My mother is organizing a fund-raising Fun Day for her local community centre, and was excited about the idea of a human fruit machine when my wife suggested it, but wanted some advice on making sure it would make a decent profit.

First some explanation of the concept. Instead of the spinning reels of a normal fruit machine, a human fruit machine comprises three (or more) volunteers, each holding a box containing a selection of fruit. A player pays the stallholder, who then gives a signal to the other volunteers, who simultaneously pick a fruit from their box and show it to the player. If the fruit are all the same, the player wins a prize.

At first glance this is a simply a nice chance to show off one's mastery of basic probability, but with real money and family honour at stake a bit of caution seems called for. As we all know, all models are wrong, and in this case it is clear that a random equiprobable selection of items from a collection is not necessarily a reliable stand-in for a human (especially a child, as has been suggested) with a box of different fruit, some probably more squashed than others.

A *useful* model of this partly human system is hard to construct, so
in the best scientific tradition it seems wise to keep the model and
simplify the system to match. Replacing the real fruit
with painted balls that can be randomised by shaking the box makes for
a system that is easier to predict statistically and also causes less
food wastage.

## Plan A

The first scenario suggested had three boxes, each with four different fruits. Players would be charged £1 a go, and would win £5 for three matching fruits or get their money back for two matching.

To calculate the probability of all three fruit matching, it doesn't matter what the first fruit is, but the others have to match it, so the probability is

\(1 \times \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}\)

The probability of two matching fruit is a bit more complicated. One way to calculate it is to start with the probability of the first two matching, but not the third:

\(1 \times \frac{1}{4} \times \frac{3}{4} = \frac{3}{16}\)

Then multiply by 3 to allow for the fact that the odd fruit out can be any of the three, giving \(\frac{9}{16}\) . In other words, a player will get their money back a bit more than half the time.

The expected payout in pounds is then

\(5 \times \frac{1}{16} + 1 \times \frac{9}{16} = \frac{14}{16}\)

It is clearer to communicate with numbers than with probabilities, as explained for example in Reckoning with Risk. In this case for every 16 games played, the stall takes in £16 and pays out on average

- £3 to each of 3 players who get three matching fruit
- £1 to each of 9 players who get two matching fruit

adding up to £14. That is the average, and it is easy to see that it only takes one extra lucky winner to wipe out that £2 profit.

## A safer bet

That doesn't seem like a surefire fund-raising scheme, so the next suggestion was to have 5 different fruits, so they are less likely to all match, not give anything back for two matching, and offer different prizes for different fruit:

- 3 bananas: £5
- 3 apples: £4
- 3 oranges: £3
- 3 pears: £2
- 3 lemons: £1

Now the probability of all three fruit matching is

\(1 \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{25}\)

and the average payout is only £3. So in 25 games we take in £25 and expect to pay on average £3 to one winner. This time we are much safer: it would take quite a few extra winners to wipe out the expected £22 profit.

The problem now, I think, is that only one person in 25 wins anything at all, which might make it less attractive and end up bringing in fewer players.

## A compromise

How can we modify the game so that more people have a little bit of success without winning back enough money to wipe out the profits. Giving back a smaller amount, say 50p, for two matching fruit would reduce the average payout, but it does have the undesirable effect that the stall would need a large float of 50p coins to start with. Once again the practicalities are not fully captured by the model.

My idea was instead of giving any money back to let the player have another turn. Since two matching fruit turn up roughly half the time, this means that a player will get on average about two turns: half the time they get one turn, a quarter of the time they get two turns:

\(1 \times \frac{1}{2} + 2 \times \frac{1}{4} + 3 \times \frac{1}{8} + \cdots = \sum\limits_{j=1}^{\infty} \frac{j}{2^j} = 2\)

Thanks to Wolfram Alpha for working out this one. So the expected payout is doubled, but remains fairly small, while the player at least gets a bit more entertainment for their money.

We could even tweak the prizes to give a small chance of a more substantial win. For example

- 3 bananas: £10
- 3 other matching fruit: £2
- 2 the same: have another go

I think the simplest way to calculate the expected payout here is to see that we only need to consider the payout in the final round, which is the first round where the result is either three matching fruit (resulting in a prize) or three different ones (no prize). So we can just calculate the expected payout for a single round given that the result is one of these two possibilities, i.e. anything except two matching. The probabilities are

- 3 bananas: \(\frac{1}{5} \times \frac{1}{5} \times \frac{1}{5} = \frac{1}{125}\)
- 3 other matching fruit: \(1 \times \frac{1}{5} \times \frac{1}{5} - \frac{1}{125} = \frac{4}{125}\)
- 3 different fruit: \(1 \times \frac{4}{5} \times \frac{3}{5} = \frac{12}{25}\)
- 2 matching: \(1 - \frac{1}{25} - \frac{12}{25} = \frac{12}{25}\)
- either 3 matching or all different: \(\frac{1}{25} + \frac{12}{25} = \frac{13}{25}\)

If we know that there were either three matching fruit or none, the probabilities given this are

- 3 bananas: \(\frac{\frac{1}{125}}{\frac{13}{25}} = \frac{1}{65}\)
- 3 other matching fruit: \(\frac{\frac{4}{125}}{\frac{13}{25}} = \frac{4}{65}\)

and the expected payout in pounds is

\(10 \times \frac{1}{65} + 2 \times \frac{4}{65} = \frac{18}{65}\)

Expressing this as numbers again for clarity, in 65 games we take in £65 and expect to pay out £18 as

- 1 prize of £10
- 4 prizes of £2

This also makes it clear that we can afford a few more lucky winners before our profit is wiped out.

## Another approach

An alternative would be to embrace the non-random element and fix the results, as Tom Leonard and friends did at their school fête. They didn't make much after paying for the fruit, but it sounds like they had fun doing it.

## The results

The Fun Day is on Monday. Sadly I won't be there to see how it works out, but I'll make sure I get a report and tell you about it here.

## Comments