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Note: Before starting, I would like to point out that I have no concrete reason as to why I need to find this, but I just wanted to know.

Here is a bit of a background

There is a 2 player game where you get a number chosen for you, from 1 to n (n included). If you have y chosen for you, then the other person has a number from 1 to y chosen. This goes on until someone has 1 chosen for them, and that person is the winner.

Now the question is to find which person, the person who has the number chosen first, or the person who has the number chosen second, has more probability of winning.

The hurdles I found were:

If I am having a number chosen from 1 to n, it is possible that I get n. If it happens, the other person has to choose from 1 to n as well, and this can happen infinitely many times for any value of n.

The probability of getting any chosen number from 1 to n is 1/n. If the number is y, then the probability of getting any chosen number from 1 to y is 1/y. But you don't know what y ends up being, and it is not feasible to make infinitely many cases

The first person also might have 1 chosen as their number in the first go itself, which is why I feel they have more probability of winning, but I am not sure how exactly do I prove that (considering it is true), and how do I get to know what the probabilities of each of them winning are.

Here is an example game. I am playing against you, I go first. n is 100.

"X: y" means "X" got the number y in their turn

I: 49

You: 42

I: 23

You: 15

I: 8

You: 3

I: 3

You: 3

I: 2

You: 1

Which concludes that you won the game. As you can see, there is a chance that there is a repetition as well.

Could anyone help me out to calculate the probabilities? Thank you!

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  • $\begingroup$ It might be worth doing the calculations for $n=2$ and then $3$ and $4$ if you can. Then guess a pattern and prove it. My initial thoughts is that the first player should have an advantage because they might get $1$ immediately and my attempts at a quick calculation seem to confirm this. $\endgroup$
    – Henry
    Commented Jul 2 at 17:03
  • $\begingroup$ It is easy for 2. I can form a GP where you get 2 for n number of times and add it. However it gets a bit tougher for 3 because for 3, I can either get a 3 "a" number of times and then get a 2 "b" number of times and "a" and "b" can take infinitely many values. So I'm a bit confused on how to proceed @Henry $\endgroup$
    – Kavin Upreti
    Commented Jul 2 at 17:20
  • $\begingroup$ If $P_n$ is the overall probability of the first player winning for given $n$ then $P_2=\frac12 \times 1 + \frac12\times(1-P_2)$ which should give the answer you have found already. Similarly $P_3=\frac13 \times 1 + \frac13\times(1-P_2)+ \frac13\times(1-P_3)$ and so on, each easily solved. $\endgroup$
    – Henry
    Commented Jul 2 at 17:24
  • $\begingroup$ Okay, I agree with the P2 one. But what if I get a 2, you get a 2, then I get another 2, then you get another 2, and then I finally get a 1? Sorry I'm not that sharp with math Also for P3, how come we're doing 1/3(1-p2)? Could you probably elaborate a bit? 😅😅 $\endgroup$
    – Kavin Upreti
    Commented Jul 2 at 18:36
  • $\begingroup$ In the $P_2$ case, either the first player gets a $1$ (probability $\frac12)$ and wins or the first player gets a $2$ (probability $\frac12)$ and the second player then has probability $P_2$ of winning i.e. the first player has a conditional probability $(1 - P_2)$ of winning. So $P_2=\frac12 \times 1 + \frac12\times(1-P_2)$ and it is $P_2$ on both sides which deals with your repetition issue; this implies $P_2=\frac23$. You can do the same for $P_3=\frac13 \times 1 + \frac13\times(1-P_2)+ \frac13\times(1-P_3)$ and generally for $P_n$. $\endgroup$
    – Henry
    Commented 2 days ago

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