# How to calculate probability of the first player winning a two player dice game, where the winner is the first to roll a total score of seven

A and B take it in turns to roll a standard die. A begins the game, and the first player to score a sum of seven or more wins. What is the probability that A wins?

Through simulation I have worked out that the probability is approximately 0.72599...

Let $P(x,n)$ represent the probability of throwing a score of $x$ or more on $n$ dice, and $Q(x,n)$ represent the probability of throwing a score of $x$ or less on $n$ dice.

I tried to work this out analytically as follows.

• Neither player can win in the first turn.
• The probability A wins in the second turn is $P(7, 2)$ (the probability neither wins in the first turn is one).
• A wins in $y$ turns by rolling seven or more on $y$ dice and neither player winning previously.
• The probability A wins in the $y$th turn is then $P(7, y)|Q(6, y-1)|Q(6, y-1)$.
• If A and B do not win on the 6th turn then A must win.

Bringing this all together, the probability of A winning is $$\sum_{i=2}^{7}P(7,i)Q(6,i-1)^2.$$

Using this sum I did not arrive at the correct probability, in fact the probability of A winning in the second or third turn is too large.

• What's the probability that $B$ wins on her first turn? Well, if $A$ gets a $i$ then $B$ needs a $7-i,\cdots 6$ so $B$ has $7-i$ winning rolls. Thus the probability that $B$ wins on her first turn is $\frac 16\times \sum_{i=1}^6\frac {7-i}6=0.583333333$. Thus your answer can not be greater than $1-.58333$. – lulu Jul 11 '17 at 11:49
The probability of having $7$ or more after $n$ turns is not independent of having $6$ or less after $n-1$ turns, so you cannot just multiply those probabilities.
Instead calculate $$\sum_i (P(7,i)-P(7,i-1))Q(6,i-1)$$