Clearly prisoners W and Z can not immediately know anything since neither of those prisoners can see any of the other prisoners. So, let’s instead focus on prisoners X and Y.Suppose we have the following scenario with the arrangement of different hat colors:

W(B) || X(B) Y(W) Z(W)

In the scenario above, prisoner X will clearly see that Y and Z both have white hats, and logically deduce that he must have a black hat since there are 2 white hats and 2 black hats all together – and he would be correct. Very simple! And this simple logic would also apply to this scenario as well:

W(W) || X(W) Y(B) Z(B)

But let’s consider another example:

W(B) || X (W) Y(B) Z(W)

In this case Z cann’t say what colour hat he is wearing .Y also knows that if the colour’s of Y and Z hats were same he would have shouted immediately.So he will guess that he and Z are wearing opposite color hats..So Y will wait till the time is about to over and then he will shout the opposite colour which Z is wearing.

1 jelly bean from the MIX jar will do the trick.The trick here is to realize that every jar is mislabeled. Therefore you know that the peanut butter jelly bean jar is not the peanut butter jelly bean jar, and the same goes for the rest. You also need to realize that it is the jar labeled p.b./grape,labelled as the MIX jar, that is your best hope. if you choose a jelly bean out of there, then you will know whether that jar is peanut butter or grape jelly jelly beans. it can’t be the mix jar because I already said that every jar is mislabeled.
Once you know that this jar is either peanut butter, or grape jelly, then you know the other jars also. If it is peanut butter, then jar 2 must be mixed because it can’t be grape (as its labelled) and it can’t be peanut butter (that’s jar 3). hence jar 1 is grape.If jar 3 is grape, then you know jar 1 must be the mix because it can’t be p.b. (as its labelled) and it can’t be grape (that’s jar 3). Hence jar 2 is peanut butter.If you pick jelly beans from jar1 or jar 2, then you would have to pick out all of the jelly beans before you knew what that jar was. This is because jar 1 and 2 could be the mix, so in order to disprove that they were the mix, you would have to pull out every jelly bean just to make sure (since there could just be one bean of the opposite flavor in there).

Of course its possible, otherwise it wouldn’t be a very interesting question. The only trick is in realizing that you want to get the two slowest people across together, because otherwise you are wasting too much time. Then once you get them across, how do you not make one of them walk back with the flashlight. Well, you just have one of the fast people already there waiting to sprint the flashlight back across.1. A & B cross. total time: 2 minutes.
C ,D |==========================| A,B(flash light)

2. B comes back. total time: 4 minutes.
C,D,B(FL)|==========================| A

3. C & D cross. total time: 14 minutes.
B |==========================| A,C,D(FL)

4. A comes back. total time: 15 minutes.
B,A(FL)|==========================| C ,D

5. A & B cross. total time: 17 minutes.
|========= =============| A,B,C,D
DONE

Here are some variants of this puzzle :
1.Minimum number of weighing required to find the fake coin when it is not known whether the fake coin is heavier or lighter.
2.Find lighter fake coin from a set of 12 identical coins.

Solution : Only two cuts required .Break the chain into three parts of length 1,2,4 .Here’s how you can pay the labrorer all 7 days satisfying the above conditions:Day 1 : we give him bar of length 1
Day 2 : we take back length1 bar and give him the bar of length 2
Day 3 : we give both the bars of length 1 and 2
Day 4 : we take back all the bars and give him bar of length 4
Day 5 : we give him bar of length 1 taking back nothing from him(4+1)
Day 6 : we take back bar of length 1 and give him bar of length 2.
Day 7 : we give him all the bars.

The prisoners need a way to communicate with each other on who when into the cell. Therefore one person is chosen as the counter.
Every time any prisoner is selected other than counter person , they follow these steps. If they have never turned on the light bulb before and the light bulb is off, they turn it on. If not, they don’t do anything.Now if Counter person is selected and the light bulb is already on, he adds one to his count and turns off the bulb. If the bulb is off, he just sits and do nothing. The day his count reaches 99, he calls the warden and tells him “Every prisoner has been in the special room at least once”.

Start both the hourglasses. The 7 minutes will be finished first. Immediately turn it upside down.When the hourglass with 11 minutes is finished (after 11 minutes).the second glass will still have sand for 3 minutes turn it upside down immediately .So now the sand will fall for 15 more minutes.(11+7-3).

The problem is ismply calculating nth fibonacci number .
Let’s write a simple function numWays(x)
int numWays(int n){
if(n<=0)return 0; if(n=1)return 1; if(n==2)return 2; else return numWays(n-1)+numWays(n-2); }

Number the bottles 1 to 1000, and write the number in binary format.
bottle 1 = 0000000001
bottle 250 = 0011111010
bottle 1000 = 1111101000
Now take your prisoner’s 1 through 10
and
Let prisoner 1 take a sip from every bottle that has a 1 in its least significant bit.
Let prisoner 10 take a sip from every bottle with a 1 in its most significant bit. etc.

Prisoner      – 10 9 8 7 6 5 4 3 2 1
Bottle 924  – 1  1  1 0 0 1 1 1 0 0
In this, bottle #924 would be sipped by 10,9,8,5,4 and 3

That way if bottle #924 was the poisoned one, only those prisoners would die.

After four weeks,
line the prisoners up in their bit order and read each living prisoner as a 0 bit and each dead prisoner as a 1 bit.

The number that you get is the bottle of wine that was poisoned.

The important thing to note here is that the piles can be of different size .Choose any 10 coins to make one pile, then turn over all 10 coins in that pile. The number of tails in this pile is now equal to the number of tails in the other pile of 90 coins. To see this, suppose there were x tails in the 10-coin pile. Since there were initially a total of 10 tails, there must be 10-x tails in the pile of 90 coins. When you flip over all 10 coins in the smaller pile, the x tails in that pile become heads and the remaining coins, which must be 10-x heads, become tails. That matches the larger pile.