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### Bubble Gum Cards

```
Date: 09/28/97 at 14:57:22
From: Rebecca Skrine
Subject: Probability

The bubble gum company decides to promote its gum by including in each
pack the photo of one of six rock music stars. Assuming that there are
equal numbers of photos of each of the six stars and that when you buy
a pack of gum your chances of getting any of the six photos are the
same, about how many packs of gum would you expect to buy to get all
six photos?
```

```
Date: 09/28/97 at 18:59:43
From: Doctor Anthony
Subject: Re: Probability

To get the first new photo you only have to buy one pack of gum.
Now the probability of getting a new photo is 5/6 with each pack you
buy. The expected number of purchases to get the second photo is
calculated from the sum of the series:

1 x prob of needing only 1 purchase + 2 x prob of needing two
purchases, and so on to infinity since in theory you could go on
getting the same old photo for evermore.

Expected puchases to get second new photo

= 1(5/6) + 2(1/6)(5/6) + 3(1/6)^2(5/6) + 4(1/6)^3(5/6) +
...........

= (5/6)[1 + 2(1/6) + 3(1/6)^2 + 4(1/6)^3 + ..........

The series inside the square bracket is of the form:

S = 1 + 2x + 3x^2 + 4x^3 + ......   where x = 1/6   Multiply S by x
xS =      x + 2x^2 + 3x^3 + ......
-----------------------------------  and subtract the two series
(1-x)S = 1 + x + x^2 + x^3 + .....    which is an infinite GP

(1-x)S =  1/(1-x)    and so  S = 1/(1-x)^2

= 1/(5/6)^2   =  (6/5)^2

And now the expected number of purchases is  (5/6)(6/5)^2

= 6/5

Having got two photos, the probability of a fresh photo is 4/6 with
each new purchase, and by an argument identical to that above, the
expected number of purchases to next new photo is 6/4. Then with
probability 3/6 of a new photo, the expected number of purchases will
be 6/3, then 6/2 and finally 6/1.

Total expected purchases to get complete set of photos is:

1 + 6/5 + 6/4 + 6/3 + 6/2 + 6/1

6[1/6 + 1/5 + 1/4 + 1/3 + 1/2 + 1/1]

= 6 x 49/20

= 14 and 7/10

So after about 15 purchases you should have a complete set of photos.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```

```
Date: 09/29/97 at 21:46:42
From: Anonymous
Subject: Re: Probability

Thanks so much for your help. Now I understand!
```
Associated Topics:
High School Probability

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