Changing The House Edge

Dear Wizard,

Thank you so much for all the valuable information on your site.

Concerning betting systems I found statements that seem contradictory to me and I would really appreaciate if you could point me "the right way".

CONs:
1) In your "Betting Systems - FAQ" you reply that betting systems can neither decrease nor increase the house edge.

PROs:
2) On your site about "Double Action Roulette" you calculate the "RETURN" of the various available bets. For simplicity I'm referring to the even money bets (probability close to 50%) here:
The return (house edge) of the double ring bets (you name them "parlay bets") is roughly twice as high as that of the single ring bets. And this holds for both the single zero wheel and the double zero wheel.

3) The "parlay" betting system(!) is a progression on winnings.
Here is how you can play a game equivalent to the double action wheel on a regular roulette wheel. Let's bet on RED in this example:
Give the wager of one unit to a friend of yours and tell him:
a) Wager one unit on RED.
b) If you win you will then have two units. Wager both on RED again.
c) Come back to me with whatever you have left over.
He will come back with either nothing (you lost 1) or with four units (you won 3).

4) On "Ask the Wizard #242" you answered "Rob from Las Vegas" with a calculation that a single parly has roughly double the advantge (based on a probability close to 50% as with RED).
Since Rob has the advantage as in general has the casino he could be considered running a casino and therfore holding the house edge on his side. Again parlay bets would increase his edge.

In cases 2) and 4) the house edge was changed,
Both cases are based on a betting system (the "parlay").
Doesn't this contradict the general statement 1) that the house edge can never be changed by a betting system?

Can you please tell me where I am wrong?

PS.: I'm not stating that betting systems can overcome a negative expectation. They cannot! But from the above I suspect that it can be worsened (progression on winnings: "parlay") or it can be brought closer to zero (progression on losses: "Martingale"; which has to be discussed further).

The wizard is right. Every bet you make it against the same house edge. You can play around with wager amounts, but ultimately, your expectation doesn't change.

With your parlay example, if you win the first, you let it ride, meaning that two wins gives you +3. Or you could lose that and you get -1. What if you lose the first one?

WW = 3
WL = -1
L = -1
L = -1

Expectation is 0.

A parlay has the ultimate effect of increasing your average wager. With a fixed percentage edge and an increased wager, the theoretical dollar loss increases. If you consider that dollar loss as a percentage of just the first bet, it is higher. But that's not a straight-across comparison; it's like saying "if I make $1000 in $10 blackjack bets, my theoretical loss is greater than if I just make a single $10 blackjack bet." It's true but not really relevant. Parlay betting does not change the theoretical percentage edge of any of the bets you're making, it just means you're betting more.

Quote: surrender88s

The wizard is right. Every bet you make it against the same house edge. You can play around with wager amounts, but ultimately, your expectation doesn't change.

With your parlay example, if you win the first, you let it ride, meaning that two wins gives you +3. Or you could lose that and you get -1. What if you lose the first one?

WW = 3
WL = -1
L = -1
L = -1

Expectation is 0.

For the house edge the outcomes have to be multiplied with their probabilities. Here for a single zero roulette wheel:
p(+3) = 18/37 * 18/37 = 324/1369
p(-1) = 1 - 324/1369 = 1045/1369 (the loss on the first roll is included here)
ev = (3 * 324)/1369 + (-1 * 1045)/1369 = 972/1369 - 1045/1369 = -73/1369 approx. -5,33%
This is the value the Wizard gives on his "Double Action Roulette" site.
The "normal" expectation of a single zero wheel would be -2,7%.

Quote: MathExtremist

A parlay has the ultimate effect of increasing your average wager. With a fixed percentage edge and an increased wager, the theoretical dollar loss increases. If you consider that dollar loss as a percentage of just the first bet, it is higher. But that's not a straight-across comparison; it's like saying "if I make $1000 in $10 blackjack bets, my theoretical loss is greater than if I just make a single $10 blackjack bet." It's true but not really relevant. Parlay betting does not change the theoretical percentage edge of any of the bets you're making, it just means you're betting more.

This is the reason why I introduced "the friend" who makes the bets.
In my humble opinion the solution is to think of games, their probabilities and returns (resulting in ev) rather than in single rolls and the intermediate wins or losses.
You hand your friend one unit and he returns the result to you with the related probabilities. You don't care about how the probabilities are generated. It's all about the probabilities and the outcomes. Please refer to the calculation in my answer to surrender88s.
Furthermore, in the two sources I referred to above (2 and 4) the Wizard clearly states the calculated advantages which are different from the "normal" values.

Quote: ThomasK

This is the reason why I introduced "the friend" who makes the bets.
In my humble opinion the solution is to think of games, their probabilities and returns (resulting in ev) rather than in single rolls and the intermediate wins or losses.
You hand your friend one unit and he returns the result to you with the related probabilities. You don't care about how the probabilities are generated. It's all about the probabilities and the outcomes. Please refer to the calculation in my answer to surrender88s.

Sure, but that's a different bet. Of course it will have a different set of probabilities and outcomes. That's not surprising. But neither does it have anything to do with "changing" the house edge on the original Red bet (or Black, Odd, Low, etc.) bet. If I fade you at 3-to-1 that you won't have two Red winners in a row, the house edge on the Red bet itself doesn't change. The per-unit edge on the Double-Red-parlay bet might be different, but I'm not sure why that's interesting

Have you ever looked into parlay cards at sports books? Or multi-race bets at horsetracks?

Quote: MathExtremist

Sure, but that's a different bet. Of course it will have a different set of probabilities and outcomes. That's not surprising. But neither does it have anything to do with "changing" the house edge on the original Red bet (or Black, Odd, Low, etc.) bet. If I fade you at 3-to-1 that you won't have two Red winners in a row, the house edge on the Red bet itself doesn't change. The per-unit edge on the Double-Red-parlay bet might be different, but I'm not sure why that's interesting

Have you ever looked into parlay cards at sports books? Or multi-race bets at horsetracks?

