Author Topic: Roulette Riddle about the house edge  (Read 2860 times)

Romn.Paras

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Roulette Riddle about the house edge
« on: April 06, 2016, 04:24:47 AM »
Hello Friends.  I have enjoyed such lively conversation from you all and I appreciate your in put.  Here is another theory I was going over the other day and thought I would share it with you.  Perhaps some of you may be able to answer this riddle I came across.

 A gentleman said that he could prove that if someone covers sufficient numbers, the house has a 100% advantage over them.  How is this possible?  He goes on to tell that if a player backs 1 dozen, the house has a 4% advantage over them.  He also goes on to tell if they back two dozen that the house has a 8% advantage over the player.

How are these scenarios possible?  I would be interested to hear your points of view on this. 
« Last Edit: April 06, 2016, 07:51:29 AM by kav »


 

kav

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Re: Roulette Riddle
« Reply #1 on: April 06, 2016, 06:15:44 AM »
Hello Romn,
Thanks for your interesting posts.
Are we talking European or American wheel?
 

kav

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Roulette Riddle about the house edge
« Reply #2 on: April 06, 2016, 07:45:17 AM »
Hi,

I think I found the answer to your riddle. And it is referring to the European single zero wheel. In American wheel the payouts are even worse.

Here is the logic.
First let's find the formula for the profits. It is:

Now we must compare our hypothetical FAIR profit with our actual profit.
I have done some preliminary calculation to chose the right bet amount for the examples to be more illustrative.
One dozen bet
Let's say we bet 48 chips (units)
Fair Profit

Actual Profit

Difference (profit lost):  4%

Two dozens bet
Let's say we bet 185 chips (units)
Fair Profit

Actual Profit

Difference (profit lost): ~8%

The situation gets worse the more numbers we bet. When we bet on 37 numbers the "profit lost" becomes 100% since we are paid less than we bet even if we win.
« Last Edit: April 06, 2016, 01:35:54 PM by kav »
 

kav

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Explanation of the profit paradox and the house advantage
« Reply #3 on: April 06, 2016, 01:11:56 PM »
Explanation of the profit paradox and the house advantage.

The funny thing is that theoretically the House edge stays the same no matter how many numbers you bet. This is because the house edge is not based on profit but on payout.

Let's see the basic Payout equations for gambling:

Now for the European wheel, the fair payout would be:

But, actually it is:


We see that the payout difference, also called House Edge, for the European wheel is always 1/37 = 2,7%
And it is independent of the numbers played and the amount of bet.
It is always the same.

Conclusion
The more numbers we play, when we win, the casino takes a larger part of our profits.
However the house edge (unfairness of payouts) remain the same.
« Last Edit: April 06, 2016, 11:14:53 PM by kav »
 

scepticus

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Re: Roulette Riddle about the house edge
« Reply #4 on: April 06, 2016, 09:32:54 PM »
He can only prove it " Mathematically" . In actual practice  " variance" determines who will win and who loses. He makes the classic " Mathematician's 'Fallacy " - confusing Expectation with Certainty.   
 

Reyth

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Re: Roulette Riddle about the house edge
« Reply #5 on: April 06, 2016, 10:57:19 PM »
LOL.  We should make a wikipedia page for:

Mathematician's Fallacy

There has to be a better title though because "Mathematician" is too general... XD

 

scepticus

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Re: Roulette Riddle about the house edge
« Reply #6 on: April 07, 2016, 08:35:15 PM »
LOL.  We should make a wikipedia page for:

Mathematician's Fallacy

There has to be a better title though because "Mathematician" is too general... XD

" Critic's Fallacy "  ?
 

Romn.Paras

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Re: Roulette Riddle about the house edge
« Reply #7 on: April 09, 2016, 03:51:09 AM »
Hello Kav.  I absolutely love your breakdown on this. Bravo my friend!  I am referring to European Roulette.  I have recently found out there are a few casinos in Las Vegas that have European Roulette tables.  You are correct Kav with your explanation.

If we bet One Dozen, the House bets us 2 to 1 that 25 numbers will beat 12. So the actual 2 to 1 on One Dozen is actually 2.08 to 1 because if we divide the House's 25 numbers to our 12, it equals 2.08 instead of two. So if we divide 2 into 2.08 that equals 96% which is a difference of the 4% . I like the mathematical breakdown that was demonstrated in your answer Kav.   Then if we back 2 dozen, the House makes US lay 2 to 1 that 24 numbers will beat 13 so the proper odds are 1.84 to 1 and we have a disadvantage of 8 %.

If we bet 1 unit on 35 numbers, we could lose 35 units if one or two numbers came out and the most we could win for that coup is 1 unit.  We were therefore laying down 35 to 1 when the real odds are 35 to 2, or 17.5 to 1.  So we were laying double the proper odds, or in other words giving the House a 100% advantage over us.

Well done Kav!   
 

scepticus

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Re: Roulette Riddle about the house edge
« Reply #8 on: April 09, 2016, 05:12:52 AM »
Twice  2,7 % is 100%  ?
I must have misunderstood your Post Romn.
 

Mike

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Re: Roulette Riddle about the house edge
« Reply #9 on: April 09, 2016, 07:20:29 AM »
He can only prove it " Mathematically" . In actual practice  " variance" determines who will win and who loses. He makes the classic " Mathematician's 'Fallacy " - confusing Expectation with Certainty.   

There is no such fallacy, it only exists in your mind. Try setting up a  Wikipedia page for "mathematician's fallacy" and see how long it lasts.  ;)

The fallacy is not recognizing that variance is just another way of saying "luck".

