Roulette Ask The Croupier

Much Ado about Nothing .

 

No that column is the probability of reaching the level before you actually make the bet at that level. If we shifted it up then we wouldn't be making the first bet on the first level 100% of the time now would we?

The actual bet at any level has a 1/37 chance to win and 36/37 chance to lose. See the two columns "Pr(win @ level)" and "Pr(Lose @ Level)" for the probability of reaching the level and then winning or losing the bet as the case may be.

You will notice in column "Pr(Lose @ level)" that the probability of getting to level 24 as a result of 23 prior losses and then placing our 24th bet and losing is 51.81%. at that point all 24 bets have lost and that agrees with what you are expecting as the probability to lose all 24 bets because my math is correct.

If I shifted it up as you incorrectly suggest then the probability of losing all 24 bets would actually decrease, and my numbers would be even better.

 

(0.78+81) × (36/37)² . So you havent multiplied 1.59 × 94.67. For clarification the E.V. values in column 14 are no relevance to the probabilty of level reached. Meaning the probability of reaching any level is independent of the amount bet. The probability of reaching any level is dependent on bet selection i.e straight-up or EC.

Column 14 is a fraction of the already "paid"/ accumulated house edge.

Expected Value is the HOUSE EDGE.
What is your total at the bottom?

In actual terms when you lose the first bet you've lost $ 29/1000. The casino has already won its HOUSE EDGE and continuing with higher progression is chasing your losses.

Column 10 (cum. bet E.V) is the cumulative sum of each bet level E.V that lost .
bet level 1 E.V = 0.78
bet level 2 E.V = 0.81
So on.

At Bet level 2 the cumulative bet E.V is 1.59 because you lost at bet 1 then at bet 2.
0.78 + 0.81 = 1.59

Its 36/37² or 36/37 × 36/37 for two levels. Thus I scrutinised column 11.

(36/37)²⁴ ???

Then multiplaction doesn't come simply to you. Column 14 is gravely miscalculated.

-$1.59 is the cum. expected value on every bet up to that point (over both levels 1 + 2). The error is you multiply the lose event (36/37) by 1 less power. a.k.a the probability of just the prior lost event level reached before the accumulated E.V.

Your formula 1.59 × (36/37) is then incorrect.
This should be 1.59 × (36/37)²

$1.59 × 97.3 = $1.54
Or 1.59 × 94.67 = $1.50

The House edge is dependent on the bet size. these values decrease from the original bet Expect Value because you are multiplying fractions together.
In roulete you are not losing these amounts I've repeatedly stated the first bet you play with a 97.3% chance of losing is slightly over the house edge of your $1000 br.

The probability of level reached at bet level 2, accumulated two Ls (36/37)². Logic would dictate the rows should be shifted up. It would correctly correspond to the current bet level.

If you want to postscript rename column 11 Pr(level reached) to the Pr(before level).
Probabilty ranges from 0 - 1. 100% is you've confirmed the choice to play the first level then you 100% place all the subsequent bets to receive the payout of the level you reach. The probability before level 24 is also 100% since you're calculating the probability that you can reaches a cum. bet E.V of $26.54. which I mock your system as its higher than half of all possible scenarios that you 51.8% you'll lose 98.2% br.

...
This is how I know you spend more time on excel than playing roulette. Your very first bet is the only opportunity the house needs to make its house edge.

29/1000 = -.029

If you reach bet level 3 . You're down -5.9% of 1000. Your progression is compounding the house edge against you.​

It's you that needs to accept you have 24 trials before you lose -98.2%. Theres 37 possible variables in roulette. Do the math.

Your if condition isnt very likely. You calculated a 51.8% overall probability to lose nearly everything by progression.

Each trial is a 97.3% expected failure.
You either continue losses trying to win everything back our accept losing the 1st bet at -2.9% br.

Correct.

 

And there you have your entire straw math argument blown away by La Partage.

 

1×(18÷37)-1×(18÷37)-0.5×(1÷37)= -.0135

50% × £1000 = £500

Eat that Pound Cake.

 

@6probability9 How on earth you get your opinion on things is bewildering.

