baccarat

Hi I have a question I believe will help my gambling addiction to baccarat and allow myself and others to perhaps ease off if they knew the rue odds.

The question is in baccarat what are the chances of say a 20 wining streak. As the way I play is by compounding half my winnings every-time I bet. Therefore to win big starting with around a $100AUD bet I would need around 20 in a row I have had 10's in a row countless amounts of times but never been on a run of 20 however I have seen these runs on tables next door to me in the casino, just I was never on one. I feel if I know the true odds It will help me to realize Its just not going to happen. Perhaps if you could give the odds of a 10 streak a 15 and 20 respectively.

Cheers

first sentence True* not rue

I guess you would just be ignoring ties:

(0.4462 / (0.4462 + 0.4586)) ^ 10 = 0.082% or 1 in 1223

For 15: 0.0023% or 1 in 42777

For 20: 0.000067% or 1 in 1'496'067

Good luck!

Quote: machine4

Hi I have a question I believe will help my gambling addiction to baccarat and allow myself and others to perhaps ease off if they knew the rue odds.

The question is in baccarat what are the chances of say a 20 wining streak. As the way I play is by compounding half my winnings every-time I bet. Therefore to win big starting with around a $100AUD bet I would need around 20 in a row I have had 10's in a row countless amounts of times but never been on a run of 20 however I have seen these runs on tables next door to me in the casino, just I was never on one. I feel if I know the true odds It will help me to realize Its just not going to happen. Perhaps if you could give the odds of a 10 streak a 15 and 20 respectively.

Cheers

Wow.

Based strictly upon this post, machine4, if I were you, I'd seriously reconsider my approach to this game.

Compounding your wins in the hopes of catching a 20-win streak and the unbelievably enormous windfall resulting from it, is, IMHO, tantamount to playing the lottery.

Nothing wrong with the utlilization of a "positive" (read: up-as-you-win) progression. It will, in fact, compound your winning streaks into some nice bankroll increases. IF you allow it to....meaning, that your positive progression is not designed to overcome tremendous odds all in the hopes of an enormous windfall. At some point....at some REALISTIC point....you need to accept a profit.

Slow your roll, machine4. Try as you might, you're not going to take home their chandeliers with you.

Much better to catch your "winning jag" and seek to terminate the session as a winner. You won't get your wished-for windfall, but, trust me on this, it'll happen a heck-of-alot more often than your desired 10, 15, and 20 winning streaks.

And, should you get really good at timing your exit strategies on both a shoe-to-shoe and a session-to-session basis, you just might be able to add up all those lesser wins (and, BTW, much lesser losses) into you're own little windfall.

Maybe you can get their chandelier.....just better to attempt it piece-by-piece.

I wish you all the very best of it.

Quote: Canyonero

I guess you would just be ignoring ties:

(0.4462 / (0.4462 + 0.4586)) ^ 10 = 0.082% or 1 in 1223

For 15: 0.0023% or 1 in 42777

For 20: 0.000067% or 1 in 1'496'067

Good luck!

Right, and since the OP mentions a gambling problem, I would wish to point out that nothing ever becomes due. The probability, for instance, of 10 consecutive winning bankers is going to be 1 in 1223 regardless of whether it is your first go, or 10 consecutive bankers has failed to happen on the last 2,000 attempts.

If you look at this:

(1-1/1223)^848 = 0.49974337097313365

That means that there is right about a 50% chance of you going 848 consecutive overall attempts without you seeing a streak of 10 bankers. However, imagine you've already gone 847 attempts without that happening, then the probability of seeing such a streak is again 1 in 1223.

I may get into some more specific numbers later on today, we'll see.

Hmm, I failed to mention I did the math for the player bet; also, I just realized I just assumed OP was betting on the player for no apparent reason...

Quote: Canyonero

Hmm, I failed to mention I did the math for the player bet; also, I just realized I just assumed OP was betting on the player for no apparent reason...

