A statistical analysis of marketing promotions.

ABSTRACT

This paper examines three types of common marketing promotions and their possible outcomes. The first of these concerns a scratch card give-away where customers can win cash or prizes simply by making purchases. The second involves a television game show where the promoter attempts to prolong viewer interest by changing the rules, but in doing so inadvertently make it easier for the contestant to win. Finally, a well-known type of competition is examined where a company includes a set of items in their product and makes it tempting for the consumer to purchase more of the product just to collect all items. In each case a statistical analysis is undertaken so that any promoter who undertakes such an activity will know what to expect.

1. INTRODUCTION

There is an endless supply of promotional ideas that organisations can use to assist them in obtaining increased sales. These can range from the simple to the complex, but in all cases a little time spent analysing the possible effects can provide big dividends and save headaches down the track. In particular, how can it be determined if a promotion has somehow 'gone wrong' and is costing a great deal more money than was first thought? Is it possible that the rules of the competition are imprecise or not what was intended? Is it conceivable that some consumers have been able to take an unfair advantage of the promotion to exploit vast amounts of money?

The only way to determine whether a promotion is running smoothly is to perform an analysis on the expected outcomes, including a calculation on the best and worst possible scenarios for the promoter. In this way it should be easy to tell about how much the promotion will cost and if it is going according to plan. The following sections consider three typical situations and give an idea of the type of statistical investigation that is required.

2. THE SCRATCH CARD PROMOTION

This type promotion endeavours to entice the customer to spend more than they otherwise would on a particular product or in an establishment. It involves a 'give-away' of cash or prizes by means of rewarding the customer with a free scratch lottery ticket provide they spend a minimum amount on other goods or services. This ticket has the possibility of the reward.

In a typical promotion, suppose that a customer has to spend $x in order to receive a free ticket and will receive one free ticket for each whole multiple of $x that is spent. Each ticket has y scratch panels that are each covered by obscuring silver paint. Under exactly k of these panels is the word WIN and under the remaining (n-k) panels are the words NO WIN. The customer must scratch exactly k of these panels and to receive their reward of $y must scratch only the k WIN panels. Otherwise they get nothing. The number of ways in which k WIN panels can be distributed among n possible panels is:

(1) Number of possible combinations = [sup.n][C.sub.k]

It is assumed that the tickets are printed such that there is an equal number of each of these combinations. For example, if N tickets are printed in total, from (1) there would be N/[sup.n][C.sub.k] of each combination printed.

Clearly for the competition to be viable, x > y/[sup.n][C.sub.k] or else the promoter is going to lose a great deal of money. The probability of a customer winning on a particular ticket is 1/[sup.n][C.sub.k] since there is no strategy available for playing.

With a reward of $y, the expected return (ER) in $ to the customer on a single ticket is:

(2) ER = y/[sup.n][C.sub.k]

so that the expected total payout (ETP) in $ by the promoter on N tickets is:

(3) ETP = N(ER) = Ny/[sup.n][C.sub.k]

The total amount spent by the customers in obtaining these N tickets is $Nx. If it can be assumed that customers might not have otherwise made the purchase to obtain these tickets, the profit (in $) on the promotion is (Nx - Ny/[sup.n][C.sub.k]) so that:

(4) Percentage profit to the promoter = 100 x {1 - [y/x([sup.n][C.sub.k])]} %

In this case the outcome is binomial (either the customer either wins $y or gets nothing), it can easily be shown that the variance 0-2 of the outcome on a single game is:

(5) [[sigma].sup.2] = [y.sup.2]([sup.n][C.sub.k] - 1] [([sup.n][C.sub.k]).sup.2]

Since there are a total of N tickets printed, from (3) and (4) the total payout for the promoter follows an approximate normal distribution with parameters:

(6) [micro] = ETP = Ny/[sup.n][C.sub.k]

(7)[[sigma].sup.2] = [Ny.sup.2]([sup.n][C.sub.k] - 1]) [([sup.n][C.sub.k]).sup.2]

The approximation to normal is better as N becomes larger, but since this would naturally be the case in any promotion it should be very close.

2.1 Example

A club runs a promotion in which they print N = 50,000 tickets, each with six covered silver panels. The customer, who must spend $5 at the gambling tables to receive a ticket, scratches exactly three panels on the card, and if they reveal the three WIN panels they receive $10. Otherwise they receive nothing.

Without doing any significant calculations, it is clear that the maximum amount (in theory) that the club would have to pay out is $10 x 50,000 = $500,000 (if all customers are correct) while the minimum is zero (if all customers are incorrect). The real payout, of course, will lie somewhere between these two values.