Again point to a specifc cell and say right there that number is incorrect and provide your value and how you calculate it.

Did you not understand the probability "Pr(Level reached)" ? That is the probability of reaching a level as explained previously in painfull detail.

There is no dollar figure in the calculation of the probabilty of reaching a level, it is a probability.

The formula is (36/37)^ (level-1)

First level (36/37)^(1-1) = (36/37)^0 = 100%
2nd level (36/37)(2-1) = (36/37)^1 = 97.3%
...
24th level (36/37)^(24-1) = (36/37)^(23) = 53.25%

Do you say this is wrong? ANSWER THE QUESTION

There is however a dollar value on the cummulative expected loss in column 14 as it is dependent on reaching that level and upon reaching that level the bet is placed and the cumulative expected value is the total house edge paid to that point.

Do not confuse expected value (house edge) with losing a bet. Wins and losses are not house edge. House edge is exactly what has been risked to that level × 1/37.

That is why we multiply the probability of reaching the level by "Cum. Bet E.V." which is the expected value (the house edge) of all bets placed up to and including that level.

​

Yes it is the cumulative house edge of all bets up to and including that level multiplied by the probability of reaching that level.

Hence it is a fraction, and surprise, surprise this is the crux of the matter. You don't seem to understand that WE STOP BETTING ON A WIN.

Why? Well we can't be paying house edge on future bets we never place when we do a get a win. unlike the moronic EC bet where like a dunce you pay it all up front because you didn't know any better.

You have to actually accept and understand we are not betting to level 24 EVERY TIME.

We CONDITIONALLY bet based on the previous bet losing. If it wins we stop betting.

You seem unable to grasp this simple concept.

So what is the probability of losing 6 bets in a row and getting to level 7 prior to putting bet 7 on the felt? (not completing level 7 just arriving at level 7).

I say it is (36/37)^6 = 84.84% which is exactly what the table says. Do you agree or disagree? ANSWER THE QUESTION

Each bet pays house edge and this is just another name for the expected value. Every bet is negative expectation due to precisely the house edge.

The expected value of each bet is clearly shown in "Bet E.V". Which you have already agreed with as you agree with the column "Cum. bet EV".

The total at the bottom is the total of the house edge of ALL bets. Make sure you understand that ALL means making 24 bets exactly. It is not the EV when making 5, 9 or 17 bets when you happen to win at those levels and STOP.

Each bet that you do not make pays no house edge.

You are confusing a loss with house edge.

Would you say when you lose your $1000 EC bet that you paid $1000 house edge? ANSWER THE QUESTION.

The house edge on the first bet is $29 × 1/37 = $0.78 it is not $29. I am afraid you are quite confused and this appears to be well over your head.

When you risk $1000 with your EC bet you stand to win $1000 and pay a chunk of house edge (2.7% or 1.35% with LP).

I am risking up to $982 to make more and pay less house edge because the series of bets can win on the earlier levels prior to bet 24.

Note the words "up to $982". You don't seem to understand that the series of bets STOPS ON A WIN.

Each bet that I do not make pays ZERO house edge.

Yes that is what my table shows. So here you say you agree with calculations for the house edge that is paid for each bet at each level and the total (cumulative) house edge up to that level.

Yet your prior statement said the house edge was $29. So which is it? ANSWER THE QUESTION.

That "typo" in my message was already followed up within minutes of my post and almost a day prior to your response.

It was actually a bait to see if you are genuine. Which you have just proved you are not.

The evidence is there for all to see. My follow-up and the spreadsheet calculations are correct.

Again which CELL do you say is wrong? Clearly provide your calculation and proposed value. ANSWER THE QUESTION.

This has been painstakingly explained a number of times but you struggle to understand the probability of a series of events.

Again this has been explained. The probability of reaching a level is the probability of reaching the level PRIOR to making the bet at that level and getting an outcome.

This is why the first level has 100% probability to be made as we always make it. Do you accept that 100% probability means always? ANSWER THE QUESTION.

What is the probability to lose the first bet? (36/37 = 97.3%) and that is why we only bet at level two 97.3% of the time.