That's what I get for trusting you! j/k

In any case, my nephew was here having a sleepover with my son last night, so we were playing video games for the better part of the night. I intend to give you a fairly specific answer this evening including probabilities and expected wins/losses based on your betting style, at least, as far as I understand it.

When you say you invest half of your profits, for example, I'm assuming you mean after paying the commission, is that so?

Yeah, just sitting here thinking about it, Banker would have to be better than 1 in 1024 to happen ten times in a row because 1 in 1024 reflects 10 in a row on a 50/50 shot...

Quote: machine4

Hi I have a question I believe will help my gambling addiction to baccarat and allow myself and others to perhaps ease off if they knew the rue odds.

The question is in baccarat what are the chances of say a 20 wining streak. As the way I play is by compounding half my winnings every-time I bet. Therefore to win big starting with around a $100AUD bet I would need around 20 in a row I have had 10's in a row countless amounts of times but never been on a run of 20 however I have seen these runs on tables next door to me in the casino, just I was never on one. I feel if I know the true odds It will help me to realize Its just not going to happen. Perhaps if you could give the odds of a 10 streak a 15 and 20 respectively.

Cheers

Okay, I will now do a full breakdown on your system because I have a few spare minutes. The one thing that I will say is that I am only going to be breaking down a ten-based system, and because it makes the math easier, I'm going to assume that half of your original bet gets added for each win, discluding the commission.

In other words, a step 1 win will result in $150 being bet on Step 2 with $45 retained.

Now, for your twenty in a row question, I'll address that only by saying that the probability of a Banker win, ignoring ties, is 0.506788499, so twenty in a row is:

(0.506788499)^20 = 0.0000012489211201029595 = 1/0.0000012489211201029595 = 1 in 800691.0796076226

Not likely.

Fortunately, your system at least allows for the potential of pulling back some wins, so let's see what happens when it happens:

Step 1

Loss: 1 - .506788499 = 0.493211501

Expected Value of Loss: 100 * .493211401 = -$49.3211401

Win: .506788499

The expected value of a Step 1 win is going to work somewhat differently, because it is going to be based only on the $45 you are pulling back with the other $50 (and the original $100 bet) moving to Step 2. You'll see that this system will reach a point where it can no longer lose money overall, so the Expected Value of a loss on a certain Step(s) will still remain positive.

Expected Value of Win: 45 * .506788499 = 22.805482455

Step 2

Loss: (.506788499 * .493211501) * -55 = -13.747465395472984

EXPLANATION OF METHOD: Okay, for there to be a Step 2, Step 1 must win. You have pulled back $45 from your Step 1 win, and your original $100 + $50 is out there for this bet, but overall, you only lose $55 if you lose here.

Win: (.506788499)^2 * ((150 * .95) -75) = 17.336334333510425

EXPLANATION OF METHOD: Okay, the first part is simply the probability of two Bankers in a row. The second part is the $150 bet, less the 5% Commission, and less $75, which is getting added for Step 3, if applicable.

Step 3

Loss: (.506788499)^2 * .493211501 * 12.5 = 1.583422125642317

EXPLANATION OF METHOD: Okay, this should be the last time an explanation is absolutely necessary. If you lose here, you have already, "Banked," $45 from the first bet and $67.50 from the second, so the ultimate result is that you still profit $12.50 on a loss here.

Win: (.506788499)^3 * ((225 * .95) -112.50) = 13.178782282562872

EXPLANATION: This is the same as Step 2, except you need three wins in a row, your total bet is $225, and you are pulling 112.50 to add to the next bet.

Step 4

Loss: (.506788499)^3 * .493211501 * 113.75 = 7.302387113272699

EXPLANATION: This is getting unusual, so I'm explaining again. You have already had three Banker wins in a row, and you multiply that by the probability of a loss. The $113.75 comes from the fact that you have banked $45, $67.5 and $101.25, which totals $213.75, but your original $100 is still, "Out," so you are losing that and profiting $113.75.

