a probability question about throwing an irregular fair die

Question

A and B throw a fair 6-sided die in turn. The die is not regular because the faces are {1,2,2,4,4,6}. The winner is whoever first gets a cumulated sum equal or greater than 10. Ask for the probability that A wins, and the expected number of throws when the game generates a winner.

Answer 1

Define the polynomial $P(x)$ by $$ P(x) = \frac{x + 2x^2 + 2x^4 + x^6}{6}. $$ This polynomial represents one throw of the die. The probability to get a sum of $s$ after $k$ throws is the coefficient of $x^s$ in $P(x)^k$. Let's denote that by $[P(x)^k]_{x^s}$. The probability that the sum reaches $10$ for the first time at time $t$ is $$w_t = \sum_{s \geq 10} [P(x)^t]_{x^s} - \sum_{s \geq 10} [P(x)^{t-1}]_{x^s};$$ in other words, $$w_t = \sum_{s < 10} [P(x)^{t-1}]_{x^s} - \sum_{s < 10} [P(x)^t]_{x^s}.$$ The probability that the sum doesn't reach $10$ at time $t$ is $$l_t = \sum_{s < 10} [P(x)^t]_{x^s}.$$ Notice that $w_t = l_{t-1} - l_t$. The probability that B wins is $$\sum_{t \geq 1} l_t w_t.$$ The probability that A wins is, similarly, $$\sum_{t \geq 1} w_t l_{t-1}.$$ The expected number of throws is thus $$\sum_{t \geq 1} 2t l_t w_t + (2t-1) w_t l_{t-1}.$$ Since $l_t = 0$ for $t \geq 10$, all these sums are finite. You can compute everything with a CAS.

Here are some results from SAGE calculations:

• A wins w.p. $64601710707175/101559956668416 \approx 0.636$.
• B wins w.p. $36958245961241/101559956668416 \approx 0.364$.
• The expected number of throws is $550136643228931/101559956668416 \approx 5.42$.