2

The question:

In a group of $n+1$ men, one of them creates a secret (I called him the Creator in my answer), and tells it to another man, every man keeps telling another man the secret (chosen randomly from the $n$ men - not including himself). all the events of telling the secret between eachother are independent.
What are the probability function of the number of times the secret has been told until it gets back..:
a) to it's creator.
b) to a man that has already heard the secret.

My Work:
a) Let $X$ be the number of times the secret has been told until it gets back to the secret's creator,
Then I'm searching for $P(X=k)$, but $k$ must be at least two since the creator cannot tell himself his own secret. So $k=2,3,4,\dots$
$P(X=2)=P(\text{Creator}\to \text{RandomMan}\to \text{Creator})= \frac{1}{n}$
$P(X=3)=P(\text{Creator}\to \text{Man}\to \text{Man}\to \text{Creator})=\frac{n-1}{n} \frac{1}{n}$ (The first man must not tell the creator so $n-1$).
$P(X=4)=(\frac{n-1}{n})^2 \cdot \frac {1}{n}$
$\dots$
$P(X=k)=(\frac{n-1}{n})^{k-2} \cdot \frac{1}{n}$.


b) let Y be the number of times the secret has been told until it gets back to a man that has already heard the secret.
$k=2,3,4,\dots,n+1$ since after $n$ times surely the secret will be told to a man that has already heard it because after $n$ times all of them will know the secret.
$P(Y=2)=P(\text{Creator}\to \text{man} \to \text{creator})=\frac{1}{n}$.
$P(Y=3)=P(\text{Creator} \to \text{man_1} \to \text{man_2} \to \text{man_1 OR Creator})=\frac{n-1}{n} \cdot \frac{2}{n}$ $P(Y=4)=P(\text{Creator}\to \text{Man_1}\to \text{Man_2}\to \text{Man_3}\to \text{Man_1 Or Man_2 Or Creator})$ $=\frac{n-1}{n} \frac{n-2}{n} \frac{3}{n}$.
$\dots$
$P(Y=k)=\frac{(n-1)(n-2)\cdots (n-k+2)}{n^{k-2}} \frac{k-1}{n}=\frac{(n-1)!}{(n-k+1)!\cdot n^{k-2}} \frac{k-1}{n}$.
Done

I would love to hear feedback if my solution is correct or not and what mistakes did I make, also I'm open to hear if any notation I wrote it somewhat incorrect since I'm new to random variables.
Thanks in advance

zhoraster
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Pwaol
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    "$P(X=2)=P(Creator->man->creator)=\frac {1}{n} \frac{1}{n}=\frac{1}{n^2}$".

    This is wrong. Actually $P(X=2) = 1/n$ (the probability is that the second randomly chosen person is the creator). $P(X=3)$ is the probability that the second person is not the creator, while the third is, i.e. $\frac{n-1}{n}\cdot \frac 1n$.

    The probabilities for $Y$ are flawed similarly. Actually, to compute its expected value, it is better to use the formula (which is valid for non-negative integer-valued random variables) $$ E[Y] = \sum_{n= 0}^\infty P(Y>n). $$

    – zhoraster Apr 24 '21 at 18:42
  • @zhoraster Thanks for pointing that out, that makes sense, about the formula, I haven't seen it before, could you tell me what I need to search to find it? (I'm pretty sure it's solvable without the formula since it's in my homework and I still didn't learn it but I do want to read about it and understand). Thanks for the help – Pwaol Apr 25 '21 at 04:57
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    It is solvable, yes. The formula just makes the solution (for $Y$) shorter. Searching for "expectation non-negative integer" yields the desired result. – zhoraster Apr 25 '21 at 07:22
  • @zhoraster Hey zhoraster, I have tried to fix both of the answer according to your comment, I would appreciate an approval if you can – Pwaol Apr 25 '21 at 12:48
  • Seems ok now, nice job! I just wonder how did I see the word "expectation" in your question, as it was never there... – zhoraster Apr 25 '21 at 12:58
  • @zhoraster Thank you! I'm not sure what you mean, I don't see any word "expectation" in my question.. anyway if you would like to write an answer maybe if you can tell me how to write the functions formally (is it just to add - otherwise, $0$?) so I can give you the bounty on the question :) – Pwaol Apr 25 '21 at 13:01
  • That's exactly my point, there's no word "expectation", but for some reason I dreamed it... Answering just to get bounty is against the principles of this community. The main purpose of the bounty is not to reward a potential answer, but to draw attention to your question, so it did serve its purpose perfectly imo. – zhoraster Apr 26 '21 at 10:05
  • @zhoraster Ohh I see, it's my first time placing a bounty, my intention was to do it since you did help me reach my answer so you might aswell get the bounty, didn't know it was against the principles. anyways Thanks alot for the all the help, it is all appreciated. – Pwaol Apr 26 '21 at 12:26

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