I'm sorry if I haven't stated my point clear enough.
I don't doubt that if you go and bet one unit on RED infinite times you will have the well known house edge.
What I'm trying to say is that if you sent your friend infinite times with that one unit to bet it in the manner described above he will come back with the payouts and the probabilities as stated.
Payouts and probabilities result in ev and this again divided by the original wager results in the advantage.
My point is that you now have a game with a different house edge (as confirmed by you) that you can play at a roulette table.
Isn't it then that you are able to play at a roulette table with a house edge different from the well known?

I'd like to try to show what I mean with (hypothetical) slot machines.
You put coins into the machine, a random event is chosen and you are payed according to the rules of the machine.
When calculating the house edge of slot machines you only care for the coin in, coin out and the corresponding probabilities.
You won't care for what happens inside: In the parlay case you wouldn't consider the single rolls, probabilities and intermediate payouts. This hiding is exactly what the double action wheel does for you. Coin in to parlay, coin out from parlay, probabilities of win and loss.
Please check these machines. I will then show that they all are based on roulette after your next post:

SLOT A
3 coins play
Payout --- probability (approx.)
6 --- 48.6%

SLOT B
3 coins play
Payout --- probability (approx.)
2 --- 38.5%
4 --- 36.5%
6 --- 11.5%

SLOT C
3 coins play
Payout --- probability (approx.)
12 --- 23.7%

SLOT D
3 coins play
Payout --- probability (approx.)
4 --- 73.6%

You have been fairly unclear.

If you take all of these, and multiply the payouts by probability, you should get the same result.

Just play blackjack and count cards. That is a betting system that CAN overcome the house edge. :)

Quote: kewlj

Just play blackjack and count cards. That is a betting system that CAN overcome the house edge. :)

So can flat betting certain games under certain conditions.

Quote: Ibeatyouraces

So can flat betting certain games under certain conditions.

Are we still talking about OVERCOMING a house edge? Or are you referring to 'certain conditions' where there is no house edge?

Quote: ThomasK

Dear Wizard,

Thank you so much for all the valuable information on your site.

Concerning betting systems I found statements that seem contradictory to me and I would really appreaciate if you could point me "the right way".

CONs:
1) In your "Betting Systems - FAQ" you reply that betting systems can neither decrease nor increase the house edge.

PROs:
2) On your site about "Double Action Roulette" you calculate the "RETURN" of the various available bets. For simplicity I'm referring to the even money bets (probability close to 50%) here:
The return (house edge) of the double ring bets (you name them "parlay bets") is roughly twice as high as that of the single ring bets. And this holds for both the single zero wheel and the double zero wheel.

3) The "parlay" betting system(!) is a progression on winnings.
Here is how you can play a game equivalent to the double action wheel on a regular roulette wheel. Let's bet on RED in this example:
Give the wager of one unit to a friend of yours and tell him:
a) Wager one unit on RED.
b) If you win you will then have two units. Wager both on RED again.
c) Come back to me with whatever you have left over.
He will come back with either nothing (you lost 1) or with four units (you won 3).

4) On "Ask the Wizard #242" you answered "Rob from Las Vegas" with a calculation that a single parly has roughly double the advantge (based on a probability close to 50% as with RED).
Since Rob has the advantage as in general has the casino he could be considered running a casino and therfore holding the house edge on his side. Again parlay bets would increase his edge.

In cases 2) and 4) the house edge was changed,
Both cases are based on a betting system (the "parlay").
Doesn't this contradict the general statement 1) that the house edge can never be changed by a betting system?

Can you please tell me where I am wrong?

PS.: I'm not stating that betting systems can overcome a negative expectation. They cannot! But from the above I suspect that it can be worsened (progression on winnings: "parlay") or it can be brought closer to zero (progression on losses: "Martingale"; which has to be discussed further).

I think your system fails at 3b. "If you win..."
Whether single zero or double zero roulette, you win red slightly less than 50% of the time (lose to black or green). You're discounting the times you lose this first bet, simply ignoring the negative effect on your bankroll.

If you do win this first bet, you lose the parlay slightly more than 50% of the time (again, on black or green). Yet if, in the less than 25% of instances you do win your parlay, you are only paid for odds of 4 for 1, even though true odds were higher than that (~4.08 for 1 on 00 tables). You lose 1 for 1, but you win slightly less (at 2 for 1) each time you win, because your odds were not 50/50 to start with. And that's the unchanging house edge in action, and where the casino makes its money.

Quote: kewlj

Are we still talking about OVERCOMING a house edge? Or are you referring to 'certain conditions' where there is no house edge?

Overcoming. HC'ing is one obvious example.

Quote: kewlj

Are we still talking about OVERCOMING a house edge? Or are you referring to 'certain conditions' where there is no house edge?

Overcoming. HC'ing is one obvious example. Flat betting will beat bj this way.

Quote: kewlj

Are we still talking about OVERCOMING a house edge? Or are you referring to 'certain conditions' where there is no house edge?

No, this is the opposite. The OP is discussing parlaying and treating the expected loss as a percentage of only the initial wager as opposed to the total action. That makes the edge look worse, not better. The expected loss increases, the denominator doesn't, so the fraction gets bigger. But it's still negative. You can't sign-flip unless you have a theoretical positive in there somewhere.

Quote: Ibeatyouraces

Overcoming. HC'ing is one obvious example. Flat betting will beat bj this way.

Well, yes, you can change the house edge by HCing. LolThat is not exactly altering the house edge by betting technique or system which is what I thought we are talking about.

Quote: kewlj

Well, yes, you can change the house edge by HCing. LolThat is not exactly altering the house edge by betting technique or system which is what I thought we are talking about.

Well...Flat betting IS a betting system ;-)

But I get what you're saying.

