A strategy better than blind chance

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Discussion Overview

The discussion revolves around a riddle involving probability and strategies for guessing the color of hats worn by participants. The problem is framed in terms of independent probabilities for each hat being black or white, and participants explore various strategies to improve the odds of winning beyond a baseline probability of 1/4 for two players.

Discussion Character

  • Exploratory
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that with independent probabilities of 50-50 for each hat, the probability of winning for two players remains 1/4 regardless of strategy.
  • Others propose strategies, such as assuming the majority color or forming a circle to guess the opposite color of the adjacent player, suggesting these could yield better odds.
  • A participant describes a strategy involving pointing at the position of the first white hat on the other player's head, claiming it results in a winning probability of 1/3.
  • There is a discussion about the possible outcomes of hat colors, with some participants arguing about the validity of excluding certain combinations like (b,b).
  • One participant mentions that the generalization for N participants leads to a winning probability of 1/(N+1), while another introduces the intriguing concept of a 1/log(N) probability, suggesting it may require a more complex strategy.
  • There is speculation about the mathematical details of achieving a better strategy, including the concept of the kth height at which all visible hats are white.

Areas of Agreement / Disagreement

Participants express differing views on the effectiveness of various strategies, and no consensus is reached regarding the best approach or the exact probabilities involved. The discussion remains unresolved with multiple competing ideas presented.

Contextual Notes

Some participants note the complexity of the riddle and the need for clearer definitions or assumptions, particularly regarding the conditions under which certain probabilities apply. There are also references to specific mathematical formulations that are not fully explored or agreed upon.

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

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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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".
 
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...
 
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.
 
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 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}.
 
asmani said:
This is an interesting riddle from here: http://www.brand.site.co.il/riddles/201607q.htm

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).
 

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