Where does the house edge come from?

I'm having a hard time figuring in a game like Three Card Poker where the house edge comes from? Sometimes I don't quite understand the things I think I understand, so I'm going to mention what I think I understand of house edges in other games, I hope people here will correct me if I'm missing the point on these, and will advise about Three Card Poker specifically.
Pai Gow - Edge comes from the house winning ties, and from commission
Blackjack - Edge comes from the player having to go first and thus having a chance to bust before the dealer even plays? And from Dealer blackjacks beating 21s?
Baccarat - Ede comes from commission?
Three card poker - Mystery!

I understand where side bets house edge comes from (bets paying off at 'incorrect' odds, like in craps...), but not in the main game... isn't it even odds whether I have a better or worse hand than the dealer?

Anyway, I'm trying to learn more about this, and would appreciate any feedback. Thanks!

If the dealer doesn't have a queen, they don't qualify. Then you only get paid on your ante bet, not the play bet.

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Quote: AndyGB

I'm having a hard time figuring in a game like Three Card Poker where the house edge comes from?

"The house edge lies in the rule that if the player folds he loses, even if the dealer doesn’t qualify."
https://wizardofodds.com/ask-the-wizard/three-card-poker/

Quote: AndyGB

Baccarat - Edge comes from commission?

For the Banker bet in regular Baccarat.
the Player bet comes from the probability of losing is greater than the probability of winning.

"The expected player return per unit wagered on the banker is .45843*.95 (added: 1 - .05; the 5% commission)+ .44615*-1 = -.01064.
The player’s loss is the casinos gain.
Thus the house edge is -1*-.01064 = .01064 = 1.064%.

Likewise the expected return on the player bet is .45843*-1 + .44615*1 = -.01228,
thus the house edge on the player bet is 1.228%."

https://wizardofodds.com/ask-the-wizard/baccarat/

Continued Good Luck

Quote: 7craps

"The house edge lies in the rule that if the player folds he loses, even if the dealer doesn’t qualify."
https://wizardofodds.com/ask-the-wizard/three-card-poker/

Not really. He can play every hand and he would still lose long-term. It's because he won't get paid on his full wager when he beats the dealer's Jack high or lower, but he will lose his entire wager if the dealer beats him with Jack high or lower.

deleted

Quote: Ibeatyouraces

True but playing every hand (blind) increases the HE.

Of course, but just showing that actual action of folding doesn't drive the house edge in 3 card poker. It's not getting paid fully on the dealer weakest hands. If the player was paid even money on all his wagers when he won and pushed ties, then it would be a fair game, obviously.

And although not especially asked, house edge in UTH comes from the poor payout of the blind bet, and the pushing of the ante bet when the dealer fails to make a pair.

On a similar note, I have always wondered how BJ came to be, whether the math was solid when it came out or came to be through trial and error by the operators.

As you probably know by now Blackjack gets most advantage to the house when both player and dealer busts as you have to go first and have already lost. Other games also use a similar "you go first" method to extract an advantage.

Consider if it was a true game, i.e. you didn't see the dealer's up-card, couldn't double and split, and the dealer played optimally, you'd play the same as the dealer. Ignoring ties, there are these possibilities:-
(i) Both hands are 21 or less - Dealer Wins.
(ii) Both hands are 21 or less - Player Wins.
(iii) Player busts - Dealer 21 or less - Dealer Wins.
(iv) Dealer busts - Player 21 or less - Player Wins.
(v) Both Player and Dealer busts - Dealer Wins.
By symmetry the chances of (i)=(ii) and (iii)=(iv).
Thus the house edge is Pr(P=Bust)*Pr(D=Bust).
In the real game this is almost compensated by Blackjacks paying 6/4 for players, Doubling and Splitting, seeing the Dealer's up-card and being able to adapt your strategy accordingly.

I would speculate Blackjack was designed from Pontoon, wanting ties to be a Stand-off; then found to have a high inherent house edge and various features added to bring the edge down. You see the same with "Push 22" or "Spanish deck" Blackjack games - a nice bonus is designed and the house edge adjusted with the "Push 22" rule, and tweaked with (say) all 21's win, Blackjack pays evens etc.


