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Say I roll a $6$ sided dice and I want to roll a $6$. what is the probability that I will have rolled the number I want after $6$ rolls?

I have been using this: $\displaystyle1-\left(1-\frac{1}{x}\right)^y$

where $x$ is the number of sides and $y$ is the amount of rolls, so it would be $\displaystyle1-\left(1-\frac{1}{6}\right)^6$ for rolling a specific number in $6$ rolls, which is $\approx66.5\%$ is this the correct way of calculating the probability of something like this, if not what is the proper way?

i'm not really sure why that formula works(if it does) so some elaboration on that would be nice.

sorry for lack of technical language

thanks in advance

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  • $\begingroup$ After exactly $6$ rolls or after at most $6$ rolls? $\endgroup$ Aug 17 '16 at 5:08
  • $\begingroup$ what i'm trying to ask is: what is the chance that by 6 rolls, you will have obtained the number that you are looking for. wether you get it on the first roll or the last, what is the probability that you will have it by the 6th roll. $\endgroup$ Aug 17 '16 at 5:18
  • $\begingroup$ it's a very good way of tackling the problem. This is a case where calculating the probability of the event you are interested in, not happening seems easier than finding the probability it does happen. $\endgroup$
    – Cato
    Aug 17 '16 at 9:21
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Simply put, your computation of the probability is correct. If the number you want (let's call it $n$) does not appear even once, every roll must have resulted in a number other than $n$. For an individual roll on an $x$-sided die, the probability of this happening is $1-\frac 1x$. After $y$ rolls, the chance of not getting $n$ in any roll must be $(1-\frac 1x)^y$. Since this accounds for every sequence off rolls in which $n$ appears and no others, the probability of seeing $n$ at least once must be $1-(1-\frac 1x)^y$.

You can also use a combinatorial argument. There are $x^y$ possible sequences of rolls, all of which are equally likely. If we stipulate $n$ cannot appear, there are only $(x-1)^y$ possible sequences. The probability of $n$ not appearing is thus $\frac{(x-1)^y}{x^y}=(1-\frac 1x)^y$, leading to the same result as before.

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I would say that it's easier to calculate the chance that you never roll this number. So that all 6 rolls are not the number you want. This chance is (5/6)^6 = 33.49%. Now to calculate the chance that you roll your number at least oncentre within 6 rolls you can take 1-(5/6)^6= 66.51%. (I don't have an approximate sign)

This is exactly the same as your equation. I just used 5/6 instead of 1-1/6.

As for the explanation. All the possibilities that exist can be added up to 100%. What we do here is we calculate the chance for all the possibilities that you won't get what you are looking for. Now if you subtract these you will be left with all the possibilities you are looking for.

The reason we sometimes use this method is because the conjugate is much easier to calculate, like in this situation.

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Split it into disjoint events, and then add up their probabilities:

  • The probability to obtain this value after exactly $\color\red1$ attempt is $\left(1-\frac16\right)^{\color\red1-1}\cdot\frac16$
  • The probability to obtain this value after exactly $\color\red2$ attempts is $\left(1-\frac16\right)^{\color\red2-1}\cdot\frac16$
  • The probability to obtain this value after exactly $\color\red3$ attempts is $\left(1-\frac16\right)^{\color\red3-1}\cdot\frac16$
  • The probability to obtain this value after exactly $\color\red4$ attempts is $\left(1-\frac16\right)^{\color\red4-1}\cdot\frac16$
  • The probability to obtain this value after exactly $\color\red5$ attempts is $\left(1-\frac16\right)^{\color\red5-1}\cdot\frac16$
  • The probability to obtain this value after exactly $\color\red6$ attempts is $\left(1-\frac16\right)^{\color\red6-1}\cdot\frac16$

Hence the probability of obtaining this value within $6$ attempts is:

$$\sum\limits_{n=1}^{6}\left(1-\frac16\right)^{n-1}\cdot\frac16\approx66.51\%$$

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    $\begingroup$ I think that by multiplying that by {1 - (1 - 1 / 6 )} / {1 - (1 - 1 / 6 )} (which is equal to 1) you can rearrange to get 1 - (1 - 1/6) ^ n $\endgroup$
    – Cato
    Aug 17 '16 at 9:21
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    $\begingroup$ @AndrewDeighton: Of course. It is the same result as in the original question. $\endgroup$ Aug 17 '16 at 9:24

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