"Spot on the Forehead" Sequel Contest

After losing the "Spot on the Forehead" contest, the two defeated Puzzle Masters complained that the winner had made a slight pause before raising his hand, thus derailing their deductive reasoning train of thought. 

And so the Grand Master vowed to set up a truly fair test to reveal the best logician among them.


He showed the three men 5 hats - two white and three black. 

Then he turned off the lights in the room and put a hat on each Puzzle Master's head. After that the old sage hid the remaining two hats, but before he could turn the lights on, one of the Masters, as chance would have it, the winner of the previous contest, announced the color of his hat. 

And he was right once again.

What color was his hat? What could have been his reasoning? 

The winner is wisest for a reason! 

Master of Logic For a Reason!

How the master was challenged?

Let's assume once again A, B and C are those logicians and C has guessed the color of own hat correctly. Here is what he must have thought - 

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"I'm assuming the grand master is conducting this test fairly denying any sort of advantage to any participant.

With that assumption, the grand master can't put 2 white and 1 black hat on heads. In that case, the person having black hat and watching 2 white hats on others' head would know the color of own hat immediately.

For fair play, he can't put 2 blacks and 1 white hat either. That will give unfair advantage to the logicians wearing black hats. Suppose A and B are wearing black and I'm wearing white hat. Now, what A (or B) would be thinking - 

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       " I'm A (or B) and I can see 1 black and 1 white hat (on head of C). If I have white 
         hat on my head then B (or A) would know color of his hat as black as there are 
         only 2 white hats available and those would be on my head and C's head.

         Moreover, 1 black and 2 white hats already eliminated as it's unfair distribution.

         That means I must be wearing black hat."

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That's how the combination of 2 black and 1 white hats also eliminated from fair play.

Hence, all of three must be wearing black hats is only fair distribution giving all of us equal chance of winning and hence I must be wearing black hat only. 

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Note : Here, C is assumed as a winner for only sake of convenience, otherwise either A or B whoever is wisest can be winner.

The Numbered Hats Test!

One teacher decided to test three of his students, Frank, Gary and Henry. The teacher took three hats, wrote on each hat an integer number greater than 0, and put the hats on the heads of the students. Each student could see the numbers written on the hats of the other two students but not the number written on his own hat.

The teacher said that one of the numbers is sum of the other two and started asking the students:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— No, I don’t.

Then the teacher started another round of questioning:

— Frank, do you know the number on your hat?

— No, I don’t.

— Gary, do you know the number on your hat?

— No, I don’t.

— Henry, do you know the number on your hat?

— Yes, it is 144.

What were the numbers which the teacher wrote on the hats?

Here are the other numbers!

Source 

Cracking Down The Numbered Hats Test

What was the test?

Even before the teacher starts asking, the student must have realized 2 facts.

1. In order to identify numbers in this case, the numbers on the hats has to be in proportion i.e. multiples of other(s). Like if one has x then other must have 2x,3x etc.

2. Two hats can't have the same number say x as in that case third student can easily guess the own number as 2x since x-x = 0 is not allowed.

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Now, if numbers on the hats were distributed as x, 2x, 3x then the student wearing hat of number 3x would have quickly responded with correct guess. That's because he can see 2 number as x and 2x on others hats and he can conclude his number as x + 2x = 3x since 
2x - x = x is invalid combination (x, x, 2x) where 2 numbers are equal.

Other way, he can think that the student with hat 2x would have guessed own number correctly if I had x on my own hat. Hence, he may conclude that the number on his hat must be 3x.

But in the case, all responded negatively in the first round of questioning. So x, 2x, 3x combination is eliminated after first round.

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That means it could be x, 3x, 4x combination of numbers on the hats.

In second round of questioning, Henry guessed his number correctly.

If he had seen 3x and 4x on other 2 hats then he wouldn't have been sure with his number whether it is x or 7x.

Similarly, he must not have seen x and 4x as in that case as well he couldn't have concluded whether his number is either 5x or 3x.

But when he sees x and 3x on other hats he can tell that his number must be 4x as 2x (x,2x,3x combination) is eliminated in previous round!

So Henry can conclude that his number must be 4x.

Since, he said his number is 144,

4x = 144

x = 36

3x = 108.

Hence, the numbers are 36, 108, 144.

Tricky Probability Puzzle of 4 Balls

I place four balls in a hat: a blue one, a white one, and two red ones. Now I draw two balls, look at them, and announce that at least one of them is red. What is the chance that the other is red?

Well, it's not 1/3!

Tricky Probability Puzzle of 4 Balls : Solution

What was the puzzle?

It's not 1/3. It would have been 1/3 if I had taken first ball out, announced it as red and then taken second ball out. But I have taken pair of ball out. So, there are 6 possible combinations.

red 1 - red 2
red 1 - white
red 2 - white
red 1 - blue
red 2 - blue
white - blue 

Out of those 6, last is invalid as I already announced the first ball is red. That leaves only 5 valid combinations.

And out of 5 possible combinations only first has desired outcome i.e. both are red balls.
Hence, there is 1/5 the chance that the other is red 

What is Color of His Hat?