Three-card poker and other games where there's an Ante and Raise mechanism work in a very similar way; even though the dealer adopts a slightly inferior strategy (e.g. plays all Qxx rather than Q64). Roughly speaking if both hands are more than Q-high it's 50:50 who wins. Similarly if one has Q-high and the other rubbish - it's 50:50 who has the good hand. The house edge comes from rubbish-rubbish, because you have to go first - the chances of the dealer qualifying are set so you're worse off if you gamble and raise on all rubbish. The compensation is achieved by paying a bonus for straights and higher.

Quote: onenickelmiracle

On a similar note, I have always wondered how BJ came to be, whether the math was solid when it came out or came to be through trial and error by the operators.

Considering the age of the game, I'd assume the latter.

Thank you to everyone who posted in this thread, it is a big help. To make sure I understand this, I have made up a (pretty lame) game akin to Beat the Dealer, except played with only 1 die on each side. I have thought of two variants with different house edge mechanisms; I know these are really simplistic examples, but I'd greatly appreciate any feedback.

Single Die BtD - Mk I
The player makes a bet (let's say $1 for the moment). The dealer rolls one die. If he rolls a 6, the player loses immediately. In all other cases, the player then rolls. If the player's value is higher than the dealer's, he gets paid 1 to 1. (Takes back his original $1 and a brand new $1 chip from the house.) If the player's value is lower, he loses his $1, and if it's the same, it's a push and nothing happens.

Single Die BtD - Mk II
The player makes a $1 bet, then rolls one die. If he rolls a 6, he wins immediately. In all other cases, the dealer then rolls. But the dealer must qualify by rolling a 2 or better. If the dealer rolls a 1 or the same value as the player, the bet is push. If the dealer rolls higher, the player loses; dealer rolls lower, player wins.

The way I figure it, the house edge in Mk I is 2.7%. Does that look right to you? Qualitatively, understanding where the house edge comes from, it is from automatically winning on a 6, and therefore not being exposed to a push

The way I figure it, the house edge in Mk II is 11.11%. Agree/disagree? Qualitatively, the house edge comes from the dealer pushing on rolls of 1 instead of losing outright.

I'm new at this, as you can tell by reading my posts, so I might be missing obvious points or not taking something into account - let me know if you see anything really glaring in the above.

Thanks!
AGB

disagree on #2. I think it's 8.3% - 12 player wins, 9 pushes, 15 dealer wins; so, 3/36 is the HE. I think you might have 6,1 as a push, but that's a player win if he rolls first and wins immediately on the 6 roll.

Quote: rdw4potus

disagree on #2. I think it's 8.3% - 12 player wins, 9 pushes, 15 dealer wins; so, 3/36 is the HE. I think you might have 6,1 as a push, but that's a player win if he rolls first and wins immediately on the 6 roll.

Oh, yes, I see that now, I did have 6,1 as a push, I filled in all pushes for Dealer-1, but you're right. I get 8.3% with that taken into account too.

Quote: AndyGB

I understand where side bets house edge comes from (bets paying off at 'incorrect' odds, like in craps...), but not in the main game... isn't it even odds whether I have a better or worse hand than the dealer?

Anyway, I'm trying to learn more about this, and would appreciate any feedback. Thanks!

OK
Simple one.
Where does the house edge come from for the basic pass line bet in Craps?
It is a two-stage bet as you know from your examples.

Do all house edges in Craps, except for the odds bets, come from paying off at 'incorrect' odds?
If yes, then what should the pass line bet pay to be a fair bet? 0% house edge

we may both have the right answer

I would like your game better without dice or a die.
Too old school using dice
nice examples

The house edge in blackjack, and most poker-based games, comes from the fact that the player must act FIRST. If both bust/fold, the player loses.

Quote: AndyGB

Baccarat - Edge comes from commission?

Nope. The commission is just a way of re-stating the payout is 20:20 for Player and 19:20 for Banker, but Banker is likely to win more often.

Its similar to Roulette where a 1:1 payout on Black or Red would be a 1.26:1 payout if the casino were some charitable fool just standing there and never making a profit.
That is why a "five dollar bet" is really 26 cents to the casino... which doesn't care if Red or Black wins.

