jon-nyc said in Puzzle time- Numbers on foreheads:

I think your solution isn't right. They each (potentially) have a different value of t.

I made a small correction and elaboration on why this is correct.

[Axtremus]

@jon-nyc said in Puzzle time- Numbers on foreheads:

The prisoners have an opportunity to conspire beforehand; how can they maximize their probability of success?

If prisoners get to choose their own number, then in the “conspire” phase all prisoners agree to one number (e.g., every prisoner chooses to paint himself with the number 0), then every one knows that his own number is 0.

[taiwan_girl]

Not sure if I understand, but to me, there is no way to make a guess any more accurate. If it is a true random number on the forhead, then there is a 10% chance of guessing correctly.

It doesnt matter if every other number is "8", your number could still be anything from 0-9. etc

What am i missing?