On a single ticket, from (1) the number of ways in which it can be scratched (n = 6, k = 3) is [sup.6][C.sub.3] = 20 and so the chance of the customer winning is 1/20 or 0.05. From (3), the expected payout from the club is therefore (y = 10) $10 x 0.05 = $0.50 per card. Since there are N = 50,000 cards, from (6) the total expected payout (in $) for the club in the promotion is:

[mu] = 50,000 x 0.50 = 25,000

while from (7) the variance is:

[[sigma].sup.2] = 50,000 x [10.sup.2] x (20 -1) = 237, 500 / [20.sup.2]

The standard deviation (in $) is therefore:

[sigma] = [square root of 237,500] = $487.34

Since the total payout follows a normal distribution, it is easy to construct a table of cumulative probabilities for the payout (to the nearest $10) that the club will have to make. These are shown in Table 1.

In fact, this very promotion recently took place in a club in Sydney, Australia. Even before all the tickets had been distributed, the total payout to customers had reached $30,000. As a result of their statistical analysis, the promoters felt something had gone terribly wrong, and indeed it had. It happened that one of the customers, by using an ultra bright light, had managed to see through the silver covering on the tickets and therefore knew exactly which squares to scratch off. It essentially cost him $5 for each ticket and was guaranteed a $10 return. In fact, he won many thousands of dollars all by himself and the club saved itself a great deal of money by immediately closing the promotion. The customer had not done anything contrary to the rules and was entitled to his winnings. But it demonstrated the practical nature of a scientific analysis of what the outcome should have been.

3. THE QUIZ SHOW PROBLEM

In this problem a sponsor wants to wring as much publicity as they can out of a promotion in which a quiz show contestant can win a valuable prize such as a car. The situation is that the show host presents to the weekly 'champion' contestant three identical looking boxes and announces that one of the boxes contains the keys to a new car while the other two boxes are empty. The host knows which box contains the keys while, of course, the contestant does not. Without any further information, the contestant has the same chance of choosing correctly if they select any box, where a success means winning the car belonging to the keys.

As a marketing ploy this can generate considerable excitement, not just from the contestant but also from the studio audience and the home viewers. Naturally, an enterprising sponsor, via the game show host, will want to prolong the excitement for as long as possible. In adopting what seems like an innocent strategy, the knowledgeable contestant can double their chances of driving home in their new car.

As it stands, the sponsor figures that it will have to part with a new car about once every three weeks, an outcome within its budget. In fact, it would not be unhappy for contestants to win on average every second week, but no more often than that. To this end, the host draws out the process by forcing the contestant to make a further decision. Once they have made their choice, the host then eliminates one of the other boxes by announcing that it does not hold the keys. There are now two remaining boxes--the one originally chosen and the remaining box not eliminated. The contestant is now asked whether they want to change their mind or stick with their original choice.

At first glance it seems that it is now a 50-50 proposition and the contestant's chance of success is the same whether they change their mind or not. But this is not the case as can be show by the use of some elementary probability theory.

Consider the problem in which there are three boxes (labelled A, B and C) from which the contestant is asked to select one. Suppose, without loss of generality, that the contestant selects Box C. The game show host, knowing in which box the key lie, reveal that the keys are not in Box A. The contestant is then given a chance to change their selection to Box B if they want to.

The question is whether the contestant should stay with Box C or switch to Box B. A first impression is that it doesn't make any difference whether Box B or Box C is now selected, since the field is now narrowed to these two boxes, each with a 50% chance of containing the keys. However, the solution is not quite that straightforward. In fact, if the contestant switches their selection to Box B, there is a two-thirds probability that they will be correct, this being their chance of success than if they stay with Box C. That is, in general, whatever box is first chosen by the contestant, their optimal strategy is to always switch their selection as soon as the host eliminates one of the two other boxes.

3.1 Verification of the solution

Consider the following two arguments expressed in different forms.

Let Pr(i) = the probability that the keys are actually in Box i, where i = A, B, or C

(a) At the outset, the probabilities for the contestant are Pr(A) = Pr(B) = Pr(C) = 1/3

Now the host reveals that Box A is empty, where it is assumed that the contestant knows that the host is aware in which box the keys are located and that the host will tell the truth. This is completely equivalent to the host putting the contents of Boxes A and B together and calling it Box B. It follows that Pr(B) = 2/3

(b) To state the argument in (a) in words, whatever selection is made by the contestant, two-thirds of the time the keys will be in one of the other two boxes. Since the host eliminates one of these two boxes, all that two-thirds chance transfers to the unchosen box, which must be the one that contains the keys. Hence by always switching their selection, the contestant will have a 2/3 chance of success. If the contestant happens to select the correct box in the first instance (which happens one-third of the time), then switching their selection would, of course, be the wrong move.

