I'll give you an answer and try to explain where it comes from and how it can be generalized to any collecting problem.
It will take on average 100 "Unknown" for a player to collect all the 26 different forms. The reason is the following :
- It takes only one "unknown" to find the first form, no problem with that.
- Suppose that you have found one form already.For each different try, you will have 1 out of 26 chances to pick the same one you already have. So the probability for it to take more than n tries to find a new one is (1/26)^n. In other terms, if we write T the number of tries, P(T>n)=(1/26)^n. This is a very usefull thing to know, because we can use it to compute the expected value of the number of days it will take to find a different form than the first one. The exepected value of T is noted E[T], and the formula is
E[T]=P(T>0)+P(T>1)+P(T>2)+P(T>3)+...
=1+(1/26)+(1/26)^2+(1/26)^3+...
(the dots mean that the sum goes on forever)
- In the same way, if you already have found k forms and want to find a new one among the 26-k that remains : there are k chances out of 26 to pick on you already have for each try. The probability for it to take more than n tries to find a new one is (k/26)^n, and the expected value is
E[T]=1+(k/26 )+(k/26)^2+(k/26)^3+....
Fortunately, we don't have to make all these calculation by hand, thanks to a very well known formula :
1+q+q^2+q^3+...+q^n=(1-q^(n+1))/(1-q)
Here, q=k/26 where k is the number of forms we have already found. As q<1, the limit of q^n as n approaches infinity is 0, so the formula becomes
1+q+q^2+q^3+...= 1/(1-q)
Let's get back to our problem : when we have already found k forms, the expected value of the number of days it will take to find a new one is 1+1/26+...=1/(1-k/26)=26/(26-k). Say you have found 13 forms out of 26, it will take an average of 26/(26-13)=2 days to find a new one.
The final answer : the expected value of the number of days it will take to find ALL of the forms is the sum of the expected value to find a (k+1)th once you have found k forms, that is
E=1+26/25+26/24+...+26/2+26/1
E=26*(1+1/2+...+1/26)
The result of this computation gives approximately 100, and this formula works for any problem of collecting n different things with equal chances of finding each one of them at each tries : it takes on average n*(1+1/2+...+1/n).
- For large values of n, a good approximation of this number is n*ln(n)
To learn more about this, you can look online for "Coupon collector's problem" ;)
I apologize in advance for any english mistake as it is not my mother tongue
That's awesome - thanks for taking the time to provide a solution :-)
I can confirm 100 days with a numerical simulation to find all 'unown's' on average !
Glad I could help! I was about to do a numerical simulation as well just to make sure I didn't make any mistake hahah
That's great! Where did you learn this? I want to know more :D