A strategy better than blind chance

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In summary: I'm not sure what you're suggesting. I was trying to suggest a strategy that would give 1/log(N) for the original problem. That is, a strategy where the probability of choosing a white hat with the kth pick is 1/k. I don't see how to get better than 1/(N+1) for the generalization.In summary, the conversation discusses a riddle involving a group of people wearing black or white hats. The probability of winning for n=2 is always 1/4, but a strategy is provided where the probability is 1/3. The conversation then moves on to discussing generalizations for larger groups, with a proposed strategy that could potentially yield a probability of
  • #1
asmani
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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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  • #2
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.
 
  • #4
Yeah, sorry I accidentally omitted the "l" in "html".
 
  • #5
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.
 
  • #6
Hey we have an infinite number of hats...
 
  • #7
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.
 
  • #8
asmani said:
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 [itex]i=1,2[/itex], let [itex]n_i[/itex] be the position of the first white hat on player [itex]i[/itex]'s head. Then have player 1 point at hat [itex]n_2[/itex] and player 2 point at hat [itex]n_1[/itex]. It is easy to see that both players win iff [itex]n_1 = n_2[/itex], which happens with probability [itex]\sum_{n=1}^{\infty}\left(\frac{1}{4}\right)^n = \frac{1}{3}[/itex].
 
  • #9
asmani said:
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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  • #10
Zafa Pi said:
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.
 
  • #11
haruspex said:
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.
My bad. Your right.
 
  • #12
haruspex said:
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.
 
  • #13
mfb said:
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.
 
  • #14
Zafa Pi said:
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.
 
  • #15
haruspex said:
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).
 
  • #16
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.
 
  • #17
The webpage has a link to the solution.

haruspex said:
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.
 
  • #18
mfb said:
The webpage has a link to the solution
I don't understand the description. Where does L come from? Have they agreed this in advance too?
 
  • #19
Yes, they agree on one L in advance. Ideally the value that is the most probable - but as there are just k options, there is certainly a value of L that gives at least 1/k probability (with the caveat discussed before).
 

1. What is "A strategy better than blind chance"?

"A strategy better than blind chance" refers to a method or plan that has been scientifically tested and proven to be more effective than random guessing or chance in achieving a desired outcome.

2. How is this strategy developed?

This strategy is developed through the scientific method, which involves formulating a hypothesis, designing and conducting experiments, collecting and analyzing data, and drawing conclusions based on the evidence.

3. Can this strategy be applied in different fields?

Yes, this strategy can be applied in various fields such as medicine, psychology, economics, and engineering. It can also be used in everyday life, such as decision-making and problem-solving.

4. What are the benefits of using a strategy better than blind chance?

The benefits of using a strategy better than blind chance include increased efficiency, accuracy, and reliability. This can lead to better outcomes and advancements in various fields.

5. Are there any limitations to this strategy?

Like any scientific approach, there may be limitations to this strategy depending on the specific application and context. It is important to carefully consider the evidence and potential biases when applying this strategy.

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