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.
Please help me to correct my logic or find an alternative way to calculate the probability.
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 $$\sum_i (P(7,i)-P(7,i-1))Q(6,i-1)$$