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Author Topic: Can the Bank have a 100% advantage?  (Read 4462 times)

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Romn.Paras

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Can the Bank have a 100% advantage?
« on: December 04, 2014, 05:02:17 AM »
Hello Friends! Here is something to think about. This is a true test in the philosophy of roulette, which I am an avid student and player.  Let me know what your thoughts are.  I have been pondering this for a little while.

We would be surprised to find how few of the people who come to play roulette have grasped the fact that the more numbers they cover, the less chance they have of winning. Some people think this is absurd.  Let me explain this theory.

If we bet 1 unit "en plein" or straight up on thirty-five numbers, we could lose thirty-five units if one or two numbers came out, and the most we could possibly win "per coup" was one unit.  We were, therefore, laying 35 to 1 when the real odds were 35 to 2 or 17.5 to 1: We were, therefore, laying double the proper odds, or in other words, giving the Bank an advantage of 100%.

When we play 8 numbers every "coup", we give the Bank an advantage of 3%, where if we play even chances that advantage would be 1.35%.

That is why my personal preference is to play even chances. I personally feel we must make as much use as possible our advantages to reduce those of the Bank to the minimum.

Some players figure that playing 2 dozens at a time is better, but this is also misleading.  By playing 2 dozens, the Bank
makes us lay 2 to 1 that 24 numbers will beat 13, whereas the proper odds are 1.84 to 1,  we consequently have a disadvantage of 8% in the odds.



 

kav

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Re: Can the Bank have a 100% advantage?
« Reply #1 on: December 04, 2014, 06:11:09 PM »
Wow!!!
Amazing topic. Did you just proved that roulette is not as well balanced as we thought?

Here is another approach. Each bet is equal, because in case of win you will be paid 2,7% less than you should for the winning bet.

Making many bets on the same time, increases your odds and decreases your payouts.

Ponder this my friend: What is better, to bet 35 numbers at once or to bet 1 number each spin for 35 spins?
According to your approach, betting for 35 spins is better. However the chance of winning if you bet 35 numbers for one spin is 35/37 - 95%, while the chance of winning if you bet 1 number for 35 spins is less than 65%.

I stay by my belief that roulette is a devilishly well balanced game.
« Last Edit: December 05, 2014, 04:28:25 AM by kav »
 

Real

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Re: Can the Bank have a 100% advantage?
« Reply #2 on: December 04, 2014, 06:32:20 PM »
Quote
ome players figure that playing 2 dozens at a time is better, but this is also misleading.  By playing 2 dozens, the Bankmakes us lay 2 to 1 that 24 numbers will beat 13, whereas the proper odds are 1.84 to 1,  we consequently have a disadvantage of 8% in the odds.


Wrong.

The house edge is the same, regardless of whether you bet one dozen, two dozens, or all of the dozens at the same time.

The outside EC has a lower house edge only because of the la patrige rule.
If you're playing roulette, and you want the best chance to win a large some of money, while playing the random game, then you want variance.  The best way to have a large amount of variance is to bet only a few numbers, straight up.

-Real
« Last Edit: December 04, 2014, 06:43:39 PM by kav »
 

Romn.Paras

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Re: Can the Bank have a 100% advantage?
« Reply #3 on: December 04, 2014, 11:47:44 PM »
I am in agreement with you Kav. The game of roulette is devilishly well balanced. I am from the school of thought that  believes that betting 1 number 35 times is a better choice of the two.  I am a firm believer that the less money on the table, the better. Kav, you are correct if we bet 35 numbers we will win 95% of the time, but what is the down side?  IF this bet loses, you have 35 units on the table, not 1 unit. What is the risk vs. reward ratio? We expect this bet of 35 numbers to win 95% of the time, but what if there is a hit back to back on one of the two numbers we didn't pick? It is possible that this event can happen and lose 70 units in 2 spins as apposed to 2 units. For me personally the risk reward ratio is too high to bet 35 numbers.  Most of us would agree that the Bank has one chance better than us in every 37 spins. This holds true if we play on one number straight up, but if we play on an even chance, the Bank has only half this advantage over us, or one chance better than us in every 74 spins, the reason being that when our stake is on an even chance, and zero appears, we do not lose our whole stake, only half of it, just like Mr. Real pointed out.  The player always has the right to take off half his stake.

We would think that if we put a unit on the first eighteen numbers straight up instead of putting eighteen units on "manqué" or low numbers 1-18, we are doing a very foolish thing; for when zero comes out, as it should do once in every 37 spins, we would lose the whole of our 18 units, whereas if had put them on 1-18, we would only lose 9 of them; in other words, the person who plays on the numbers is quite sure to lose, on average, 9 units an hour more than the player who plays 1-18.

