"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.

Spot On The Forehead!

Three Masters of Logic wanted to find out who was the wisest among them. So they turned to their Grand Master, asking to resolve their dispute.

"Easy," the old sage said. "I will blindfold you and paint either red, or blue dot on each man's forehead. When I take your blindfolds off, if you see at least one red dot, raise your hand. The one, who guesses the color of the dot on his forehead first, wins."


And so it was said, and so it was done. 

The Grand Master blindfolded the three contestants and painted red dots on every one. 

When he took their blindfolds off, all three men raised their hands as the rules required, and sat in silence pondering.

Finally, one of them said: "I have a red dot on my forehead."

How did he guess? 

And this is how his logical brain responded! 


Similar kind of puzzles are - 

The Greek Philosophers  

Real Test Of Genius  

Logical Response By Master of Logic!

What was the test?

Let A, B and C be the names of three logicians and C be the logician who correctly guessed the color of dot on forehead. 

Now, this could be the C's logic behind his correct guess - 

"If I had blue dot on my forehead then A and B must had raised hands after looking red dots on the foreheads of other. In case, what A (or B) would have thought? His logic would be -

    "If C is with the blue dot then B (or A) must have raised hand after noticing 
    red dot on my forehead, hence I must have red dot."

So A (or B) would have successfully guessed color of dot on own forehead easily.

But neither A or B not responding that means I must have red dot on my forehead!"  

The logician who guess it correctly could be either A or B not necessarily be C; here it is assumed C is wisest for the sake of convenience.

Different Kind of Dice Game!

Timothy and Urban are playing a game with two six-sided dices. The dice are unusual: Rather than bearing a number, each face is painted either red  or blue.

The two take turns throwing the dice. Timothy wins if the two top faces are the same color, and Urban wins if they’re different. Their chances of winning are equal.

The first die has 5 red faces and 1 blue face. What are the colors on the second die?

The second die must have 'these' colors!

Die Needed For Different Kind of Dice Game!

What was the different in dice game?

Throwing two six-sided dice produces 36 possible outcomes. Since both Timothy and Urban have equal chances of winning, there are 18 outcomes where top faces are of same color.

Let's assume there are 'x' red faces and (6-x) blue faces on the other dice.

Remember, the first die has 5 red faces and 1 blue face. Then there are 5x ways by which top faces are red & 1(6-x) ways by which they both are blue. But as deduced above, there are 18 such outcomes in total where faces of dice matches color.

Therefore,

5x + 1(6-x) = 18

4x = 12

x = 3

Hence, other die have 3 red colored & 3 blue colored faces.

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 

Cut The Blue Cube Puzzle

A solid, four-inch cube of wood is coated with blue paint on all six sides.

Then the cube is cut into smaller one-inch cubes. These new one-inch cubes will have either three blue sides, two blue sides, one blue side, or no blue sides. How many of each will there be?

Here is solution of the puzzle! 

Cut The Blue Cube Puzzle : Solution

What is the puzzle?

Apart from the 8 cubes at the center all 4 x 4 x 4 - 8 = 56 will have some paint on ones side at least. See below the 1/4 th cube is taken out.

The cubes at the 8 corners will have blue paint on three sides.

The cubes between corner cubes along 12 edges of big cube will have 2 sides painted. That is 12 x 2 = 24 cubes will painted with blue on 2 sides.

And 4 center cubes on each of 6 faces (left, right, top, bottom, front, back) will have only 1 side painted with blue. That is , there are 6 x 4 = 24 cubes having paint on one side only.

To conclude, out of 56 painted cubes,

24 cubes have paint on 1 side,

24 cubes painted with 2 sides,

8 are painted with three sides.

"Square,Square; Which Color?"

A square tabletop measures 3n × 3n. Each unit square is either red or blue. Each red square that doesn’t lie at the edge of the table has exactly five blue squares among its eight neighbors. Each blue square that doesn’t lie at the edge of the table has exactly four red squares among its eight neighbors. How many squares of each color make up the tabletop?

Here is correct way to count those!

Counting Colorful Squares!

How squares are arranged?

The tabletop measures 3n × 3n, so we can divide it evenly into n² ( 3 × 3) squares that together tile the surface completely.

Let's consider a piece of square of size 3 x 3. For each such unit of 3 x 3 -

1. If the center of the square is red square, then there are 5 blue squares and 3 red squares surrounded with it. 

2. If the center is blue square, then there are 4 blue and 4 red squares surrounding that square. 

In any case, for 3 x 3 = 9 squares, there are 5 blue and 4 red squares. 

Therefore, for tabletop of 3n x 3n, there will be 5n² blue squares and 4n² red squares.  

Count The Number of Kids

Fourteen of the kids in the class are girls. Eight of the kids wear blue shirts. Two of the kids are neither girls or wear a blue shirt. If five of the kids are girls who wear blue shirts, how many kids are in the class?

Find The Count Here!
 

Count of Number of Kids.

 Read the problem first!

 Let's recollect all the data given in the question.

1.Fourteen of the kids in the class are girls.

2. Eight of the kids wear blue shirts. 

3.Two of the kids are neither girls or wear a blue shirt.

4.Five of the kids are girls who wear blue shirts.

Now, (3) clearly suggests that there are 2 boys who are not wearing blue shirts. 

From (5) and (2), it's clear that 3 of 8 kids who are wearing blue shirts are boys as 5 of them are girls.

Above 2 conclusions indicates that there are total 5 boys in the class; 2 not wearing blue shirts and 3 wearing blue shirts.

But as per (1), there are 14 girls in a class. 

Hence, total 14 + 5 = 19 kids are there in a class.

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).