Program manager Interview Questions

Everyone is wrong. You stay with your first pick, the odds becoming slanted greatly against the other door.

Interview Candidate and Anonymous are right. This is also known as the Monte Hall problem

Your choice splits the doors in two sets. Set A contains the door you selected, and the probability that is a prize behind this door is 1/3. The set B contains all remaining doors, and the probability that the winning door is somewhere in there is 2/3. By removing one door, which all have the success probability of zero because there's coal behind them, from set B, only one door remains in B, but the overall probability for success in set B is still 2/3. Therefore you must switch.

I can't believe that some of the answers up here actually state that you should switch. -------------------------- Before the host showed you that door#3 was hiding coal, your chances to pick the right door where 1/3. Now that you know that coal is behind door#3, you only know that 1 of the 2 remaining doors leads to coal while the other leads to the prize. While your probability of making the right choice has increased from 1/3 to 1/2, this probability still applies to both of the doors. You can't differentiate them. Seriously.

Tarik, You are correct that the right choice has increased from 1/3 to 1/2, but you have to account for variable change in this instance. Think of the same problem, but with different numbers. 100 doors, same conditions. You pick 1 door, with a 1/100 chance of being correct. The game show opens 98 doors with coal (because he knows where the prize is). He now offers you the chance to keep your door or switch to the last remaining door. In this Because of variable change you will always be better off switching doors.

Scratch that, the right choice hadn't increased from 1/3 to 1/2. It's late and I have no idea why I typed that.

Really??? I can see why they ask the question if so many people have so little understanding of probability. There are NOW two doors to choose from and there is one prize. DAAAAA It doesn't matter how many there were to begin with or which one of the two you originally chose. There are now two doors and each one has an equal chance of containing the prize so your odds of success do not change at all regardless if you stay with your original choice of switch to the other. The odds of your original door being the correct one were 1/3 when there were 3 option but the odds of your door being the correct one CHANGE to 1/2 when you eliminate one of the options. To say the probability of your door being correct are still 1/3 is to say that your odds of success are still one third if you were to pick door 3 even though you know it does not have the prize. As an interviewer I could have easily eliminated the majority of the applicants above.

What about sticking to your original decision and not deviating without knowing the facts(Data). i will stick with my decision and might fail, sure but at list i will not make decision on just the 1/3 and 1/2 number game.

Averee has given the best explanation. Before anyone writes any further comment or contradicts what he says, do yourself a favor, go on youtube and watch the video on monty hall problem - you will not object to averee's solution after that.

What everyone who is answering "stick to original door" is missing the host's knowledge of which door has the prize and his actions after the first door is chosen. Host's action puts the odds of prize behind the switched door as 0.5. However, the original door still has a 0.33 probability of the prize. Yes I know it is a difficult concept to grasp - so you all need to watch the monty hall problem on youtube.

Switching is the correct answer.

Thank you Akhil. I'd like to add that the youtube video "Monty Hall Problem for Dummies - Numberphile" explained it very well.

PS. Watch it twice.

You must switch the door. Imagine the same situation with 100 doors and you will understand.

Rikesh's answer is not correct. cuz 1+6= 7. But with his implementation, you would use 1, 6 to represent 16th also. You can argue that you could use 6,1 to represent the 16th. However, I think MS wants the more "geek" way of the solution, which is one I upvoted the base 6 solution.

How about... Dice 1: each pip equals 1 day: 1 pip =1 day, 2 pips = 2 days, etc. Dice 2 each pip equals 5 days: 1 pip = 5 days, 2 pips = 10 days 3 pips = 15 days etc. Die 2 is always to the left of Die 1. For days less than or equal to 5 use 1 die. When counting a day that is a multiple of 5 (days 5,10,15 etc.) count day by putting Die to 5 pips and Die 2 to the remaining number of days.

You are on Desert.. ! Why don't write the count down/up number on it. and cover using the dice ? Just a Approach keeping basis of PMO PM: Analytical thinking, – it’s thinking about the parts or elements of a situation. PMO: Synthetical thinking – it’s thinking about how those parts or elements work together.

Given a Tea Powder, Milk, Gas Stove and two match stick to prepare Tea in less than 60seconds. Approach: Using the two match stick keeping crossed at Right angles and done "T" Its about "what can be done to achieve goal" not "what is given to achieve the goal"

Why use dice when you can use the island to help you?

It should be 6x-y!.. consider the dice to be x and y. Irrespective of the order they are placed in.. x is always the lower (or equal) dice value, so it's easy to identify x and y! Day 1: 6x1 - 5 (dice values:1,5) Day 2: 6x1 - 4 (dice values: 1, 4) Day 3: 6x1 - 3 (dice values: 1,3) Day 4: 6x1 - 2 (dice values: 1,2) Day 5: 6x1 - 1 (dice values: 1,1) Day 6: 6x2 - 6 (dice values: 2,6) . . .

Draw numbers in the sand with the dice.

Basically 2 dice has 6 sides so for Dice 1: mark 0 1 2 4 6 8 Dice 2: mark 1 2 3 5 7 9 And those two can make up all days of the month Or you can map 1-6 with those numbers for each dice

Using the dice you can count to a maximum of 66 combined. When the plane flies by on the 1st month's 30th day, each day after that at the same approximate time in the day you can increase the value of the dice by 1. For example plain flies by on 30th day of 1st month. the next day you wake up and assuming plane flew by at dawn you place the side with the 1 facing up. Each dawn after that you increase to 2, 3, 4, 5, 6-- after six you would use both die, and make some indication to your self saying that 1+6= 7, 2+6=8... and so on after you reach the 4+6=10th day it would be easier to use the die to show "1""1"-> 11th day, etc. That way eventually, you can make a calendar that tells you the day of the month.