Luck runs out, but the house edge is constant, which is why you need a real edge.
 

scepticus

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Re: Roulette Riddle about the house edge
« Reply #10 on: April 09, 2016, 12:14:11 PM »
He can only prove it " Mathematically" . In actual practice  " variance" determines who will win and who loses. He makes the classic " Mathematician's 'Fallacy " - confusing Expectation with Certainty.   

There is no such fallacy, it only exists in your mind. Try setting up a  Wikipedia page for "mathematician's fallacy" and see how long it lasts.  ;)

The fallacy is not recognizing that variance is just another way of saying "luck".

Luck runs out, but the house edge is constant, which is why you need a real edge.

All life is "Luck" . We are all here because of luck though, perhaps , you actually chose to exist. Our upbringing ,our health etc.  we owe to luck. So don't scorn "Luck".
Casinos depend on Luck- that all too many people bet haphazardly rather than with method. Your views on gambling are rather simplistic -that the HE rules .Gambling  is all about  trying to beat "The Odds" so your continuing repetition that we cannot beat the HE is boring - and in a gambling  website too! Get REAL Mike.
btw. I said that my Parlayed two dozen DID give the player a "real edge" and you did not dispute that. I now give you the opportunity to do so.
PROBABILTY IS ABOUT EXPECTATION- NOT CERTAINTY  !
p.s. Anything for the Grand National ?
 

kav

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Re: Roulette Riddle about the house edge
« Reply #11 on: April 09, 2016, 12:45:55 PM »
I think what is interesting here is the
If we bet One Dozen, the House bets us 2 to 1 that 25 numbers will beat 12. So the actual 2 to 1 on One Dozen is actually 2.08 to 1 because if we divide the House's 25 numbers to our 12, it equals 2.08 instead of two. So if we divide 2 into 2.08 that equals 96% which is a difference of the 4% . I like the mathematical breakdown that was demonstrated in your answer Kav.   Then if we back 2 dozen, the House makes US lay 2 to 1 that 24 numbers will beat 13 so the proper odds are 1.84 to 1 and we have a disadvantage of 8 %.

I think this is the interesting subject of this thread.
Please focus on this topic, this is quite specific.
 

scepticus

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Re: Roulette Riddle about the house edge
« Reply #12 on: April 09, 2016, 01:03:07 PM »
I think what is interesting here is the
If we bet One Dozen, the House bets us 2 to 1 that 25 numbers will beat 12. So the actual 2 to 1 on One Dozen is actually 2.08 to 1 because if we divide the House's 25 numbers to our 12, it equals 2.08 instead of two. So if we divide 2 into 2.08 that equals 96% which is a difference of the 4% . I like the mathematical breakdown that was demonstrated in your answer Kav.   Then if we back 2 dozen, the House makes US lay 2 to 1 that 24 numbers will beat 13 so the proper odds are 1.84 to 1 and we have a disadvantage of 8 %.

I think this is the interesting subject of this thread.
Please focus on this topic, this is quite specific.

But how does this give the House an edge  of 100 % ?
 

kav

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Re: Roulette Riddle about the house edge
« Reply #13 on: April 09, 2016, 03:03:33 PM »
Please read again my profit post and payout post in this topic to understand better what is happening here.
None really claims that the house edge is 100%. The "house edge" is constant and well known: 2,7%


But there is more than this.
Scepticus this whole discussion is about what would be FAIR the casino to pay us in case we win and what it actually pays us. And how this percentage changes depending on the amount of numbers we bet.
Ronm said:
Quote
If we bet 1 unit on 35 numbers, we could lose 35 units if one or two numbers came out and the most we could win for that coup is 1 unit.  We were therefore laying down 35 to 1 when the real odds are 35 to 2, or 17.5 to 1.  So we were laying double the proper odds, or in other words giving the House a 100% advantage over us.

Kav said:
Quote
The situation gets worse the more numbers we bet. When we bet on 37 numbers the "profit lost" becomes 100% since we are paid less than we bet even if we win.

 
« Last Edit: April 09, 2016, 03:26:30 PM by kav »
 

scepticus

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Re: Roulette Riddle about the house edge
« Reply #14 on: April 09, 2016, 03:56:12 PM »
Please read again my profit post and payout post in this topic to understand better what is happening here.
None really claims that the house edge is 100%. The "house edge" is constant and well known: 2,7%


But there is more than this.
Scepticus this whole discussion is about what would be FAIR the casino to pay us in case we win and what it actually pays us. And how this percentage changes depending on the amount of numbers we bet.
Ronm said:
Quote
If we bet 1 unit on 35 numbers, we could lose 35 units if one or two numbers came out and the most we could win for that coup is 1 unit.  We were therefore laying down 35 to 1 when the real odds are 35 to 2, or 17.5 to 1.  So we were laying double the proper odds, or in other words giving the House a 100% advantage over us.

Kav said:
Quote
The situation gets worse the more numbers we bet. When we bet on 37 numbers the "profit lost" becomes 100% since we are paid less than we bet even if we win.
So   we   bet
1 number = 2.7%
2 numbers = 5.4%
3 numbers = 8.1%
4 numbers =10.8%
5 numbers = 13.5%
etc,
17 numbers=45.9%
18numbers =48.6%
19numbers=51.35
35 numbers =94.5%
36 numbers =97.2%
37 numbers =100 %
 So  it is inadvisable to bet more than 18 numbers.
and ,preferably , no more than 17.
And we didn't already know this ?
« Last Edit: April 09, 2016, 04:11:16 PM by scepticus »