Sean finds a coloured ticket inside the wrapper of a bar of chocolate. A message says that the card comes in four different colours and encourages him to collect all four. Assuming there is an equal chance of getting any one of the colours, what is the expected number of chocolate bars Sean must buy to get all four?

Answer: This one proved a bit too chewy for most readers, but the answer is 8.33, or nine bars. Extremely well done to: **Stephan Paischer**, head of product management special products, Semperit AG Holding, Vienna, Austria; **Federico Fiorini**, managing director, MAI Italia srl, Isola Rizza (VR), Italy; **Michele Girardi**, Scame Mastaf Spa, Suisio, Italy – and everyone else who had a go.

*Solutions*

*Our top trio came up with some really neat answers:*

*Stephan Paischer*

For the first colour it’s easy – one box (4/4) as any colour will fit; for the second it’s just 3 out of 4 that will give a suitable color (4/3) as one is already selected from the first round; for the third 2 of 4 (4/2) and for the last one 1 of 4 boxes (4/1).

So 4/4+4/3+4/2+4/1 = 8.33

*Federico Fiorini*

Sean has collected the first colour from the first bar. Then he’ll have 75% chance to get a different colour from the second one (3 new colours out of 4 colours available) then with a simple proportion he can expect a new colour in 1/0.75=1.33 times.

After that he’ll have 50% chance to get a different colour from the following one (2 new colours out of 4 colours available) then he can expect a new colour in 2 chocolate bars. Once he got the third he’ll have 25% chance to get the remaining colour (1 out of 4) then he will need to eat 4 chocolate bars to get the final one.

Total expected number of boxes = 1 + 1.33 + 2 + 4 = 8.33

That said… there is always that lucky scenario where he will get all 4 colours in 4 boxes only BUT also the very unlikely but not impossible chance he will have to try infinite chocolate bars…

*Michele Girardi*

The exact number is 8.33, as explained below .

The expected number of boxes is the sum of the number multiplied the probability

4*P(4)+5*P(5)+6*P(6) + .....

Let's call the colours A, B, C, D

For N=4 , let's suppose that the last colour to appear is A the probability that A doesn't appear in the first 3 boxes is (3/4)^3, the probability for A to appear in the fourth box is 1/4 But not all the BCD combinations in the first 3 boxes are useful, since one of the colours may be missing, for example C in BBD the total number of cases is 3^3 = 27 the bad ones are :

- the ones with missing B , i.e all the terns based on C and D , 2^3= 8

- the ones with missing C , i.e all the terns based on B and D , 8

- the ones with missing D , i.e all the terns based on B and C , 8

the total is 24, but we must subtract 3, since the combinations BBB,CCC,DDD are counted twice The number of good combinations is 27-21 = 6 the fraction of good combinations is (27-21)/27 or 1-21/27 So the probability P(4, A last) = (3/4)^3* ( 1-21/27)*1/4= 0,023 this can be repeated for B,C and D, so P(4) = 4*0,023 = 0.094 the general formula is P(N)= (3/4)^(N-1) * ( 1 - (2^(N-1)*3-3) / 3^(N-1)

* 1/4 *4

The sum 4*P(4)+5*P(5)+ .....+ N*P(N) converges to 8.3

N P(N) N*P(N) cumulative

4 0,094 0,3750 0,4

5 0,141 0,7031 1,1

6 0,146 0,8789 2,0

7 0,132 0,9229 2,9

8 0,110 0,8818 3,8

9 0,088 0,7960 4,6

10 0,069 0,6924 5,3

11 0,053 0,5873 5,8

12 0,041 0,4893 6,3

13 0,031 0,4023 6,7

14 0,023 0,3275 7,1

15 0,018 0,2645 7,3

16 0,013 0,2124 7,5

17 0,010 0,1696 7,7

18 0,007 0,1349 7,8

19 0,006 0,1069 7,9

20 0,004 0,0845 8,0

21 0,003 0,0665 8,1

22 0,002 0,0523 8,1

23 0,002 0,0410 8,2

24 0,001 0,0321 8,2

25 0,001 0,0251 8,2

26 0,001 0,0196 8,3

27 0,001 0,0152 8,3

28 0,000 0,0119 8,3

29 0,000 0,0092 8,3

30 0,000 0,0071 8,3

31 0,000 0,0055 8,3

32 0,000 0,0043 8,3

33 0,000 0,0033 8,3

34 0,000 0,0026 8,3

New, maybe easier, teaser on Monday.