16 Consecutive Banks in Baccarat

What are the odds of 16 consecutive banks in eight deck baccarat?

see Murphy's law, odds are you will go broke

about 1 in 260,000

Odds that this will happen to someone somewhere over a period of a few years and they will post on this board: very good.

Quote: rdw4potus

about 1 in 260,000

Thanks, that's what I got, a little over 261,000 to 1. Guy had it this morning and won £58,000. No outlay stake, won by playing game and being on the shoe.

Follow the dragon...

http://www.worldgamingmag.com/en/gaming/gaming-insights/item/70-fate-in-the-cards-understanding-baccarat-trends

That's nuts... this morning I dealt 11 consecutive players in 8-deck bacc... unfortunately for one of our customers it cost him $1500 in about 2 minutes when he jumped in to the shoe after seeing 6 straight Player wins and bet on the Banker $100, $200, $300, $300, $600, all losses.

At my casino we very rarely see big action (our table max is $300) like that during the day... I couldn't believe it didn't bounce once for him!

Quote: kijani

That's nuts... this morning I dealt 11 consecutive players in 8-deck bacc... unfortunately for one of our customers it cost him $1500 in about 2 minutes when he jumped in to the shoe after seeing 6 straight Player wins and bet on the Banker $100, $200, $300, $300, $600, all losses.

At my casino we very rarely see big action (our table max is $300) like that during the day... I couldn't believe it didn't bounce once for him!

I guess this is why Stoneynv's system for martingaling a player/banker streak doesn't start until win #13. He could have won on the 16 banker streak if he had been awake to see it though. :) (FMI see the betting systems area)

Quote: skrbornevrymin

I guess this is why Stoneynv's system for martingaling a player/banker streak doesn't start until win #13.

Never heard of it, but how long would I have to sit there making minimum bets and staring at the dealer's tits waiting for a string of 13 to take place?

I often see punters doing their absolute cobblers taking a Martingale-like approach to baccarat. Not to be recommended.

Hello,

1st post on here :)

Agree with TheSweeney's answer about the odds of 261k but if the question were changed and ties were ignored the odds of pulling off a 16 timer on the Bank I get as 52,700 to 1 and change.

Interestingly Genting now a new table game side bet in the UK where the jackpot is for just that - 16 Bank wins.

so .506^16 = 54,149 to 1 of it happening (bank winning - omit ties)

the odds of it hitting again? about 50%

Banker:

(0.458597)^16 = 0.000003827360320856526 = 1/0.000003827360320856526 = 1 in 261276.68057555915

Yay, an easy one!!!

Paytable:

Eight Consecutive: 100

Nine Consecutive: 150

Ten Consecutive: 250

Eleven Consecutive: 350

Twelve Consecutive: 500

Thirteen Consecutive: 1000

Fourteen Consecutive: 2000

Fifteen Consecutive: Base of 2500

Sixteen Consecutive: Base of 10000

Probability of Loss and EV of Loss

(1 - 0.458597 ) + (0.458597 * (1-0.458597)) + (0.458597^2 * (1-0.458597)) + (0.458597^3 * (1-0.458597)) + (0.458597^4 * (1-0.458597)) + (0.458597^5 * (1-0.458597)) + (0.458597^6 * (1-0.458597)) + (0.458597^7 * (1-0.458597)) = 0.9980436359436812

EV of Loss = 0.9980436359436812 * -1 = -0.9980436359436812

Probability of Win(s) and EV of Win(s)

(0.458597)^8 * 100 = 0.19563640563188966

(0.458597)^9 * 150 = 0.13457740307035154

(0.458597)^10 * 250 = 0.10286132219309001

(0.458597)^11 * 350 = 0.06604065128329828

(0.458597)^12 * 500 = 0.043265777937952485

(0.458597)^13 * 1000 = 0.03968311193002239

(0.458597)^14 * 2000 = 0.03639711216354496

(0.458597)^15 * 2500 = 0.020864508058581532

(0.458597)^16 * 10000 = 0.03827360320856526

Here comes the pain!