Quote: ThomasK

On "Ask the Wizard #242" you answered "Rob from Las Vegas" with a calculation that a single parly has roughly double the advantge (based on a probability close to 50% as with RED).
Since Rob has the advantage as in general has the casino he could be considered running a casino and therfore holding the house edge on his side. Again parlay bets would increase his edge.

In cases 2) and 4) the house edge was changed,
Both cases are based on a betting system (the "parlay").
Doesn't this contradict the general statement 1) that the house edge can never be changed by a betting system?

I thought I would provide ask-the-W #242

Quote:

Suppose there are two football games that I feel have a player advantage. Let’s say each has a 55% chance of winning, and I have to lay 110. Which is more profitable, to bet the games straight up or as a single parlay?

Rob from Las Vegas

Good question. Straight up, the advantage per wager is 0.55×(10/11) - 0.45 = 0.05. As a parlay, the advantage is (0.55)2×((21/11)2-1)-(1-(0.55)2) = 10.25%. So, it would seem the parlay is the way to go to maximize advantage.

However, the variance is greater as a parlay. If you are following the Kelly Criterion, then you will have to protect your bankroll for the parlay with a smaller wager. In this example, the optimal Kelly bet straight up is 5.48% of bankroll if the two games overlap, 5.50% if you first game ends before you bet on the second game, and 3.88% for the parlay. Multiplying the wager times the advantage, we get 0.00275 straight up (based on a 5.50% advantage) and 0.00397 for the parlay. Thus, the parlay results in the greater profit.

I considered the general case for this kind of question, looking also at 3-team and 4-team parlays and money line wagers. Assuming a small advantage for all bets, as a rule of thumb, if the probability of each event winning is less than 33%, then you should bet straight up. If each probability is between 33% and 52%, then you should do a 2-team parlay. If each probability is between 52% and 64%, then you should do a 3-team parlay. If each probability is greater than 64%, then you should do a 4-team parlay. If you are doing straight up wagers, then you are about equally well off doing 2-team or 3-team parlays, again assuming you have an advantage to begin with.

I should stress that if you are a recreational gambler going against a house edge (what sports bettor will admit to that?), then betting straight up minimizes the house advantage.

Since I couldn't do the math myself, I have trouble evaluating this. I also know zilch about parlaying sports bets; but it does seem to me [looking at this] there is something standard about the way sports parlays are done so that the payoff odds are increased? Note that I am using a question mark. If so, it explains how the HE can change too.

In any case, it is relevant that the player has advantage to start with.

Quote: beachbumbabs

I think your system fails at 3b. "If you win..."
Whether single zero or double zero roulette, you win red slightly less than 50% of the time (lose to black or green). You're discounting the times you lose this first bet, simply ignoring the negative effect on your bankroll.

If you do win this first bet, you lose the parlay slightly more than 50% of the time (again, on black or green). Yet if, in the less than 25% of instances you do win your parlay, you are only paid for odds of 4 for 1, even though true odds were higher than that (~4.08 for 1 on 00 tables). You lose 1 for 1, but you win slightly less (at 2 for 1) each time you win, because your odds were not 50/50 to start with. And that's the unchanging house edge in action, and where the casino makes its money.

3b is the conditional state if and only if you won the first roll.
Losing on the first roll is covered by the sequence 3a - 3c.

Assuming a single zero wheel:
losing on the first roll: 19/37 (game over, lost 1, total -1)
winning on the first roll: 18/37 (staying in the game)
winning on the first roll and losing on the second roll: 18/37 * 19/37 (game over, lost 1, total -1)
winning on the first roll and winning on the second roll: 18/37 * 18/37 (game over, won 3, total +3)

(-1)*19/37 + 18/37*((-1)*19/37 + 3*18/37) =
-19/37 - 342/1369 + 3*324/1369 =
-703/1369 - 342/1369 + 972/1369 =
-73/1369 approx. -5.33%

Which again is the value the Wizard calculated for the Double Action Roulette.

Consider a standard double-zero roulette table. 50c bet on black has edge -2/38 = -5.26%. 50c bet on red has the same edge.

Now consider the playing strategy "Make a 50c black bet and a 50c red bet, and keep doing that until a 0 or 00 spins." In other words, you're eventually going to lose $1. That $1 loss is 100% of the initial $1 wager.

But the house edge for red or black didn't change, it's still -5.26%. In fact, you can calculate how long it is likely to take to lose that $1. It is, not surprisingly, 38/2 (19) spins. -2/38 * 38/2 = -1. In other words, your expected loss for a single $1 roulette bet is $0.0526. Your expected loss for 19 $1 roulette bets is $1.

Quote: MathExtremist

No, this is the opposite. The OP is discussing parlaying and treating the expected loss as a percentage of only the initial wager as opposed to the total action. That makes the edge look worse, not better. The expected loss increases, the denominator doesn't, so the fraction gets bigger. But it's still negative. You can't sign-flip unless you have a theoretical positive in there somewhere.

I really like your expression "total action" because that is exactly what I'm trying to discuss. I introduced "the friend" who does the actual betting exactly for the reason to hide the single rolls from you as a player of a game (not of rolls) and direct your attention to the game ("parlay") where you "put one unit in" and get the according result.

In the case of the Double Action Roulette the special wheel hides this sequence of rolls from you and for you.

So my point is that if you think in terms of games instead of rolls all the intermediate results should be irrelevant for the calculation of the house edge of this particular game. And this should be the reason why it should be appropriate to calculate only relative to the initial wager.

You are absolutely right that the parlay game makes the edge (look) worse.

(For a first glimpse of a better edge please refer back to my slot machine "D" I posted earlier. Since I'm interested in your opinion of my view of the "parlay" game I will discuss SLOT D at a later time.)

Quote: surrender88s

You have been fairly unclear.

If you take all of these, and multiply the payouts by probability, you should get the same result.

