Why my random experiment has a log normal distribution?

Main Question or Discussion Point

Hi,

I am confused with the results of a seemingly simple simulation that is generating a log normally distributed output. Please see the attached results file.

Simulation: I have built a Scratch program that randomly picks six letters from a group of six letters (A, B, C, D, E & E). The program displays the order in which the letters have been picked. I am interested in finding out how many iterations the program takes to get a specific order of letters (say, EEDCBA).

I repeated this experiment 100 times and I was surprised to see log normally distributed results. I was hoping to see a normal distribution with an average of 360.

Can someone please explain what is going on?

Thanks,

Attachments

• 26.5 KB Views: 122

Related Set Theory, Logic, Probability, Statistics News on Phys.org
The actual distribution should be geometric with p=1/6^6 (i.e. counting the number of failures before a success).

I guess a smallish sample (size 100) would superficially resemble lognormal.

I assume you understand, that you cannot actually get neither lognormal nor normal distribution, as they are continuous, and your r.v. is discrete.

If I understood your description, then your r.v. is just "the number of failures, before first success", where success is getting "EEDCBA" and trials are independent, right? In this case what you should get is the geometric distribution.

P.S. I can't open your excel file, so can't give you details of what it's doing wrong

Thanks folks.

For those who are not able to open my excel file, I have attached a text file with my results.

If I understood your description, then your r.v. is just "the number of failures, before first success", where success is getting "EEDCBA" and trials are independent, right?
That is correct.

Also, I got the 360 as follows: Prob of getting E in the first place = 2/6, prob of getting the second E in the second place = 1/5, prob of getting D in the third place = 1/4...and so on.

probability of getting EEDCBA = 2/6 * 1/5* 1/4* 1/3 * 1/2 * 1 = 1/360
How is this number related to the distribution? Is it the mean of the distribution?

Attachments

• 505 bytes Views: 200
Last edited:
Thanks folks.
How is this number related to the distribution? Is it the mean of the distribution?

Also, if p is the probability of success, then 1-p is probability of failure. In order to get (first) success on k-th trial (iteration), you need to fail k-1 times in a row and then have a success, thus $$P(X=k) = (1-p)^{k-1}p,$$ which is exactly the pmf of geometric distribution.