If I used 1 extra power as you incorrectly say I should then the probability of making the first bet would be (36/37)^1 = 36/37 = 97.3%

Clearly you are wrong as we ALWAYS make the first bet 100% of the time not 97.3% of the time. Do you agree? ANSWER THE QUESTION.

The second level is ONLY played if the first bet losses. (36/37)^(2-1) = (36/37)^1 = 36/37 = 97.3%

Exactly what the table shows. Do you agree? ANSWER THE QUESTION.

Similarly getting to level 7 means we have lost 6 times (not 7 times). And guess what (36/37)^6 = 84.84%

Again, exactly what the table shows. Do you agree? ANSWER THE QUESTION.

Be thankfull I am not doing what you suggest as my house edge would be even less. Again you don't understand the basics or the implications of what you are suggesting as it only improves my numbers but it would still be incorrect and I am 100% about mathematical correctness.

Do you agree that losing the straight up bet 6 times in a row has a probability of 84.84% and that the table clearly shows we reach level 7 84.84% of the time? ANSWER THE QUESTION

​

Losing a bet twice means we are now at level 3.

The probability to get to level 2 is (36/37)^(2-1) = (36/37)^1= 36/37 = 97.3%

The cummulative house edge paid when we make the bet at level 2 is the house edge of bet 1 and the house edge of bet 2. Which is $0.78+$0.81 = $1.59

What is the probability of getting to level 2? We just calculated that and it is based on losing the first bet which is 36/37 = 97.3%

Great so now we can multiply them together to get the expected value for that level of the progression:

$1.58 × 97.3% = $1.55​

Again you struggle to understand the probability of a sequence of events and you even suggest that I should be paying even less house edge: $1.58 × 94.67% = $1.50

Last time I looked $1.55 is more house edge than $1.50.

Do you really wish to insist that my house edge should be the lower $1.50? ANSWER THE QUESTION.

Even when your incorrect math benefits my position I will still reject it as it is incorrect.

Which house edge is correct for the second bet level $1.55 or $1.50? ANSWER THE QUESTION

There you go again saying the house edge is now $29 again. You are confused over the difference between a loss, a win and the house edge.

Again using your erroneous belief, your first and only EC bet must therefore be a $1000 house edge? I say it is 1/37 of $1000. ANSWER THE QUESTION.

Even with your absurd logic, $29 is still less "house edge" than $1000.

Can you count?

We start at level 1 and always make a bet 100% of the time.
Getting to level 2 happens if bet 1 loses. (1 bet lost)

THEN a bet is made which has an expected value.​

Getting to level 3 happens if bet 2 loses (2 bets lost)

THEN a bet is made which has an expected value.​

...
Getting to level 24 happens if bet 23 loses (23 lost bets)

THEN a bet is made that has an expected value.​

And we stop.

Did you notice the conditional logic that we stop betting on a win? ANSWER THE QUESTION.

No. We CONDITIONALLY make each bet when the bet before loses. If the bet wins we never bet the higher levels and can never pay house edge on bets we don't make.

Pretty simple to understand but your blindness to the concept is quite literally stupifying.

You wouldn't know the first thing about anything I do. You barely know how to count to 24.

... and there you go again confusing loss vs a win vs the house edge.

Now I'm convinced you're a troll.

House edge is based on amount risked. We have not risked $1000 on our $29 bet. We risked $29. We know the house edge on the bet is 1/37th of $29 or $0.78 or 78 cents.

Let's consider your logic with your EC bet:

1000/1000 = -1

which is -100% edge or $1000 with your loopy logic.

A reasonable and honest person would say it is amount risked × 1/37
​

You seem to forget the bet selection is about risking up to $1000 to make $1000. My approach reduces the house edge vs the EC bet.

It is a pity you can't grasp the idea to stop betting after a win and therfore reduce the average house edge paid.

It seems you prefer to pay a fully compounded lump sum house edge with premium added in for good measure rather than the discounted option.

I did the math and it stands correct. The expected value (house edge) on the series of bets is lower than that an EC with LP.

As a comparison to risking $1000 on an EC the series bet EV is -$8.10 or -0.81% whilst yours is sky high at -1.35%

You have 1 bet to lose 100% and pay sky high EC with LP house edge which you think is God's gift.

