Given a 12 sided dice, you roll the dice repeatedly until the cumulative sum is odd. What is the number of the cumulative sum you can have with the highest probability?
[Math] 12 side dice. Roll to first odd cumulative sum. What is the most likely odd number to appear
probability
Related Solutions
Note that there are a total of $6 \times 6 \times 6 = 216$ options. The possible sums are from $3$ to $18$.
Also note that the distribution has to be symmetric since if a roll gives $x,y,z$ adding to $n$, then $7-x,7-y,7-z$ adds to $21-n$.
Hence, the number of ways of getting $n$ is same as the number of ways of getting $21-n$. We would hence expect the maximum to occur for $10$ and $11$.
Number of ways to get $3$ is $1$ i.e. $\dbinom{2}{2}$, which is the same as the number of ways to get $21-3=18$.
Number of ways to get $4$ is $3$ i.e. $\dbinom{3}{2}$, which is the same as the number of ways to get $21-4=17$.
Number of ways to get $5$ is $6$ i.e. $\dbinom{4}{2}$, which is the same as the number of ways to get $21-5=16$.
Number of ways to get $6$ is $10$ i.e. $\dbinom{5}{2}$, which is the same as the number of ways to get $21-6=15$.
Number of ways to get $7$ is $15$ i.e. $\dbinom{6}{2}$, which is the same as the number of ways to get $21-7=14$.
Number of ways to get $8$ is $21$ i.e. $\dbinom{7}{2}$, which is the same as the number of ways to get $21-8=13$.
Number of ways to get $9$ is $25$ i.e. $\dbinom{8}{2}-3$, which is the same as the number of ways to get $21-9=12$.
Number of ways to get $10$ is $27$ i.e. $\dbinom{9}{2} - 3 - 6$, which is the same as the number of ways to get $21-10=11$.
Now by symmetry you can get the number of ways to get the sum between $11$ and $18$.
Hence, the distribution peaks at $10$ and $11$ as expected.
As a sanity check, we have $2 \left( 1 + 3 + 6 +10 +15 + 21 + 25 + 27\right) = 216$.
Below is the distribution of the number of times a number occurs as a sum of three dices, where the $X$-axis the sum of the three dice and $Y$-axis is the number of times the sum occurs.
I tried several strategies, and the best seems to be to reroll on $9$ or less and stop with $10$ or more, which gives an expected payout of $\frac{293}{27}=10.8518518518519$ just above $\$10.85$
Reroll on $8$ or less
Let's say we reroll on $8$ or less and stay with $9$ or greater. The probabilty of rolling $9$ or greater is $$ \frac{6+6+6+5+4+3+2+1}{60}=\frac{33}{60} $$ so on average, it takes $\frac{60}{33}$ rolls to get $9$ or greater paying on average $\frac{27}{33}$ dollars to get there. The expected payout of $9$ or greater is $$ \frac{6\cdot9+6\cdot10+6\cdot11+5\cdot12+4\cdot13+3\cdot14+2\cdot15+1\cdot16}{6+6+6+5+4+3+2+1}=\frac{380}{33} $$ Thus, the expected value of this strategy is $\frac{380}{33}-\frac{27}{39}=\frac{353}{33}=10.6969696969697$
Reroll on $9$ or less
Let's say we reroll on $9$ or less and stay with $10$ or greater. The probabilty of rolling $10$ or greater is $$ \frac{6+6+5+4+3+2+1}{60}=\frac{27}{60} $$ so on average, it takes $\frac{60}{27}$ rolls to get $10$ or greater paying on average $\frac{33}{27}$ dollars to get there. The expected payout of $10$ or greater is $$ \frac{6\cdot10+6\cdot11+5\cdot12+4\cdot13+3\cdot14+2\cdot15+1\cdot16}{6+6+5+4+3+2+1}=\frac{326}{27} $$ Thus, the expected value of this strategy is $\frac{326}{27}-\frac{33}{27}=\frac{293}{27}=10.8518518518519$
Reroll on $10$ or less
Let's say we reroll on $10$ or less and stay with $11$ or greater. The probabilty of rolling $11$ or greater is $$ \frac{6+5+4+3+2+1}{60}=\frac{21}{60} $$ so on average, it takes $\frac{60}{21}$ rolls to get $11$ or greater paying on average $\frac{39}{21}$ dollars to get there. The expected payout of $11$ or greater is $$ \frac{6\cdot11+5\cdot12+4\cdot13+3\cdot14+2\cdot15+1\cdot16}{6+5+4+3+2+1}=\frac{266}{21} $$ Thus, the expected value of this strategy is $\frac{266}{21}-\frac{39}{21}=\frac{227}{21}=10.8095238095238$
Best Answer
Eleven.
For $s$ odd let $S_s$ be the set of tuples $(a_1,\ldots, a_k)$ with $k\ge 1$, $a_k\in\{1,3,5,7,9,11\}$ and all other $a_i\in\{2,4,6,8,10\}$ and $a_1+\ldots +a_k=s$. Define $|(a_1,\ldots,a_k)|:=k$. The probability of ending at sum $s$ is $$P(s) =\sum_{\alpha\in S_s}12^{-|\alpha|}.$$ If $s<11$ we define an injection $f\colon S_s\to S_{11}$ by $$(a_1,\ldots,a_k)\mapsto (a_1,\ldots,a_{k-1},a_k+2)$$ (noting that $s<11$ implies $a_k<11$). Since $(2,2,2,2,2,1)$ is not in the image of this map and $|f(\alpha)|=|\alpha|$ we conclude $P(11)>P(s)$.
If $s>11$ then observe that $$g\colon S_s\to\bigcup_{k=1}^6S_{s-2k},\\ (a_1,\ldots,a_m)\mapsto(a_1,\ldots,a_{m-1}-1)$$ is a bijection. As $|g(\alpha)|=|\alpha|-1$, we have $$ P(s)=\frac1{12}\sum_{k=1}^6 P(s-2k)$$ and by induction $P(s)\le \frac12P(11)$ for $s>11$.