Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose you toss three fair six-sided dice simultaneously until you observe a six. So if none of the three dice are a six, then you toss them simultaneously again. Find the probability that the first occurrence of a six requires n tosses.

Since P(a six first toss) =$1/6+1/6+1/6=1/2$ shouldn't it be $n/2$ after $n$ tosses?

share|improve this question
2  
The probability of a 6 on $any$ toss is $p$, which you have to calculate (hint: it's 1-(prob. of no 6)). Now, think of the Geometric distribution. –  David Mitra Feb 2 '12 at 18:19
    
@David: Good hint(s). –  Did Feb 2 '12 at 18:27

2 Answers 2

up vote 3 down vote accepted

The probability of getting a six on the first toss is not $1/2$. If the reasoning that led you that conclusion were valid, you could argue that with six dice you’d be certain to get a six on the first toss, which clearly isn’t the case.

The easiest way to calculate the correct probability of getting a six on the first toss is to calculate the probability of not getting a six and subtract that from $1$. In order not to get a six, you must roll a not-$6$ on each of the three dice. The probability of rolling a not-$6$ on one die is $5/6$, and the dice rolls are independent, so the probability of rolling a not-$6$ on all three dice is $(5/6)(5/6)(5/6)=125/216$, and the probability of getting at least one six is therefore $1-125/216=91/216$, or a little over $0.42$.

Now in order to get a six for the first time on the $n$-th toss, you must get no sixes on each of the first $n-1$ tosses, and then you must get at least one six on the $n$-th toss. The probability of getting no six on one toss is, as we just saw, $125/216$, so the probability of getting no six on each of $n-1$ tosses is $$\left(\frac{125}{216}\right)^{n-1}\;.$$ This needs to be followed by a successful toss, an event whose probability is $91/216$, so the probability of getting your first six on the $n$-th toss is $$\left(\frac{125}{216}\right)^{n-1}\left(\frac{91}{216}\right)=\frac{91\cdot125^{n-1}}{216^n}\;.$$

share|improve this answer

The first toss:

$$P(\text{no 6}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$$

We generalize. The chance you throw a six after $n$ turns is:

$$P (\text{six in }n\text{th throw}) = \left(\frac{125}{216}\right)^{n-1} \frac{91}{216} = \frac{91 \cdot 125^{n-1}}{216^n}$$

share|improve this answer
    
I saw it as soon as I posted. Edited –  Hidde Feb 2 '12 at 18:32

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.