# Dice game keep rolling [closed]

There is a dice game where both players roll a fair dice once. If player A rolls a 1, then he keeps rolling. And in the end, if the score is tied, player B wins. What is the probability of player A winning?

## closed as off-topic by Namaste, Shailesh, Adrian Keister, Xander Henderson, StrantsAug 28 '18 at 0:35

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• This seems a bit unclear to me. Are you basically saying that player A will get one of the numbers 2, 3, 4, 5, 6; and B will get one of the numbers 1, 2, 3, 4, 5, 6; higher score wins; B wins if the score is tied? – paw88789 Aug 27 '18 at 16:50
• B wins if the score is tied. Other outcomes, A wins. – user587649 Aug 27 '18 at 16:54
• What is unclear is how a person's score is decided. "If player A rolls a 1, then he keeps rolling" might be interpreted as his final score is the sum of all rolls of the dice that he has thrown, so he could potentially have a score of 1000 if he happens to roll 995 ones in a row followed by a five. It is also unclear if when the first die is a $1$ is allowed to roll again and the second die is also a $1$ whether he must keep the result of the second die or if he can continually reroll until getting something other than a $1$. – JMoravitz Aug 27 '18 at 17:02
• A's final score is the first roll that is not 1. – user587649 Aug 27 '18 at 17:05
• Assuming each player's score is merely the final result shown on a die and the first player may reroll indefinitely until getting something other than a $1$, although I discourage "counting by hand" this problem is small and easy enough that you can begin counting by hand (and hopefully spot the pattern in order to not have to do it ever again). You can make yourself a $5\times 6$ grid with rows corresponding to $A$'s final result and columns corresponding to $B$'s. Count how many entries correspond to $A$'s score being strictly higher than $B$'s. (triangles can make counting faster) – JMoravitz Aug 27 '18 at 17:06

$\begin{array}{|cccccc}\hline \color{blue}{2>1}&2\leq 2&2\leq 3&2\leq 4&2\leq 5&2\leq 6\\ \color{blue}{3>1}&\color{blue}{3>2}&3\leq 3&3\leq 4&3\leq 5&3\leq 6\\ \vdots\\ \color{blue}{6>1}&\dots&&\dots&\color{blue}{6>5}&6\leq 6\end{array}$
We count how many "good" possibilities there are. In the first row there is one, in the second row there are two, etc... on up until the last row which has five good possibilities giving a total of $1+2+3+4+5=\frac{5\cdot 6}{2}=15$ scenarios in which $A$ wins.
Notice that each of the outcomes in the grid are equally likely to occur. (This might be somewhat challenging to a beginner, but the hand-wavy explanation is that since $A$ is rerolling whenever he gets a $1$, he might as well have rolled a "five-sided die" instead which only has numbers $2$ through $6$.)
$$Pr(A~\text{wins})=\frac{15}{30}=\frac{1}{2}$$