Based on some heuristic calculations I did, it seems that the probability of escape with this plan is exactly 4⁄10.
Interesting. Do you agree that every number is reached by the z function defined above, infinite number of times?
And yet, every single time z != sleeping_round? In the 60 percent of this Sleeping Beauty imprisonments?
Even if the condition x>y is replaced by something like x>y+sqrt(y) or whatever formula, you can’t go above 50%?
Extraordinary. Might be possible, though.
You clearly have a function N->N where eventually every natural number is a value of this function f, but f(n)!=n for all n.
That would be easier if it would be f(n)>>n almost always. But sometimes is bigger, sometimes is smaller.
Do you agree that every number is reached by the z function defined above, infinite number of times?
Yes, definitely.
Yes. I proved it.
Well, on average we have f(n)=n for one n, but there’s a 50% chance the guy won’t ask us on that round.
Based on some heuristic calculations I did, it seems that the probability of escape with this plan is exactly 4⁄10.
Interesting. Do you agree that every number is reached by the z function defined above, infinite number of times?
And yet, every single time z != sleeping_round? In the 60 percent of this Sleeping Beauty imprisonments?
Even if the condition x>y is replaced by something like x>y+sqrt(y) or whatever formula, you can’t go above 50%?
Extraordinary. Might be possible, though.
You clearly have a function N->N where eventually every natural number is a value of this function f, but f(n)!=n for all n.
That would be easier if it would be f(n)>>n almost always. But sometimes is bigger, sometimes is smaller.
Yes, definitely.
Yes. I proved it.
Well, on average we have f(n)=n for one n, but there’s a 50% chance the guy won’t ask us on that round.