House edge calc for Let It Ride poker

I believe I found an error in the wizard's calculation of the house edge for Let It Ride poker (given as 3.51%).

I agree with the initial calculation of 3.51%. However, one game involves betting 1, 2 or 3 chips, not just 1. The expected value of the number of chips bet in one game is 1.2321.

This means that the house edge is actually 3.51% divided by 1.2321, which is 2.845%.

Another way to think about this is, if you were playing video poker $1 denomination bet 5, and the average prize is $4.949 per hand, the difference is $0.051. This does not mean the house edge is 5.1%. The house edge is 5.1% divided by 5 (or 1.02%) because you are betting 5 $1's per hand, not just one $1 per hand.

There are a few places that allow you to only bet on the $ symbol.

Quote: KevinAA

I believe I found an error in the wizard's calculation of the house edge for Let It Ride poker (given as 3.51%).

I agree with the initial calculation of 3.51%. However, one game involves betting 1, 2 or 3 chips, not just 1. The expected value of the number of chips bet in one game is 1.2321.

This means that the house edge is actually 3.51% divided by 1.2321, which is 2.845%.

Another way to think about this is, if you were playing video poker $1 denomination bet 5, and the average prize is $4.949 per hand, the difference is $0.051. This does not mean the house edge is 5.1%. The house edge is 5.1% divided by 5 (or 1.02%) because you are betting 5 $1's per hand, not just one $1 per hand.

You're actually making the Wizard's argument for him, on a calculation he calls Element of Risk. There is some controversy about which figure to use among table game designers, distributors, and clients.

The House Edge calculation was developed to compare games with no decision point. You placed a bet on craps, roulette, baccarat, or blackjack, and then something happened, and you won or lost. There was no opportunity to place more money on the table (or cut your losses) once you had any knowledge of how valuable that hand, spin, or roll might be. Even if you placed more than one bet on the table, say in roulette or craps, it happened before any action.

Then came Carnival games. Caribbean Stud was pretty much the first, a poker derivative with a decision point. You placed a bet, as above (for these games called the ante). Then you saw your hand, plus 1 of the dealer's cards. THEN you either folded or bet 2x your original bet. So you were risking some amount more than your original bet to play, on average. But the HE formula would only count the amount of the original bet in calculation. Apples to oranges.

Let It Ride has the same issue, in reverse. The HE, which has a very narrow definition, is only calculated from the amount of the single required bet, the $ circle (the one you cannot withdraw). But, as you point out, on average the player risks 1.2x the ante, because they start with 3x ante, then with optimal play will sometimes let those additional bets ride, with 2 decision points available to leave their bets up or pull each back.

The Element of Risk returns apples to apples comparison to dissimilar games, just as you show, dividing the ante (1) by the average bet amount under optimal strategy, then multiplying it by the HE. There is widespread but not universal agreement that this is a valid and/or necessary correction. Some, not all, casinos use it to figure their choices for games offered, and/or factor it into their calculation of your play (ATD, your theoretical value to them).

I'm sure the Wizard will appreciate your independent finding that agrees with his. :)

Yep. EoR is 2.85%

https://wizardofodds.com/gambling/house-edge/

Okay, I understand. Thanks for the replies!

The "Element of Risk" is worse than meaningless -it actually creates the wrong impression. And its not a replacement for House Edge.

1. Why House Edge is meaningful. When I step up to a table and place a bet, I want to know what fraction of my bet I'm going to lose on average. House Edge relates to an initial decision to play the ding dong game in the first place. Every subsequent decision has its own EV or House Edge or "Player Edge".

2. Illustration that Element of Risk is not meaningful. To illustrate how silly the "Element of Risk" is, let me provide two different ways of implementing a pay-out table that rewards a royal flush at 100:1:

Method a) player makes an initial bet of $10, gets a royal flush and is awarded $1000
Method b) player makes an initial bet of $10, gets a royal flush and is required to increase his bet to $1000 at which time he is paid-off at 1:1.

Method (b) would have a lower element of risk because the rules of the game required you to "wager" $1000. But it is not less risky than if the game had used method (a).

3. Why games with low Element of Risk are actually MORE risky. Many first-time players are horrified when they first make a $10 initial bet on Mississippi Stud and then find that they need to wager as much as $40 on a speculative hand to find out whether they lose or win. That's not a favorable game aspect that needs to be labeled as a low "Element of Risk." Instead, the Wizard should state that the maximum possible loss per hand is 4X what you initially bet and 4X what the casino says is your "minimum bet". For most people, Miss Stud is a far more risky game-with a higher risk of ruin - because the rules require that they put even more money at risk as the only alternative to folding.