Let me break this down a little further.  Real, you make a very interesting observation. I understand your philosophy about the house edge being the same. I will say that I respectfully disagree about the columns and dozens bets in that they are the same. Here is my opinion on the this matter.

It seems to me, and I maybe wrong, but if we back any given dozen, the Bank has a 4% advantage over us and if we back 2 dozen for the same spin, the Bank has a 8% advantage over us.  If we back only one dozen the Bank only bets us 2 to 1 that 25 numbers will beat 12, whereas the proper odds are 2.08 to 1. In the same way if we back 2 dozens, the Bank makes the player lay 2 to 1 that 24 numbers will beat 13, whereas the proper odds are only 1.84 to 1, and we have a disadvantage of 8% in the odds.

I do enjoy the philosophy and discussions that roulette can provide us with. It makes the game that much more interesting and fun to play.

 

Real

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Re: Can the Bank have a 100% advantage?
« Reply #4 on: December 05, 2014, 02:26:37 AM »
Quote
It seems to me, and I maybe wrong, but if we back any given dozen, the Bank has a 4% advantage over us and if we back 2 dozen for the same spin, the Bank has a 8% advantage over us.  If we back only one dozen the Bank only bets us 2 to 1 that 25 numbers will beat 12, whereas the proper odds are 2.08 to 1. In the same way if we back 2 dozens, the Bank makes the player lay 2 to 1 that 24 numbers will beat 13, whereas the proper odds are only 1.84 to 1, and we have a disadvantage of 8% in the odds. I do enjoy the philosophy and discussions that roulette can provide us with. It makes the game that much more interesting and fun to play.

Wrong.  You're not understanding the math.

If you bet on three columns on the single zero wheel, then you're essentially betting on 36 numbers.  The probability of winning is 36/37 for a payoff of only 35 to 1. The house edge is therefore 2.7% on this bet.

If you bet two columns on the single zero wheel, then you're essentially betting on 24 numbers.  The probability of winning is 24/37 for a payoff of only 35 to 1.  The house edge is therefore 2.7% on this bet.
If you bet one column on the single zero wheel, then you're essentially betting on 12 numbers.  The probability of winning is 12/37 for a payoff of only 35 to 1. The house edge is therefore 2.7% on this bet.

Sorry, but there's no disputing the math.

-Real
« Last Edit: December 05, 2014, 04:35:12 AM by kav »
 

kav

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Re: Can the Bank have a 100% advantage?
« Reply #5 on: December 05, 2014, 05:08:20 AM »
Hi Romn,

In the article about the house edge it is explained why the house edge is always the same. There is no dispute to it. (see attached image) However I do agree with you that different bets have different qualities and risk/reward.

Let's analyze the different qualities of the two options I posted above:
35 numbers at once vs. 1 number for 35 spins

In the first case the probability of winning is 95%.
The reward in case of winning is 1 unit.

In the second case the probability of winning is  1-(36/37)^35 = less than 65%. However the possible reward is higher, because it is possible that you win in more than one spin. or that you win in the first spin or... etc.

So which way of betting is better? None. It all depends on your strategy and what you want to achieve.

This is a very important issue. If one understands this, this can be a solid foundation to move further:
There are no advantageous bets in the sense of changing the house edge. The player disadvantage is the same all the time, no matter what or how he bets.
On the other hand, not all the bets are the same. Different bets have different risk/reward qualities and can be considered  suitable or unsuitable depending on our overall strategy, bankroll etc.

[PS: we ignore the "la partage/en prison" rule and the "first five" in American roulette, as these are special cases and will complicate the discussion]
« Last Edit: December 05, 2014, 05:13:45 AM by kav »
 

dobbelsteen

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Re: Can the Bank have a 100% advantage?
« Reply #6 on: December 05, 2014, 11:36:02 AM »
 I have explained this item in my topic “Who is the smartest player, the one who plays flat one unit on dozen 1 and the one who plays a unit on dozen 2 and 3”. My conclusion was the number one.
 

dobbelsteen

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Re: Can the Bank have a 100% advantage?
« Reply #7 on: December 05, 2014, 12:07:22 PM »
In my house casino, I see very often gamblers who covers 37 numbers. This is the same by wagering on red and black. The kick is, every spin gives a payout. It is not easy to convince such player that there is an important difference between 18 units on the red numbers and 18 units on the red EC.
 

