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The first result is the roll of a 20 sided die The second result is an independent roll of a 20 sided die that has its roll multiplied by 2

What is the probability that the first result is greater than or equal to the the second result?

I thought it would be easier to calculate the probability of failure (that the second is result is strictly greater than the first) and then from there derive the probability of success and so I started by assuming that a roll of $11$ or higher on the second die is an instant failure ($\dfrac{10}{20}$). On a similar note a roll of $1$ on the first die would result in instant failure ($\dfrac{1}{20}$). Then we could use a summation to calculate the number of times a roll on the first die would beat a roll on the second die. $$\dfrac{10}{20}\times\dfrac{19}{20}\sum_{n=2}^{20}\dfrac{1}{19}\times(1-\dfrac{1}{2}\times\dfrac{n}{10})$$ I derived the above equation from the fact that the only times the first roll has a chance of success is if the result of the second roll is $1-10$ which has a probability of $\dfrac{10}{20}$. In addition the first roll cannot be a $1$ which has a probability of $\dfrac{19}{20}$. The final condition is that the roll of the first die is strictly less than $2\times$ the roll of the second die, so we create a summation that traverses all of the possible rolls on the first die from $2-20$. The odds of each roll on the first die is $\dfrac{1}{19}$ and the odds that it is strictly less than a roll on the second die is $1-\dfrac{1}{2} \times \dfrac{n}{20}$. In other words it is $1 - $ the number of times the first roll would be greater than or equal to the second roll. This brings the full equation to:

$$\dfrac{10}{20} + \dfrac{1}{20} + \dfrac{10}{20}\times\dfrac{19}{20}\sum_{n=2}^{20}\dfrac{1}{19}\times(1-\dfrac{1}{2}\times\dfrac{n}{10})$$

With the above equation I obtained the probability 0.76375, which again is the probability that the second roll is strictly greater than the first, but which we can use to obtain the probability that the first roll will be greater than or equal to the second as such: $1 - 0.76375 = 0.23625$ But when I map out the correct number of victories for the first roll over the second I get:

|---------------------|------------------|---------------------|------------------|
|      First Roll     |   # Victories    |      First Roll     |   # Victories    |
|---------------------|------------------|---------------------|------------------|
|          1          |        0         |          11         |        5         |
|---------------------|------------------|---------------------|------------------|
|          2          |        1         |          12         |        6         |
|---------------------|------------------|---------------------|------------------|
|          3          |        1         |          13         |        6         |
|---------------------|------------------|---------------------|------------------|
|          4          |        2         |          14         |        7         |
|---------------------|------------------|---------------------|------------------|
|          5          |        2         |          15         |        7         |
|---------------------|------------------|---------------------|------------------|
|          6          |        3         |          16         |        8         |
|---------------------|------------------|---------------------|------------------|
|          7          |        3         |          17         |        8         |
|---------------------|------------------|---------------------|------------------|
|          8          |        4         |          18         |        9         |
|---------------------|------------------|---------------------|------------------|
|          9          |        4         |          19         |        9         |
|---------------------|------------------|---------------------|------------------|
|          10         |        5         |          20         |        10        |
|---------------------|------------------|---------------------|------------------|

The average of the above being: $\dfrac{victories}{total} = \dfrac{100}{400} = 0.25$

So the equation I derived was close but that doesn't account for much in mathematics so any help that can be given would be greatly appreciated, thank you in advance.

As a side, is there any chance that the derived formula can be constructed so that it is applicable to any factor for the second roll (for example $\times 5$ instead of $\times 2$)?

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  • $\begingroup$ The first problem I see is that when you add $\frac{1}{20}$ to account for the automatic loss of rolling a 1, you should actually add half of that, $\frac{1}{40}$. Because you already counted the space where the doubled die rolls an 11-20 and your $\frac{1}{20}$ space has half of its space overlapping it. (Imagine a 20x20 grid) $\endgroup$ – turkeyhundt Jan 5 at 6:27
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Your equation needs a couple edits, as I mentioned in comments above. $$\dfrac{10}{20} + \dfrac{1}{20} + \dfrac{10}{20}\times\dfrac{19}{20}\sum_{n=2}^{20}\dfrac{1}{19}\times(1-\dfrac{1}{2}\times\dfrac{n}{10})$$ First, the $\dfrac{1}{20}$ should be changed to $\dfrac{1}{20}\times\dfrac{1}{2}$ because you only want the events where the first die is a 1 and the second die is 1-10 (you already counted 11-20).