In Godfather terms the casino "wets its beak" when you win...and takes all your money when you lose.

The casino's wet beak (just enough to pay the light bill) makes millionaires out of casino owners.

Quote: FleaStiff

Its similar to Roulette where a 1:1 payout on Black or Red would be a 1.26:1 payout
if the casino were some charitable fool just standing there and never making a profit.

I think you were going too fast.
If the casino paid 1.26 to my 1 bet, I have a very nice edge. Yes?

I think you mean 1.11 to 1 (20/18) to 1 = 0% HA

anything less than a 20/18 payoff = you got shorted
"Make your bets"

i was under the influence that all games were 50/50

you either win or you dont...

Quote: nezbit

i was under the influence that all games were 50/50

you either win or you dont...

Sure, like the lottery:-)

Quote: nezbit

i was under the influence that all games were 50/50

you either win or you dont...

"under the influence" could be a dangerous thing especially in a casino
For the Roulette example for "red" to win on a 00wheel
18 ways to win and 20 ways to lose

for this bet to be a 50/50 shot, there should be 19 ways to win and 19 ways to lose
19/19 = 1
50/50 = 1
19/19 = 50/50

20/18 can never = 1
but it does = 1 (2/18) = 1 (1/9) = 1.111...

Quote: 7craps

OK
Simple one.
Where does the house edge come from for the basic pass line bet in Craps?
It is a two-stage bet as you know from your examples.

Do all house edges in Craps, except for the odds bets, come from paying off at 'incorrect' odds?
If yes, then what should the pass line bet pay to be a fair bet? 0% house edge
...

Ok, I'll take it as a quiz, let's see if I am getting this at all. Another way of asking your question would be "How do you get craps to be a 0 edge game, that everyone just stands around playing to try to get lucky on." I know the odds bets carry 0 edge, but you don't get the same true odds payout on the flat bet portion. So first instinct is to say pay the PL bet at 2:1 for a 4/10, 3:2 for a 5/9 or 6:5 for a 6/8. Those payouts would reflect the true odds of the thing happening (for the 2nd phase of craps) which should be how you get to an even money, no-vig setup.

But, then you'd have major problems on the come-out roll. (This goes back to a funny conversation I had with a friend who insisted you could take down your PL bet after a point was established...). I think you could make naturals pay 1:2. Then you'd win 8 times for a win of 4, and lose 4 times for a loss of 4, zero sum game. Or you could make the odds of winning/losing your passline bet 50-50. So you could change 11s from winners to losers, which means you'd win only six times (six 7s) and lose six times (2s, 3s, 11s and 12s). So that's my answer: adjust the Come out portion to either make it an even chance of winning (fewer winning rolls) or adjust the payout there to halve it, to cut the 33% player edge to zero. On the Point portion of the game, adjust the payout on the flat part of the bet to reflect the true odds.

Sound good?

I don't think that was the question 7craps was asking. Try answering this:

Right now, the passline pays 1-to-1 when it wins. Suppose instead that the passline paid N-to-M when it wins. Find N and M such that the passline has no house or player edge. Don't change the rules of the game, just change the payout.

Quote: MathExtremist

I don't think that was the question 7craps was asking. Try answering this:

Right now, the passline pays 1-to-1 when it wins. Suppose instead that the passline paid N-to-M when it wins. Find N and M such that the passline has no house or player edge. Don't change the rules of the game, just change the payout.

Ok thanks. My approach to this is as follows:

House edge is EV/amt wagered. Amount wagered is $1 (let's say) so House edge reduces to EV. EV can be calculated by (amt won * number of times won) - (amt lost * number of times lost). So figure that the passline bet wins or loses only on 2, 3, 7, 11, 12. Total possible dice outcomes that resolve the passline bet is 12. So we're looking at a number out-of-12. There are 8 wins out of 12 rolls, and 4 losses. All of which is to say:

8x/12 - 4/12 = 0
(8x/12) = 4/12
8x=4
x = .5.

So for every $1 wagered on the passline, a payout of $.50 for naturals would make the expectation and therefore the edge 0. I guess you'd call that paying 1:2, or laying 2 to win 1.

For the point phase of the game, pay the passline bet off at the same odds you use for the odds bet.

