Any sum of 2 dice with equal probability - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T14:59:05Z http://mathoverflow.net/feeds/question/41310 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability Any sum of 2 dice with equal probability jakab922 2010-10-06T18:28:17Z 2010-10-11T07:02:24Z <p>The question is the following: Can one create two nonidentical loaded 6-sided dice such that when one throws with both dice and sums their values the probability of any sum (from 2 to 12) is the same. I said nonidentical because its easy to verify that with identical loaded dice its not possible.</p> <p>Formally: Let's say that $q_{i}$ is the probability that we throw $i$ on the first die and $p_{i}$ is the same for the second die. $p_{i},q_{i} \in [0,1]$ for all $i \in 1\ldots 6$. The question is that with these constraints are there $q_{i}$s and $p_{i}$s that satisfy the following equations: <br>$q_{1} \cdot p_{1} = \frac{1}{11}$ <br>$q_{1} \cdot p_{2} + q_{2} \cdot p_{1} = \frac{1}{11}$ <br>$q_{1} \cdot p_{3} + q_{2} \cdot p_{2} + q_{3} \cdot p_{1} = \frac{1}{11}$ <br>$q_{1} \cdot p_{4} + q_{2} \cdot p_{3} + q_{3} \cdot p_{2} + q_{4} \cdot p_{1} = \frac{1}{11}$ <br>$q_{1} \cdot p_{5} + q_{2} \cdot p_{4} + q_{3} \cdot p_{3} + q_{4} \cdot p_{2} + q_{5} \cdot p_{1} = \frac{1}{11}$ <br>$q_{1} \cdot p_{6} + q_{2} \cdot p_{5} + q_{3} \cdot p_{4} + q_{4} \cdot p_{3} + q_{5} \cdot p_{2} + q_{6} \cdot p_{1} = \frac{1}{11}$ <br>$q_{2} \cdot p_{6} + q_{3} \cdot p_{5} + q_{4} \cdot p_{4} + q_{5} \cdot p_{3} + q_{6} \cdot p_{2} = \frac{1}{11}$ <br>$q_{3} \cdot p_{6} + q_{4} \cdot p_{5} + q_{5} \cdot p_{4} + q_{6} \cdot p_{3} = \frac{1}{11}$ <br>$q_{4} \cdot p_{6} + q_{5} \cdot p_{5} + q_{6} \cdot p_{4} = \frac{1}{11}$ <br>$q_{5} \cdot p_{6} + q_{6} \cdot p_{5} = \frac{1}{11}$ <br>$q_{6} \cdot p_{6} = \frac{1}{11}$ <br></p> <p>I don't really now how to start with this. Any suggestions are welcome.</p> http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41311#41311 Answer by Robin Chapman for Any sum of 2 dice with equal probability Robin Chapman 2010-10-06T18:39:55Z 2010-10-06T18:39:55Z <p>You can write this as a single polynomial equation $$p(x)q(x)=\frac1{11}(x^2+x^3+\cdots+x^{12})$$ where $p(x)=p_1x+p_2x^2+\cdots+p_6x^6$ and similarly for $q(x)$. So this reduces to the question of factorizing $(x^2+\cdots+x^{12})/11$ where the factors satisfy some extra conditions (coefficients positive, $p(1)=1$ etc.).</p> <p>This is a standard method (generating functions).</p> http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41313#41313 Answer by Michael Lugo for Any sum of 2 dice with equal probability Michael Lugo 2010-10-06T18:45:06Z 2010-10-06T21:42:54Z <p>You can write a polynomial that encodes the probabilities for each die:</p> <p>$$P(x) = p_1 x^1 + p_2 x^2 + p_3 x^3 + p_4 x^4 + p_5 x^5 + p_6 x^6$$</p> <p>and similarly</p> <p>$$Q(x) = q_1 x^1 + q_2 x^2 + q_3 x^3 + q_4 x^4 + q_5 x^5 + q_6 x^6.$$</p> <p>Then the coefficient of $x^n$ in $P(x) Q(x)$ is exactly the probability that the sum of your two dice is $n$. As Robin Chapman points out, you want to know if it's possible to have</p> <p>$$P(x) Q(x) = (x^2 + \cdots + x^{12})/11$$</p> <p>where $P$ and $Q$ are both sixth-degree polynomials with positive coefficients and zero constant term. </p> <p>For simplicity, I'll let $p(x) = P(x)/x, q(x) = Q(x)/x$. Then we want</p> <p>$$p(x) q(x) = (1 + \cdots + x^{10})/11$$</p> <p>where $p$ and $q$ are now fifth-degree polynomials. We can rewrite the right-hand side to get</p> <p>$$p(x) q(x) = {(x^{11}-1) \over 11(x-1)}$$</p> <p>or </p> <p>$$11 (x-1) p(x) q(x) = x^{11} - 1.$$</p> <p>The roots of the right-hand side are the eleventh roots of unity. Therefore the roots of $p$ must be five of the eleventh roots of unity which aren't equal to one, and the roots of $q$ must be the other five. </p> <p>But the coefficients of $p$ and $q$ are real, which means that their roots must occur in complex conjugate pairs. So $p$ and $q$ must have even degree! Since five is not even, this is impossible.</p> <p>(This proof would work if you replace six-sided dice with any even-sided dice. I suspect that what you want is impossible for odd-sided dice, as well, but this particular proof doesn't work.)</p> http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41328#41328 Answer by Aaron Meyerowitz for Any sum of 2 dice with equal probability Aaron Meyerowitz 2010-10-06T20:46:44Z 2010-10-06T20:52:37Z <p><strong>sorry</strong> misread as identical to standard case. Delete this if you know how.</p> <p>We want $p(x)q(x)$ to be the same as $r(x)^2$ where $r(x)$ encodes a standard die. Put in that $r(x)$ and factor.</p> http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41333#41333 Answer by Steven Heston for Any sum of 2 dice with equal probability Steven Heston 2010-10-06T21:14:06Z 2010-10-06T21:14:06Z <p>You can't even solve this with two-sided dice. Consider two dice with probabilities p and q of rolling 1, and probabilities (1-p) and (1-q) of rolling 2. The probability of rolling a sum of 2 is pq, and the probability of rolling a sum of 4 is (1-p)<em>(1-q). These are equal only if p=(1-q). Hence they are equal to one third only if p satisfies the quadratic equation p</em>(1-p) = 1/3. Since this has no real roots, it cannot be done. This logic extends to multisided dice.</p> http://mathoverflow.net/questions/41310/any-sum-of-2-dice-with-equal-probability/41753#41753 Answer by Michael Lugo for Any sum of 2 dice with equal probability Michael Lugo 2010-10-11T07:02:24Z 2010-10-11T07:02:24Z <p>Here is an alternate solution, which I ran across while looking through Jim Pitman's undergraduate probability text. (It's problem 3.1.19.)</p> <p>Let $S$ be the sum of numbers obtained by rolling two dice,, and assume $P(S=2)=P(S=12) = 1/11$. Then</p> <p>$P(S=7) \ge p_1 q_6 + p_6 q_1 = P(S=2) {q_6 \over q_1} + P(S=12) {q_1 \over q_6}$</p> <p>and so $P(S=7) \ge 1/11 (q_1/q_6 + q_6/q_1)$. The second factor here is at least two, so $P(S=7) \ge 2/11$. </p>