need help with a probability problem

This post was edited by 2wo1ne on Nov 17 2014 09:59pm

Post it up.

There are two players and an N sided die. They take turns rolling the die. The player that does not improve the value from the previous roll loses. What is the probability that the first player wins?

This post was edited by 2wo1ne on Nov 17 2014 10:08pm

50%?

intuitively, the more you roll, the more likely you are to lose. since player 1 rolls before player 2, player 1 is more likely to lose. so i'm not sure where you got 50% from.

This post was edited by carteblanche on Nov 17 2014 10:21pm

So in order for the first player to win, the second player has to roll lower than the first player. If the first player wins, then the game will be lost on an even roll. Calculate the probability of a good roll on every odd roll and a bad number on every even roll and add them all up (i.e. prob of 2 rolls + 4 rolls + 6 rolls + 8 rolls....will eventually be zero). This is a function of N.

So player one rolls. This problem gets fairly tricky here because you don't know what they rolled. Let's say they roll x, and x is some integer between 1 and n. The probability of the second person losing on the next roll would be (x)/n, while the probability of them not losing is (n-x)/n.

We keep the probability of player 2 losing on roll 2, which is x/n.

Player 2 rolls a y and doesn't lose. The probability of player 1 rolling a non-losing number is...(n-y)/n. Player 1 rolls z. Then player 2 must lose on roll 4, and the probability of that happening is z/n. So to get to player 2 losing on exactly roll 4 we get (n-x)(n-y)(z)/n^3

Don't know how this converges. I would say that your problem is not answerable as prompted. Every roll's probability is based on all previous roll values, which are unknown and running probability on every possibility is stupid unless N is small, like 6.

Calculating player 1's chance of winning on a 6 sided die is fairly simple, albeit still tedious.



For a 6 sided die:
Player 1 rolls a 6 first roll and wins: 1/6
Player 1 rolls a 5 first roll, player 2 must roll a 1-5, 5/36
Player 1 rolls a 4 first roll, player 2 must roll a 1-4, 4/36
Player 1 rolls a 3 first roll, player 2 must roll a 1-3, 3/36
Player 1 rolls a 2 first roll, player 2 must roll a 1-2, 2/36
Player 1 rolls a 1 first roll, player 2 must roll a 1, 1/36
Sum up all of these above and we get the probability of player 1 winning to be: appx 58.33% (21/36)
It appears to be something along the lines of [n+n-1+n-2+n-3+n-4........]/n^2 which SHOULD be n(n+1)/2n^2 Ans.
There are a lot of terms after this, this is just 2 probabilities although there are a lot of permutations that must be considered. The probability is really low (1/36^2 so we can neglect it all).
Player 1 rolls a 3 first roll, player 2 rolls a 4, player 1 rolls a 5, player 2 rolls a 1-5. 5/36^2
Player 1 rolls a 1 first roll, p2 rolls 2, p1 rolls 3, p2 rolls 4, p1 rolls 5, p2 rolls 1-5. 5/36^3

This post was edited by Dontrunaway on Nov 17 2014 11:19pm

Thanks a lot! ill see if i can work out a concrete answer with this. Again i appreciate it!

Take a look at this if you are interested in an indepth explanation

http://cr4.globalspec.com/blogentry/21768/Roll-of-the-Dice-Newsletter-Challenge-January-2013

Oh well, I tried.

there's 2 players.
50% chance to win and lose here brah
since there's 2 choice for who wins





This post was edited by Uguu on Nov 18 2014 12:43am