Romn.Paras

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Re: Can the Bank have a 100% advantage?
« Reply #8 on: December 06, 2014, 02:14:12 AM »
 12/37 is not 2.7%. It is 32.4%. The payout is not 35 to 1 on outside bets, it is 2 to1 or 1 to 1 respectfully.
« Last Edit: December 07, 2014, 07:49:37 PM by Romn.Paras »
 

Romn.Paras

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Re: Can the Bank have a 100% advantage?
« Reply #9 on: December 06, 2014, 05:23:59 AM »
Thank you Kav, Real, and Dobbel for clearing that up for me. Sorry for the confusion. I misread an article and it confused me.  It mentioned the house edge in European Roulette, American Roulette and the Five Number bet which the house has a 7.89% edge.  I was confused that the house edge could change because of the 7.89% Five Number Bet. I was trying to figure out how this bet is different and gives the house a higher edge if the house edge always remains the same.   Here is part of what I read. The example was an American Roulette wheel.

The presence of the green squares on the roulette wheel and on the table is technically the only house edge. Outside bets will always lose when a single or double zero comes up. However, the house also has an edge on inside bets because the pay outs are always set at 35 to 1 when you mathematically have a 1 out of 38 (1 out of 37 for French/European roulette) chance at winning a straight bet on a single number. [To demonstrate the house edge on inside bets, imagine placing straight $1 wagers on all inside numbers (including 0 and 00) to assure a win: you would only get back $36, having spent $38.] The only exceptions are the five numbers bet where the house edge is considerably higher (7.89% on an American wheel), and the "even money" bets in some European games where the house edge is halved because only half the stake is lost when a zero comes up.

The house edge should not be confused with the "hold". The hold is the average percentage of the money originally brought to the table that the player loses before he leaves—the actual "win" amount for the casino. The Casino Control Commission in Atlantic City releases a monthly report showing the win/hold amounts for each casino. The average win/hold for double zero wheels is between 21–30%, significantly more than the 5.26% house edge. This reflects the fact that the player is churning the same money over and over again. A 23.6% hold, for example, would imply that, on average, the player bets the total he brought to the table five times, as 23.6% is approximately equal to 100% - (100% - 5.26%)^5. For example, a player with $100 making $10 bets on red (which has a near 50/50 chance of winning) is highly unlikely to lose all his money after only 10 bets, and will most likely continue to bet until he has lost all of his money or decides to leave. A player making $10 bets on a single number (with only 1/38 chance of success) with a $100 bankroll is far more likely to lose all of his money after only 10 bets.

Kav, I was figuring in the La Partage/ En Prison as well as the First Five American. Again, sorry for the confusion my friends.  Definitely a fun topic. 

I am going to say that I had a wonderful conversation with Kav about this topic and I want to share it with you all.  I understand that If we bet 1 dozen, it is 12/37 or 2.7% house advantage.  But the Bank is betting us 2 to 1 that 25 numbers including 0, will beat 12 numbers because 25/12=2.08.to 1. If we divide what the bank is giving us, which is 2 divided by the actual odds which is 2.08, because we are getting paid for 24 numbers instead of 25,  that gives us a 4% disadvantage. Meaning that we are getting paid 4% less on 1 dozen than we should by the actual odds. The same goes for 2 dozen in the sense of WE bet the bank 2 to 1 that 24 numbers will beat 13, or 24/13 which =1.84. So if we divide the actual odds 1.84 on 2 dozen divided by 2 that gives us an 8% disadvantage. Meaning the bank is paying us 8% less than the actual odds we are given because we are betting 24 numbers beats 13 as apposed to 24 numbers beating 12 which we are getting paid for.   
 

Romn.Paras

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Re: Can the Bank have a 100% advantage?
« Reply #10 on: December 07, 2014, 03:14:04 AM »
I am figuring the math. I will post a more detailed description soon.

« Last Edit: December 07, 2014, 03:19:41 AM by Romn.Paras »
 

Romn.Paras

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Re: Can the Bank have a 100% advantage?
« Reply #11 on: December 07, 2014, 03:18:42 AM »
Wrong.  You're not understanding the math.

If you bet on three columns on the single zero wheel, then you're essentially betting on 36 numbers.  The probability of winning is 36/37 for a payoff of only 35 to 1. The house edge is therefore 2.7% on this bet.

If you bet two columns on the single zero wheel, then you're essentially betting on 24 numbers.  The probability of winning is 24/37 for a payoff of only 35 to 1.  The house edge is therefore 2.7% on this bet.
If you bet one column on the single zero wheel, then you're essentially betting on 12 numbers.  The probability of winning is 12/37 for a payoff of only 35 to 1. The house edge is therefore 2.7% on this bet.

Sorry, but there's no disputing the math.