Does it sound like I know what I'm talking about?

I figured that is what you would you would come up with.
Take each win state and make a different payoff. it does make the payoff simple.

Currently as you know, the pass line bet pays even money or 1 to 1 on both possible win outcomes.
The cor and point round.

I think many players would question your payoff on a bet of $100 and you only pay them $50 for the cor.
Does not look fair!
They would say OK to the point round payoffs but that would not make up for the cor... even tho your method does work.

as you know the formula for EV (for simple win/lose bets) is simply
EV = P(win)*$bet*PF - P(loss)*$bet*LF
where PF = the payoff factor is (some bets pay 1 to 1 ; others 2 to 1; some 1 to 2; some 19 to 20 and so on)
LF = the loss factor. Most wagers that = 1. (you lose what you bet)
this can be left out to keep things simple or have it equal 1 as in most cases it is.

If we make EV in the formula 0 then we can solve for the PF
the win and lose probabilities can be determined (or most cases are easily known) as well as the $ bet
that leaves the only unknown to be the PF

examples

see if it works for you and then do the pass line again.
We want the pass line to pay the same either for the cor or the point round
to keep it an even money bet (bet $X and win $X) and to see if it is even practical

So far
A for extra effort for the pass line payoffs for a fair game.

Quote: AndyGB

Ok thanks. My approach to this is as follows:

House edge is EV/amt wagered. Amount wagered is $1 (let's say) so House edge reduces to EV. EV can be calculated by (amt won * number of times won) - (amt lost * number of times lost). So figure that the passline bet wins or loses only on 2, 3, 7, 11, 12. Total possible dice outcomes that resolve the passline bet is 12. So we're looking at a number out-of-12. There are 8 wins out of 12 rolls, and 4 losses. All of which is to say:

8x/12 - 4/12 = 0
(8x/12) = 4/12
8x=4
x = .5.

So for every $1 wagered on the passline, a payout of $.50 for naturals would make the expectation and therefore the edge 0. I guess you'd call that paying 1:2, or laying 2 to win 1.

For the point phase of the game, pay the passline bet off at the same odds you use for the odds bet.

Does it sound like I know what I'm talking about?

That's not actually how the passline works -- there aren't different payouts during different phases. The point, at least of my question, was to determine what a single payout amount on passline wins would need to be in order to eliminate the edge. Paying different amounts for different winning outcomes is answering a different question.

Okay, thanks again for the feedback on this. I'm going to post this in several parts, mostly because of being at work, where work keeps interfering...

To adjust the PL payouts so we don't have to wangle around with sometimes paying it one way and sometimes another way, the key thing to know is How many times does the bettor win the passline bet. Seems straightforward:

The odds of winning the passline bet overall are the odds of hitting a natural, plus the odds of hitting a box number and then making the box number.

The result of doing the above, which I'll work on next, should be checkable. You can calculate the exact opposite, the odds of hitting a craps plus the odds of hitting a box number and then missing the box number, which should also work out to 1-(the answer to the first method).

Knowing the odds of winning the passline bet over both phases should give the payout you could set to to make it 0 ev.

More in a little bit.

The Wizard can help too
https://wizardofodds.com/games/craps/appendix/1/

simple threads like this one may end up getting 20,000 views in the next few years.
It does happen
Others will follow and we all may learn something new... even if it hurts

So far I get the following:

Odds of a natural are (8/36)
Odds of hitting a 4: (3/36) Odds of making a 4: (1/3) [same goes for 10]
Odds of hitting a 5: (4/36) Odds of making a 5: (2/5) [same goes for 9]
Odds of hitting a 6: (5/36) Odds of making a 6: (5/11) [same goes for 8]

The odds of two things happening is the product of their probabilities (two 'heads' = 1/2 * 1/2) so I'm multiplying each box number, the odds of hitting it originally times the odds of making it to get the odds of hitting and making it. Since we're adding each set of numbers twice I'll simplify a bit below:

(8/36) + [2*(1/12) * (1/3)] + [2*(1/9)*(2/5)] + [2*(5/36)*(5/11)]

With some good old fashioned fraction simplification and then division on the calculator I get
Probability of the passline bet winning in any way at .4928

Check this doing the losing side:

Odds of a craps are (4/36)
Odds of hitting a 4: (3/36) Odds of Missing a 4: (2/3) [same goes for 10]
Odds of hitting a 5: (4/36) Odds of Missing a 5: (3/5) [same goes for 9]
Odds of hitting a 6: (5/36) Odds of Missing a 6: (6/11) [same goes for 8]


(4/36) + [2*(1/12) * (2/3)] + [2*(1/9)*(3/5)] + [2*(5/36)*(6/11)]

With some good old fashioned fraction simplification and then division on the calculator I get
Probability of the passline bet losing in any way at .50715.

