Calculating house advantage percentage from expected value?

Hey guys,

I see the house advantage for certain rules expressed as a percentage in a lot of places. Like, one set of rules gives the house an edge of 2% or whatever.

Let's say I've worked out an EV for a game and it's -0.04, meaning that every hand I play, I will lose 4 cents on average per dollar bet.

How do I turn that into a percentage?

I guess I don't really understand what the percentage means?

Thanks!

As you already know by now EV and HE is the same thing.

If your EV calculation is -0.04, that means the HE is 4%. You simply multiply the EV calculation by 100 to get the % or just move the . two spots to the right.

I hope I'm not giving you bad advice here as math is like only my 2nd strongest attribute :)

Thanks for the quick reply.

Instinctively I'm not sure if this is quite right though, since EV can swing from -1 to +1, i.e., there's a range of 2, whereas a percentage can only go from 0 to 100%, i.e., a range of 1.

But I guess the house edge could be negative too, in which case the percentage could be negative and also swing from -100% to 100%.

Okay, then that makes sense I guess. :)

EV is dependent on the size of the bet. The house advantage is what it is.

Quote: motrek

Thanks for the quick reply.

Instinctively I'm not sure if this is quite right though, since EV can swing from -1 to +1, i.e., there's a range of 2, whereas a percentage can only go from 0 to 100%, i.e., a range of 1.

But I guess the house edge could be negative too, in which case the percentage could be negative and also swing from -100% to 100%.

Okay, then that makes sense I guess. :)

Perhaps you are confusing EV with probability? The two are very much different.

Sorry, I know EVs depend on the bet size, but I am just working with EVs assuming bets of 1.0.

This is what confuses me between EVs and percentages, one seems to be an amount and the other seems to be a probability.

I can't make sense of the "probability" though... the probability of what?? There are a lot of possible scenarios and outcomes when playing blackjack, you might split 4 times and end up winning 4x your original bet, I know what that means for an EV but how can that be factored into a "house edge" percentage?

Quote: motrek

Sorry, I know EVs depend on the bet size, but I am just working with EVs assuming bets of 1.0.

This is what confuses me between EVs and percentages, one seems to be an amount and the other seems to be a probability.

I can't make sense of the "probability" though... the probability of what?? There are a lot of possible scenarios and outcomes when playing blackjack, you might split 4 times and end up winning 4x your original bet, I know what that means for an EV but how can that be factored into a "house edge" percentage?

As I mentioned before EV and HE is exactly the same thing. There are simply different formulas using different things to calculate each.

In the EV calculation you use probability as well as the payout. EV = (netpay)*probability + (netpay)*probability.

The answer is expressed as a figure (it should be a negative otherwise the house has a disadvantage). Figures are not as easy to express as what percentages are so in order to make it easy to understand we convert the answer into a %. That % is what is commonly refered to as HE.

I hope I'm explaining myself accurately as I'm not too good at math.

What are you looking at accomplishing?

EDIT: You ask the probability of what.....Seems to me that perhaps you want to start there? Try and figure out what probabilities are and how they partain to a casino table game. I suggest Bob Hannum's and Anthony Cabot's Casino Math. It does quite a good job of explaining how it all works!

I am trying to run a simulation of a few different strategies and I'm calculating the EVs (based on bets of 1.0) and I'm curious how that translated to the "house edge" percentages that I've seen elsewhere online.

I have no reason to doubt you, instinctively I just wasn't sure if they were the same thing or not. Thanks for the replies. :)

Some people make the following mistake. The odds of wining a pass line bet are 49.29292929%. They mistakenly assume the house edge is 0.70707071%. Actually the house edge is 2*0.70707071%.

You can talk about expected value that has nothing to do with house edge. For instance you can say, what is the EV of the number of spins before I lose a red/black bet 6 times in a row? The EV is 97 spins for American roulette, and 110 spins for European roulette.

Quote: motrek

I am trying to run a simulation of a few different strategies and I'm calculating the EVs (based on bets of 1.0) and I'm curious how that translated to the "house edge" percentages that I've seen elsewhere online.

I have no reason to doubt you, instinctively I just wasn't sure if they were the same thing or not. Thanks for the replies. :)

You might find THIS THREAD useful; same and closely related discussion in further detail, including plain-language discussion of calculating these things.

Quote: pacomartin

The odds of wining a pass line bet are 49.29292929%. They mistakenly assume the house edge is 0.70707071%. Actually the house edge is 2*0.70707071%.

The House Edge is the amount, on average in the long run, that the house wins per (for arguments sake) $100. Sometimes this can be worked out absolutely by considering all the possible outcomes and seeing how much the player gets back. Other times, such as on track betting or NFL-type wagers, it can only be based by looking at the odds being offered.

In a simple example say you play Heads/Tails with $1 paying 95c. If you toss the coin 100 times, on average 50 will be heads and 50 will be tails (don't get into the discussion about what actually happens when you do it, in this case on average half will be heads and half won't, thus 50 in 100 will be). You lose $1 x50 and win 95c x50, so lose $50-$47.50 = $2.50 per $100. This gives a House Edge of 2.5%. Another way to look at it is you should have received $2 ($1 stake + $1 win) but only received $1.95 which is 97.5% of what you should have got.

In the craps example 49.293% wins $2 ($1 stake + $1 win) and the rest receive nothing back, total return = 98.586 so the loss per 100 is 1.414.

In a very simple fruit machine you wager $1 - for each 100 spins on average, 30 wins are cherries paying $2, 5 wins are bars paying $5, 1 jackpot pays $10, total paid out = 30x2+5x5+1x10 = $95. So the payback is 95% or the house edge is 5%.