Are you asking what's the highest number I can choose which guarantees I get to 2 (chance greater than 33% of getting a 1) in 4 moves or less?
OK. Let me explain what I mean further.
An alternative way of asking for the same thing would be: What's the maximum number of n I can select to ensure that there is a 33% chance or less that the game will last 4 rounds or less.
(or to put it yet another way, the game lasts 5 rounds or more at least 66% of the time).
For example, let's say n is equal to 2.
In this case, the only way the game will last 5 rounds or more is if roll n returns 2,2,2,2. This means 1 in 16 times, or 6.25% of the time. This is much smaller than 33%.
Now let's say n is equal to 3.
In this case, the game will still last 4 rounds or less quite often. I mean there is a 33% chance of it ending on round 1 if you get 1. If not, you may get 2 which will increase its chances of ending quicker. Or you may get 3 which will give you three more tries.
Regardless the chance is greater than 6.25% from n=2, but smaller than 33%
Now let's say n is equal to 1000000
You can see that the game will very often last more than 4 rounds. Definitely more than 33% of the time.
So what's the maximum value n can take for the chance to be 33% or less?
Is that clearer now?
This does not halp me at all! I already knew all this
Sorry, was typing. Just wanted to post the problem first.
Does the above make it clearer of what the problem is?
Also, just wanted to note that I do not have an answer. Just some observations on how to approach and a quick and dirty way of getting an approximation. But I'm happy to share any insights. And if I find some spare time, I might give it a proper try too.