-snip-
The second one yields 0.9*0.9 and so on...
The formula for the EV after n consecutive plays of the totality of the bonus is: 1*0.9^n (i.e. bonus*(1 - house edge)^n) which obviously tends towards 0 as n grows to infinity.
I don't think this is correct just because it tends towards 0 and it is an easy formula.
If I give you 1 BTC and tell you that you have to wager 10 BTC at a 10% house edge, the EV of that bonus is exactly 0, as you calculate above.
To say more than that we will need to know the terms of the bonus.
For example, with some bonuses I am allowed to "cancel" it at any time. But does cancelling a bonus remove the full 1 BTC from my balance, or just the amount I didn't already lose of the bonus?
RHaver should be right. Bonuses mess things up. You could lose bonus before the wagering is completed (so no house edge lost on next bets which takes average EV up).
Cancelling gets the balance at deposit minus loss if there is any. In this case there is no deposit so 0.
Ok, let's say we're playing on a dice site with a 1% house edge.
We have a bonus balance of 1,000 with the following terms.
We must choose one of three bets to make for an amount between 1 and 1,000
After the first bet - we can not change it. We have to make that same bet over and over and over.
Options
A) 1.01x
B) 2x
C) 750x
(all with 1% HE as seen on just-dice and ignoring any potentially exploitable rounding issues)
There are only two possible outcomes : A) meet the wagering requirement and cash out balance or B) go broke
Let's say in one scenario there was a 10x playthrough, and another with 100x.
750x is the better. Even if there is no bonus.
You should read this:
https://bitcointalk.org/index.php?topic=939776.0 (RHaver, dooglus, blockage)