Trials until First Success: Challenging Probability Problem


On the average, how many times must a die be thrown until one gets a 6?


Solution:

Let pp be the probability of a 6 on a given trial. Then the probabilities of success for the first time on each trial are (let q=1pq = 1 - p):

TrialProbability of success on trial
1p p
2pq pq
3pq2 pq ^2
..
..
..

The sum of the probabilities is

p+pq+pq2+=p(1+q+q2+)=p1q=pp=1\begin{align*} p+pq+pq^2+\ldots & = p\left( 1+q+q^2+\ldots \right) \\ \\ & = \frac{p}{1-q} \\ \\ & = \frac{p}{p} \\ \\ & = 1 \end{align*}

The mean number of trials, mm, is by definition,

m=p+2pq+3pq2+4pq3+m = p + 2pq + 3pq^2 + 4pq^3+ \ldots

Note that our usual trick for summing a geometric series works:

qm=pq+2pq2+3pq3+qm = pq + 2pq^2+3pq^3 + \ldots

Subtracting the second expression from the first gives

mqm=p+pq+pq2+m-qm=p+pq+pq^2+\ldots

or

m(1q)=1m\left( 1-q \right) = 1

Consequently,

mp=1mp=1

and

m=1/pm=1/p

We see that p=1/6p=1/6, and so m=6m=6.

On the average, a die must be thrown 6 times until one gets a 6.


Advertisements
Advertisements