Well, that exactly is the point I try to discuss in this thread:

SLOT A
(-3)*51.4% + 3*48.6% = -0.084

SLOT B
(-3)*13.5% + (-1)*38.5% + 1*36.5% +3*11.5% = -0.08

SLOT C
(-3)*76.3% + 9*23.7% = -0.156

SLOT D
(-3)*26.4% + 1*73.6% = -0.056

Remark: SLOT A and SLOT B actually have the identical ev and differ here only because I intended this as a teaser and gave only approximate probabilities.
I will show you the exact math at a later time when the "parlay" discussion will be resolved.

Quote: ThomasK

So my point is that if you think in terms of games instead of rolls all the intermediate results should be irrelevant for the calculation of the house edge of this particular game. And this should be the reason why it should be appropriate to calculate only relative to the initial wager.

You are absolutely right that the parlay game makes the edge (look) worse.

Like I said, it's a different bet. Go back and look at my roulette "play until you lose" example. If you always lose a bet eventually, the edge is -100%. But it's clearly not the same bet as one spin on red. You can obviously win that one.

Quote: MathExtremist

Like I said, it's a different bet. Go back and look at my roulette "play until you lose" example. If you always lose a bet eventually, the edge is -100%. But it's clearly not the same bet as one spin on red. You can obviously win that one.

You call it "a different bet" I call it "the game".
In my humble opinion calculating ev and house edge serves the purpose to find out which bets have a lower disadvantage than others. And then choose those for your play. So if there's the simple bet on RED and "a different bet" one would compare both evs or house edges and decides to play the more favorable. In this case both would be played at a roulette table.

Isn't it like choosing

• basic strategy only or card counting on a single deck blackjack
• a 8/5 or a 9/6 jacks or better
• pass or don't pass at a craps table
• ...

In this case you may choose at a single zero roulette table between

• betting on red (-2.7%)
• single parlay betting on red (-5.33%)

Quote: ThomasK

...Isn't it like choosing

• basic strategy only or card counting on a single deck blackjack
• a 8/5 or a 9/6 jacks or better
• pass or don't pass at a craps table
• ...

In this case you may choose at a single zero roulette table between

• betting on red (-2.7%)
• single parlay betting on red (-5.33%)

The former has different payoffs. Blackjack odds change. The latter doesn't.

Quote: ThomasK

In this case you may choose at a single zero roulette table between

• betting on red (-2.7%)
• single parlay betting on red (-5.33%)

And

• betting on red and black until a zero appears (-100%)

I don't follow your point, though. Obviously if you bet more (which you do when you parlay) then your theoretical dollar loss increases. Why is that relevant to the initial bet? Here's an example where things break down:

a) Bet $1 red.
b) If win, parlay that and add $146 for a total of $148.

Total expected action on red is
100% * $1 +
18/37 * $148
= $73
Theoretical edge on red is -1/37 * $73 = -197.3%.

So you expect to lose nearly 200% relative to your initial bet with that strategy. But you can't lose 200% of a bet. It's easy to be misled by percentages. If you want to compare losses from one strategy to another, use Expected Loss Per Hour (ELPH).

https://wizardofvegas.com/forum/gambling/other-games/15219-is-the-element-of-risk-a-useful-concept/#post272469
https://wizardofvegas.com/forum/gambling/other-games/15219-is-the-element-of-risk-a-useful-concept/#post273521

ThomasK seems like a smart guy that doesn't quite speak the gambling lingo. And I also don't follow his point.

Betting systems do not change the house edge. The confusion seems to be caused by how the house edge is defined.

Suppose there is a sports bet with a 48% chance of winning and pays even money.

Betting one game at a time the house edge is .48 - .52 = 4.00%.

If you do a two-team parlay (both with 48% chance of winning) then the house edge is 4*.48^2 - 1 = 7.84%.

If you call the parlay a "betting system" did it change the house edge? No. Because with a betting system the money runs through your hands as you play. Had you planned to parlay your winnings by hand with the two-team parlay, then the total expected action would have been .48*3 + .52*1 = 1.96 units. The expected loss divided by the expected action is 7.84%/1.96 = 4.00%, the same as the house edge on a single game.

Bottom line -- with a game of luck, the ratio of expected loss to expected action is always the same. Not only can't a betting system beat that house edge, it can't even dent it. I see others have already said the same thing. Listen to those who did.

Quote: Wizard

Bottom line -- with a game of luck, the ratio of expected loss to expected action is always the same. Not only can't a betting system beat that house edge, it can't even dent it. I see others have already said the same thing. Listen to those who did.

Well, what about card counting and wagering more on positive expectation hands and less on negative expectation hands? Do you not consider that a 'betting system'?

Or you could take it a step further and implement wonging only strategy where you are only wagering on positive expectation rounds....That is really changing the house edge, no? Would that not be a betting strategy that changes the house edge (to the player)

Quote: kewlj

Well, what about card counting and wagering more on positive expectation hands and less on negative expectation hands? Do you not consider that a 'betting system'?

Or you could take it a step further and implement wonging only strategy where you are only wagering on positive expectation rounds....That is really changing the house edge, no?

As you say, blackjack is a game where the instantaneous edge changes from hand to hand. You're not changing the house edge on any single hand, but you *are* changing the average edge on your total action by weighting your action heavier when it's positive and lighter (or zero) when it's not.

On a game like roulette or craps, where the edge is constant for a given wager, none of that applies. Unless you can control the dice or clock the wheel...

Quote: Wizard

Betting systems do not change the house edge. The confusion seems to be caused by how the house edge is defined.

Suppose there is a sports bet with a 48% chance of winning and pays even money.

Betting one game at a time the house edge is .48 - .52 = 4.00%.

If you do a two-team parlay (both with 48% chance of winning) then the house edge is 4*.48^2 - 1 = 7.84%.

If you call the parlay a "betting system" did it change the house edge? No. Because with a betting system the money runs through your hands as you play. Had you planned to parlay your winnings by hand with the two-team parlay, then the total expected action would have been .48*3 + .52*1 = 1.96 units. The expected loss divided by the expected action is 7.84%/1.96 = 4.00%, the same as the house edge on a single game.