-Real

I have a question. I sat down and figured the math.  I did 12/37 and it is not 2.7%, it is 32.4% for one dozen.  If there was no zero, it would be 33.34% on 1 dozen. The pay out on a dozen is 2 to 1, not 35 to 1.    I think that the 2.7% house edge applies to the inside bets. On the outside bets, the house edge is figured into the pay outs.  Should we look at the house edge in a different light? My theory is that the house edge of 2.7% applies to the inside bets and there is a different house edge on the outside bets.

For example, if we bet 1 dozen, we get paid 2 to 1 for our bet.  The truth is that we should be getting paid 2.08 to 1 because we are betting against 25 numbers instead of 24 because 25/12=2.08.  In actuality we are getting paid for betting against 24 not 25 numbers.   We do not have a 35 to 1 pay out on outside bets, so I am thinking that 2.7 percent house edge is on the inside numbers because whatever we bet on the inside, the pay out is minus - 2.7%  of what we win. There is another house edge that is  figured into the outside bets. 

For 2 dozen , 24/37=64.8% instead of 66.67% if there was no house edge.  Now when we bet 2 dozen, we are betting that 24 numbers will beat 13 numbers for a payout of 1.85 to 1 instead of the 2 to 1 we are expected to get because 24/13=1.85.  Once again, the house edge is figured into the outside bets and we are getting paid 8% less than we should for this bet according to the math.

If we bet All 3 dozens or Columns it is 32.4% *3= 97.3%. So the house advantage is 2.7% because of zero.

So I stand by my theory that we are at a 4% disadvantage for 1 dozen because the house edge is figured into the outside bet, and we are at an 8% disadvantage for 2 dozen because the house edge is figured into the outside bet here as well.  So it is in my opinion that there is a house advantage of 2.7% for inside bets, and there is a different house edge on outside bets depending on which outside bet we choose.
[/quote]

Kav, the example you highlighted for me was really informative about the house edge. On the even chance example, I figured the math and 19/18=1.055 or 105.5% not 102.7%. I think there might be a different edge on even numbers as well. I am still checking and working on this. I will let you know my findings. I find this quite fascinating.

If I approach even chances in the same manner, the truth is that we bet 18 numbers will beat 19 numbers not 37. That comes out to 94.7%, so the house edge is 5.25%.  If we got paid the way we are suppose to for this bet it would be 1.055 instead of 1. So if we bet 1 unit we should win 1.055, not 1. so 1/1.055=94.7 and again we come up with 5.25%.  If we bet both even chances it is 36/37 and that is 97.3% so the house edge is 2.7% when we bet both even chances at the same time. Also if we factor in the La Patrage/ En Prison rule on Even Chances, then the house edge is 1.4% on even chances.  So if we take 18/37 = 48.6.% Then we are betting against 19/37 because we INCLUDE zero in the 19 numbers. That comes out to 19/37=51.4% . Now we subtract the difference 51.4%-48.6%= the 2.7% house advantage. When we add the La Patrage/ En Prison it makes the edge 1.4%.  But in theory we are STILL at a disadvantage because we are only getting paid 1 to 1 odds instead of receiving  1.055 to 1 odds because 19/18=1.055. Adding this to the equation gives us a total of 5.25% disadvantage because what we get paid 1 vs. what we Should be paid 1.055 = 1/1.055=94.7%. We are getting paid 94.7% of what we should get paid. So 100%-94.7%=5.25%. Then IF we factor the La Patrage/ En Prison rule in this way, the house edge becomes 2.7% on Even Chances.  So in actuality there are 3 possible outcomes playing "even" chances and we can lose 2 out of the 3 chances. We lose if it hits the opposite chance and we also lose as well if it lands on zero.  I think the confusion lies in that on the outside bets we are using the inside bet edge and odds and applying them to the outside as well. The outside bets are not 35 to 1, they are 2 to1 or 1 to 1 and we have to calculate them as such.  I am still pondering this theory.
« Last Edit: December 07, 2014, 07:43:58 PM by Romn.Paras »
 

dobbelsteen

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Re: Can the Bank have a 100% advantage?
« Reply #12 on: December 08, 2014, 12:56:52 PM »
Every number of the FR has a hitchance of 2.7%. A dozen or all 12 random numbers has a hit chance of 2.7X12=32.4%

A sixline has a chance of 6*2.7=16.2% .

zero-5-32-21-36 has a hitchance of 5*2.7=13.5%. Etc. Don`t mix up payouts, profit percentages and hitpercentage.

The house edge of a mixed bet ,you can compute with a true sample of 37 spins. The HE should be between 1.35 and 2.7!!