Trevor
Trevor
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December 1st, 2014 at 9:55:14 AM permalink
Hi Everyone,

This is probably a daft question but here goes anyway. When the house edge differs from the element of risk or % RTP, doesn't this just mean that the house edge was calculated incorrectly?

Cheers,

Trev
CrystalMath
CrystalMath
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December 1st, 2014 at 10:23:23 AM permalink
No. They are just different metrics.

I prefer using house edge, which is the percent of the initial bet that the casino will win, on average.

The element of risk is the percent of the total average bet that the casino will win, on average. Let's say that the house edge of a game is 1% and the player will place an extra wager, such as a double down or split, 20% of the time (average wager 1.2 credits). Then, the element of risk is 1%/1.2 = 0.83%.
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Mosca
Mosca
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December 1st, 2014 at 11:27:05 AM permalink
The way I've understood it, the element of risk takes into account that some games let you bet more when you have an advantage, or less when you don't. For example, in Ultimate Texas Hold'em, you can bet 4x the original ante when you are holding A-A. Or, if you have 8-4-2 unsuited in Three Card Poker, you can fold rather than raise. Part of your bet is made with a little bit of strategy due to extra knowledge, offering a better chance overall.

Is my understanding wrong, based on CrystalMath's explanation?
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CrystalMath
CrystalMath
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December 1st, 2014 at 12:12:32 PM permalink
Quote: Mosca

The way I've understood it, the element of risk takes into account that some games let you bet more when you have an advantage, or less when you don't. For example, in Ultimate Texas Hold'em, you can bet 4x the original ante when you are holding A-A. Or, if you have 8-4-2 unsuited in Three Card Poker, you can fold rather than raise. Part of your bet is made with a little bit of strategy due to extra knowledge, offering a better chance overall.

Is my understanding wrong, based on CrystalMath's explanation?



You are correct. In my example, I said that you would bet more for a double or a split, which is similar to your examples of betting more when it is advantageous.
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mrsuit31
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December 1st, 2014 at 12:21:05 PM permalink
Quote: CrystalMath

You are correct. In my example, I said that you would bet more for a double or a split, which is similar to your examples of betting more when it is advantageous.



EOR is very misleading to players sometimes. because of what the previous two posters have said. Sometimes people try and sneak EOR through as HA making a 5% HA game look like a 1.something% game...

But like they expressed above, its merely just a matter of either regarding ante wagers alone or the avg bet in total.
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Trevor
Trevor
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December 2nd, 2014 at 1:48:26 AM permalink
Quote: mrsuit31

EOR is very misleading to players sometimes. because of what the previous two posters have said. Sometimes people try and sneak EOR through as HA making a 5% HA game look like a 1.something% game...

But like they expressed above, its merely just a matter of either regarding ante wagers alone or the avg bet in total.



Is EOR misleading? If the house actually took a smaller % of my total bet how can this be misleading? Is it because i am forced to make a second bet to play the game?
kubikulann
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December 2nd, 2014 at 8:30:59 AM permalink
Quote: Trevor

Is EOR misleading? If the house actually took a smaller % of my total bet how can this be misleading? Is it because i am forced to make a second bet to play the game?

I wouldn't say "misleading". As always, it all depends on what you want to measure.

I like to use a game like Caribbean Stud to show the problem. (I think Texas Hold Up is similar).
You place an ante (say 10), you get your cards and you choose either to fold or to play, paying twice the ante (20, making 30 in total).
Typically, House Edge is the expected value divided by 10. Element of Risk is the expected value divided by a weighted average of 10's and 30's.

But actually you would not engage in the game if you are not able to put 30 on the table. One may view the game as requiring an initial bet of 30, so some people (me, in particular) would measure the expected value divided by 30.

There. You have three different measures for one same game. Why? Because they are based on "initial bet", which is ambiguously defined in such games.
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CrystalMath
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December 2nd, 2014 at 8:48:42 AM permalink
I agree with kubikulann. It is only misleading if you don't know what you're looking at.

For the casino, they must know the house edge, which is the total expected loss divided by the initial bet. That way, when tracking your bets, they just need to observe your initial bet and can calculate points/comps based on that.

For the player, they should know the house edge and total bet required to play (as in kubikulann's example).

To me, I only use the house edge to calculate an expected loss. In 3CP, for example, the house edge is about 3.3%, so I would expect to lose 33 cents per $10 initial bet. Of course, I must also know that I need $20 total to be in the game. I don't care that the EOR is 2%, because then I also need to remember that the total average bet is 1.67.
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