There are 5 man in a row.

They all have Either White color or Black Color Hat in their heads.

two of them have white color and two other have black color.

They can look at others Hat but cant see their own and also cant talk to each other.

how many know the color of their own Hat and how???

"two of them have white color and two other have black color" -where is the fifth one? anyways it doesn't matter coz' i am gonna tell them who's wearing what color of hat !!

All of them know the color of their hats because they all have a good memory to remember it from the time they wore them.

OR

No one knows them because someone else put the hats on their head.

HAhahaha...one of them has to be correct.

khurrukka haat le hat nikalera here bhai halcha ni . kina sodhnu paryo. were they all blind as well ?

2 have white, 2 have black, suppose the fifth one has black
then, the guys with white hats can tell that each of them have a white hat. the guys with black hats cannot tell what color hats they have.

Suppose fifth one has white
then, the guys with black hats can tell, and the ones with white hats cannot.

Four of them know what color hat they are wearing. As all four of them can see either 3 of the others are wearing white or 3 of the other are wearing black, thus he is wearing the hat that falls in the one of the other two.


Hurraaay

Sorry guys, Its 2. My Bad

All of them know the color of their hats. This is assuming that all five of them know whether a person was able to tell the color of his/her hat or not. Here's how:

Notation: 1B = the first person has a black hat.

Condition 1:
If the fifth hat is a WHITE one
Then the condition will be something like: 1B, 2B, 3W, 4W, 5W (Here, the order of the hats does not matter. )

Now, when 1 looks at other people, he can only see one black hat. So, he instantly knows that his MUST have a black one too. Same logic with 2 too; So 2 also knows that he has a black hat. When 3 looks at other people, he can see two white hats and two black hats. But he knows that if he had a black hat, then 1 and 2 would not be able to tell the color of their hats. But since 1 and 2 know the color of their hats, 3 must have a white hat. Same logic with 4 and 5 too, and hence they know that they have white hats too.

Condition 2:
If the fifth hat is a BLACK one.
Its the similar logic. We will then have condition like: 1W, 2W, 3B, 4B, 5B.

Here's another puzzle that I posted on the other thread but did not get any attempts.
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Among 10 coins which look exactly alike, two are fake coins. All real coins weigh the same. Both fake coins weigh the same, but they weigh less than the real coins. You have a beam balance. How do you find the fake coins using a balance not more than 5 times?

Fake Coins Puzzle:
A, B, C, D, E, F, G, H, I, J

Divide this up in two grps

First Use of Balance: Weigh ABCDE vs FGHIJ
Two possibilities, weghts are the same and weights are different.

1. If Weights are the same for the two grps, ie. ABCDE = FGHIJ
There is a fake coin on each group.
For each grp; ABCDE and FGHIJ
make a sub grp of two coins X 2 and 1 coin, eg. AB, CD and E for ABCDE

Second Use of Balance: Weigh AB vs CD, if they weigh the same E is fake.
if AB < CD, then A or B is fake
Third Use of Balance: Weigh A vs. B, whichever is less, it's fake.

Repeat the same for FGHIJK. The max use of balance is 5.

2. If ABCDE < FGHIJ then
ABCDE has both fake coins.
Divide into sub grp of 2 coins X 2 and a single coin, like AB, CD and E

Second Use of Balance: Weigh AB vs CD, if they are both equal, both of them have one fake each.

Then weight A vs B and C vs D. Smaller weights are fakes. (Total number of balance use 4)

If AB < CD, then AB must contain at least one fake, weigh A vs B. If A=B, then they are both fake. (total number of balance use = 3)

If A
Weight C vs D. If they are equal, E is fake. (Total number of balance use 4). If C< D them C is fake. (Total Number of balance use 4)

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Million Dollars,
Good try, but in your step 1 you have to be able to differentiate BOTH fake coins in less than 5 attempts not just one.

*in not more than 5 attempts

My bad...u r right!