Bottom line -- with a game of luck, the ratio of expected loss to expected action is always the same. Not only can't a betting system beat that house edge, it can't even dent it. I see others have already said the same thing. Listen to those who did.

May I please first verify that my understanding of "parlay" is correct?
I wager on the first bet and then I let ride the pay of the first bet which consists of my initial wager plus the winning amount (only if I win the first bet, of course).

Now: You mention the "total action" as MathExtremist did earlier.
My question still is: Why do I have to calculate using the "total action"?
I only brought a single unit to start betting from home and I go back home with whatever the result is: Either that one unit is lost (no matter whether on the first bet or on the second) or I bring home the four units which would give me a plus of three units.

This would be exactly what I experienced if I stayed at home, gave that one unit to a friend, and sent her to play the game. She'd bring back home the same amounts (either nothing or four) and with the same probabilities you showed above. If I didn't ask her what exactly she did I had no idea at all about the total action.

Perhaps I now can make my motivation clear:
If I can show that there is at least one more house edge than the well known one
it is not unlikely that there exist more (if not an infinite number of) house edges and
there certainly will be some worse than the well known but also some better than that well known one.
And now I am able to choose the lower or even the lowest of them.
Isn't that what advantage play is all about?

Quote: ThomasK

Perhaps I now can make my motivation clear:
If I can show that there is at least one more house edge than the well known one
it is not unlikely that there exist more (if not an infinite number of) house edges and
there certainly will be some worse than the well known but also some better than that well known one.
And now I am able to choose the lower or even the lowest of them.
Isn't that what advantage play is all about?

I highlighted the part that's wrong. If you're increasing the expected dollar loss and leaving the initial wager size the same, the fraction can only increase (become more negative). You'll never find a lower expected loss from a parlay system as opposed to making a single bet.

Quote: MathExtremist

Quote: ThomasK

Perhaps I now can make my motivation clear:
If I can show that there is at least one more house edge than the well known one
it is not unlikely that there exist more (if not an infinite number of) house edges and
there certainly will be some worse than the well known but also some better than that well known one.
And now I am able to choose the lower or even the lowest of them.
Isn't that what advantage play is all about?

I highlighted the part that's wrong. If you're increasing the expected dollar loss and leaving the initial wager size the same, the fraction can only increase (become more negative). You'll never find a lower expected loss from a parlay system as opposed to making a single bet.

Yes I am aware that parlay worsens the house edge (in the sense I'm discussing here).
Back some posts I offered the SLOT machines A, B, C, and D. In response to surrender888s' remark I calculated the expected values of all four.
(Can anyone please confirm that the equations are carried out correctly?)

I use slot machines here because it hides the total action from the player (and I won't have to bother my friend running around ...)

SLOT D seems to be of interest for advantage play.
Here is the exact calculation
ev: (-3)*19^2/37^2 + 1*(18/37 + 19/37*18/37) = -75/1369 approx. -0.0548
Divided by the 3 initial units the hous edge is approx. -1.8%.
This is the result of applying a two step Martingale.

I give here the short descriptions of the machines:

SLOT A
3 coins on RED, single roll: -2.7%(!)

SLOT B
3 consecutive rolls on RED, 1 unit each: -2.7%(!)

SLOT C
3 coins parlay on RED: -5.3%

SLOT D
two step Martingale (1 coin on first roll, two coins on second roll if first roll was a loser): -1.8%

(Sorry I haven't managed yet to do the formatting of the detailed table containing all intermediate steps and results.)

Quote: Wizard

Bottom line -- with a game of luck, the ratio of expected loss to expected action is always the same. Not only can't a betting system beat that house edge, it can't even dent it. I see others have already said the same thing. Listen to those who did.

I'm feeling extra patient and need a distraction from work, so here's why you are wrong with your house edge calc on "slot" D.

Slot D is not a game where you wager 3 coins and have a ~73% chance of winning 4.

It's a game where you bet 1 coin and have a ~48% chance of winning 1, and a ~52% chance of playing another game for 2 coins. That second game has a ~48% chance of winning 2 coins.

If you step up to a roulette wheel with 2 coins in your pocket, and bet 1 coin, the house edge is based on your bet, or your action. The house edge is not lower just because you have more money in your pocket.

Does this make sense? I hope some of this is sinking in, I feel like we've been very nice and helpful here. I don't know why I keep posting in this thread, heh...

Quote: surrender88s

Quote: Wizard

Bottom line -- with a game of luck, the ratio of expected loss to expected action is always the same. Not only can't a betting system beat that house edge, it can't even dent it. I see others have already said the same thing. Listen to those who did.

I find the idea quite intriguing and therefore wanted to discuss it with you here and perhaps have my calculations and the idea itself verified by the Wizard.
If it worked you could all go out and advantage play roulette.
And since the strategy(?) is a betting strategy(?) you could even apply it to your favorite game and give the strategy you already use an extra kick.

Please spare your benevolence. I think you're on to something here. All the casinos are fools, offering their beatable roulette wheels to the masses. ThomasK, you and you alone should take these casinos for all their worth using your betting systems. Leave us doubters and nay-sayers behind. Good luck and have fun! :-)

Quote: surrender88s

Slot D is not a game where you wager 3 coins and have a ~73% chance of winning 4.

It's a game where you bet 1 coin and have a ~48% chance of winning 1, and a ~52% chance of playing another game for 2 coins. That second game has a ~48% chance of winning 2 coins.

You're trying to manipulate SLOT D!
SLOT D wants 3 coins in (as SLOTs A, B, and C do) and is going to pay 4 coins out, approx. 73% of the games.
That first single coin will sit there in the machine until two more are inserted. And only then the game will start.

But we're coming closer to my point ...

You keep saying "advantage play" but in order to have an overall advantage, you need to have a positive EV at some point. That is never true with roulette. Playing games with the denominator won't change the sign of the fraction. The edge is still negative regardless of how much you squeeze and knead it.

