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During the holiday season I found time to reconnect with family over board games. One evening we played a game that required each player to have their own dice. Unfortunately we ran out of 6 sided die. During my turn of sitting out, I found an 8 sided die, and wondered how I could modify the game to rejoin with a dice of n sides by summing scores:

Evidently, I need equal expected values. Therefore, $\mathbb{E}[6s] = 3.5$ and $\mathbb{E}[6s] = 4.5$. The LCM being 31.5, meaning I need to throw 9 and 7 times respectively to have a 50-50 distribution.

Now I attempted to generalize the question without fixing a probability:

What is the probability of player A winning summing $\alpha$ throws of an $n_a$ sided dice, vs. player B summing $\beta$ throws of an $n_b$ sided dice.

The denominator representing all possible outcomes: $(n_a)^\alpha \cdot (n_b)^\beta$

I have made attempts to solve for the numerator, the number of outcomes with A winning, but have not found an answer I am intuitively confident in. Please enlighten me on a correct approach of finding this.

CFM
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  • you will need to use a computer for any of your calculations, so just plot both distributions and compare them using some CAS or programming language. Also you can use this site. – Masacroso Jan 02 '21 at 10:29
  • Is there no short closed-form solution to the problem? – CFM Jan 02 '21 at 11:46
  • no, there is the formula for the distribution of the sum of dice. You can approximate these probabilities using normal distributions, but at thi end of the day, from one approach or other, you will a computer to calculate the probabilities – Masacroso Jan 02 '21 at 11:59
  • @CFM In your game you was able to use the next strategy: you throw a dice and if you see a number from 1 to 6 then you take it, otherwise you throw a dice one more time. – Botnakov N. Jan 03 '21 at 09:19

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