Win: (.506788499)^4 * ((337.5 * .95) - 168.75) = 10.018282937441748

Step 5

Loss: (.506788499)^4 * .493211501 * 265.625 = 8.64189816927075

Win: (.506788499)^5 * ((506.25 * .95) - 253.00) = 7.619904583386087

NOTE: I got sick of the decimals, so now you're carrying $253 over for the next bet.

Step 6

Loss: (.506788499)^5 * .493211501 * 493.5625 = 8.137839178764201

Win: (.506788499)^6 * ((759.25 * .95) - 380 = 5.782037247767017

Step 7

Loss: (.506788499)^6 * .493211501 * 834.85 = 6.975930572318382

Win: (.506788499)^7 * ((1139.25 * .95) - 570) = 4.398463703865881

Step 8

Loss: (.506788499)^7 * .493211501 * 1347.1375 = 5.704694269351155

Win: (.506788499)^8 * ((1709.25 * .95) - 855) = 3.345186360279326

Step 9

Loss: (.506788499)^8 * .493211501 * 2115.9250 = 4.540957831896467

Win: (.506788499)^9 * ((2564.25 * .95) - 1282) = 2.5448411326706837

Step 10

Loss: (.506788499)^9 * .493211501 * 3269.9625 = 3.556450118500153

Win: (.506788499)^10 * ((3846.25 * .95) + 3269.9625) = 7.737814102714492

EXPLANATION: Okay, you're keeping the money you've already pulled off the table, but it is not being bet, so it is not subject to being commissioned twice. You figure out the amount of the bet won AFTER the commission, then add it to the money pulled off on that side of the equation, and then multiply by the probability of ten Bankers in a row, which is the left side of the equation.

CONCLUSION

This is going to seem strange, but the amounts for, "Win," do not count towards the expected value, except for on the tenth step, because they are already being accounted for in the, "Loss," sections, which is when the system terminates. The system terminates only on a loss or after ten consecutive wins.

Therefore, to determine the expected value, you simply sum up all of the loss columns + the last win column and the result, and going backwards we get:

7.737814102714192 + 3.556450118500153 + 4.540957831896467 + 5.704694269351155 + 6.975930572318382 + 8.137839178764201 + 8.64189816927075 + 7.302387113272699 + 1.583422125642317 - 13.747465395472984 - 49.3211501 = -8.887222013742672

Thus, you will lose $8.89, approximately, per attempt.

The Expected Loss exists as a function of the House Edge and the Average Bet per system run, thus the Average Bet total per system run---DISCLUDING TIES will be:

8.887222013742672/.010579 = $840.0814834807328 or $840.08---DISCLUDES TIES

The probability of the system being profitable is simple, if the Banker wins twice in a row, the system is profitable regardless of what else happens, thus:

(0.506788499)^2 = 0.256834582718673

The probability of the system losing is simply:

1 - 0.256834582718673 = 0.743165417281327

The system will lose on every three out of four attempts, approximately. Like most (if not all) Positive Progression systems, the system is capable of tremendous wins and the losses tend to be limited. In this case, the system can only lose either $100 on Step 1 or it can lose $55 (overall) on Step 2.

If you wish to modify the system to stop on Steps 3-9, then you can use the formula as in Step 10 to determine the new Expected Value. To do this, simply remove the subtraction of half of the bet on the right, and replace it with adding the total amount banked at that point. After you have done that, add up the Expected Values of all losses and then that of the final win.

If you choose to do the above, you will notice that the overall Expected Loss will be lower per attempt. That's simply a function of the fact that the average bet will decrease the less steps you do and will increase the more steps you do. To get the average bet, simply divide the expected loss (expressed positively) by the House Edge.

I may do that later as a proof of the above numbers to prove the average bet, but I'm out of time for right now.

Gambling Problem

If you really do have a Gambling Problem, do the following, in order:

1.) Quit gambling.

2.) If you cannot quit gambling, get help, it's out there, there are many websites and toll-free numbers where you can find support. If you want to quit, you will quit, if you do not want to quit, then you won't quit.

3.) If it turns out you do not want to quit, then play EZ Bacc so this kind of thing will be easier for me to do next time, no bullshit Commission to worry about.