You cannot beat roulette with accounting tricks. You can make the negative edge almost as small as you want by cooking the books with the denominator but it doesn't change the expected loss. Each dollar you wager in roulette has a theoretical cost of about 5.26c. There is nothing you can do about that, at all. If your system wagers an average of 4 dollars over the course of some complex betting pattern, the average dollar loss will be about 21c. You can artificially deflate a ratio by, for example, dividing 21c by a million dollars. That's no different than saying "100x odds on craps is the best bet on the table because the house edge is 0.0002%" But that's misleading. Odds bets have zero cost at all. The line bet has a 1.41% edge, so a $5 line bet costs 7c. That 7c theoretical cost doesn't change when you pile on more odds.

Quote: MathExtremist

You keep saying "advantage play" but in order to have an overall advantage, you need to have a positive EV at some point. That is never true with roulette. Playing games with the denominator won't change the sign of the fraction. The edge is still negative regardless of how much you squeeze and knead it.

You cannot beat roulette with accounting tricks. You can make the negative edge almost as small as you want by cooking the books with the denominator but it doesn't change the expected loss. Each dollar you wager in roulette has a theoretical cost of about 5.26c. There is nothing you can do about that, at all. If your system wagers an average of 4 dollars over the course of some complex betting pattern, the average dollar loss will be about 21c. You can artificially deflate a ratio by, for example, dividing 21c by a million dollars. That's no different than saying "100x odds on craps is the best bet on the table because the house edge is 0.0002%" But that's misleading. Odds bets have zero cost at all. The line bet has a 1.41% edge, so a $5 line bet costs 7c. That 7c theoretical cost doesn't change when you pile on more odds.

I admit that I may have used the term "advantage play" inappropriately but in the good faith that everything that makes you "a more knowledgeable gambler"¹⁾ could be subsumed under it. I'm sorry for that mistake.
I never claimed to have a positive expected value and I stated that in my initial post and also during the discussion:

Quote: ThomasK

#post469907: ...PS.: I'm not stating that betting systems can overcome a negative expectation. They cannot! ...
#post470058: ... I don't doubt that if you go and bet one unit on RED infinite times you will have the well known house edge. ...

What I would like to contribute is a way of losing less. Just like the advise to play single zero roulette instead of double zero, betting the don't pass in craps instead of the pass line, or playing blackjack along proper basic strategy instead of intuitive decisions.

Talking about the expected loss is exactly the point that I'm trying to make but with a certain twist:

Quote: Don Schlitz. "The Gambler"

You never count our money when you're sittin' at the table. There'll be time enough for countin' when the dealin's done.

Say, I go to a casino playing at a single zero roulette wheel. I have $300 with me and play exactly 100 times wagering $3 each. I repeat this kind of session forever. After each session when I get home checking my bankroll I will find that I lost different amounts depending on the way I chose to play:

• If I bet $3 on RED 100 times, on average I lose $8.10.
• If I bet $1 each on RED three consecutive rolls and do that 100 times, on average I lose $8.10.
• If I bet $3 single parlaying on RED 100 times, on average I lose $16.00
• If I use the $3 for a two step Martingale on RED, as described before, and do that 100 times, on average I lose $5.48

Summary:

• I play roulette.
• I can lose more or less of my money.
• The math as shown before helps me to decide which of the ways eats more or less of my bankroll.

__________
¹⁾ Bob Dancer. Richard W. Munchkin. "Gambling With an Edge"

Quote: ThomasK

What I would like to contribute is a way of losing less.
Talking about the expected loss is exactly the point that I'm trying to make but with a certain twist:

Okay, but the only way you can have a lower theoretical loss is to have a lower total action. The house edge on roulette is inviolate: $1/37 per dollar wagered in the single zero game, discounting any French rules like la partage.

Quote:

Say, I go to a casino playing at a single zero roulette wheel. I have $300 with me and play exactly 100 times wagering $3 each. I repeat this kind of session forever. After each session when I get home checking my bankroll I will find that I lost different amounts depending on the way I chose to play:

• If I bet $3 on RED 100 times, on average I lose $8.10.
• If I bet $1 each on RED three consecutive rolls and do that 100 times, on average I lose $8.10.
• If I bet $3 single parlaying on RED 100 times, on average I lose $16.00
• If I use the $3 for a two step Martingale on RED, as described before, and do that 100 times, on average I lose $5.48

This isn't entirely accurate. Your first bullet is correct; 3*100*1/37 = $8.10. In your second bullet, you appear to be saying that "three spins = 1 play" so 100 plays = 300 spins. If so, you again have $300 in action. In your 3rd example, you have wagered significantly more than $300 in action. In your 4th example, you have wagered significantly less.

In other words, your premise that you "play exactly 100 times wagering $3 each" is not true in either of your last two examples. You are not changing the expected loss relative to a fixed wager amount of $300. You are instead changing the expected wager amount, and because you have changed the wager amount, the expected loss that results from the unvarying house edge times that wager amount also changes. I would redirect your inquiry, and rephrase it, to investigate the shape of the distributions of various roulette betting systems vis-a-vis wager size and win amount. This has been done before but it's a worthwhile effort to reproduce if you want to get your hands dirty. (see Ken Elliott's paper: http://www.conjelco.com/downloads/elliott-paper.pdf)

Quote: ThomasK

Talking about the expected loss is exactly the point that I'm trying to make but with a certain twist:

huh?
no twist
just a different way to increase or decrease the average bet

i agree with ME that you have not done anything to change the house edge
is that what you are trying to prove?


just a reminder the house edge comes directly from the fact the casino short pays on any win

Quote: ThomasK

Say, I go to a casino playing at a single zero roulette wheel. I have $300 with me and play exactly 100 times wagering $3 each. I repeat this kind of session forever. After each session when I get home checking my bankroll I will find that I lost different amounts depending on the way I chose to play:

• If I bet $3 single parlaying on RED 100 times, on average I lose $16.00

i agree on the avg loss
ok
you lose more because the avg bet is more than 3, i say
no HE is changed as expected