Base EV & House Edge

0.19563640563188966 + 0.13457740307035154 + 0.10286132219309001 + 0.06604065128329828 + 0.043265777937952485 + 0.03968311193002239 + 0.03639711216354496 + 0.020864508058581532 + 0.03827360320856526 - 0.9980436359436812 = -0.32044374046638513

Base House Edge of 32.044374046638513%!!!

Advantage Point

As with any uncapped Progressive, there will be an advantage point. Results Banker^15 and Banker^16 must equal the House Edge, then add those amounts to the base and you have your break-even point.

0.32044374046638513 = ((0.458597^15) * .25x) + ((0.458597^16) * x)

0.32044374046638513 = (0.000008345803223432613 * .25x) + (0.000003827360320856526 * x)

x = 54185.65686

PROOF

(0.19563640563188966 + 0.13457740307035154 + 0.10286132219309001 + 0.06604065128329828 + 0.043265777937952485 + 0.03968311193002239 + 0.03639711216354496) + (0.458597^15 * (64185.65686/4)) + (0.458597^16 * 64185.65686) - 0.9980436359436812 = x

0.6184617842101493 + 0.1339202154800819 + 0.24566163623407647 - 0.9980436359436812 = x

0.9980436359243077 - 0.9980436359436812 = x

x = -1.9373502802011444e-11

Errors, rounding, you know the drill.

AP

Play at a jackpot of $64,185.66, or better.

Quote: Mission146

Play at a jackpot of $64,185.66, or better.

It's capped at $60,000, so you can't. Just noticed that. The best EV at $60,000 is as follows:

0.19563640563188966 + 0.13457740307035154 + 0.10286132219309001 + 0.06604065128329828 + 0.043265777937952485 + 0.03968311193002239 + 0.03639711216354496 + (0.458597^15 * (60000/4)) + (0.458597^16 * 60000) - 0.9980436359436812 = -0.024753184130651084

That's still a House Edge of 2.4753184130651084%, optimal conditions.

The bet is ridiculous, and the cap of $60,000 disgusts me.

Let's say the jackpot increases indefinitely, but a win can only be worth up to $60,000, the 25% of Max Amount can be based on a JP of over 60K, though:

First calculate the Base ER for $60,000:


(0.458597)^16 * 60000 = 0.22964161925139154

(0.458597)^15 * 15000 = 0.1251870483514892

0.19563640563188966 + 0.13457740307035154 + 0.10286132219309001 + 0.06604065128329828 + 0.043265777937952485 + 0.03968311193002239 + 0.03639711216354496 + 0.22964161925139154 + 0.1251870483514892 - 0.9980436359436812 = -0.024753184130651084

Okay, so we need to compensate with:

0.024753184130651084 = (0.000008345803223432613 * x)

x = 2965.943896

The jackpot would need to be (60,000 + (2965.943896 *4)) = 71863.775584

So, then:

(0.19563640563188966 + 0.13457740307035154 + 0.10286132219309001 + 0.06604065128329828 + 0.043265777937952485 + 0.03968311193002239 + 0.03639711216354496) + (0.458597^15 * (71863.775584/4)) + (0.458597^16 * 60000) - 0.9980436359436812 = -2.8940183582903955e-12

Errors, rounding.

AP

$71863.78, or better.

Quote: THESWEENEY

I often see punters doing their absolute cobblers taking a Martingale-like approach to baccarat. Not to be recommended.

I would agree that if things are going against you in a two option game, its just better to either walk or switch than it is to keep this " double up to catch up " thing going. Too often: you don't catch up and if you do happen to catch up... that is just about all you do: catch up, not get ahead.

Greetings!

Okay, so the first thing that I want to do is apologize to Genting Casino, if the website below is to be believed, the bet is not a bet at all...it's free.

It will still be a sufficiently easy problem, and will result in a reduction to the overall House Edge. How it works is, basically, they pass the shoe around, and if the player dealing deals the requisite number of Bankers, he wins. If the player deals a Player/Tie, then he passes the shoe to the next player who gets to deal.

I don't know whether or not you can wong out, so the Optimal Strategy (if you can't) would be to play with as few people as possible at the table, (so you can deal as often as possible) and do bet nothing but Banker at Table Minimum. The reason for Table Minimum is because you're looking to cut into the House Edge as much as possible, and possibly turn it into a Player Advantage, but the payouts are fixed, so the minimum bet will balance the scales faster.

http://www.ukcasinotablegames.info/puntoinstantbaccaratbonus.html

I just need some additional verification. I'm probably going to call Genting and figure out the Math on this. More to come.

