Average amount of die rolls to roll a specific number

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onelegout
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Hi,
I know very little about maths. I'm struggling to get my head around a question I was asked today, and I was wondering whether someone could explain it to me.

Question:
If a very large group of kids took it in turns to roll a die again and again until it landed on '1', on average how many rolls would they take each.

The question assumes a true and fair 6-sided die numbered 1 to 6, and I take 'very large' to mean 'large enough to be statistically viable'.

I am told the answer to this is '6'.

I have trouble understanding why the answer is 6 and not 3.5.

The way I look at it, the probability of rolling a 1 is 1/6. This means that probability states that the average child should roll a 1 in the first 6 attempts.

This also means that the 1 has an equal chance of appearing on each of the first 6 rolls.

To me that means that it should average out at 3.5 rolls.

I understand that it is perfectly possible however unlikely that the die will be rolled many times without landing on a 1, for example a child could roll it 10,000 times before it landed on a 1. This is extremely unlikely, but it IS possible.

What I want to know, is how I apply this improbable but possible situation to answering the question.

I'm sorry if this all seems a little jumbled or if I've got the wrong end of the stick completely, but I have little more than a basic GCSE maths education (which I failed dismally!)

Thanks,
Henry
 
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If a 1 were rolled every 3.5 times, it becomes impossible for each number (1-6) to appear with equal probability.
 
Welcome to PF!

Hi Henry! Welcome to PF! :smile:

Suppose the die is rolled 6,000 times altogether …

how many 1s would you expect, and how many times would a new child take over the throwing? :wink:

(incidentally, the usual way of solving this is the series 1/6 ∑ (5/6)n :biggrin:)
 
The answer is 6/# of kids, because there is a 1/6 chance of getting a 1. Which means, on average, for every six rolls of the die, a 1 should come up once.
 
The answer involves the definition of "average". For a specific sample of things, average has one meaning. For a "random variable", it has another and it is usually called the "mean" of the random variable instead of the "average".

The mean of a random variable with a discrete number of possible values (such as 1,2,3,..) is defined as a sum of terms. Each term is the product of a value (such as 3) times the probability that the value occurs.

If we let the random variable X be the number of thows up to and including the first throw that produces a 6, then X can take on the values 1,2,3,4,5,6,7,8,... etc.

Each value has a certain probability. For example the probability that X = 3 is the probability that the first two throws are not-6 and the third throw is a 6. So the term corresponding to that outcome is (3)(5/6)(5/6)(1/6).

So (as tiny-tim indicates) computing the mean requires computing an infinite sum, which begins

(1)(1/6) + 2(5/6)(1/6) + 3(5/6)(5/6)(1/6) + ...

This general type of random variable is called a "geometric random variable".
http://en.wikipedia.org/wiki/Geometric_distribution