Topic: Calculating the EV of a Casino Bonus (Read 837 times)

I finally get it, thanks, guys.  I've been wondering about this for years (literally) and it's not something one can figure out with google.  Now I know why.

Dooglus' bad strategy is not something I had ever considered and pretty much changed the way I look at things:


Haven't looked too closely at Pocketdice.  Mostly motivated by the way all those coingaming and softswiss type sites like to act as if their 100%+ first-time deposit bonuses are so generous that they have to hold cashouts hostage for player dox to make sure nobody is "abusing their bonuses."  Offering that same bonus for every deposit all the time would probably only cost them money because they wouldn't be able to cancel as many withdraws.

Does anyone know where I could find some software to run simulations of simple slot type games?  For example, If I could run a million simulations of a game that either paid out 0x 2x 20x or 200x with a 5% HE with limits and a basic betting strategy, I could figure out a reasonable number to put on different bonuses. 

I think most of the people running these sites are pretty oblivious and just do what has become standard.  Would be nice to enlighten them, or something. 

Bonus 2x wagering required, No deposit, Dice 50% chance. All-in.
Bonus amount withdrawable after completing the requirement.


There are 3 possible outcomes:
1 Win Win (25% probability) Balance : 2.6
2 Win Lose (25% probability) Balance : 0.8
3 Lose (50% probability) Balance : 0


EV : 2.6*0.25+0.8*0.25 = 0.85.


1, 0.9, 0.85, ........

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We are looking for the average EV and the best possible EV?

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Nope. You would need to know the variance of the game you are playing, just finding that is difficult.

Are you referring to Pocketdice? Their cashback is +EV, all the rest is -EV.

I don't think this is correct just because it tends towards 0 and it is an easy formula.




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.





750x is the better. Even if there is no bonus.

You should read this: https://bitcointalksearch.org/topic/challenge-whats-the-best-way-to-win-1-btc-with-1-btc-939776 (RHaver, dooglus, blockage)

Edit: didn't see dooglus' post, gonna read it later.

I can't wrap my head around this idea that equity in a vacume (not influenced by human traits like time or emotion) can be influenced by variance.  


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.

I guess you've never taken a betcoin.tm bonus!

But assuming that's the case, that you get to play your bonus without touching your deposit, and the bonus is cancelled as soon as you bust the bonus amount, then the EV of the bonus would depend upon the odds, house edge, and stake sizing. You would want to pick a game and a play style which maximizes the variance (since your EV is the average of just the positive results)

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Suppose we put the 1 BTC on a 0.9% chance of 100x payout, and if we win, we put 9 BTC on another 0.9% chance of 100x payout. Those are both 10% house edge bets, and result in us wagering 10 BTC if we win the first bet.

We have a 0.991 chance of busting the bonus: bonus value = 0 BTC
We have a 0.009 chance of winning the first bet (leaving us with 100 BTC), and then:
    We have a 0.991 chance of losing the 2nd bet: bonus value = 100 - 9 = 91 BTC
    We have a 0.009 chance of winning the 2nd bet: bonus value = 91 + 900 = 991 BTC

So the expected value of the bonus is 0.009 * (0.991 * 91 + 0.009 * 991) = 0.8919 BTC

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To try to pick a 'bad' strategy, what if we bet 1 BTC with a 90% chance of a 1x payout ten times in a row... Those are also 10% house edge bets.

We have a 0.9 ** 10 = 0.3486784401 probability of winning all ten bets, and ending up with our original 1 BTC. That's an EV of 0.3486 BTC.

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Can we do worse than 0.3486 BTC somehow while only risking 10 BTC? Or better than 0.9 BTC?

Suppose we put the 1 BTC on a 0.09% chance of 1000x payout, and if we win, we put 9 BTC on another 0.09% chance of 1000x payout. Those are both 10% house edge bets, and result in us wagering 10 BTC if we win the first bet.

We have a 0.9991 chance of busting the bonus: bonus value = 0 BTC
We have a 0.0009 chance of winning the first bet (leaving us with 1000 BTC), and then:
    We have a 0.9991 chance of losing the 2nd bet: bonus value = 1000 - 9 = 991 BTC
    We have a 0.0009 chance of winning the 2nd bet: bonus value = 991 + 9000 = 9991 BTC

So the expected value of the bonus is 0.0009 * (0.9991 * 991 + 0.0009 * 9991) = 0.89919 BTC

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This looks like it approaches an EV of 0.9 BTC as the payout multiplier increases.

I think there's an idea is when you "bust" your bonus, it's over. You can never lose. Your worst case is making 0 btc profit, but you can make money from the bonus, ergo the EV needs to be >0

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?

Maybe my crappy diagram will help make it easier to understand:




The blue curve is controlled by your playing strategy. The higher variance it is, the flatter it is. The lower variance it is, the steeper it is.  So as you can see in this example, the flatter the probability distribution is, the more of the curve contributes to our expected value.


So perhaps the ideal way to maximize our EV is to do a single bet at 100000000000000000000000000x multiplier, and then if we win (slightly less than 1 in that big number) we can easily satisfy our play-through requirements like nothing.


Now would mean, the EV of bonus is always  0.9 btc *regardless* of play through requirements. (Of course no casino lets you bet at an arbitrary payout, though. So the problem needs to be more constrained )

If you want a clear answer, you should state precisely all the details about the situation you want to analyze.

For example, what fraction of the bonus must be gambled at each round? Depending on it, the answer will change dramatically.
Assuming, you have to gamble it entirely at each round, then the fraction of the bonus you end up after a single round is 100% - 10% = 90% = 0.9
Starting with 1BTC, you get 1 * 0.9 = 0.9BTC after one round.
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.
It is never 0 for any finite n, but in practice the EV will fall under one satoshi which will be rounded as 0.