Okay, here we go!

The methodology behind this is still going to be fairly simple and straightforward. The first thing that we have to do is determine the expected loss of each individual bet on Banker (optimal strategy) that has to be made, starting with that of one banker bet.

Remember, ties break the banker streak on this and the shoe gets passed to the next player for the deal, so as a result, we're going to be determining our probabilities on the premise that we only want the Banker to win. In other words, we're going to use the correct expected loss for Banker, but probability-wise, we'll be using the Banker's actual winning percentage and counting ties as a loss, with respect to this part of the bet.

In other words, most of the work that I did yesterday is still applicable to this, the only thing that I need to do is determine the expected loss on all of the Banker bets, by probability of the Banker not being the result, and get the expected losses from those. We will then tally up the expected value, based on all possible results of this bet, on the Base Pays, and then apply the formula from yesterday to determine our advantage point.

Expected Loss, Consecutive Bankers 0-7

Okay, so the Expected Loss on the Banker bet will be reflected as .010579, because I'm assuming an eight-deck shoe. I'm also assuming a minimum bet of 100 pounds, because I haven't received a response from Genting (it hasn't been that long since I requested one).

Now, we have to figure out how many hands we are likely to play that are subject to this expected loss. The tie, as we know, breaks the Banker's streak, but the overall House Edge includes the probability of a tie, so we only need our probabilities of the Bankers from yesterday.

Here we go:

Zero Bankers:

(100 * .010579) = 1.0579

One Banker:

(200 * .010579) = 2.1158

Two Bankers:

(300 * .010579) = 3.1737

Three Bankers:

(400 * .010579) = 4.2316

Four Bankers:

(500 * .010579) = 5.2895

Five Bankers:

(600 * .010579) = 6.3474

Six Bankers:

(700 * .010579) = 7.4053

Seven Bankers:

(800 * .010579) = 8.4632

The Streak Begins

You will recall that you get paid based on where the streak ends, so we're going to use the expected loss of Eight or more Bankers in a row (as appropriate) and add it to the Expected Values of getting paid on the bonus to achieve a new Expected Value of those consecutive Bankers.

Eight Bankers:

(900 * .010579) - 0.19563640563188966 = 9.325463594368111

Nine Bankers:

(1000 * .010579) - 0.13457740307035154 = 10.444422596929648

Ten Bankers:

(1100 * .010579) - 0.10286132219309001 = 11.53403867780691

Eleven Bankers:

(1200 * .010579) - 0.06604065128329828 = 12.6287593487167

Twelve Bankers:

(1300 * .010579) - 0.043265777937952485 = 13.709434222062046

Thirteen Bankers:

(1400 * .010579) - 0.03968311193002239 = 14.770916888069976

Fourteen Bankers:

(1500 * .010579) - 0.03639711216354496 = 15.832102887836454

Fifteen Bankers:

(1600 * .010579) - 0.020864508058581532 = 16.905535491941418

Sixteen Bankers:

(1700 * .010579) - 0.03827360320856526 = 17.946026396791435

Expected Loss

Okay, we have our expected losses for each individual occurrence, so now we have to multiply those expected losses by the probability of that occurrence actually happening, then we will sum that up for a total expected loss in pursuance of this bet.

We will also determine the probability of placing a certain number of pounds in action for each run, and we will sum that up. When we have tallied our total expected losses, we will then be able to compare those to the number of dollars bet in order to get an Effective House Edge.