3 + (18/37 *6) = 3 + 108/37 = 219/37 = avg bet for system (5.9189189189189189189189189189189)
and you do this 100 times = 21,900/37 total wagered = A (591.89189189189189189189189189189)
A * HE (1/37) = 21,900 / 1,369 = expected loss = B (15.997078159240321402483564645727)
HE = B/A = 1/37
as expected

not even a dent in the house edge
because the payoff on a win does not change
next

Quote: ThomasK

• If I use the $3 for a two step Martingale on RED, as described before, and do that 100 times, on average I lose $5.48

yes, i agree on the average loss

now you have a much lower average bet for the system you could choose
see
1 + (19/37 *2) = 1 + 38/37 = 75/37 = avg bet for system (2.027027027027027027027027027027)
and you do this 100 times = 7,500/37 total wagered = A (202.7027027027027027027027027027)
A * HE (1/37) = 7,500 / 1,369 = expected loss = B (5.4784514243973703433162892622352)
HE = B/A = 1/37

again as expected
sweet!

Quote: ThomasK

Summary:

• I play roulette.
• I can lose more or less of my money.
• The math as shown before helps me to decide which of the ways eats more or less of my bankroll.

so, in other words,
bet more (total action) = lose more
bet less = lose less

this was fun

thanks!

Quote: MathExtremist

Okay, but the only way you can have a lower theoretical loss is to have a lower total action. The house edge on roulette is inviolate: $1/37 per dollar wagered in the single zero game, discounting any French rules like la partage.

It seems I use the term "house edge" inappropriately, is it?
Because I am not completely sure and since the "house edge" is derived from the expected value I intentionally constructed the examples with an identical coin in so that it would be possible to compare the games by their expected value, too.

I'd like to think of the expected value as a pure mathematical equation where I am free to do equivalent transformations. Then I try to match the equations with some real world experiments (sort of reverse engineering). Afterwards I'd like to claim that the experiments are as equivalent as are the equations that describe them. (Perhaps there is the flaw in my thinking.)

equivalent equations	type of gamble
(-3)*1045/1369 + 9*324/1369	SLOT C: 3 coins in; 12 coins out approx. 23.7% of the games
(-3)*(703/1369 + 342/1369) + 9*324/1369	{intermediate step}
(-3)*(19/37 + (18/37)*(19/37)) + 9*((18/37)*(18/37))	$3 parlay on RED: 19/37 - losing $3 on first roll - game over; total -$3 of my own money (18/37)*(19/37) - winning on the first roll and letting ride on the second roll but losing on the second roll; total -$3 of my own money (18/37)*(18/37) - winning on the first roll and letting ride on the second roll and finally winning; total +$9 (payout $12 minus my own $3)

equivalent equations	type of gamble
(-3)*361/1369 + 1*1008/1369	SLOT D: 3 coins in; 4 coins out approx. 73.6% of the games
(-3)*361/1369 + 1*((666/1369 + 342/1369))	{intermediate step}
(-3)*((19/37)*(19/37)) + 1*(18/37 + (19/37)*(18/37))	$3 Martingale on RED: (19/37)*(19/37) - losing $1 on first roll and losing $2 on second roll; total -$3 of my own money (18/37) - winning $1 on the first roll - game over; total+$1 (payout $2 minus my own $1) (19/37)*(18/37) - losing $1 on the first roll but winning $2 on the second; total +$1 (payout $4 minus my $2 and minus my $1 of the first roll)

The equations in each table calculate the expected values of the slot machines and those of the play at the roulette table.
Since the equations within one table are equivalent I tend to conclude that the corresponding games are equivalent as well.
If the expected value of SLOT C is more negative than the expected value of SLOT D I would choose SLOT D for play.
If the roulette plays were equivalent to their corresponding slot machines isn't it then that I should choose the two-roll-Martingale?

Therefore your explanation below perfectly meets my question. Why do I have to think about the action at the table? Why may I not think in terms of "coin in" and "coin out" and corresponding probabilities?

Quote: MathExtremist

This isn't entirely accurate. Your first bullet is correct; 3*100*1/37 = $8.10. In your second bullet, you appear to be saying that "three spins = 1 play" so 100 plays = 300 spins. If so, you again have $300 in action. In your 3rd example, you have wagered significantly more than $300 in action. In your 4th example, you have wagered significantly less.

In other words, your premise that you "play exactly 100 times wagering $3 each" is not true in either of your last two examples. You are not changing the expected loss relative to a fixed wager amount of $300. You are instead changing the expected wager amount, and because you have changed the wager amount, the expected loss that results from the unvarying house edge times that wager amount also changes.

In the case of the parlay: If I should win on the first roll I'd receive extra temporary $3 from the house that I am supposed to wager on the second roll. Perhaps here is the misunderstanding: The rule that I have to obey says that I'm not allowed to leave the table at that point!
That is what the slot machine does. And by the way the same the Double Action Roulette wheel does.
(I'm sorry that I assumed that this extra rule would be implicitly clear from the combination of slot machine and roulette play. Mea culpa!)

For the Martingale the rule says that I am not allowed to start playing if I am not willing to wager $3. That means I am not allowed to leave the table after losing $1 on the first roll.
This again is what the slot machine does: The game won't start before I put all three coins in.
And thus I cannot avoid losing two of my three coins by some "early exit".

Assuming that this hidden extra rule caused the misunderstanding:
Is it correct when I am calling it the "house edge" provided I am playing after this rule at the table?

Quote: MathExtremist

I would redirect your inquiry, and rephrase it, to investigate the shape of the distributions of various roulette betting systems vis-a-vis wager size and win amount. This has been done before but it's a worthwhile effort to reproduce if you want to get your hands dirty. (see Ken Elliott's paper: http://www.conjelco.com/downloads/elliott-paper.pdf)

Thank you for that reference. It will take some time for me to go through it.
Interrestingly enough it starts with craps play where this extra rule can also be applied: If I were going to play on a 3-4-5X table with full odds a corresponding (or equivalent) slot machine would require 6 coins in to be prepared for an established point of 6 or 8 with 5 coins odds bet. Nevertheless, in case I were for example a frontline winner, 5 of these coins wouldn't have been used.
That is similar to the Martingale example where I win on the first roll and the other two coins are not "used".

