A mind-flexer ! August 9, 2009
Posted by ilabs in Puzzled.1 comment so far
Suppose you are seated in a dark room and on the table in front of you are placed ten coins – 3 heads up and 7 tails up.
The task @ hand is to separate these coins into two piles such that both piles have the same number of heads up coins.
Instructions : The room is dark. So obviously you cant see if a coin is heads up / tails up. Neither can you tell the difference by touching the coin. But to make your task a tad simpler – flipping the coins is always allowed.
That’s it. All set. Let’s flex that mind !
Solution:
Separate the coins into two piles of 3 and 7 each. We obviously don’t know of the number of heads / tails in each pile.
Take the smaller pile of 3 coins. Flip all the coins in this pile.
Now, both the piles will have the same number of heads.
How? It’s like this…
Let’s say the pile with 3 coins has ‘h’ heads. The total number of heads-up coins being three, the pile with 7 coins will have ‘3-h’ heads up. Hence, number of heads in the 3 coins pile is ‘h’ and number of coins in the 7 coins pile is ‘3-h’.
Now, take the smaller pile of 3 coins. Here, number of heads-up coins is ‘h’ and number of tails up coins is ‘3-h’ (because total number of coins in the pile is 3 and if ‘h’ of them are heads-up, only ‘3-h’ of them can be tails-up). When you flip all the coins here, all the coins which were previously heads become tails and vice versa. Hence, now the number of tails up coins here would be ‘h’. Therefore, number of heads up coins would be ‘3-h’.
Now, both piles have same number of heads up coins i.e. ‘3-h’.
Hence, problem solved !!
The Gabbar Kalia puzzle – Russian Roulette July 25, 2008
Posted by ilabs in Puzzled.Tags: puzzle, trivia
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Here is an adaptation of the famous puzzle – Russian Roulette.
A scene from the movie ‘Sholay’.
‘Gabbar Singh’ (the dacoit) is furious at his accomplice ‘Kalia’. Gabbar loads his revolver with exactly two bullets in continuous slots (of the six possible slots) and points the muzzle @ Kalia’s head. Gabbar says ‘I am going to kill you’. And saying this, triggers the revolver.
But no bullet is fired. Kalia is alive and also a lot confused.
Gabbar, amused by the turn of fate, gives Kalia an option, ‘There are exactly 2 bullets in here and I am going to take only one more shot. You have a choice – 1. Either I spin the cylinder (of the revolver) again and then take a shot. 2. Or I go ahead and take another shot. Whatever now happens is your destiny. And your destiny is in your hands. What do you want me to do – Spin the cylinder and shoot OR Take another shot?’
Considering that Kalia really wants to live another day, what option should he be suggesting?
(I came across the russian roulette puzzle online. Also read in a book called ‘How would you move Mount Fuji?’ by William Poundstone. You could also read about the Russian Roulette on wikipedia http://en.wikipedia.org/wiki/Russian_roulette)
Answer
Option 2: Go ahead and take another shot.
Case 1: Revolve the cylinder again and shoot.
In this case, the no of cases where Kalia will survive is 4 (if any of the 4 empty slots gets selected again). Total no of slots is 6 (4 empty + 2 with bullet each).
Hence, Probability of Survival, P(S) = Favourable cases / Total no of cases = 4/6 = 66.66 %
Case 2: Take another shot.
In this case one empty slot has already been exhausted (as the first shot did not fire any bullet). Now to survive, the next shot must also land up on one of the remaining empty slots. Remaining empty slots = 3. Hence, favourable cases = 3.
From the total no of possible slots that can be considered, one empty slot has already been exhausted.
Also the bullets are loaded continuously. Hence, the second continuous bullet slot also has to be excluded. Because the only way that the second bullet slot can now be selected is if the first shot would have landed on the first bullet shot. But that cannot be the case, since the first shot did not fire any bullet.
Hence Total no of possible slots = 6 – 1 (empty slot that was selected for the first shot) – 1 (the second continuous bullet slot which cannot be selected) = 4
Hence, Probability of Survival, P(S) = Favourable Cases / Total Possible Cases = 3/4 = 75%
This case has a higher Probability of Survival.
Hence, Kalia must select this option.
Find the heavier ball… July 22, 2008
Posted by ilabs in Puzzled.Tags: puzzle, trivia
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There are 8 iron balls. All are of equal weight except one. And you are given a weighing balance. What is the minimum no of times you need to use the weighing balance to find the heavier ball?
Ans : 2 times
Three times is what is the most common response you’ll come across.
But you can spot the heavier one by using the balance just two times.
Keep three balls on each side of the balance.
Case 1: Both the sides are equally balanced. Which would mean that heavier ball is one of the two left out. Now put each of those two on either side of the balance. One of them would be heavier and the corresponding side would be lowered.
So, for case 1, you can spot the heavier ball by using the balance only two times.
Case 2: Both the sides are not equally balanced. This would mean that the heavier ball is one amongst the three balls, for which the balance is lowered. Next task is to spot which of the three is the heavier one.
So for this, take any two of the three and place them on the scale on either side.
If both of them are of same weight (i.e the balance stays balanced), the heavier one is the third ball which has been kept out.
Else if the heavier ball is one of these two, the corresponding side will be lowered.
So, in case 2 also, we can spot the heavier ball in just two attempts.
Hence, minimum no of times you would need to use the balance is 2.
