# Probability - Tales Game

A dice game played by two players is like this: each player throw two dice and sum theier results; that is the number of points the player scored. Whoever scores more, wins.

One additional detail is that if the numbers of both dices of a player are equal, the player can roll the two dice again and the sum of these points will be added to the previous sum - and so on, indefinitly.

  A) A player has k points. Calculate his probability of victory.
B) A group of friends decided to play the same game with n players.
Find the winning probability for a player who scored k points.


I've tried for some time to do this, but it seems impossible to me. I don't know much of this kind of probability. Does anyone know a way to solve this?

-
try to add some info as to what you have tried and where the question is from. Copy-Pasting the question is generally not a good way to get the answer you need. – nbubis Oct 16 '12 at 19:52
A player's result is the sum of a series of doubles ($\frac 56$ of the time none) followed by a non-double throw. You can calculate the probability distribution of the non-double throw by deleting some cases from the usual 36 of two dice. Calculating the expectation of the series of doubles is not hard, it is only 1.4 points. I don't see a neat way to get the exact distribution, but the generating function experts can probably do so. Then you can add the two distributions to get the distribution of a full turn. – Ross Millikan Apr 25 '13 at 0:18

When you throw 2 dice, the sum is one of the numbers $2,3,\dots,12$. Do you know the probability of each of these sums? If you do, then you can figure out the probability of winning if you get $k$; it's the probability that your opponent gets a number less than $k$, plus half the probability that your opponent gets exactly $k$.
EDIT: For example, suppose you roll a 4. You win if your opponent rolls a 2 or a 3; the total probability of that is 1/12 (do you know why?). With probability 1/12 your opponent rolls a 4, and in that case, your probability of winning is 1/2, since you are essentially back to square one, and it is a symmetric (thus, fair) game. So your total probability of winning if you throw a 4 is $$(1/12)+(1/2)(1/12)=1/8$$
Sure. Suppose you are one of 7 players, and you roll a 4. You win if each of the other 6 throw 3 or less --- that's $(1/12)^6$. If they all throw 4 or less (probability $(1/6)^6$), and at least one of them throws exactly 4, then your rules don't specify whether they all get to throw again, or just the ones who threw 4 get to throw again. – Gerry Myerson Oct 6 '12 at 23:41