Ive been lurking the forum for a while now. No i dont love anything and i didnt invented martingale ;)

Im playing a dice game at the moment where you can adjust winchance and the payout.

I tried everything that came to my head and always went into the - at the end but i got an idea where i like to hear your thoughts.

First some informations on the dice game:

Rolls 0-9999 2,5% House Edge

To my Idea

Roll on x1.2 payout with 81,25% winchance winroll <8125.

If i lose i would switch to payout x5.2 winroll >8124

I did some simulations with a few million and billion rolls as an example 1 million:

x1: 123500 ( 247.0 % )

x2: 23091 ( 46.182 % )

x3: 4415 ( 8.83 % )

x4: 760 ( 1.52 % )

x5: 155 ( 0.31 % )

x6: 35 ( 0.07 % )

x7: 4 ( 0.008 % )

Thats how often i would lose/lose in a row. x2 as an example is a 2 times in a row lose.

It doesnt matter how high i go in the simulated rolls the numbers are the same with a few differences in it.

So if i hit the ~23091 2 losing trains with the 5.2 payout roll on the second lose i would reduce the toal loss by about ~100k if i bet with an inital of 1.

Am i stupid and there is a big mistake in my thoughts or could this work in the long run?

ps: English is not my nature language if i did any mistakes excuse me pls.

And sry for the wall of text

A number from 0 to 9999 is chosen at random

There are two possible bets:

Betting that the number is 0-8124 has an 81.25% success rate and pays 1-5

Betting that the number is 8125-9999 has a 18.75% success rate and pays 21-5

Each bet has an HE of 2.5%

The "strategy" is to bet on the 1-5 bet, and if it loses, switch to the 21-5 bet.

What I don't understand is, what is the actual "system" - do you keep switching back and forth between the two bets?

And what do the percents in parentheses mean?

Also, does 2x mean the number of times you lost at least 2 pairs of bets, or the number of times you lost exactly 2 pairs of bets but won on the third?

And do you always bet the same amount each time? Winning a 1.2x bet other than the first one isn't going to make up for the losses in the previous bets.

Its the number i lost exactly 2bets in a row.

By switching o 5.2x payout i would win on the second loss.

And yea its always the same amount that i bet.

the 1.2x bet gives me the base profit, the 5.2x bet just reduces the loss i would have when betting always the same number.

https://wizardofvegas.com/forum/gambling/craps/24843-charting-a-table/#post509859

But we talk about 144000 rolls per day where the Law of large numbers should apply if my thoughts are not wrong

Quote:AlkaniosBut we talk about 144000 rolls per day where the Law of large numbers should apply if my thoughts are not wrong

The "Law of Large Numbers" does not work that way. It applies to fractions, not to specific counts.

Here is an example:

Toss a coin 1,000,000 times. 490,000 times, it comes up heads, so heads came up 49% of the time.

Toss it another 1,000,000 times; this time, it came up heads 495,000 times. The total percentage of heads is now 985,000 / 2,000,000 = 49.25%, and it's getting closer to 50%, but there were still more tails than heads in the second set of 1,000,000 tosses. All that the Law of Large Numbers says is, the total percentage "should" get closer and closer to 50%.

Let's take a look at your original problem. Since bets are paid off "to 5", I will assume each bet is 5.

The probability of winning your first bet is 0.8125, or 13/16; you are ahead 1 if it wins.

The probability of losing your first bet but winning your second is 3/16 x 3/16 = 9/256; you are ahead 16 (lose 5 on the first bet, win 21 on the second) if it does.

The probability of losing both bets is 3/16 x 13/16 = 39/256; you are behind 10.

Each "step" (1 or 2 bets) has an expected value of (13/16 x 1) + (9/256 x 16) + (39/256 x (-10)) = -38/256, although remember this is based on a bet of 5.

The probability of losing N consecutive pairs of bets is (39/256)

^{N}.

The approximate number of times of times you should lose at least N pairs of bets in a row out of 1 million steps is:

2 times or more - 23,209

3 times or more - 3536

4 times or more - 539

5 times or more - 82

6 times or more - 13

7 times or more - 2

As I said earlier, one significant problem with this is, unlike the Martingale or D'Alembert systems, it is possible to reach a "reset" point and still be behind since the last reset point. If you lose two bets and then win the third, your third bet only won 1, which doesn't cover the 10 you lost in the previous 2. Once you get past four losses in a row, even a 21-5 bet win won't cover the losses.

But either i understand something wrong or you did. In your calculation i would always switch after 1 win or 1 lose, but i only switch the payout after a lose.

So i dont bet in a pair of bets.

Bet 5 until i lose and then switch arround to the x5.2 Payout.And switch back after that,doesnt matter if i won or lost that roll

Or am i wrong and the calculation includes that?

Quote:AlkaniosThanks alot for the Math ThatDonGuy.

But either i understand something wrong or you did. In your calculation i would always switch after 1 win or 1 lose, but i only switch the payout after a lose.

So i dont bet in a pair of bets.

Bet 5 until i lose and then switch arround to the x5.2 Payout.And switch back after that,doesnt matter if i won or lost that roll

Or am i wrong and the calculation includes that?

I think I misunderstood your method.

What I was doing was, you make x1.2 bets until you lose, then you make one x5.2 bet, then you switch back to x1.2 even if your x5.2 bet won.

Is this correct:

You make x1.2 bets until you lose, then you make x5.2 bets until you lose again, then you make x1.2 bets until you lose again, and so on.

I did some rolls in the past 5 hours and so far im +170 profit with an inital of 1.

in a fair game where you are against a 2.5% house edgeQuote:AlkaniosFirst some informations on the dice game:

2,5% House Edge

<snip>

could this work in the long run?

simple answer: no

no one could ever beat that game in the long run

no matter what you bet and how you do it.

<<< >>>

that is way way way worse than the 1% dice games I have seen.

of course if there is cash back with losses that could be different

for a while

have fun playing

Sally

Quote:AlkaniosAhh no you were right. x1.2 until lose then one x5.2 and switch back.

I did some rolls in the past 5 hours and so far im +170 profit with an inital of 1.

How many rolls is "some"?

Most systems "work" because, in the long run, they match a large chance of winning a small amount with a small chance of losing a large amount that, in most cases, would wipe out all of the wins up to that point. If you're lucky enough to avoid the big loss, at least long enough so your wins will cover it, than it will look like the system works.

Otherwise i wouldnt even mention it ;)

Im writing a script at the moment that automates the process.

So ill provide some more information on alot of rolls in the next days.

a simple Excel sheet can do this tooQuote:Alkaniossome are something like 15.000 did that for 10 hours straight.

I simulated 1000 players doing your "system" for 15,000 "rolls" each

as a simple test

the 1.2x bet after each roll

either

-1 on a loss or the bankroll gain 0.2 on a win

the 5.2x bet made

only after a 1.2x loss

that results in a -1 loss or a 4.2 bankroll gain

then back to the 1.2x bet

results:

avg session = -447.547

best session ended at -3.80

worst was -828.20

no way to beat that very high edge of 2.5%

you must understand things differently

good luck

"LUCY!"