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1

In 2e, each point of a character’s ability score gives a +5% to the chance of success. In 3e, every two points of a character’s ability score gives a +5% to the chance of success. Which is a fundamental difference. If you want scores to count the way they do in 2e but you want roll-over like 3e, you’d want to use score - 10 as the modifier to the roll. ...


3

With 2 coins, there's an 11/36 chance that at least one die is a 1. With 3 coins, there's a 20/36 = 5/9 chance that at least one die is a 1 or 2. With 4 coins, there's a 27/36 = 3/4 chance that at least one die is a 1-3. With 5 coins, there's an 32/36 = 8/9 chance that at least one die is a 1-4. The expected number of steps is just the inverse of the ...


6

Here's some Python to lend some experimental results to the theoretical results others have added: import random def die_roll(): dice = [random.randint(1, 6), random.randint(1, 6)] return dice def play_game(starting_coins): coins = starting_coins rolls = 0 while coins < 6: dice = die_roll() rolls += 1 if ...


4

About 7½ rolls This is a pretty simple problem, because your system only stores a tiny bit of state -- the number of coins. Once we know the state, we know the behavior of your system from that point onward. Which means we can break your system down into simple problems, then add them up to find the average you want. So we'll solve the simplest ...


41

Answers Below are the expected number of steps/rolls to get to 6 coins: Steps To Go From 2 Coins to 6 Coins = 7.52631 Steps To Go From 3 Coins to 6 Coins = 4.25833 Steps To Go From 4 Coins to 6 Coins = 2.45833 Steps To Go From 5 Coins to 6 Coins = 1.125 Solution The best way to solve this problem in general is not Anydice, but with a tool called Markov ...


11

You can do this analytically using geometric distributions. Anydice runs into problems with these because they are infinite and computer programs don't like infinite loops. The probability distribution to get below the number of coins you currently have is a geometric distribution and we can use anydice to get the probabilities of each of these from n=2 to ...



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