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14

Advantage. If one of the d20s hits you hit once. If both of the d20s hits you hit once. Separate. If one of the d20s hits you hit once. If both of the d20s hit you hit twice. Hope this helps and answers the question


40

It depends on your situation (typically, getting Sneak Attack is key) Let's get something out of the way first: in the simple analysis, attacking twice is usually much better than attacking once with advantage. In both situations, you roll two d20s: when you attack twice, you hit each time one rolls a hit. On a single attack with advantage, you hit if either ...


13

Polynomial-time method without custom functions Here's an example of a +5 modifier against DC 10, 15, and 20. AnyDice link. set "explode depth" to 100 M: 5 output 3 + [explode d20+M < 10] + [explode d20+M < 15] + [explode d20+M < 20] The x-axis is the number of tries, and the y-axis is the chance of taking at most that many tries to ...


14

Note: While this answer works, HighDiceRoller's solution is a lot simpler, faster and more elegant. I encourage anyone reading this to upvote it. Here's a simple recursive AnyDice function that should hopefully do what you want, and allow fairly easy modification to explore variations of this mechanic: function: roll S:n successes after K:n on DIE:d in N:n ...


4

The pigeonhole principle The probabilistic generalisation of the pigeonhole principle tells us that for \$n\$ pigeons randomly assigned to \$m\$ pigeonholes with equal probability, the chance that one hole will contain at least 2 pigeons is: $$ 1-{{(m)_n}\over{m^n}} $$ where \$(m)_n\$ is the falling factorial. For us, our pigeons are the number of dice and ...


4

You're going to have a fairly limited range where success (of some level) is possible, but isn't guaranteed. For 1d6, it's obviously impossible to roll doubles. 7d6 guarantees at least one double by the pigeonhole principle. So only from 2d6 to 6d6 do you have any uncertainty at all. Because my conditional probability skills have atrophied, I just brute-...


1

For a numerical answer, we can use the AnyDice code I just recently wrote for another answer, which calculates the highest number of identical values in a roll of dice: function: highest number of matches in ROLL:s { MAX: 0 loop I over ROLL { COUNT: ROLL = I if MAX < COUNT { MAX: COUNT } } result: MAX } output [highest number of matches ...


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