New answers tagged

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 ...


1

If the goal is to keep having fewer dice from being advantageous for avoiding fumbles, this change mostly succeeds. The pessimal pool size (fumble-wise) under the original mechanics ranges from three to seven; with the new mechanics it is consistently two, except when the target is very high (a 9 or a 10). There are some other effects, however: the ...


0

While there are certainly more robust answers on this page already, a friend told me that a simple Expected Value equation can illustrate the difference at a more "base level". This doesn't include any of the additional on hit effects (such as sneak attack, +x damage, etc.) that will certainly exacerbate the problem. It also doesn't incorporate AC which ...


0

Conclusion For a fighter with critical hits on 19-20, attacking with Advantage gives a 40% increase in damage for AC 13 foes, 64% for AC 18 foes, 86% for AC 23 foes. Attacking with the +2 bonus the GM is using in place of Advantage, these numbers are 14%, 22% and 48% respectively. In other words, a +2 Attack Bonus is one third to one half as effective as ...


-2

I realized, as I was going online to relearn probability to the point where I could be useful to you, that I wasn't sure if your DM does things the 3.5 way and only lets you crit if you roll a 19 or 20 naturally. If so, then your crit rate (success AND failure) remains unchanged, no matter the stats, and all you get is bonus to-hit. Average roll on a d8 is ...


9

Assumptions I'm going to be making the following assumptions, based on what you've already provided: 3rd level (since you get a crit on a 19 or 20) 16 Strength (no ASI to bump up to 18) Fighting a CR 3 creature (for base math) Average damage is 7.50 (4.50 from the die +3 Str mod) Average crit damage is 12.00 (4.50 per die +3 Str mod) Attack bonus is +5 ...


11

Everyone is making it hard on themselves - first ignore the modifier and use anydice to calculate output [lowest 1 of 2d20] - [highest 1 of 2d20] and display it as at most, giving this: # % -19 0.95 -18 2.75 -17 5.32 -16 8.54 -15 12.35 -14 16.64 -13 21.33 -12 26.35 -11 31.60 -10 37.02 -9 42.53 -8 48.05 -7 53.52 ...


10

Here's an AnyDice program to do what you want: function: greater A:n or B:n{ if A > B {result: 1} if A < B {result: -1} result: 0 } loop MOD over {-15..15}{ output [greater [highest 1 of 2d20]+MOD or [lowest 1 of 2d20]] named "[MOD]" } Setting the data view to "Transposed" shows the chances to lose as "-1", a tie as "0" and the chances to ...


2

First considering without modifiers When you have advantage the probability of getting a result of X can be governed by the equation P(x)=(2x-1)/400 When you have disadvantage the probability of getting a result of X can be governed by the equation P(o)=(41-2o)/400 The overall Probability of winning when you roll x, with advantage vs disadvantage, that ...


19

This is sort of brute-force, but the computers nowadays...: >>> all = [(i,j,k,l) for i in range(20) for j in range(20) for k in range(20) for l in range(20)] >>> p = lambda X: 1.0*len([t for t in all if max(t[0],t[1]) + X > min(t[2],t[3])])/len(all) >>> print '\n'.join("%3d: %.4f" % (X, p(X)) for X in range(-15,16)) -15: 0.0854 ...



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