However, a popular alternative argument yielding a different solution is often advanced along the following lines. Once the host has eliminated Box A, there are only two remaining boxes B and C which are now equally likely to contain the keys. That is, Pr(B) = Pr(C) = 1/2, so that it doesn't make any difference whether the contestant switched their selection or not. (This is often the favoured stance of contestants who like to stay with their first choice regardless since they may feel that the host is simply trying to talk them out of winning the car.)

In fact, this latter argument is incorrect, since we must consider the conditional probabilities. This means we should actually be trying to find Pr(B|Host eliminates Box A) and Pr(C|Host eliminates Box A). It is assumed that, if the contestant selects Box C and is correct, then it is equally likely that the host will eliminate either Box A or Box B and that the contestant knows this.

The unconditional prior probability that the host will eliminate Box A is found from:

Pr(Host eliminates Box A)

= [SIGMA] Pr(Host eliminates Box A I Keys are in Box i) x Pr(Keys are in Box 0

= (0 x 1/3) + (1 x 1/3) + (1/2 x 1/3) = 1/2

Also,

Pr(Keys are in Box C and Host eliminates Box A)

= Pr(Host eliminates Box A I Keys are in Box C) x Pr(Keys are in Box C)

= 1/2 x 1/3 = 1/6

Then Pr(Keys are in Box C | Host eliminates Box A)

= Pr(Keys are in Box C and Host eliminates Box A)/ Pr(Host eliminates Box A)

= 1/6/1/2 = 1/3

It follows that Pr(Keys are in Box B I Host eliminates Box A) = 1 - 2/3 = 1/3 and so the arguments in (a) and (b) are correct.

The above discussion suggests that, if the contestants are made aware of the above arguments, then the sponsor will be giving away a car about two thirds of the time, double what they would have done if the host had not given a hint and about 30% more often than if contestants equally decided to change their choice.

4. THE CEREAL BOX COMPETITION

This is a special but common type of scenario in which a company has a promotion that is designed to make the customer purchase more of their product than they otherwise might. Although this can be aimed specifically at children, it really applies to all persons. The basic premise is that the company issues a 'set' of different items and places one of the items in boxes of their product. The consumer does not know which of the items in the set they will get until they purchase the product and open the packaging. The question of interest here is what is the expected number of items that must be purchased before the complete set is obtained? This situation is sometimes known as the 'cereal box problem' (since the items are often a set of toys found in a packet of cereal) and the aim here is to analyse it in the general and then demonstrate by using specific examples.

Suppose that the company decides to issue a set that consists of n different items. These items are distributed equally among packets of its product and so that if a single packet is purchased there is a 1/n chance that it will contain a particular item. The key here is that the statistical distribution applicable is the geometric distribution, this being the one that deals with the expected time until the first success.

Formally, if there is a series of independent trials of process in which the probability of a 'success' at any one trial is a constant p, then the time to achieving the first 'success' is given by the geometric distribution with parameter values:

(8) [mu] = 1/p

(9) [[sigma].sup.2] = (1 - p)/p

In our problem, the first packet purchased will always contain one of the items. For each subsequent packet, there is a (n-1)/n chance that the item found will be different to the first, and so from (8) the expected number of packets that must be purchased to find a different item is n/(n-1). Once two items in the set are obtained, there is now a (n-2)/n chance that a subsequent packet will obtain a different item from the first two items. In a similar way, from (8) the expected number of packets that must be purchased to find a different item is n/(n-2).

The expected total number of packets, E(n), that must be purchased to obtain the entire set of n items is given by:

E(n) = n/n + n/(n-1) + n/(n-2) + n/(n-3) + .... n/2 + n/1

(10) = n [n.summation over (i = 1)]1/i

It follows from (10) that:

E(n+1) = (n+1) [n+1.summation over (i = 1)]1/i

(11) = [(n+1)/n]E(n) + 1

The recursive relationship in (11) makes it easy to calculate values of E(n) for various values of n. Since, for example, E(1) = 1, then E(2) = 2E(1) + 1 = 3 and E(3) = 1.5E(2) + 1 = 5.5. For larger values of n we can use Euler's approximation for the partial sum of a harmonic series. That is:

(12)[n.summation over (i = 1)1/i [approximately equal to] [log.sub.e]n + 0.5772

Substituting (12)into (10), an approximate value of E(n)is given by:

(13) E(n) = n[[log.sub.e]n + 0.5772]

The values in (13) can be easily found using a pocket calculator. To test the accuracy of the approximation, Table 2 shows the values given by (13) for n = 1 to 10 and a comparison made with the exact values.

Table 2 shows that for values of n greater than 2 the approximation if very good. Using (13) we can easily find excellent approximate values of E(n) for large values of n. For example, E(20) [approximately equal to] 71 (exact is 71.95), E(50) [approximately equal to] 224 (exact is 225.0) and E(100) [approximately equal to] 518 (exact is 518.7).