We are talking about bonus without any deposit, right?

Assuming a 50% win or lose game.

Suppose you get a 1BTC bonus, and you have to wager 1 time, you will lose 10% of that to the house, so the Expected Value of the bonus is 0.9BTC

Suppose you have to wager 2 times, the EV would be 2.6*0.25+0.8*0.25= 0.85BTC

3 times, the EV would be 0.125*3.4+0.125*1.6+0.0625*1.44= 0.715BTC

and so on 

Maybe I should of made this thread self moderated.

Sorry, that was a dick thing to say.

Luck really has nothing to do with the question though. 

You forget luck, which is the most important role in gambling. Once I got two casino bonus(2 mbtc and 3mbtc chip) from surprise box in betcoin, I passed 50X playthrough, but most cases, I can't complete it, the percentage I can pass the 50X playthrough is less than 5%.

In real life betting, if we deposit 100 times of one btc for 50X rollover requirement bonus, I afraid they will be ALL busted. I don't have so big bankroll to try, cause once I claimed tons of satoshibet no deposit bonus to try 35X rollover, I rarely pass it, only cashed out twice.

Remember: casino bonus is the most profitable way to make money, from casino degree, the bonus is using to waste our bitcoin.

High variance betting (both via bet size and game selection) is pretty clearly the optimal strategy for clearing high playthrough bonuses (and also lossback promotions), but I thought it was mostly due to our value of time and the fact that walking away with 1% of the bonus is the same to most as walking away with 0 and going bust on the first bet 999 times for ever one time we 900x our money is preferable.

I don't see how increasing our variance should change out EV in a vacuum, but if it does - when does it start?



Whether we lay 1:10, flip a coin or take a 100:1 shot, A $100 bonus with 1x playthrough always = $100 - House Edge  

How did you get this?  I think 1000x seems way high, but still curious.


Btw, I'm sure I was wrong, but i put my formula into excel and it came out to 97x before our original $100 was worth less than 1 penny.  Theoretically even with infinite playthrough, our value would be >0.  



How do you calculate this?

It depends on variance, not just house edge. So you would need to know the total number, chances of and payouts of each event.



Exactly. It can even get positive.

^ Disregard my answer. I believe the question is not solvable with the information you gave. My logic was based upon someone betting with 50-50 odds of winning (e.g. 0.8x multiplier) and always betting MIN(playerBalance, amountRequiredToHitWageringReq), however now that I dig deeper it appears that the EV changes depending how the player bets. (Or that is wrong, which would also make my original answer wrong). So either way, I was wrong



It's very counter-intuitive, but I believe the riskier bets you take, the higher your EV will actually be.

For instance, imagine you have to wager 100 BTC @ 10% house edge. The normal expected value of doing that is -10 BTC, and the less-risky bets you make, the closer your actual result will be to the expected value. But in this case, we only have 1 BTC to wager with, so the less risky bets we do, the closer we are to certain doom (busting our 1 BTC). e.g. If we attempting to hitting the wagering requirement by doing a million tiny bets, the variance would be so small that we would always bust, and we would have 0 EV.


But if we make some extremely risky bets, our variance is huge. And the benefit in this case, is our worst result is making 0 money. e.g. our EV is actually the   so its our goal to increase the variance.




Disclaimer: it's late, wait for someone like blockage or dooglus to answer this, as they're a lot more competent than me.

Thank you very much for the calculation of deposit bonus distribution, as i was also having some doubts but now i am clear , that is why they are giving 100% bonus as they know before a user clears the deposit bonus requirement he will be in full loss of atleast 70% of loss.

That is why i dont go for deposit bonus in casino and dice, ya but in poker and sports betting this type of bonus we can reach the requirement and we can win the bonus, i have tried twice and have earned it.

Very interesting question!


I think it's 0.563 bitcoin.



Yeah, I think you're doing it wrong. But it's late, I've had a few beers and I'm sleepy so I very well might be wrong. I'll try provide some working tomorrow if you're interested.

For what it's worth I get:
1x roll over: 0.90 BTC
10x roll over:  0.563 BTC
100x roll over: 0.315 BTC
1000x roll over: 0.196 BTC


Which seems strangely high.  But I guess the important thing to note, is a player can never lose money, as your worst case is losing the free 1 BTC bonus. So the EV is the average of a bunch of weighted >= 0 numbers.

 In a real casino, it might be a 100% deposit bonus ... which means although you get some "free" money you have to gamble like crazy to the point you risk your actual deposit (and can have a loss), to the point that the "free" bonus money might cause you more harm than good!

Does anyone know of an easy was to calculate the true EV of a bonus considering the house edge and playthrough requirement?

For example, lets say we receive a BTC1 bonus to play a game with a 10% HE and 10x playthrough.  

We can expect to lose BTC1 for every BTC10 wagered at this game, BTC10 is the amount we have to wager, but that doesn't make the EV of our bonus = 0

I think the calculation for the Value of the bonus looks something like this:

Starting BR = BTC1

BTC1 x .1 = .1
BTC1 - .1 =BTC0.9

Bonus Value with 1x Required = BTC.9

BTC0.9 x .1 = .09
BTC0.9 - .09 = BTC0.81

Bonus Value with 2x Required =  BTC0.81

BTC0.81 x .1 = .081
BTC0.81 - 0.81 = BTC0.729

Bonus Value with 3x Required = BTC0.729

Etc.

Am I doing it wrong?
If I'm doing it right, it there an easy way to calculate Bonus Value with 100x or more?

(Variance and deposit/no deposit are not relevant.)