# Why my random experiment has a log normal distribution?

• musicgold
In summary, The conversation discusses a simulation that randomly picks six letters and displays the order in which they are picked. The goal is to find out how many iterations it takes to get a specific order of letters. The results were found to be log normally distributed, but it was expected to be normally distributed with an average of 360. However, upon further discussion, it was determined that the actual distribution should be geometric with a probability of success of 1/360. The average number of iterations should be 360 as well. The conversation also mentions the use of a Bernoulli trial and the probability of success being related to the mean of the geometric distribution. The source for more information on this topic is suggested to be the Wikipedia article on

#### musicgold

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

• names.xls
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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

• Data.txt
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Last edited:
musicgold said:
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.