return Math.floor(((100 * e) - h) / (e - h));
That's the only really interesting line. We can ignore the "0x crash 1-in-101 games" part, because that is only used to fund the 1% bonus that it paid out on the other 100-in-101 games. It has no effect on the house edge at all.
We can even lose the Math.floor bit, because that just makes sure the result has no more than 2 decimal places. And divide the top by 100, because that function is returning the crash point times 100 (so it's an integer).
So the crash point is simply:
e - h/100
---------
e - h
e is a big number, but it doesn't really matter what it is (it's 2^52)
h is a random number between 0 and e-1 inclusive
Let's look at three cases:
1. If h is at its minimum, 0, the crash point is 1x
The house edge is 0%. (We can ignore the 1-in-101 0x crashes, because on the other 100-in-101 games we get a 1% bonus, assuming we're the only player, or that all the other players are also playing at a 1x cashout).
2. If h is half way through its range, e/2, the crash point is 1.99x
So we have a 50% chance of winning 1.99x. That gives an RTP of 50%*1.99 = 99.5% and so a
house edge of 0.5%3. If h is at its maximum, e-1, the crash point is 0.99e+0.01 (ie. very high, since e is a big number)
So we have a 100/e% chance of winning 0.99e+0.01. That gives an RTP of (99e+1)/e% ~= 99% and so a
house edge of a tiny bit under 1%.
So you can see that the house edge ranges from 0% to 1% as your payout multiplier ranges from 1x to high numbers.
i just don't know how i would go about calculating the house edge for a certain multiplier
The payout multiplier for any particular h is (from the code you pasted):
m = (e - h/100) / (e-h) [1]
and the probability p of hitting that payout is:
p = (e-h)/e [2]
h = e-ep
the RTP for that h is [1]*[2]*100:
(e - h/100) / (e-h) * (e-h)/e * 100
= (e - h/100) * 100/e
= 100 - h/e
and so the house edge is simply:
edge = 100 - RTP
= h/e
= (e-ep)/e
= 1-p
Now you want the house edge for a given multiplier, m. Let's find h in terms of m:
m = (e - h/100) / (e-h)
(e-h)m = e - h/100
em - hm = e - h/100
em - e = hm - h/100
e(m-1) = h(m-0.01)
h = e(m-1)/(m-0.01)
And so the house edge:
edge = h/e
= e(m-1)/(m-0.01)
= (m-1)/(m-0.01)
And there's your answer. The house edge for a probability of p is 1-p, and for a multiplier of m is (m-1)/(m-0.01)
Examples:
the house edge when you have a 0.5 probability of winning is 1 - p = 1-0.5 = 0.5%
the house edge for a multiplier of 2x is (m-1)/(m-0.01) = (2-1)/(2-0.01) = 0.50251256%
Edit: if all you want to do is change the house edge so that it goes from 0% to N% instead of 0% to 1%, I think you only need to replace the 100 in the Math.floor line by 100/N. Don't change the insta-crash logic unless you want to unbalance the bonus system. Looks like you already changed it to insta-crash every 51 games (without changing the corresponding comment), but that's a bad idea. People won't want to play a game that insta-crashes so often. Adjust the edge using the Math.floor line instead.
Example:
>>> e = 2**52
>>> h = e * 0.75
>>> 100 - (e - h/100) / (e-h) * 25 # the bustabit setting - house edge is 0.75% when we have 25% chance of winning
0.75
>>> 100 - (e - h/20) / (e-h) * 25 # increase the house edge by a factor of 5 - house edge is 3.75% when we have 25% chance of winning
3.75
What would the code look like if you wanted a set house edge which is not scaling?