Actually, with M.Stud and following basic strategy (No AP play), you can lose 8x. Example being dealt KQ suited. Bet 1x to see 3rd street. If that card and 4th street give you straight flush or royal draws or higher you'll 3x the rest of the way and if you brick 5th street, you lose 8x. Example hand:

KQ spades
3rd street, J spades
4th street, A spades
5th street, 5 clubs

Obviously you'd 3x any 4 card flush as well.

Quote: gordonm888

The "Element of Risk" is worse than meaningless -it actually creates the wrong impression. And its not a replacement for House Edge.

1. Why House Edge is meaningful. When I step up to a table and place a bet, I want to know what fraction of my bet I'm going to lose on average. House Edge relates to an initial decision to play the ding dong game in the first place. Every subsequent decision has its own EV or House Edge or "Player Edge".

2. Illustration that Element of Risk is not meaningful. To illustrate how silly the "Element of Risk" is, let me provide two different ways of implementing a pay-out table that rewards a royal flush at 100:1:

Method a) player makes an initial bet of $10, gets a royal flush and is awarded $1000
Method b) player makes an initial bet of $10, gets a royal flush and is required to increase his bet to $1000 at which time he is paid-off at 1:1.

Method (b) would have a lower element of risk because the rules of the game required you to "wager" $1000. But it is not less risky than if the game had used method (a).

3. Why games with low Element of Risk are actually MORE risky. Many first-time players are horrified when they first make a $10 initial bet on Mississippi Stud and then find that they need to wager as much as $40 on a speculative hand to find out whether they lose or win. That's not a favorable game aspect that needs to be labeled as a low "Element of Risk." Instead, the Wizard should state that the maximum possible loss per hand is 4X what you initially bet and 4X what the casino says is your "minimum bet". For most people, Miss Stud is a far more risky game-with a higher risk of ruin - because the rules require that they put even more money at risk as the only alternative to folding.

1. Your calculation doesn't show how much you will lose, it shows how much you will forfeit by folding. You also can lose (and often do) even with a hand good enough to bet the play. The optimal play calculation for EOR takes that into account.

2. That's a non-pertinent example b. It doesn't happen that way in any game. The closest to your example might be in games where the dealer must qualify, and then your hands are an exact copy and you push. But that's a different matter.

3. On MSstud, the total possible loss is 10x (including your initial bet), not 4x.1x-3x-3x-3x is possible and not uncommon, as you bet after receiving your 2 card hand, seeing the 3rd, and seeing the 4th. The 5th card is a disaster more often than not, and all bets will lose. But the math says that you win them often enough that the odds paid will offset those losses. Very high volatility, that game requires very aggressive play, and a little luck. But HE is most misleading on this game. With all winning bets paid at odds, HE severely underestimates the house exposure, and the player potential. EOR does not.

Quote: beachbumbabs

3. On MSstud, the total possible loss is 10x (including your initial bet)

A hole carder or edge softer or some other AP type play might lose 10x, but not a proper BS player will not . They should only bet 3x (and most don't even do that) after seeing their own two cards if they are a pair or better. If they are 2's-5's, and that pair doesn't connect on 3rd street, they now 1x 4th street. 6's or higher will never lose.

Here is an example from Three Card Poker.

You are playing one-on-one against the dealer. Basic Strategy is to BET on Q64 or higher and FOLD on Q63 and lower. The House edge is about 3.83% (for a common pay table) and the element of risk is 2.28%.

Now let's say that you decide to play with a strategy that is BET on Q63 and FOLD on Q62 and lower. Naturally, with Q63 your EV on the BET vs FOLD decision is about -1% for the BET choice, But that is the strategy you take.

What is the impact of always BETTING on Q63 vs the conventional (optimal strategy)?

• The House Edge becomes larger than 3.83%
• The Element of Risk becomes smaller than 2.28%

That's right! By adding a negative EV wager to the Three Card Poker strategy you can reduce the Element of Risk! Bwaah-hah-hah.

Why? Because when you BET on Q63 the House Edge is only about -1% which is lower than the House Edge of -3.83% on your initial bet. Thus, the "average House Edge on all the money you wager" -which is the Element of Risk metric - goes down.

Also, by the way, when you add "BET on Q63" to your 3CP strategy, your risk of ruin goes up. So, Element of Risk goes down while House Edge and Risk of Ruin goes up. Don't you see a problem with that?

