mathematicalchances

[Vimto]

A company gives away a collectors card in every packet of its product. It could be a "gum packet", a "cigarette packet", a tea packet or whatever. The company distributes them fairly and randomly keeping no numbers in short supply: let's say there are 50 cards to collect a complete set. What is the number of packets of the product that a customer must buy in order to have an even chance of completing a full set. I would appreciate the maths involved in working out the answer to this problem if they are explained in an idiot type way as my mathematics are way back in the 1960's and a little rusty but I did reach as far as ballistics if that gives a clue to my standard.

[Physicsgirl]

working on the other but his chances of doing it with only 50 buys are (50!) divided by 50 to the power of 50. This is no use to you whatsoever.
By even chance, do you mean 1 in 2?
P.S. 50!= 50 factorial = 50*49*48*47.....etc.

[Physicsgirl]

that's 1 in 257 million million million

[GetOverIt]

I've found this - to sum it up its says

For my sets of 50 Player's Motor Car cards the answer is 224.5, and I should expect to have had to buy 225 cards to make up my set of 50.

http://plus.maths.org/issue37/outerspace/index .html

[Vimto]

Thank you "physics", "GetOverIt" has found exactly what I wanted. If you go to the site he mentions you will see what I meant plus a very elegant solution to the problem. I can follow the maths for the part I required but start to lose it later on. Thank you both very much.