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.
a probability question about throwing an irregular fair die
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0@Yuval, can you give an e$x$ample? Also, analytically, are there any convenient method please? thanks. – 2011-01-27
1 Answers
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 $550136643228931/101559956668416 \approx 5.42$.
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0CAS is shorthand for Computer Algebra System. One freeware CAS is SAGE (http://www.sagemath.org). – 2011-01-27