Quote: MathExtremist

Consider a standard double-zero roulette table. 50c bet on black has edge -2/38 = -5.26%. 50c bet on red has the same edge.

Now consider the playing strategy "Make a 50c black bet and a 50c red bet, and keep doing that until a 0 or 00 spins." In other words, you're eventually going to lose $1. That $1 loss is 100% of the initial $1 wager.

But the house edge for red or black didn't change, it's still -5.26%. In fact, you can calculate how long it is likely to take to lose that $1. It is, not surprisingly, 38/2 (19) spins. -2/38 * 38/2 = -1. In other words, your expected loss for a single $1 roulette bet is $0.0526. Your expected loss for 19 $1 roulette bets is $1.

This of course also is my favorite way of picturing the house edge in roulette.

I like to take it even one step further and bet 18*50c on red and 18*50c on black which doesn't change the principles of calculating the house edge.
What I also can do then and is perfectly equivalent with respect to the house edge is to bet 50c straight on each of the 18 red numbers and 50c straight on each of the 18 black numbers.
If I now add 50c on zero and 50c on double-zero I have covered all numbers on the table and I will lose this $1 on each of the rolls.

The side effect of this is that I now won't experience a random experiment any more.

• If I am the only player versus the dealer at the table even the dealer (representing the house) will experience a non-random situation where he only spins the wheel to determine which 50c pieces to rake in and to which of the 50c piece to pay the 35 pieces of winnings to.
• If I am one among many players at the table the situation seems to be that
  
  • I will still experience the non-random experiment paying $1 for each roll.
  • All the other players playing the way they usually do will experience a random game with its 5.23% house edge.
  • The dealer (representing the house) will also still experience his well known random game with 5.23% house edge.

For me this seems to be an indication that a single player can actually experience the game of roulette in a different way than the house does:

• Covering the whole table results in a non-random situation.
• Playing in a way that simulates the payouts and corresponding probabilities of slot machines (as discussed before) should lead to a different house edge for the player - the house edge of the slot machine simulated.

There is a further point to be considered:
Playing 50c on red and simultaneously 50c on black never generates any winnings and therefore no progression can be constructed based on this model.
Only when I start betting 35 numbers or fewer I will win extra money with the corresponding probability.
And only then it is meaningful to "let ride" for a progression on winnings or "increase the wager" for a compensation in a progression on losses.
Since this model "red and black" cannot be applied to these multi-roll games any more, in my opinion this is another indication that a player might be able to experience a different house edge.

Algorithm for a two step Martingale.
Sitting at the roulette table assume two rolls: The player and the bettor.
As the player you own all the chips. As the bettor you don't own any chips and you depend on (yourself) the player.

1. (You) The player hand three chips to (youself) the bettor.
2. (You) The bettor bets one chip on RED.
3. IF RED wins (you) the bettor gives back the now four chips to (yourself) the player. GOTO 1.
4. ELSE (you) the bettor bets two chips on RED.
5. IF RED wins (you) the bettor gives back the now four chips to (yourself) the player. GOTO 1.
6. ELSE (you) the bettor has lost all three chips and cannot give anything back to (yourself) the player. GOTO 1.

You the player only sees either 4 chips coming back or all three chips lost.
The corresponding expected value (and the house edge?) has been calculated before and is smaller than the well known for roulette.

Quote: mustangsally

huh?
no twist
just a different way to increase or decrease the average bet

i agree with ME that you have not done anything to change the house edge
is that what you are trying to prove?


just a reminder the house edge comes directly from the fact the casino short pays on any win

[...]

That is exactly my question:
Why do I have to consider the "average bet" or the "total action"?
If I have got two random experiments that are mathematically equivalent with respect to money wagered, payout and their probabilities (see before) and one hides the particular details about how the random events are derived (slot machine) but the other shows every single step (multi roll game at a roulette table) why should then both expected values and house edges be different? Shouldn't they be equivalent as well?

Quote: mustangsally

[...]

Quote: ThomasK

Summary:

• I play roulette.
• I can lose more or less of my money.
• The math as shown before helps me to decide which of the ways eats more or less of my bankroll.

so, in other words,
bet more (total action) = lose more
bet less = lose less

this was fun

thanks!

If I only consider the "coin in" ($3) I won't wager more or less but always the same amount.
But by the math of the games I am free to choose

• A game where I lose $8.10 in the long run, e.g. $3 on RED, e.g. 3 times $1 on RED.
• A game where I lose $16.00 in the long run, e.g. $3 single parlay on RED.
• A game where I lose $5.48 in the long run, e.g. $3 in a two step Martingale on RED.

Quote: ThomasK

If I only consider the "coin in" ($3) I won't wager more or less but always the same amount.

That's like saying "2 + 3 = 2 because I only consider the first value." It's wrong to consider it that way. Your parlay does not have $3 in coin in (handle) and neither does your martingale. You can "consider" whatever you want but it doesn't actually make it correct.

But I don't even know what your point is. It's not like you believe you can decrease the expected loss by using any of these strategies. You certainly don't believe you can beat the game.

Edit: Yes I do know what your point is. If the martingale and the parlay were contract bets, they would have different probability distributions. Your mistake is thinking that a combination of bets, where later bets are conditional (that is, they may not occur at all) is equivalent to a contract bet that goes for longer than one outcome. In your parlay, you are not putting $3 at risk if you lose the first bet. In the martingale, you are not putting $3 at risk if you win the first bet. You only wager $3 *some of the time*. If these were contract bets and you actually did wager $3 all of the time, as in your slot game example, the way you're calculating the return would be correct.