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Can anyone tell me what are the odds that stage 4 will be reached?:

Stage 1: roll a 20 sided die results must be 13 or lower

Stage 2: roll a 20 sided die results must be 13 or lower

Stage 3: roll 2 separate 20 sided dice taking lowest dice and ignoring higher dice, results must be 13 or lower

Stage 4: roll 2 separate 20 sided dice taking lowest dice and ignoring higher dice, results must be 13 or lower

Source

I would like to learn the formula for adding consecutive successive dice rolls, but I cant find it anywhere on the net. This has to do with a dungeons and dragons game.

Thanks

Thank you for replies. I really appreciate it. This problem has been on my mind for a long time and I couldn't find any good answers online.

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  • $\begingroup$ Nice! D&D is good fun. Try anydice.com for a really nice dice probability calculator for anytime in the future you need this type of thing. I think the probability of rolling 13 or less with 2d20 is 87.75%. Subject to double checking from someone better at probability than myself. $\endgroup$
    – Old mate
    Apr 23 '15 at 22:42
  • $\begingroup$ Just curious why you refer to adding rolls. None of what you describe has any actual addition. $\endgroup$
    – Brian Tung
    Apr 23 '15 at 23:42
  • $\begingroup$ On your EDIT: Stage $1$ is right, at 65 percent. Stage $2$ is correct, if you mean the cumulative probability of passing stage $1$ and then also stage $2$. After that, your probabilities are not correct. I have given the correct probabilities in my answer. $\endgroup$
    – Brian Tung
    Apr 24 '15 at 0:11
  • $\begingroup$ By the way, that's a First Edition zombie, isn't it? $\endgroup$
    – Brian Tung
    Apr 24 '15 at 0:17
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Be careful of the distinction between rolling below a number, and rolling that number or lower. As @calculus noted, the probability of rolling $13$ or lower is $13/20 = 65$ percent, not $60$ percent.

One will roll $n$ or higher at normal odds with probability $(21-n)/20$. One rolls $n$ or higher at advantage with probability $1-(n-1)^2/400$. And one rolls $n$ or higher at disadvantage with probability $(21-n)^2/400$. For $n = 13$, this yields $8/20 = 40$ percent, $1-144/400 = 256/400 = 64$ percent, and $64/400 = 16$ percent, respectively. I'm not sure where the linked table got $63.9$ percent.

For your problem, your phases are equivalent to failing to roll a $14$ at normal (first two phases) and then failing to roll a $14$ at disadvantage (second two phases). Those probabilities are $1-(21-14)/20 = 1-7/20 = 13/20 = 65$ percent, and $1-(21-14)^2/400 = 1-49/400 = 351/400 = 87.75$ percent, respectively.

To compute the probability of joint events (e.g., one must pass phases $1$, $2$, and $3$ to reach phrase $4$), one multiplies the probabilities associated with each phase. That is, the probability of reaching phase $4$ is the product of the probabilities of passing phases $1$ through $3$; the probability of passing phase $4$ is the product of the probabilities of passing phases $1$ through $4$.

In other words, you would multiply $(13/20)(13/20)(351/400) \doteq 37.074$ percent probability of reaching your phase $4$ (i.e., failing the first three rolls), and $(13/20)(13/20)(351/400)(351/400) \doteq 32.533$ percent probability of passing through phase $4$ (i.e., failing all four rolls).

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  • $\begingroup$ Ah so clear thank you very much!!!! $\endgroup$
    – Dunlop
    Apr 23 '15 at 23:55
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The probability for stage 1 and 2 is $p_1= p_2=\frac{13}{20}\cdot 100\%=65\%$. To calculate the other two probabilities you can make a $20 \times 20$-table. $\begin{array}{|c|c|c|c|c|c|c|} \hline & 1 & 2 & 3 & 4 & \ldots & 20 \\ \hline 1 & & & & & & \\ \hline 2 & & & & & & \\ \hline 3 & & & & & & \\ \hline 4 & & & & & & \\ \hline \vdots & & & & & \\ \hline 20 & & & & & & \\ \hline \end{array} $

Then you can mark the cells, which fullfill the condition in stage 3 and 4 and count them.

Let y be the number of marked cells. The probability for stage 3 and 4 is $p_3=p_4=\frac{y}{400}$

400 in the denominator are the number of all cells (=20x20)

The chance of reaching stage 4 is $p_1\cdot p_2\cdot p_3\cdot p_4$

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  • $\begingroup$ I did the 20x20 chart and found 169 possible rolls that fullfill the condition (both dice being below 13. So the formula would be (13/20)x(13/20)x(169/400)x(169/400)=0.0754 or 7.5% of the time the condition will be fulfilled after rolling d20, d20, 2d20, 2d20 with no result over 13? $\endgroup$
    – Dunlop
    Apr 23 '15 at 23:29
  • $\begingroup$ @Dunlop The probability for $p_3$ and $p_4$ is not right. Both dices have to be $\leq 13$ Thus you do NOT have to mark the cells where both dices are $\geq 14$. The number of NOT marked cells is 7x7=49, because the numbers of cells from 14 to 20 is $7(=20-14+1)$. Thus the number of marked cells is 400-49=351. And $p_3=p_4=\frac{351}{400}$ $\endgroup$
    – callculus42
    Apr 23 '15 at 23:49
  • $\begingroup$ Thanks, I see where I messed up on the chart. I didn't mark the results with where one dice was greater than 13. thus my return was 169 instead of 351. $\endgroup$
    – Dunlop
    Apr 24 '15 at 0:13
  • $\begingroup$ Nice, that you get it now. You are welcome. $\endgroup$
    – callculus42
    Apr 24 '15 at 0:31

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