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'.
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?
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'.
Where are you getting 0.5?
fair game, probability of greater or equal 11= probability of less than 11.
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" ?
I think this question is ok.
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
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...
U can do it as
1-(33/36)^n=0.5
which is same, and you get Ans n=8
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
first I accept BGM Ans, but you should read the question it says "How many rolls would be required to make it a fair game?".
yes I read it and what I say is that it does not depend on the number of rolls, the reason is given above.
regards
rogojel
You can say Probability of equal and greater 11=probability of less than 11
3/36 = 33/36
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.
I don't have emotions and sometimes that makes me very sad.
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