# Short Math Question

Im doing a small cost benefit analysis out of my own interest but I havent learned (or dont remember) how to go about it properly. How would you go about solving a problem similar to this?

I have a system like the following:
at level 0, you pay \$100, with a 97% chance to get to level 1
at level 1, you pay \$100, with a 95% chance to get to level 2
at level 2, you pay \$100, with a 93% chance to get to level 3
at level 3, you pay \$140, with a 91% chance to get to level 4
at level 4, you pay \$140, with a 89% chance to get to level 5
at level 5, you pay \$140, with a 87% chance to get to level 6
at level 6, you pay \$180, with a 85% chance to get to level 7
Assume the pattern continues. If you fail, you go back to level 0

At any point you can opt to pay 4x the price which will change the situation so that if you fail you get to stay where you were and try again. At what point is it better to start paying the higher price?

Right now I have made a small chart of % chance to reach a certain level, but I am not sure where to go from here if this would even be useful.

If I understand correctly the idea is that if he fails and drops back to zero, he spend the money to get back up to the current level and at what point will it be cheaper to pay the insurance rather than risk having to build your way up again.

To start with work out the total price it costs to get to a give level iteratively

Let PN be the total price to get to level N, SN be the success rate for going from N to N+1 (0.97 for N=0)
CN be the cost to try to go from N to N+1 (100 for N=0, 140 for N=3)

The expected number of trials before a success is given by 1/SN

If you don’t buy the insurance,
PN+1 = (PN+CN )/(SN)

Since each run you have to pay to get back to your starting point.

If you do buy the insurance, then

PN+1 = PN+(4*CN )/(SN)

You don’t have to repay to get back to the start for each run, but you have to pay 4* the advance cost for each run.

So from this it looks like you are better off buying the insurance if
PN*(1-SN)>3*CN

Do you get to try again just the once, or is it essentially guaranteed advancement?

Too late to Edit: