Adding probability of multiple dice rolls 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. 
 A: 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).
A: 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  $p_1\cdot p_2\cdot p_3\cdot p_4$