(1.0579) * (1-.458597) = 0.5727502337000001

(1.0579*2) * ((.458597) * (1-.458597)) = 0.525323077848238

(1.0579*3) * ((.458597^2) * (1 - .458597)) = 0.3613673812979525

(1.0579*4) * ((.458597^3) * (1-.458597)) = 0.22096266261479616

(1.0579*5) * ((.458597^4) * (1-.458597)) = 0.1266660177339471

(1.0579*6) * ((.458597^5) * (1-.458597)) = 0.06970638688168192

(1.0579*7) * ((.458597^6) * (1-.458597)) = 0.03729499655557512

(1.0579*8) * ((.458597^7) * (1-.458597)) = 0.019546712611882386

9.325463594368111 * ((.458597^8) * (1-.458597)) = 0.009877357298151085

10.444422596929648 * ((.458597^9) * (1-.458597)) = 0.005073246659639646

11.53403867780691 * ((.458597^10) * (1-.458597)) = 0.0025692960853329165

12.6287593487167 * ((.458597^11) * (1-.458597)) = 0.0012901037827425716

13.709434222062046 * ((.458597^12) * (1-.458597)) = 0.0006422656606820383

14.770916888069976 * ((.458597^13) * (1-.458597)) = 0.0003173465888115973

15.832102887836454 * ((.458597^14) * (1-.458597)) = 0.00015598979698166148

16.905535491941418 * ((.458597^15) * (1-.458597)) = 0.00007638669685781084

17.946026396791435 * (.458597^16) = 0.00006868590934812334

0.5727502337000001 + 0.525323077848238 + 0.3613673812979525 + 0.22096266261479616 + 0.1266660177339471 + 0.06970638688168192 + 0.03729499655557512 + 0.019546712611882386 + 0.009877357298151085 + 0.005073246659639646 + 0.0025692960853329165 + 0.0012901037827425716 + 0.0006422656606820383 + 0.0003173465888115973 + 0.00015598979698166148 + 0.00007638669685781084 + 0.00006868590934812334 = 1.953688147722621

Therefore, the ultimate expected loss is 1.953688147722621 Pounds.

We will now determine the expected total bet.

100 * (1-.458597) = 100

200 * ((.458597) * (1-.458597)) = 49.657158318200004

300 * ((.458597^2) * (1-.458597)) = 34.15893574987735

400 * ((.458597^3) * (1-.458597)) = 20.88691394411534

500 * ((.458597^4) * (1-.458597)) = 11.973345092536824

600 * ((.458597^5) * (1-.458597)) = 6.589128167282532

700 * ((.458597^6) * (1-.458597)) = 3.525380145153145

800 * ((.458597^7) * (1-.458597)) = 1.8476900096306252

900 * ((.458597^8) * (1-.458597)) = 0.9532632322648978

1000 * ((.458597^9) * (1-.458597)) = 0.4857373983633169

1100 * ((.458597^10) * (1-.458597)) = 0.24503348504494426

1200 * ((.458597^11) * (1-.458597)) = 0.12258722306307959

1300 * ((.458597^12) * (1-.458597)) = 0.060902977129647366

1400 * ((.458597^13) * (1-.458597)) = 0.030078378187549884

1500 * ((.458597^14) * (1-.458597)) = 0.0147791292875098

1600 * ((.458597^15) * (1-.458597)) = 0.007229508644121739

1700 * (.458597^16) = 0.006506512545456093

54.14030000000001 + 49.657158318200004 + 34.15893574987735 + 20.88691394411534 + 11.973345092536824 + 6.589128167282532 + 3.525380145153145 + 1.8476900096306252 + 0.9532632322648978 + 0.4857373983633169 + 0.24503348504494426 + 0.12258722306307959 + 0.060902977129647366 + 0.030078378187549884 + 0.0147791292875098 + 0.007229508644121739 + 0.006506512545456093 = 184.70496927132635

Therefore, the expected total bet is 184.70496927132635 Pounds.

Effective House Edge

The Effective House Edge will simply be the Expected Loss divided by the total amount bet, under the Base Pays, which will be:

1.953688147722621/184.70496927132635 = 0.010577344807939133

It comes as no surprise that the effect on the House Edge is minimal, it reduces it by:

.010579 - 0.010577344807939133 = 0.0000016551920608669223, to be exact.

Advantage Play

Can't happen.

The only possible advantage play would be one where the value of the increase to the Progressives was sufficient to overcome the expected loss:

1.953688147722621 = (0.000008345803223432613 * .25x) + (0.000003827360320856526 * x)

x = 330360.2543

You'd have to play at a Progressive point of 340,360.26, or better.

