Roulette Ask The Croupier

& if you decide to do just that, play EC+LP once in a whole lifetime,
all you've achieved is getting ½ of your stake back;
mind that with that now you'd have to 'win two spins in a row' to make a profit
(first, to equalize back at ±0, & second, to walk away as a winner.

So your pronounced strategy on 'EC+LP for one spin only & walk away' -- is invalidated in its entirety.

 

Column 14 "cum. Level E.V." (the one you inserted.)

I'm confident the values in this column are false.
Explain how you even calculated from bet levels 2-24 in the cumulative. Level E.V. ?
The individual values are fabricated to add up a lower sum of bet E.V. + the statistical average of all your losses, doesn't aid your system. Bet 1 is the most crucial level of your system. If you lose it 97.3% likely to. The casino has already taken it's house edge from your starting 1000.
Shall we continue?

The house edge on any wager on any level is independent of whether you lose or win the previous level. The expected value of a wager straight-up is expected to lose -2.7%. You stated that basic logic should tell you (straight-up) bet $ize × 1/37 = Expected Value.

Add it up.



Column 10 Cum. Bet E. V. Is the accurate accumulation of the house edge when you lose up to 24 levels which is 51.8% likely.
Likes wise 26.54 ÷ 24 doesn't aid

The total house edge winning is $26.54. That is the cumulative expected value of all your progression that proved betting (straight-up) fractions of 1000u will still result with a negative E.V -2.7%.

Your starting bankroll of 1000, assuming you lose(51.8% chance) to level 24. The total capital lost is $982 the remainder $18 that you haven't bet would be the remainder of $27.02 - $26.54.

 

Comparing this to the very similar rule of En Prisone:

you even have the advantage of, at Zero appearing & your staked position being locked,
doing one more spin only, so at a higher probability, to potentially walk away as a winner.

So, your promotion of LP over En Prison, is making you a false prophet ..
where the boot imprint on your face still stands.

 

Can you Quote me when/re I ever said that?

 

You have no concept of math or logic.

You quoted yourself.

 

-TwoUp wants to have his cake and eat it.
51.8% chance of losing $982.

 

If you want to get technical your probability of win at level 24 (48.2%) is still lower than any EC = 48.65%.

Your high-risk to gain is fantasy in the face of a negative expected value game.

 

The values should be shifted one row Up. To include 51.81%

 

So after figuring all your calculations what are you guys betting on?

 

48.65% of the members don't have the resources to play La Partage online, let alone in a casino.

 

... you don't even know your primary colours.

Please refrain from littering this thread with nonsense games in your head.

 

Let's first clearly establish the calculation to column 14:

"column 10" x "column 11"​

Which is:

"Cum. Bet E.V" × "Pr(Level reached)"
​

You have indicated that you agree with the "Cum. Bet E.V." calculations. So everything up to and including column 10 you agree with as that is the result of all the prior columns.

Column 14 simply multiplies the accumulated Bet EV (which is the house edge paid on every bet up to that point) by the probability of actually getting to that level.

So you can bet at level 2 until you lose level 1, and you can't bet at level 3 you have first lost at levels 1 and level 2.

Column 11 "Pr(Level Reached)" calculates the probability of reaching a level. As you can see we always reach the first level (100%) meaning we always make the first bet. We only reach the next level when the bet loses.

Each bet has a 36/37 probability to lose.

37/36 = 97.23% for level two
To reach level 3 we have to have lost twice. The probability of losing twice is 37/36×37/36 = 94.67%

The formula is multiply the previous probability by 36/37 as that is always the probability of losing the bet. Right at the bottom we see the probability of losing 24 times in a row is 51.8%.

The general formula for computing the probability of back to back series losses on a single bet is (36/37)^level. Same method as calculating the probability to lose a coin toss three times in a row (1/2)^3 = 1/2×1/2×1/2 = 1/8

So column 14 is just column 10 multiplied by column 11. Which is the cumulative loss of each independent bet multiplied by the probability of reaching that level.

A simple multiplication will verify the numbers are correct.

No they are not fabricated unless by that you mean calculated.

The casino cannot take $1000 worth of house edge until bet 24 which is 24 losses in a row. We have 24 opportunities to pay less house edge on a win.

The "Cum. Bet E.V" column is exactly that. It is each bets EV summed and not taking into account the probability of reaching that level.

We do not keep playing the progression if we win a bet. We don't blindly always go to bet 24. We stop on a win, having made in excess of $1000 profit which is the target.

Because the level we progress to IS DEPENDENT on losing all of the prior bets the house edge we pay is therefore conditional. The "Pr(Level reached)" provides the probability of reaching a level as already described.

At this point you have to accept that we do not bet every level if we win on a prior level.

You also have to accept that the probability of a win is never zero on the prior levels.

I provide the expected probability of witnessing a win and a loss in columns 12 and 13 for informational purposes. Again it is a simple multiplication of:

Pr(Level reached) x 1/37 for a win
Pr(Level reached) x 36/37 for a loss
​

More precisely when you lose EXACTLY 24 times, not up to 24, EXACTLY 24 times and not 1 time less.

Not sure what you meant to say there.

That is the cumulative house edge NOT considering any prior win. It is the total house edge paid when ALL 24 bets lose.

The $18 has not been risked, only money risked is subject to house edge.

 

Should say:

36/37 = 97.23%​

To reach level 3 we have to have lost twice. The probability of losing twice is 36/37×36/37 = 94.67%​

 

Which is cheaper than a 51.4% chance of losing $1000.

51.8% × $982 = $508.76
51.4% × $1000 = $514

Last time I checked $514 is a greater loss than $508.76

I like eating cake.

 

Well the principle here is that you are always better to use a higher odds wager in roulette with a series of smaller bets and progressively risk your money to the house edge as opposed to paying it all in a lump sum with an EC.

 

It seems like we're comparing apples with oranges, but I'm with TwoUp on the straight up.

 

I know, & everyone sees -- its obvious what I did .. no need to blabb about that;

now rather than evading address the written instead --
my quote quoted by me, & mg quote quoted by you.
Of you gonna pussyup, as expected ..most probably.

Then even better, add another spin .. there's plenty of extra room towards the whole 1000u.

 

All the best of the best do that.
Losers like you chicken out.

 

So will all the fractions .. or "every & each no-hit non-Zero spins" in the continuous EC+LP play, what!?
"result with the negative E.V -2.7%., or pay the house edge of 2.7%.

Just the same.