A strategy better than blind chance

• A
This is an interesting riddle from here: http://www.brand.site.co.il/riddles/201607q.htm [Broken]

I'm having difficulty understanding the problem. If each hat is black/white with 50-50 probability, independent of the colors of other hats, then the probability of winning for n=2 is always 1/4, no matter what strategy they plan. Can you please provide an example strategy, in which the probability is not 1/4?

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RUber
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I am having difficulty accessing your link. Could you copy some of the pertinent text?
Usually these sorts of problems involve grouping to increase your odds, but I would have to see the riddle to know for sure.

Yeah, sorry I accidentally omitted the "l" in "html".

RUber
Homework Helper
Got it, thank you.

One possible strategy I can think of is for everyone to assume they are in the majority.
If the true proportion is 60:40 in favor of white, then most people would see that white is more common. If everyone says white, then 60% of the people are correct.
Unfortunately, this strategy also guarantees failure, unless everyone is wearing the same color hat, since success is defined as everyone guessing correctly.

So, what if everyone were to form a circle? Then you could look left and just guess that your hat is the opposite color compared to the person on your left.
For n=2, you have the following cases:
white, white = fail
white, black = success
black, black = fail
black, white = success.
This would at least give you 50/50 odds of winning for n=2.

As far as working out the large number options, I have not gone that far yet.

Hey we have an infinite number of hats...

RUber
Homework Helper
Sorry, I was looking at the guess your color variation with one hat.
You are talking about the "pick a white hat" option?
That seems to be about the same in theory, but a lot more complicated to draw out the specific cases.

Can you please provide an example strategy, in which the probability is not 1/4?

Sure. Both players will almost surely be wearing at least one white hat. For $i=1,2$, let $n_i$ be the position of the first white hat on player $i$'s head. Then have player 1 point at hat $n_2$ and player 2 point at hat $n_1$. It is easy to see that both players win iff $n_1 = n_2$, which happens with probability $\sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^n = \frac{1}{3}$.

This is an interesting riddle from here: http://www.brand.site.co.il/riddles/201607q.htm [Broken]

I'm having difficulty understanding the problem. If each hat is black/white with 50-50 probability, independent of the colors of other hats, then the probability of winning for n=2 is always 1/4, no matter what strategy they plan. Can you please provide an example strategy, in which the probability is not 1/4?
An easier statement is to adopt post #8 strategy and notice that there 3 equally likely possible outcomes (b,w), (w,b), and (w,w). (b,b) is excluded. Thus 1/3.

I don't think your post should be labeled A, but rather B

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haruspex
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there 3 equally likely possible outcomes (b,w), (w,b), and (w,w). (b,b) is excluded
That's not quite right. BB is possible, e.g. the stacks are BWB... and BBW...
With this ingenious algorithm, the probabilities are equally likely WW, BB, BW. If it's not WW then one of them has the earlier first white. That player will equally likely pick a B or W from his own stack, but the other player is guaranteed to pick a B from her own.

That's not quite right. BB is possible, e.g. the stacks are BWB... and BBW...
With this ingenious algorithm, the probabilities are equally likely WW, BB, BW. If it's not WW then one of them has the earlier first white. That player will equally likely pick a B or W from his own stack, but the other player is guaranteed to pick a B from her own.

mfb
Mentor
That player will equally likely pick a B or W from his own stack, but the other player is guaranteed to pick a B from her own.
And that is a key point. Each player still has 1/2 probability to pick a white hat, but the choices are now correlated.

With 1/3 probability they both pick a white hat, with 1/3 probability they get black+white (or white+black), with 1/3 probability they both pick a black hat.

The generalization to N participants gives 1/(N+1) probability to win. The 1/log(N) in the puzzle is interesting - it looks specific enough to suggest a solution, but it needs some better strategy.

Edit: Indeed. And there is one.

The generalization to N participants gives 1/(N+1) probability to win. The 1/log(N) in the puzzle is interesting - it looks specific enough to suggest a solution, but it needs some better strategy.

Edit: Indeed. And there is one.
OK, I get the 1/(N+1). Are you going to share the 1/log(N)? I find it amazing.

haruspex
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OK, I get the 1/(N+1). Are you going to share the 1/log(N)? I find it amazing.
My guess is that it involves picking the kth height at which all the visible hats are white, k tending to infinity. But I've not attempted the algebra yet.

My guess is that it involves picking the kth height at which all the visible hats are white, k tending to infinity. But I've not attempted the algebra yet.
Is the kth height the same as the minimum height where you see all white? If so then you get probability 1/(N+1).

haruspex
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No, it's the kth height at which you see all white.
I.e., of all the heights at which you see all white, the kth instance.

mfb
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The webpage has a link to the solution.

My guess is that it involves picking the kth height at which all the visible hats are white, k tending to infinity. But I've not attempted the algebra yet.
That would not give a result better than 1/(N+1) I think.

haruspex