Unfortunately, that can't happen because the Maximum Payout is only $60,000, which is fine, because this is still a great deal for the player. Let's look at the expected loss assuming an increase to the Progressive of 50,000:

1.953688147722621 - ((0.000008345803223432613 * 12500) + (0.000003827360320856526 * 50000)) = 1.657997591386887

The average total bet of 184.70496927132635 would remain the same, so the effective House Edge would be:

1.657997591386887/184.70496927132635 = 0.008976464455330032 = .8976464455330032%

Again, I await information about the Table Minimum, and will adjust accordingly. The Effective House Edge is mainly going to be a function of Expected Loss, so if the expected loss comes down enough on the Base Game, there could be an advantage.

Theoretical Advantage Point

If we again use the maximum Progressive increase with additional value of:

((0.000008345803223432613 * 12500) + (0.000003827360320856526 * 50000)) = 0.29569055633573393

It would overcome an Expected Loss of 0.29569055633573393 on the whole thing.

Just using the regular House Edge for the Banker Bet at Baccarat, this expected loss would reflect an average bet of:

.29569055633573393/.010579 = 27.950709550594002

Which is roughly 28 pounds. This, of course, does not factor in the value of those increases. If the Table Minimum is 25 pounds, or less, we may have an advantage point. I eagerly await information concerning the minimum bet.

Okay, so I called Genting Casino, Manchester, and I have the following to offer:

1.) The player does not have to be banking the Table in order to qualify. The dealer simply passes a token around to indicate which player is, "Dealing," and if the requisite amounts of consecutive bankers come at that time, the player wins the appropriate bonus.

2.) The table minimums are either 5 pounds or 10 pounds, as of ten minutes ago, they had two at five and one at ten.

Now, figuring out the average bet necessary for this will be easy, because it is simply going to be the bet of 5 pounds multiplied by 1.8470496927132635 which is simply the earlier average bet, divided by the base bet of 100, so the multiplier remains the same.

5 * 1.8470496927132635 = 9.235248463566318

The expected loss on a 5 pound bet is 5 * .010579 = 0.052895 pounds on the Base Game of Baccarat.

Expected Loss

The Expected Losses will be the same as before, the only difference being that we have to divide the Base Expected Loss we were using by 20.

Zero Bankers:

((100 * .010579)/20) = 0.052895000000000004

One Banker:

((200 * .010579)/20) = 0.10579000000000001

Two Bankers:

((300 * .010579)/20) = 0.158685

Three Bankers:

((400 * .010579)/20) = 0.21158000000000002

Four Bankers:

((500 * .010579)/20) = 0.264475

Five Bankers:

((600 * .010579)/20) = 0.31737

Six Bankers:

((700 * .010579)/20) = 0.37026499999999995

Seven Bankers:

((800 * .010579)/20) = 0.42316000000000003

Eight Bankers:

((900 * .010579)/20) - 0.19563640563188966 = 0.28041859436811034

Nine Bankers:

((1000 * .010579)/20) - 0.13457740307035154 = 0.3943725969296485

Ten Bankers:

((1100 * .010579)/20) - 0.10286132219309001 = 0.47898367780691003

Eleven Bankers:

((1200 * .010579)/20) - 0.06604065128329828 = 0.5686993487167017

Twelve Bankers:

((1300 * .010579)/20 - 0.043265777937952485 = 0.6443692220620475

Thirteen Bankers:

((1400 * .010579)/20) - 0.03968311193002239 = 0.7008468880699775

Fourteen Bankers:

((1500 * .010579)/20) - 0.03639711216354496 = 0.757027887836455

Fifteen Bankers:

((1600 * .010579)/20) - 0.020864508058581532 = 0.8254554919414185

Sixteen Bankers:

((1700 * .010579)/20) - 0.03827360320856526 = 0.8609413967914349

Once again, we must take the expected loss for each occurrence and multiply it by the probability of that occurrence happening:

(.052895) * (1-.458597) = 0.028637511685000002

(.052895*2) * ((.458597) * (1-.458597)) = 0.02626615389241189

(.052895*3) * ((.458597^2) * (1 - .458597)) = 0.018068369064897623

(.052895*4) * ((.458597^3) * (1-.458597)) = 0.011048133130739807

(.052895*5) * ((.458597^4) * (1-.458597)) = 0.0063333008866973545

(.052895*6) * ((.458597^5) * (1-.458597)) = 0.003485319344084095

(.052895*7) * ((.458597^6) * (1-.458597)) = 0.0018647498277787556

(.052895*8) * ((.458597^7) * (1-.458597)) = 0.0009773356305941192

0.28041859436811034 * ((.458597^8) * (1-.458597)) = 0.00029701415072724905

0.3943725969296485 * ((.458597^9) * (1-.458597)) = 0.0001915615192183925

0.47898367780691003 * ((.458597^10) * (1-.458597)) = 0.00010669730895697445

0.5686993487167017 * ((.458597^11) * (1-.458597)) = 0.00005809606159746866

0.6443692220620475 * ((.458597^12) * (1-.458597)) = 0.0000301876953802258

0.7008468880699775 * ((.458597^13) * (1-.458597)) = 0.00001505738410781159

0.757027887836455 * ((.458597^14) * (1-.458597)) = 0.000007458808685723624

0.8254554919414185 * ((.458597^15) * (1-.458597)) = 0.0000037297735089551553

0.8609413967914349 * (.458597^16) = 0.0000032951329406623315

The total expected loss is:

0.028637511685000002 + 0.02626615389241189 + 0.018068369064897623 + 0.011048133130739807 + 0.0063333008866973545 + 0.003485319344084095 + 0.0018647498277787556 + 0.0009773356305941192 + 0.00029701415072724905 + 0.0001915615192183925 + 0.00010669730895697445 + 0.00005809606159746866 + 0.0000301876953802258 + 0.00001505738410781159 + 0.000007458808685723624 + 0.0000037297735089551553 + 0.0000032951329406623315 = 0.09739397129732712

Effective House Edge

The effective House Edge, with no meter increase, is:

0.09739397129732712/9.235248463566318 = 0.010545896158782618

Again, right off the top, the decrease to the House Edge is minimal:

.010579 - 0.010545896158782618 = 0.000033103841217381816

But, this can be AP'ed.

Advantage Play

Easy money, the value of the Progressive increase must simply overcome the Expected Loss:

0.09739397129732712 = (0.000008345803223432613 * .25x) + (0.000003827360320856526 * x)

x = 16468.90122

Good times, bet the minimum and play when the Jackpot is 26468.91, or better.

Not baccarat but about 10 years ago I won 16 hands in a row on Pai Gow Times (discounting half a dozen or so ties that occurred during). On betting relatively smelly but won around $3k.

Quote: Mission146

But, this can be AP'ed.

Advantage Play

Easy money, the value of the Progressive increase must simply overcome the Expected Loss:

0.09739397129732712 = (0.000008345803223432613 * .25x) + (0.000003827360320856526 * x)

x = 16468.90122

Good times, bet the minimum and play when the Jackpot is 26468.91, or better.

Good analysis.

I'm not sure about the assumption of independence though. Winning bank hands are associated with bank favourable cards. Each winning hand will deplete the pack of those bank favourable cards, reducing win expectation fractionally.

This is a very small effect to be sure, indeed very few people know it exists at all, but when you are talking about compound effects over a sixteen hand streak it becomes significant.

Btw A card-counter, for example using Thorp's linear system in the Mathematics Of Gambling, would be able to improve on the results you listed above without much effort.

Quote: GBV

Good analysis.

I'm not sure about the assumption of independence though. Winning bank hands are associated with bank favourable cards. Each winning hand will deplete the pack of those bank favourable cards, reducing win expectation fractionally.

This is a very small effect to be sure, indeed very few people know it exists at all, but when you are talking about compound effects over a sixteen hand streak it becomes significant.

Btw A card-counter, for example using Thorp's linear system in the Mathematics Of Gambling, would be able to improve on the results you listed above without much effort.

Thank you for the compliment, I'm about to confirm my results by running a probability analysis using the actual Expected Values of every possible result. In fact, that's probably the way I should have done it to begin with, but if I end up being wrong, at least I'll be able to figure out what jackpot you would need.

I also agree with you about the EoR, but the problem is, I suck, so I don't know how to do simulations.

