What is your most memorable interview question or experience?

[Beavis]

well I've heard this one at least twice:



You and I play a game.  We each have 1 die.  I win if i roll a 4.  You win if you roll a 5.  If I go first, what is the probability that I win?

[Corey]

At the risk of embarrassing myself with a terribly poor answer, here is my solution...



For the first iteration...

P(I win) = P(Roll 4) = 1/6

P(You Can Roll) = !P(Roll 4) = 5/6

P(You win) = P(you can roll) ^ P(Roll 5) = !P(Roll 4) ^ P(Roll 5) = 5/36

P(2nd Iteration) = P(you can roll) ^ !P(Roll 5) = !P(Roll 4) * !P(Roll 5) = 25/36



Second iteration...

P(I Win #2) = P(2nd Iteration) ^ P(Roll 4)

P(You Can Roll #2) = P(2nd Iteration) ^ !P(Roll 4)

P(You Win) = P(You Can Roll #2) ^ P(Roll 5)

P(3rd Iteration) = P(You Can Roll #2) ^ !P(Roll 5)



etc ...



So we get ...



P(You Can Roll #n) = P(Nth Iteration) ^ !P(Roll 4)

P(Nth Iteration) = P(You Can Roll #(n-1)) ^ !P(Roll 5)





P(I Win #n) = P(Nth Iteration) ^ P(Roll 4)

= P(You Can Roll #(n-1)) ^ !P(Roll 5) ^ P(Roll 4)

= P((n-1)th Iteration) ^ !P(Roll 4) ^ !P(Roll 5) ^ P(Roll 4)

= P(You Can Roll #(n-2)) ^ !P(Roll 5) ^ !P(Roll 4) ^ !P(Roll 5) ^ P(Roll 4)

...

P(I Win #n) = ((!P(Roll 4) ^ !P(Roll 5))^(n-1)) ^ P(Roll 4)

= (25/36)^(n-1) * 1/6



Is your probability of winning the nth round...



I think? Maybe? Someone help?

[proton]

My guess:



Cum. probability of win after infinite rolls = 1 / (6-25/36)

[tickminator]

Embarrassing if wrong as well:



I tested both solutions - result for n=1 can't be?! Why does it matter if it goes to infinity? Every new round the probability repeats itself - doesn't it??



So my solutions without fancy formulas



P(I win) = 1/6



P (I definitely lose) = 5/6 * 1/6 = 5/36



1/6 + 5/36 = 11/36



So that this solution = 1 multiply by 36/11



P(I win before Player 2) 1/6  * 36/11 = 6/11 (>.5 so may be right)



At an AC with limited time and no paper to write on I would be pretty screwed even if it's correct

[rowdyroddypiper]

You and I play a game.  We each have 1 die.  I win if i roll a 4.  You win if you roll a 5.  If I go first, what is the probability that I win?



I think you are over thinking this everyone.  No where does it speak about repeated rolls or a series of turns.  It doesn't say you roll until you win or anything like that. 



The simple answer would be 1/6.  Just my thought.

[sammyzee]

[img]/User%20Files/4006/Latex-Equation-7679.gif[/img]

[tristanreid]

For those of you who like the verbose descriptions, here's why sammyzee is correct (under the interpretation of roll-til-win):



Just count up all the situations under which you win. You only win on odd rolls, so for any odd roll it's as below, where indentation stands for the next roll:



1/6 you win



+ 5/6 you didn't win



       *  5/6 he didn't win on next roll



               * 1/6 you win on the next roll



                +5/6 you didn't win



                        *... etc



-t.

[DrTarr]

Man, I need to do some of those -t brain exercises.  I kept looking at sammyzee's thinking no Confused , that is not what the verbose answer is saying , it's that plus the initial 1/6 on the first roll.  Duh? we are going from n=0, not n=1.  Now if I never have an interview where they ask me the dice question - I just wasted 15 minutes of my life........ Smiley



I should just stick with the simple answer - its 50-50, either you win or I win.

[Corey]

Looks like I was right on the money (lucky guess, really)...just didn't sum from 0 to infinity. Good little brain teaser!

[cygnet]

p = 1/6 + 5/6(1-p)