Probability Calculus Puzzle?

I have a real life probaiblity puzzle, derived from a video game (Super Smash Brothers Melee) that I don't know how to approach (at least not with brute force simulations...)

This game has about 125 trophies you can 'win', little 3D models of Nintendo characters and misc memorabilia. You can collect these in the normal course of gameplay, albeit slowly, or you can play a little slot machine they've setup, and I can't figure out what the optimal slot machine strategy is if I want to win 1 of each different kind of trophy.

You win coins by playing the regular modes of the game, and you can then use these coins in the slot machine. You 'deposit' a coin, and you get a random trophy, but possibly one you've already received. (The game displays what your % chance of winning a previously unseen trophy is; since this decreased by .8 every time I won a trophy, I did the math and figured there were probably 125 trophies, was my reasoning correct given my assumptions?)

So here's the catch: you have the option of putting in extra coins for each slot pull, and each coin increases your chance of winning by 5%...(I guess they might have to round up or down in some cases.)

So with all that, what's the optimal strategy for winning every different trophy with the minimum amount of coins? It seems that for the first part, when you haven't won many trophies, it makes sense to do one coin pulls, since chances are good that you'll be seeing a lot of new trophies. At the other border condition, it may make sense to put in many coins, since if it's down to, say, a 1% chance of seeing a new trophy, you'd have to (on average) put in 100 singleton coins to win a new trophy, but you can put in 10 coins and have 50/50 odds, or 20 coins and almost definately get it. So what is the ration or pattern for number of unseen trophies to number of coins it makes sense to insert?

Question as spoiler space..
http://kisrael.com/mortal

I agree that there should be about 125 trophies.

Lets say at some point you have already gained x different trophies, so you have 125-x left. The probability you win (get a new trophy) with one coin is (125-x)/125. So with n coins, the probability you win is (125-x)/125 + 0.05n + 0.05. The expected number of attempts that you will have to make before getting a new trophy is 1/this. So the expected cost is n/((125-x)/125 + 0.05n + 0.05). Using a long division and getting rid of decimals we get cost of 20 + (160x - 19000)/(50n - 8x + 950). Note that the denominator is always positive, as x<125 and n>1. If the numerator is negative, ie 160x < 19000 ie x <= 118, we want the denominator as small as possible to decrease the cost. So we insert only one coin. If the numerator is positive, ie 160x > 19000 ie x >= 119 we want the denominator as large as possible, so we insert as many coins as we can.

So with 118 or less trophies, put in one coin, else put in as many as possible.

You want to maximize the quantity (# of trophies won)/(# coins spent). Call this tpc for trophies per coin. If your initial probability of winning is p (after 1 coin) and your probability of winning is p + (q - 1)/20 for q total coins, then you want to maximize TPC(q) = [p + (q-1)/20]/q = (p-1/20)/q + 1. Differentiate w.r.t. q and set equal to zero:

(1/20-p)/q^2 = 0.

Now this equality never holds except in the limit that q goes to infinity (or p =1/20 when it's always true), but it's still instructive. When p > 1/20, the derivative is negative and adding coins only reduces tpc. When p < 1/20, the derivative is positive and adding coins always increases tpc (sort of.. see below). And when p = 1/20 it makes no difference. So the strategy seems to be that you bet only one coin until you've collected almost all the trophies (down to a 5% chance of winning) and then for all future plays you dump in as many coins as it takes to get to a 100% chance of winning (sort of.. see below).

There's a slight detail I've neglected which is that the formula for the probability of winning for q total coins is wrong. In particular, this probability is upper-bounded at 1. This means that when you're at a probability of > 95% you have to check to see whether paying the extra coin to top-off at 100% is beneficial. In general this will depend on how close you are to a multiple of 5% (it seems to make little sense to pay to go from 99.8% to 100%) and how many coins have already been bet (when you've already bet a lot of coins, betting one more is less damaging to your TPC