Easy | Probability |

We have a weighted coin which shows a Head with probability p, (0.5<p<1). How do we get a fair toss from this? That is, how do we toss this coin in such a way that we can have probability of winning = loosing = 50%?

Hint

Clearly we cannot have a fair toss in a single flip of this coin. But by tossing this coin 2 times, we can assign the outputs to "win" or "loose", such that we have a 50% chance for both.

Answer

Toss 2 times, mapping HT to "win", TH to "loose" and repeat the process otherwise.

Solution

Toss the coin twice. If consecutive Heads-Tails appears (HT), we "win". In case of (TH), we "loose". If (TT) or (HH) appears, repeat the process. Probability of infinite repetition (p*p*.....) + (1-p)*(1-p)*.... = 0 + 0 = 0;

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Probability of H-T and T-H is equal hence it's fair now.

I proposed a faster method, "lets keep tossing the coin to form a sequence of H's & T's . I win if HT appears before TH" . Was I bluffing?

Actually yes, the probability of HT before TH is just p in second game.

Source: Common

Enable Like and Comment Easy | Probability |

Spiderman has two girlfriends, Mary Jane & Gwen Stacy. After every mission, he rushes to the central subway. Since spidey is a nice man, much impartial, he takes which-ever train arrives first. From subway, one series go towards Mary's place, and another series move towards Stacy. Trains from either series appear every 10 minutes. Also, Peter Parker sticks with the train which arrives first.

But somehow, he notices that he is spending 9 times more dates with Mary Jane than Stacy. Can you explain why?

But somehow, he notices that he is spending 9 times more dates with Mary Jane than Stacy. Can you explain why?

Hint

Do you think ratio should be 50:50? Do you think the gap between each train's arrival is 5 minutes? Can a train arrive always earlier than the other and still be at every 10 minutes gap?

Solution

Train to Mary Jane's place comes at say times 0 and 10, while the train to Stacy's place comes at times 1 and 11. So 9/10 times, train to Mary Jane's place is appearing earlier.

Easy | Strategy |

An Egg breaks only if dropped from above a threshold floor, within this 100 story building. Every time you drop the egg, it is counted an attempt. You are given two eggs, find an algorithm to deduce the threshold floor, with minimum number of attempts in worst case!

Hint

If we had only 1 egg, we would go linearly from 1 to 100. Having an extra egg gives an opportunity to jump (skip some floors from testing). When 1st egg breaks, second egg moves linearly. Think why first egg should not move in constant jumps, but rather decreasing jumps! This will give 14 steps in the worst case.

Answer

It can be done in 14 steps in the worst case.

Solution

A solution for minimum steps in worst case is the smallest integer greater than or equal to the positive solution of n(n+1)/2=100...which gives 14....

start at 14th floor, if egg breaks start linearly from 1, if it does not break drop the egg from 14+13 = 27th floor, and so on....

My friend was asked only one puzzle in his interview, "3 eggs". He got the job.

start at 14th floor, if egg breaks start linearly from 1, if it does not break drop the egg from 14+13 = 27th floor, and so on....

My friend was asked only one puzzle in his interview, "3 eggs". He got the job.

Source: Quant Interview

Enable Like and Comment Easy | General |

Without explicitly calculating, find out which is bigger: e^Pi or Pi^e?

Answer

e^Pi

Solution

This is a common placement test question. Since e^x = 1 + x…., we have e^x>1+x. Now, conveneintly choosing x = (Pi / e -1) and solving, we get e^Pi > Pi^e.

Another way is to observe for f(x) = x^(1/x), differentiating: f'(x) = f(x)*(1/x^2)*(1- ln x); f is decreasing for x > e so pi^(1/pi) < e^(1/e); hence e^(pi)> pi^(e)

Another approach is to assume e^Pi > Pi^e <=> Pi*lne > e*ln Pi <=> lne/e > lnPi/Pi. Now, notice that f(x) = lnx/x is decreasing function since f'(x) is negative in range (e<x<pi), hence above assumption is true!

Another way is to observe for f(x) = x^(1/x), differentiating: f'(x) = f(x)*(1/x^2)*(1- ln x); f is decreasing for x > e so pi^(1/pi) < e^(1/e); hence e^(pi)> pi^(e)

Another approach is to assume e^Pi > Pi^e <=> Pi*lne > e*ln Pi <=> lne/e > lnPi/Pi. Now, notice that f(x) = lnx/x is decreasing function since f'(x) is negative in range (e<x<pi), hence above assumption is true!

Easy | General |

100 pirates stand in a circle. They start shooting alternately in a cycle such that 1st pirate shoots 2nd, 3rd shoots 4th and so on. They continue in circle till only one pirate is left. Which position should Captain Jack Sparrow stand to survive?

Solution

This is not a puzzle at all, this is apti/maths question, but has become quite common. You can calculate for a N=100, 73rd position is safe. For a general N, find largest m such that 2^m < N. Take t = N-2^m. Then Safe position is 2*t + 1. Please learn this formula. No need to prove it using strong induction. Proof: http://en.wikipedia.org/wiki/Josephus_problem#k.3D2

Source: Written Tests

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