Then, for the summation, you do not need the $\dfrac{19}{20}$. You just need to sum up the events you are counting and then multiply by the probability of that happening. Which is the $\dfrac{1}{19}$. Further, you can then pull that $\dfrac{1}{19}$ out of the summation to get a final equation of: $$\dfrac{10}{20} + \dfrac{1}{40} + \dfrac{10}{20}\times\dfrac{1}{19}\sum_{n=2}^{20}(1-\dfrac{1}{2}\times\dfrac{n}{10})$$ Which should equal the desired answer of 75%

Finally, I think a better way to write the summation would be $\dfrac{1}{20}\sum_{n=1}^{20}(10-floor(\frac{n}{2}))$ to calculate the whole space for when the second die rolls a 1-10. And not worry about the $\frac{1}{40}$ as its own component. Giving an equation for the probability of the 2x die being greater as: $$\dfrac{10}{20} + \dfrac{1}{400}\sum_{n=1}^{20}(10-floor(\frac{n}{2}))$$ You could also think of the equation in terms of outcomes instead of probabilities. There are 200 ways the second die is an automatic winner $$20\times10=200$$ And there are 100 ways the second die is a winner when it is a 10 through 20 $$\sum_{n=1}^{20}(10-floor(\frac{n}{2}))=100$$ for a total of 300 wins for the second die out of 400 total outcomes.

For your question about generalizing the number $M$ that you are multiplying the second die by, if $M$ evenly divides 20 (M=2, 4, 5, 10), then you can say that the probability that the second die is greater $$\frac{M-1}{M}+\frac{1}{400}\sum_{n=1}^{20}(\frac{20}{M}-floor(\frac{n}{M}))$$

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  • $\begingroup$ Brilliant, the solution you provided is not only correct, but believably correct, which is the best kind because it means I understand it! You are also correct in that the final equation you provided is probably more readable but as far as I'm aware you can't simplify the summation if it contains a floor operation within it, at least I'm not aware of a summation identity that could perform such a task. So if I was to use actually solve the equation it would be easier if I could simplify it using the identity $\sum_{n=m}^{k}n = \dfrac{(k - m + 1)(k + m)}{2}$ $\endgroup$ – Ryoku Jan 5 at 7:06
  • $\begingroup$ Considering the space created by 2 twenty sided die rolls, the probabilities that are generated by these problems will always be in increments of $\dfrac{1}{400}=0.0025$. If I wanted to increase the "precision" for lack of a better word, could I implement the above in such a way that we are comparing two infinite sided dice where the second result is $x2$ as in the initial problem? If I were to attempt to would I need to integrate over infinity? Or can I avoid using an integral and just continue using summations?My gut feeling is that I would have to take the limit of the equation at infinity $\endgroup$ – Ryoku Jan 5 at 7:10
  • $\begingroup$ Maybe parameterize the number of sides and then take the limit of the equation at infinity with respect to the number of sides. $\endgroup$ – Ryoku Jan 5 at 7:14
  • $\begingroup$ Yeah, I was thinking about your continuous function and was thinking that it was working because the remainders kind of cancelled out. I'm not sure how to do it, as these comparisons, by definition, create a step function when the resolution is the 20 faces of the die. But I think you are thinking the right way with the infinity-sided die. $\endgroup$ – turkeyhundt Jan 5 at 7:23
  • $\begingroup$ An infinite sided die might work... But I can't help but feel that its conceptually the same as simply taking an integral instead of a summation. The first two values of the equation $\dfrac{10}{20}+\dfrac{1}{20}$ are generated from the ratio between the factors of both rolls, in this case 1 and 2. So even if we left integer space we should be able to generate ratios from factors of 1 and 1.5 to be $\dfrac{2}{3}+\dfrac{3}{40}$ = $\dfrac{2}{3}+\dfrac{1}{120}$.... Thus we have freed ourselves from the space of 20 sided dice in the first part of the equation $\endgroup$ – Ryoku Jan 5 at 7:49

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