As well as calculating the average time to collect all coupons, it is also useful to determine the variation of this time. To do this, we consider the variance of the geometric distribution given in (9) and apply it to the case where n coupons are to be collected. The variance in the sum of the time taken is the sum of the variances of the individual times. This yields:

(14) Variance of total time = [[sigma].sub.2]/ = [1 - (n-1)/n]/(n-1)/n + [1 - (n-2)/n]/(n-2)/n +.... + [1 - (1/n)]/ 1/n

= 1/(n-1) + 2/(n-2) + 3/(n-3) +.... + (n-1)/1

= [n-1.summation over (i = 1)i/(n - i)

From (14), the variance [[sigma].sub.2] and hence the value of the standard deviation [sigma] (by taking the square root) can easily be found. The values of [mu] and [sigma] for a selection of values of n is shown in Table 3.

The values in Table 3 can be used to calculate the approximate probability that the number of boxes that will have to purchased to collect all n coupons will be less than or exceed a specified value. This involves using the Central Limit Theorem which suggests that the total number of boxes that must be purchased follows an approximate normal distribution with the mean and standard deviation values shown in Table 3. The approximation works better for large values of n (say, at least 20) but still gives a rough estimate for lower values. A 95% confidence interval for the number of boxes can be found by taking 1.96 standard deviations either side of the mean. The results are shown (to the nearest integer) in Table 4.

For example, from Table 4 it follows that if 10 coupons are to be collected, while the average number of boxes that will need to be purchased is about 29, you would be lucky to get them all in less than 21 boxes and unlucky if it took you more than 38 boxes. For further discussions on this type of problem, see Litwiller and Duncan (1992), Maul and Berry (1992) and Wilkins (1999).

5. REMARKS

This paper has considered only three types of promotion but of course there are many more. In most cases the analysis required does not involve complex statistics but it is still essential that it be carried out to obtain a full understating of the possible consequences. There are a number of excellent publications that deal with general problems such as these, including Eastman and Wyndham (1999) and Haigh (2003).

TABLE 1 CUMULATIVE PROBABILITIES OF PAYOUTS FOR THE
CLUB PROMOTION IN THE EXAMPLE

Amount of    Probability the club will
payout ($)    pay out less than this
                      amount

24,380                 0.10
24,590                 0.20
24,740                 0.30
24,880                 0.40
25,000                 0.50
25,120                 0.60
25,260                 0.70
25,410                 0.80
25,620                 0.90
25,800                 0.95
26,140                 0.99

TABLE 2: COMPARISON OF APPROXIMATE AND ACTUAL VALUES
OF E(n) FOR n = 1 TO 10

n    Approximate value of   Actual value of
             E(n)                E(n)

 1            0.6                 1.0
 2            2.5                 3.0
 3            5.0                 5.5
 4            7.9                 8.3
 5           10.9                11.4
 6           14.2                14.7
 7           17.7                18.2
 8           21.3                21.7
 9           25.0                25.5
10           28.8                29.3

TABLE 3: THE MEAN AND STANDARD DEVIATION OF THE NUMBER OF
PACKETS THAT MUST BE BOUGHT FOR SEVERAL VALUES OF n

n     [micro]   [sigma]

5       11.4      2.53
10      29.3      4.32
20      72.0      7.21
30     119.8      9.48
50     225.0     13.23

TABLE 4: UPPER AND LOWER BOUNDS ON THE NUMBER OF PACKETS
THAT MUST BE BOUGHT FOR SEVERAL VALUES OF n

n    Lower bound     Average     Upper bound

5         6            11             16
10       21            29             38
20       58            72             86
30       101          120            138
50       199          225            251

REFERENCES

Eastman, R. and Wyndham, J., Why do buses come in threes?: the hidden mathematics of everyday life. Robson Books, London, 1999.

Haigh, J., Taking chances: winning with probability. Oxford University Press, Oxford, 2003.

Litwiller, B.H. and Duncan, D.R., "Prizes in cereal boxes: an application of probability", School Science and Mathematics, 92(4), 1992, 193-195.

Maull, W.S., and Berry, J., "Modelling the coupon collector's problem" Teaching Statistics, 19(2), 1997, 43-46.

Wilkins, J.L.M., "Cereal box problem revisited", School Science and Mathematics, 99(3), 1999, 193-195.

Author Profile

Professor John S. Croucher earned his Ph.D. in Operations Research at the University of Minnesota in 1973. He is currently a professor of management at the Macquarie Graduate School of Management at Macquarie University in Sydney, Australia. John is the author of 14 books on statistics, mathematics, crime, humour and management along with over 75 research papers in his fields of expertise. He is also famous for his extensive media appearances, including 8 years as a TV presenter on football telecasts.