need help understanding interview question

r4qx

New Member
#1
Hello folks,

I had a weird question which I am confused about. I was wondering if you anyone could help me or point me in the right direction (ie. what i need to review to answer these types of questions).

Two dice are rolled, you win $X when you roll greater than or equal to 11. otherwise, you lose $Y dollars. How many rolls would be required to make it a fair game?


At first I am thinking, nothing would make it a fair game, but the interviewer insisted there was an answer.. what am I missing here?
 
#2
Probability of greater than or equal to 11=3/36
suppose $X=10, $Y=5
So your expected win =10*3/36
your expected loss =-5*3/36
How many rolls mean you have to find 'n'
Simply solve this for 'n'
(3/36)^n=0.5.
hint take log and than find 'n'.
 

BGM

TS Contributor
#5
The winning probability is \( \frac {2} {6^2} + \frac {1} {6^2} = \frac {1} {12} \)

It is a fair game if the expected outcome is \( 0 \)

The expected outcome (per roll) is

\( \frac {x} {12} - \frac {11y} {12} \)

So for any \( x, y \) satisfying \( x = 11y \) would be a fair game.

Maybe there is some wording problem (or extra information), not sure what the question is asking for.
 

r4qx

New Member
#6
I am still a bit confused.

I understand the 3/36^n , but not quite understanding why you set it equal to 0.5? is 0.5 basically saying "set it to probability of 50% and solve for y" ?
 
#8
if coin is fair why we can say it fair, because probability of head and tail is equal or 0.5.
there is small difference between your first and second question, you have to under stand that
 

r4qx

New Member
#9
Question is asking how many rolls would make the game fair... I thought it was a trick question... I still think it is.. considering when I do log(3/36) 0.5 I get 0.27894294... so thats less than half a turn.. so the game can never be fair...
 

rogojel

TS Contributor
#11
I think, if the game is fair it has to be fair for any number of rolls. I mean there is no way the game could be unfair for one roll, i.e. I would have a high chance of winning, but would be fair for , say, three rolls. This would be equivalent with the dice having some kind of memory and remembering that I am now at my second or third roll, which is absurd.

You can set up the game to be fair for one, and consequently any number of rolls by adjusting the X and Y values as BGM showed how.

regards
rogojel
 

Dason

Ambassador to the humans
#15
Azeem I really think you are misunderstanding what it means for something to be a fair game. It does not mean that there is a 50% chance of winning and 50% chance of losing as you apparently seem to think. It means that the expected amount that you win is equal to 0.

For example - we play a game where you roll a single fair die and if it lands on a 6 I'll give you 5 dollars but if it lands on anything else you give me $1. Clearly I have a higher probability of winning the game than you do but you have the possibility to win more money. If we calculate the expected value for how much each person will win it comes out to be $0. So this is a fair game.
 
#16
Dason I accept your explanation, but please read the original question. As you say I lose $1 in first roll, I think question is now how many times we play more such that my lose goes to $0, vice versa. You all are focus on over all expected gain should be zero, yes I accept it because for fair game it is necessary, keep in mind if you and me win same Dollars it is also a fair game, thats why I put probability=0.5, The original question was "How many rolls would be required to make it a fair game?", If you have other solution please share it, It will be useful for discussion.
 
#17
The winning probability is \( \frac {2} {6^2} + \frac {1} {6^2} = \frac {1} {12} \)

It is a fair game if the expected outcome is \( 0 \)

The expected outcome (per roll) is

\( \frac {x} {12} - \frac {11y} {12} \)

So for any \( x, y \) satisfying \( x = 11y \) would be a fair game.

Maybe there is some wording problem (or extra information), not sure what the question is asking for.

This makes sense, but the question is HOW MANY throws would make it a fair game... Using Azeem's answer it's roughly ~8.. but I am not sure if thats the correct way of going about it.


I understand the x/12 - 11y/12 such that it = 0 would make it fair.. but what if there are X values and Y values that don't match? What relationship am I missing here??
 

Dason

Ambassador to the humans
#18
Is it supposed to be that if you get a 11 or higher on any of the rolls then you win, if not then you lose? If so then for certain values of X and Y there could be some fair games for certain values of n...
 
#19
Is it supposed to be that if you get a 11 or higher on any of the rolls then you win, if not then you lose? If so then for certain values of X and Y there could be some fair games for certain values of n...
Yes 11 or higher you win, otherwise you lose. and the question was phrased as "so many many rolls do you think would make it a fair game"
 

Dason

Ambassador to the humans
#20
I think you misunderstood my question. I was trying to figure out what the original question could mean. I was conjecturing that it might be saying that instead of just doing a single roll to attempt to achieve the success condition you get multiple rolls?

For example maybe you get 3 rolls to try to get 11 or greater and if you get 11 or greater on ANY of those rolls then you win, otherwise you lose. Not that you play the game 3 separate times and each time you either win or lose.

If it's that first case I mentioned then there might be a solution. If it's the second case then there isn't a solution unless X and Y are chosen properly.