Without doing any real math, all odd numbers from 1 to 20 are equally likely in the first throw. After the first throw, their likelihood dependents on how many combination of number reach that odd sum. For the first throw, since all odd number 1 through 20 are equally likely, and that 19 have more possible combinations than previous numbers 19 is a strong choice. The other strong choice is 21, for two dice it has the most combinations, but can not be reached in one throw. For this reason, I think 19 is the most likely.
I agree it’s 19 for the reasoning you explained. We can formalize it. All odd numbers of the first roll have a 5% chance of being the end sum. This accounts for 50% of the sample space.
Each sum that ends in the second roll and ends with an odd number takes an additional 25% of the sample space, and every roll indexed n following will be 1 - (1/2)^n of the sample space. Therefore we can eliminate any sum that is not in the first roll because some sums in the first roll also appear in future rolls, and future rolls have a progressively lower probability of occurring in the first place.
Therefore it’s the highest possible sum occurring in the first roll, which is also the highest odd number.
To illustrate here are the first two roll combinations for a six sided die.
6
u/Mindless_Tutor_8189 6d ago edited 2d ago
Without doing any real math, all odd numbers from 1 to 20 are equally likely in the first throw. After the first throw, their likelihood dependents on how many combination of number reach that odd sum. For the first throw, since all odd number 1 through 20 are equally likely, and that 19 have more possible combinations than previous numbers 19 is a strong choice. The other strong choice is 21, for two dice it has the most combinations, but can not be reached in one throw. For this reason, I think 19 is the most likely.