I also agree with you about the potential for card counting the Banker bet, but the problem is, you'd have to be able to wong-in. You'll also notice that my analysis conveniently assumes that the individual playing is the only player. At a table of two (or more) people, you'd also have to eat the expected loss on all other hands you would play when not acting as, "Dealer." Whether or not they'd let you only bet when the button is on you, I have no idea.

I'm also a bit surprised this doesn't exist as an independent side bet, low hit rate would kill it, probably. The paytable would need to be more generous, and you'd probably have to start the pays at four (or so) consecutive Bankers.

It's going to have to wait until tomorrow, I'm tired.

I know I'm tired because there is no way the Base paytable for this results in a Player Edge, but that's what I'm getting. I think my problem is that I'm not really sure how to treat the Tie...

Quote: Mission146

I also agree with you about the EoR, but the problem is, I suck, so I don't know how to do simulations.

Yeah, I might screw this up too, but considering I have never heard of anyone even trying to count Baccarat, the EOR is a lot less than BJ. Mainly because payouts are very flat, unlike BJ, where BJ pays 3 to 2.

Quote: Mission146

(math for the baccarat side bet)

Somewhere along the way, my results disagree with yours. I made the educated guess that the prizes are on a "for one" basis as opposed to a "to one" basis. If so, I come up with an RTP of 38.441% (a house edge of 61.559%) at reset, an RTP of 63.2262% (house edge of 36.7738%) with a £60,000 jackpot, and a break-even jackpot amount of £134,184.89.

Consecutive Banker Wins	Probability	Prize	Return
16	0.0000038274	10,000	0.038274
15	0.0000045185	2,500	0.011296
14	0.0000098529	2,000	0.019706
13	0.0000214848	1,000	0.021485
12	0.0000468489	500	0.023424
11	0.0001021570	350	0.035755
10	0.0002227596	250	0.055690
9	0.0004857410	150	0.072861
8	0.0010591884	100	0.105919
7 or less	0.9980436215	0	0.000000
Totals	1.0000000000		0.384410

Consecutive Banker Wins	Probability	Prize	Return
16	0.0000038274	60,000	0.229645
15	0.0000045185	15,000	0.067778
14	0.0000098529	2,000	0.019706
13	0.0000214848	1,000	0.021485
12	0.0000468489	500	0.023424
11	0.0001021570	350	0.035755
10	0.0002227596	250	0.055690
9	0.0004857410	150	0.072861
8	0.0010591884	100	0.105919
7 or less	0.9980436215	0	0.000000
Totals	1.0000000000		0.632262

Consecutive Banker Wins	Probability	Prize	Return
16	0.0000038274	134,184.89	0.513581
15	0.0000045185	33,546.23	0.151579
14	0.0000098529	2,000	0.019706
13	0.0000214848	1,000	0.021485
12	0.0000468489	500	0.023424
11	0.0001021570	350	0.035755
10	0.0002227596	250	0.055690
9	0.0004857410	150	0.072861
8	0.0010591884	100	0.105919
7 or less	0.9980436215	0	0.000000
Totals	1.0000000000		1.000000

On average, you'd have to place the bet 511 times before it finally wins something. Few side bets have such a lousy RTP and hit frequency.

Quote: tringlomane

Yeah, I might screw this up too, but considering I have never heard of anyone even trying to count Baccarat, the EOR is a lot less than BJ. Mainly because payouts are very flat, unlike BJ, where BJ pays 3 to 2.

The EOR's for baccarat are roughly nine times smaller than that of BJ, but in this particular case those effects will be compounded over the sixteen wins.

Quote: JB

Somewhere along the way, my results disagree with yours. I made the educated guess that the prizes are on a "for one" basis as opposed to a "to one" basis. If so, I come up with an RTP of 38.441% (a house edge of 61.559%) at reset, an RTP of 63.2262% (house edge of 36.7738%) with a £60,000 jackpot, and a break-even jackpot amount of £134,184.89.

On average, you'd have to place the bet 511 times before it finally wins something. Few side bets have such a lousy RTP and hit frequency.

That's where our difference is, whether we are treating it as, "For One," or, "To One."

