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Fairground spinner game
By Clare Nicholson (P1732) on Wednesday, December 15, 1999 - 10:02 pm:

There is a stall at the fairground and it costs 25p to spin a spinner which has 8 equal sections. On 3 of them you get 0p if you land on it, on 2 of them you get 10p if you land of it, on 1 of them you get 25p, one 50p,and one £1.If in one night the spinner gets spinned 960 times what will be the expected profit or loss for the stall holder?

CN

By Richard Samworth (Rjs57) on Thursday, December 16, 1999 - 07:45 pm:

Clare,

This sounds like homework to me, so I'd better not spoil your fun by doing the whole thing for you! But here's the general idea:
Think about one spin. Your expected winnings are given by the formula

Expected winnings = p1×x1 + p2×x2 + p3×x3 + p4×x4 + p5×x5,

where x1,x2,x3,x4 and x5 are your winnings if you land in each of the five different areas, and p1,p2,p3,p4 and p5 are the corresponding probabilities of landing in these areas.
You are given x1,x2,x3,x4 and x5 in the question, and should be able to work out p1,p2,p3,p4 and p5 quite easily.
So your expected profit for one spin (which equals the stall holder's expected loss for one spin) is given by:

Expected profit = Expected winnings - 25.
(In this formula, the expected winnings should be left in pence).

Now, note that your expected profit over 960 spins is 960 multiplied by your expected profit for one spin. This should give you your final answer.

Please write back if any of this is unclear.

Richard

By Clare Nicholson (P1732) on Friday, December 17, 1999 - 08:03 pm:

Richard,

I worked this one out after I sent my email but I think I did it in a different way. The probability of landing on one section is 1 over 8. So in eight spins he/she would have landed on each one.

If you divide 960 by 8 you get 120. So you times the total cost of all the numbers on the spinner by 120 add you get the cost he gives away.

This is £1.95. We can call this X.

You take X away from the cost of each spin times 8.

This equals £2.00. We can call this Y.
So if you take X away from Y you get £0.05 (Z)
So Y-X=Z
Z*120=£6 (the answer).

I worked it out but I don't understand totally about your p1's and x1's please explain. Thanks for all your help.

CN

PS: This was misunderstood Classwork not homework.

By Richard Samworth (Rjs57) on Friday, December 24, 1999 - 03:23 pm:

Clare,

Sorry for taking such a long time to get back to you. You do indeed get the right answer using your method, but it's probably not the easiest way to do the problem. Maybe I should explain a little about expectation...

Let's suppose you roll a dice, and let X be the value obtained. So X can take any integer value between 1 and 6, and takes each value with probability 1/6. X is an example of a random variable, which you can think of as a variable which takes a specified set of values with given probabilities. Note that the probabilities don't all have to be the same (think of a random variable representing the sum of the scores on each of two dice - it's more likely that you'll get a score of 7 than one of 12).

In your fairground problem, if X represents the amount won by the spinner on one spin, X can take the following values:
The value 0 with probability 3/8
The value 10 with probability 2/8 = 1/4
The value 25 with probability 1/8
The value 50 with probability 1/8
The value 100 with probability 1/8

Note that I have totally defined the random variable, by specifying its possible values and the corresponding probabilities. The values 0,10,25,50 and 100 are what I called x1,x2,x3,x4 and x5 in my previous message, and the probabilities 3/8,2/8,1/8,1/8 and 1/8 are what I called p1,p2,p3,p4 and p5 before.

Returning to the general situation...
When a random variable X takes a finite number of values x1,x2,...,xn with probabilities p1,p2,...,pn, we define the expected value of X, which we write as E(X), by

E(X) = p1×x1 + p2×x2 + ... + pn×xn.

This is called the expected value because if we observed X on many different occasions (for example if we span the spinner many times), then we would 'expect' the average value of X to be close to its expected value! Try to think about why this is the case.

In the fairground problem, X does take only finitely many different values (5, in fact!), so the formula for the expectation of X given above applies:

E(X) = (3/8)×0 + (2/8)×10 + (1/8)×25 + (1/8)×50 + (1/8)×100 = 195/8.

This is the spinner's expected winnings for one spin, so if we take off 25, we get his/her expected profit/loss:
195/8 - 25 = -5/8 (i.e. a loss of 5/8)

Now multiply by 960 for the number of spins:

-5/8 × 960 = -600.

So the spinner's expected loss (i.e. the stall holder's expected profit) is 600p = £6, which is the answer you had.

I hope that by doing the problem this way, I have shown you that it is easy to use the formula for expectation in other (possibly more complicated) situations, and that you now understand my notation of x's for the values on a spin and p's for the probabilities.

Please write back if you'd like me to explain anything in more detail.

Richard