There is a basket full of hats. 3 of them are white and 2 of them are black. There are 3 men Tom, Tim, and Jim. They each take a hat out of the basket and put it on their heads without seeing the hat they selected or the hats the other men selected.

The men arrange themselves so Tom can see Tim and Jim’s hats, Tim can see Jim’s hat, and Jim can’t see anyone’s hat.

Tom is asked what color his hat is and he says he doesn’t know.

Tim is asked the same question, and he also doesn’t know.

Finally, Jim is asked the question, and he does know.

What color is his hat?

Know color of his hat! 

'THIS' is The Color of His Hat!

What was the challenge?

Since there are only 2 black hats if Tom had saw 2 black hats (on heads of Tim & Jim) then he would have realized that he must be wearing white hat. Since, he says he doesn't know color of his hat that mean he can see either 2 white  hats or 1 black & 1 white hat on heads of other 2.

Now when Tim is saying he doesn't know color of his hat that means Jim must not be wearing black hat. If Jim had black hat then Tim would know that his color of hat is white but can't be black again as in that case Tom would have identified color of his hat in first attempt.

Hence, Jim must be wearing white hat! 

Three Hat Colors Puzzle

A team of three people decide on a strategy for playing the following game.  

Each player walks into a room.  On the way in, a fair coin is tossed for each player, deciding that player’s hat color, either red or blue.  Each player can see the hat colors of the other two players, but cannot see her own hat color.  

After inspecting each other’s hat colors, each player decides on a response, one of: “I have a red hat”, “I had a blue hat”, or “I pass”.  The responses are recorded, but the responses are not shared until every player has recorded her response.  

The team wins if at least one player responds with a color and every color response correctly describes the hat color of the player making the response.  In other words, the team loses if either everyone responds with “I pass” or someone responds with a color that is different from her hat color.

What strategy should one use to maximize the team’s expected chance of winning?

These could be the strategies to maximize the chances of winning!

Three Colors Hats Puzzle - Solution

What's the puzzle? 

There can be two strategies to maximize the chances of winning in the game.

STRATEGY - 1 :   

There are 8 different possible combinations of three color hats on the heads of 3 people. If we assume red is represented by 0 & blue by 1 then those 8 combinations are - 

Here only 2 combinations are there where all are wearing either red or blue hats. That is 2/8 = 25% combinations where all are wearing hat of same color and 6/8 = 75% combinations where either 1 is wearing the different colored hat than the other 2.  In short, at least 2 will be wearing either red or blue in 75% of combinations.

Now for any possible combination, there will be 2 hats of the same color (either blue or red). The one who sees the same color of hats on heads of other two should tell the opposite color as there are 75% such combinations. That will certainly increase the chances of winning to 75%.

STRATEGY 2 :    

Interestingly, here 3 responses from each member of team are possible viz RED (R), BLUE (B) and PASS (P). And every member can see 3 possible combinations of hats on the heads of other 2 which are as 2 RED (2R),  2 BLUE (2B) and 1RED:1BLUE (RB). See below.

Let's think as instructor of this team. We need to cover up all the possible 8 combinations in form of responses in the above table. 

For every possible combination, at least 1 response need to be correct to ensure win. But out of 9 above, 3 responses of 'PASS' are eliminated as they won't be counted as correct responses. So we are left with only 6. Let's see how we can do it.

First of let's take case of 2R. There are 2 responses where A sees 2 RED hats (000,100). We can't make sure A's response correct in both cases. So let A's response for this case be R. So whenever this 000 combination will appear A's response will secure win.

After covering up 000, let's cover up 001. For that, C's response should be B whenever she sees 2 red hats on other 2. And only left response P would be assigned to B.

So far,we have covered up these 2 combination via above responses.  

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Now, let's take a case of 2B. A can see 2B hats whenever there 011 or 111 appears. Since A's R response is already used previously, let B be her response in the case. So the combination 111 will be covered up with A's response.  

B can see 2B hats in case of 101 or 111. Since 111 is already covered above, to cover up 101, B should say R whenever she sees 2 BLUE hats on the heads of other 2. With this only response left for C in case of 2B is P.

 
With these responses, we have covers of so far,

 
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After filling remaining 1 possible response in response table for every team member in case of 1 RED and 1 BLUE hat, 

 B's response as BLUE in this case will ensure win whenever 011 or 110 combination will appear. Similarly, C's response as a RED will secure win whenever 010 or 100 appears as a combination.

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In this way, there will be at least 1 response correct for every possible 8 combinations. This strategy will give us 100% chances of winning this game!

The above table shows who is going to respond correctly for the given combination ( the block of combination & correct response are painted with the same background color).
 
SIMPLE LOGIC : 

The same strategy can be summarized with very simple logic. 

There must be someone to say RED whenever she sees 2 RED hats; someone should say BLUE and remaining one should say PASS. Similarly, one has to say BLUE; other should say RED & third one should say PASS whenever 2 BLUE hats are seen. Same logic to be followed in case of 1 RED and  1 BLUE hats seen. But while doing this, we need to make sure responses are well distributed & not repeated by single member of team (See table below).