Unfortunately, I'm sorry about the work you did, but all of that is irrelevant. As I posted later on, Instant Baccarat Bonus is actually not a side bet, it's free, so my RTP numbers should be correct because they aren't even taking anything.

If you're so inclined, would you like to check my later posts and see if they make sense?

Quote: Mission146

Instant Baccarat Bonus is actually not a side bet, it's free, so my RTP numbers should be correct because they aren't even taking anything.

Aha, now it makes sense. At £100/hand, it reduces the house edge by 0.3844% when the jackpot is £10,000 and by 0.6323% when the jackpot is £60,000. Betting more than £100/hand will result in a proportionately smaller reduction in the house edge.

Quote: JB

Aha, now it makes sense. At £100/hand, it reduces the house edge by 0.3844% when the jackpot is £10,000 and by 0.6323% when the jackpot is £60,000. Betting more than £100/hand will result in a proportionately smaller reduction in the house edge.

Right, and the Table Minimum is £5.

.010579 - 0.010577344807939133 = 0.0000016551920608669223, to be exact.

.010579 * (1 - .003844) = 0.010538334324

I ended up with a House Edge of .010577344807939133 above on the base paytable. I guess mine was fairly close.

I thought I saw somewhere that the table minimum was £100, but now I don't know where I got that from. My figures above were a reduction in the house edge on the banker, player, or tie, if the bet was £100. At a £5 bet I think it would have a player advantage, but I'm not sure and I'm giving up on it.

Quote: THESWEENEY

What are the odds of 16 consecutive banks in eight deck baccarat?

I've seen 17 consecutive banks followed by 14 consecutive players. But I work in a casino at the table full time so that's not unusual considering I sit through about 20 shoes a day.

Just to clear this up, egalite's " as they are called in the UK" break the streak, the button stays with the current player. Casinos due to paranoia no longer give the shoes to the player to handle, rather a white button goes in a clock-wise direction after every punto result. It is not only Genting in the UK that offer this free "lucky banco" side bet in the UK, the Rank chain also has it, but at a lower payout. It has been reported that a lucky gentleman in chinatown birmingham had the button when 16 Banco's struck and got paid approx £58k just for being in the correct seat.

Another way to lower the EV, surely would be to bet only when the button came around to you. What some of us do, is after 7 banks, put £50 on the punto and £5 on the tie, or £25 on the Punto, £5 on Tie after 6 banks then the former. It becomes a no-lose proposition, because should those bets lose, you are getting paid £100 from the bank run regardless.....

It was only last Wen, the phenomenal happened.

Chinese guy two spots to my left has the button (not the shoe, they no longer give the players the shoe any more), there is a 9 Banco streak for which he gets paid £150 for nothing, then there is a 9-9 Tie (of all things) and another Banker. So everybody is thinking, if it wasn't for the Tie breaking the streak of 9, he would have got £250 instead of £150, little did we know. The Banco streaked for 12 more times. It was amazing to watch, basically every time hand the Punto was drawing dead with a really low score.

So the lucky Chinese guy ended up being paid twice, £150 for the B9 and anther £500 for the B12 streak, not bad for a freebie bet. Genting offers a much bigger bonus, however to get your hands on those, you need a B16 streak. In over 10000 shoes of Baccarat I've played in my life-time, I've only seen run of one-side of 22 hands ONCE, which was last week.

Although I did hear tales of something similar in Adelaide 2004 on the Player side, that run had many Ties in-between and punters were literally fighting to get money on the table and the tray had to be replenished three times within the shoe, this was followed by a Bank result, then another 17 Player streak. The 8 deck 70+ hand shoe finished with about 11 banks in total, however in this particular situation due to certain circumstances, it is my belief that perhaps under-hand shenanigans may have been in play, instigated by the casino which then back-fired, this is best laid to rest and was the first thing that sprang to my mind with the 22 hand sequence and no Punto, although last week was unfounded.

Not too sure - but I do know that I introduced a casino funded Progressive whereby an amount is entered into a Jackpot for every coup dealt (no play no cost - lots of play who cares about the cost) and in the space of 18 months my game generated in excess of £5 000 000 gross profit for the casino

Why do I not believe this ?