Does this check out to the Losing probability being equal to 1 minus the winning? It does, within a matter of a couple ten-thousandths on the calculator.

Next: summing up.

Quote: 7craps

see if it works for you and then do the pass line again.
We want the pass line to pay the same either for the cor or the point round
to keep it an even money bet (bet $X and win $X) and to see if it is even practical

So far
A for extra effort for the pass line payoffs for a fair game.

So, overall the fair "anytime" payout (0 EV, not changing for which phase of the game you're in) for the PL bet is: (.5070/.4928). Using the Wizard's fractions to make it cleaner, should be something like (251/244). Or to lop a couple of decimal places off, ~103/100. Sanity checking this, to make sure it's right side up, the longer shot pays more. So "PL wins" is a slight dog, so it pays a slight premium.

Okay! So what was the question? :)

P (Win)
Natural Win 7 0.166666667 990
Natural Win 11 0.055555556 330
Point=4 0.027777778 165 55
Point=5 0.044444444 264 66
Point=6 0.063131313 375 75
Point=8 0.063131313 375
Point=9 0.044444444 264
Point=10 0.027777778 165
2928 1.414 141% 5856 Need to,pay 84
Craps=2 0.027777778 165
Craps=3 0.055555556 330
Craps=12 0.027777778 165 1.363 636% 5859
Point loses=4 0.055555556 330
Point loses=5 0.066666667 396
Point loses=6 0.075757576 450
Point loses=8 0.075757576 450
Point loses=9 0.066666667 396
Point loses=10 0.055555556 330
3012 -1.414 141% 6024
5940
(Sorry if above doesn't line up) but basically if you eliminate fractions (but multiplying each figure by 5940)
then it's easier to see that the Win Line pays 5856 (2928 times you pay the original bet PLUS the win) and
where the bar is 12 the Don't Win pays 5859.
This means for the win line you'd have to pay 84 more over the 5940 outcomes. One way of getting close is to pay 2/1 if make
point with 3-3 (which adds 75), another way is to pay 6/4 if a point of 10 is made (which adds 82.5).
I think only paying 1/2 for 7s and paying 2/1 winning 9s 10s and (5-3) works, but it's open to biassed
dice.

Quote: AndyGB

So, overall the fair "anytime" payout (0 EV, not changing for which phase of the game you're in) for the PL bet is: (.5070/.4928). Using the Wizard's fractions to make it cleaner, should be something like (251/244). Or to lop a couple of decimal places off, ~103/100. Sanity checking this, to make sure it's right side up, the longer shot pays more. So "PL wins" is a slight dog, so it pays a slight premium.

Okay! So what was the question? :)

I forgot about this one.
yes.
So paying $(251/244) would make the pass line bet a fair one.
But not practical unless you bet something like 244 and the payoff is 251
now you can see the short pay of $7 per 244 bet (7/495)

The last poster comes close to making the pass line a fair bet,

At least you showed how this is done for any that follow
Thanks

Here's another way of looking at it.
Nat Win 7.............. 990 0.5 1485
Nat Win 11............ 330 . . 1 660
Point 4 made......... 165 . . 1 330
Point 5 made......... 264 . . 1 528
Point 6 made......... 375 . . 1 750
Point 8 made [4-4].. 75 . . 1 150
Point 8 made [5-3] 150 . . 2 450
Point 8 made [6-2] 150 . . 1 300
Point 9 made [5-4] 132 . . 2 396
Point 9 made [6-3] 132 . . 2 396
Point 10made [6-4] 110 .. 2 330
Point 10made [5-5]. 55 . . 2 165
....................................... 5940