The series bet risks less, wins a higher amount and pays lower house edge $8.10 on average as per the math.

It's equivalent to an EC. In expected value terms is smashes your EC bet. And expected value is ALL THAT MATTERS.

Yes each trial is 36/37 probability to lose (97.3%) String 24 of them together and you have an equivalent probability to losing and winning as an EC.

Which has the lowest house edge?

Your first (and only) bet risks 100% of br. and pays full house edge (2.7% or 1.35% for LP). Fool you.

The series bet has the mathematically and logically proven lower house edge in absolute terms. It achieves this by progressively risking the $1000 as any win means you don't have to risk the remainder and therefore can't pay house edge on money you never put on the felt.

Now ANSWER THE QUESTIONs.

 

The irony when you can't even look at the numbers and claim a hollow victory. Like when your horse "won" and your ticket actually had the losing horse.

The LP bet lost more:

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

51.8% × $982 = $508.76 (my series bet)
51.4% × $1000 = $514 (your retarded EC bet)

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

Can you do subtraction and see that 514 is $5.24 more than the series loss? ANSWER THE QUESTION.

 

That cake is far too expensive. But you go ahead, stuff your face.

I prefer the more efficient, higher yeilding petit fours cake. I only have to eat as much as I need and it's only $8.10 vs your overpriced $13.50 cake.

 

The probability you get to level 2 is (36/37)² or (36/37) × (1/37).

You are calculating the prior formula because you accumulated two Ls. Therefore
(0.78 + 0.81) = 1.59 × (36/37)^(2-1) is wrong!

 

Again with your incorrect -$8.10 average. Average is the sum of set values ÷ by the total values.
So you must include the cumulative loss at bet level 24. That is the average 51.8% event when you're left with $18 r.

 

Can you do subtraction and see that 500 is 8.76 less than your 508.76 series loss? ANSWER THE QUESTION.

-50% × 1000 =?

 

•Assuming I use a $1000 br and bet $27 on E.C. without La Partage.

•Assuming you start with a $1000 br and bet $29 on a straight-up.

The probability that you lose immediately to the house edge is higher than EC.

 

Regard yourself as caught, again.

 

How it is possible to lose twice before I bet at level 2? We only make the second bet when the first bet loses.

Losing twice means we now need to bet at level 3. It means we both reached level 2 and then lost at level 2.

The first bet we always make 100% of the time.

The second bet we only make when the first bet loses. The level probability (36/37)^1 is 36/37.

Can you not understand that oh so simple concept?

Do you really believe we lost two bets BEFORE we make bet number 2? How is this possible when we only made one bet?

The probability to lose bet #1 is 36/37.

The probability to lose bet #1 and THEN lose bet #2 is 36/37 × 36/37 = 94.67%. this takes us to level 3.

The probability to lose bet #1 and then win bet #2 is 36/37×1/37= 2.63%. We stop and the probability to go to level 3 is zero (never).

This is like teaching a child to count.

Let's say I am betting on a coin flip for heads and I stop when I get heads. I always have to flip the coin at least once so the first flip is 100%.

The probability of flipping the coin a second time is 1/2. It is NOT 1/2×1/2 = 1/4.

You just seem incapable of understanding the basic idea that you only make the second attempt IF and ONLY IF the first atttempt fails.

You alsos eem to be confusing the level probability with the outcome of the bet at that level losing when the probability is actually about the prior bets losing. That is why it is Pr(Level reached) and not Pr(Lose @ Level) which I also include but which you don't seem to understand.

You are beyond learning.

 

You will never understand house edge vs loss. But one more time.

House edge is based on $ risked.

The risk is my case is $29 and pay 78c win or lose.

You risk $1000 and pay house edge win or lose $27 or $13.50 for LP.

 

I have to say that you @6probability9 have perhaps the lowest IQ of anyone I have interacted with on a forum and believe me that is low bar to get under.

 

You @6probability9 need to ANSWER THE QUESTIONS.

They are simple direct questions that are easy to answer will will easily show if you are being genuine.

Now put up or shut up.