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2

There's another way of looking at this, which works particularly well with a ten-sided die. Think of the single-digit numbers from 0-9. How many of them have a 1? One, out of ten, or 10%. Now, out of all of the combinations of two digits, from 00-99, how many have any 1s? Well, there's 01, plus all ten of the numbers from 10-19, and then 21, 31, 41, 51, 61,...


1

Whenever you have an probability experiment where there are exactly two outcomes and you can assign a probably to the one of the outcomes, it's called a binomial experiment: True, false, yes, no, black, white, etc. In this case, you get a one or you don't. The number of dice would constitute the number of trials, since it wouldn't matter if you rolled a ...


8

Think of that way: if you roll a die, there's a 10% chance of it being a 1. If you roll two dice, it's the same as if you first rolled one of them (since both rolls are independent of each other), then the other. Say you make the experiment 100 times (and equate occurrences with percents). With a single die, 10 times you lose, 90 times you win. With two ...


4

Putting together a layman's answer/example. To further simplify, I will use the example of a coin flip to emphasize the results. Given a single coin, you have a 50% change of flipping a "heads". Given a second coin (or flipping the same coin a second time), you now have a separate and independent 50% chance for each coin flip. With two coin flips you now ...


61

Yes it does. Your instinct is right. The more dice, the more likely you are to roll some 1s. If I'm reading you right you're just interested in whether any 1s appear in your rolled pool. It may not seem obvious but the easiest way to think of this is to model the probability of rolling all 2s-through-10s. That is $$P(\text{not }1)=\frac 9 {10} = 0.9$$ ...


10

Yes, the odds of rolling at least one of anything increase as the number of dice goes up. I don't know the actual function to determine how those odds increase, but as an example, here's an Anydice graph showing a 65.13% chance of getting at least one 1 on 10d10: http://anydice.com/program/8b09 and only 27.10% chance of getting at least one 1 on 3d10: http:/...


0

Probability The answers provided effectively cover the probability for every result, 1 through 20, for advantage/disadvantage with 2d20. For completeness, the probabilities follow: Expected Value When rolling 2d20, and keeping the Maximum value from each of the 400 permutations, the expected value is 13.825. By contrast, the expected value when you ...



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