It's a minefield trying to decide whether the better measure for a game is HE or EoR. Personally I prefer to know how much it costs for me to play a hand (or spin etc) at (say) a $5 table. In that way, combined with the number of hands per hour, one can compare the cost of playing various games. It is a side effect that I may have to stump up more, in some cases much more, just to play the game - but that is a different issue.

Even that doesn't solve the problem for games such as Craps - this is because some bets are resolved every roll of the dice and others take multiple rolls. For instance if you make a $5 Pass bet (or Don't Pass, Come, Don't Come) you will on average lose about 1.4%. However it can take several rolls before the bet either wins or loses. On average it costs 2.095c per roll (for a $5 bet).

Compare this with a $6 bet on (say) 6. The House Edge is complicated because do you consider the loss per roll of the dice or per outcome (i.e. either a 6 or 7 are rolled). Also sometimes the bet isn't "working" (on the come out roll) so it actually costs less in the long term than a Pass bet! The House Edge is slightly higher (1.5%) but the average cost per roll is 1.955c per roll (for a $6 bet).

I do think that House Edge could be defined in a different way for those table games where it is necessary to place two or more bets. Like for Ultimate Texas Hold'em, it is mandatory to make an Ante Bet and a "Blind" Bet that is the same size as the ante bet. I would have no problem defining the UTH House Edge to be 1.1% rather than 2.2% because your initial bet really is two units.

I came up with a solution to this. The expected value of playing any game is sum(prize*count)/sum(bet*count). If you are playing a game with flat betting, you can reduce this to sum((prize/bet)*count)/sum(count).

For Let It Ride Poker, using the Wizard's table, we have:

outcome prize prob amt won*count bet*count
Royal Flush 3,003 1.53908E-06 240240 240
Royal Flush 2,002 0 0 0
Royal Flush 1,001 0 0 0
Straight Flush 603 6.77194E-06 212256 1056
Straight Flush 402 7.07975E-06 147936 736
Straight Flush 201 0 0 0
Four of a kind 153 0.000151445 1204416 23616
Four of a kind 102 8.86508E-05 470016 9216
Four of a kind 51 0 0 0
Full House 36 0.000642719 1202688 100224
Full House 24 0.000797858 995328 82944
Full House 12 0 0 0
Flush 27 0.000192539 270216 30024
Flush 18 0.001772863 1658736 184304
Flush 9 0 0 0
Straight 18 8.58805E-05 80352 13392
Straight 12 0.001705605 1063872 177312
Straight 6 0.002133161 665280 110880
Three of a kind 12 0.007013575 4374720 1093680
Three of a kind 8 0.006328839 2631744 657936
Three of a kind 4 0.007786037 1618848 404712
Two pair 9 0.011045341 5167152 1722384
Two pair 6 0.013135562 4096656 1365552
Two pair 3 0.023358112 3642408 1214136
Tens or better 6 0.049474328 15429816 7714908
Tens or better 4 0.050934989 10590240 5295120
Tens or better 2 0.062117232 6457608 3228804
Loser 0 0.001021101 0 159228
Loser 0 0.018042294 0 1875648
Loser 0 0.742156478 0 38576700

Of the 51,979,200 outcomes, we have sum(prize*count)=62,220,528. The sum(bet*count)=64,042,752. Divide the two and you get 0.971546757, 1- that is 0.028453243.

Therefore, the house edge for this game is, without a doubt, 2.845%. This means that if you wager hundreds of thousands of hands, you can expect to lose 0.028453243*amt wagered, with little variance (because of the law of large numbers). For the casino, with players playing hundreds of thousands of hands (enough to be in "law of large numbers land"), they will earn 0.028453243*amt wagered.

Quote: KevinAA

...Of the 51,979,200 outcomes, we have sum(prize*count)=62,220,528. The sum(bet*count)=64,042,752. Divide the two and you get 0.971546757, 1- that is 0.028453243.

Therefore, the house edge for this game is, without a doubt, 2.845%. This means that if you wager hundreds of thousands of hands, you can expect to lose 0.028453243*amt wagered...

You can see the definition of "House Edge" and "Element of Risk" on this page https://wizardofodds.com/gambling/house-edge/ and the figure for LIR as 3.51% which corresponds to (using your figures above) the loss (64042752-62220528) divided by the number of outcomes (or initial wagers) (51979200). Also the Element of Risk there agrees with your figure which is the loss divided by the total amount wagered.

I read that page before but I guess it didn't "click" with me. Oh well there's nothing wrong with replicating the Wizard's calculation.

For many years I had always been calculating "house edge" as 1-expected value of $prize per $bet on flat betting games and only recently decided to try it on Let It Ride.