dice probability

how do you calculate the probability of rolling a particular number on a dice on a particular roll. For instance you want a six but the first roll is not, the second roll is not but third roll is a six.

3 Answers

Assuming I understand what you want: this is called Geometric probability. In this case the probability not to toss $6$ is $\frac{5}{6}$. Assuming the dice is fair and and rolls are independent all you need is $$P(X=k)=(1-p)^{k-1}p$$ where $1-p=\frac{5}{6}$.

You have $5$ choices for the first and second roll but $1$ choice for the third roll, and $6^3$ outcomes so the probability is $(5\times 5\times 1)/6^3$. In general this becomes $5^k/6^n$ where $n$ is the number of tosses and $n-k$ the number of required $6$ faces.

• so is this correct? 25./(6.**3). I used an algorithm that used random choice to calculate the probability and got 0.116051 but the answer was supposed to be in reduced fraction so I thought my logic was wrong somehow but the result of 25./(6.**3)= 0.11574... Am I still missing something? Commented Apr 6, 2014 at 18:05
• Well the values are really close. The other answers are also saying its $5^2/6^3$. Commented Apr 6, 2014 at 18:13
• I must have interpreted the question incorrectly somehow. I got it wrong but could not figure out why! Commented Apr 6, 2014 at 18:14
• Can you explain this algorithm you are using, and why you are certain it gives the correct result? Commented Apr 6, 2014 at 18:16
• @PadraicCunningham How about you write a program that lists all the possible outputs and selects the ones whose first $2$ entries are not $6$ while the $3$rd is, it should not be too hard. Commented Apr 6, 2014 at 18:21

The probability of not getting a number $a$ on a six-sided dice is $\mathrm{Pr}(a)=\dfrac{5}{6}$. The probability of getting $a$ in the first roll is $1-\mathrm{Pr}(a)$. However, see that whether or not you get $a$ in the first roll does not affect the second roll, i.e., rolling a dice is an event independent of the previous roll(s). Thus, the probability of getting $a$ in the $n$th roll is equivalent to getting $a$ in the first roll, i.e., $1-\mathrm{Pr}(a)$. (However, if you want the probability of getting $a$ in all $n$ rolls, the probability is $1-(\mathrm{Pr}(a))^n$.)

• The problem wanted the probability of not getting it on the first two rolls but getting it on the third roll exactly. As I said in another comment I used an algorithm to calculate the P over 1000000 trials but as it was supposed to be in reduced fraction I could not use my result to answer but using the logic how I see it here I don't get an answer where I can use reduced fraction either. I think I have been out of school for too long! Commented Apr 6, 2014 at 18:13