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01-27-2015 , 02:19 PM

#51

Didace

Carpal \'Tunnel

Join Date: Nov 2009 Posts: 19,888

Quote:

Originally Posted by RickySteve

Poker players are so adorable.

Are you suggesting that math is a sham?

Quote

01-27-2015 , 10:42 PM

#52

RickySteve

banned

Join Date: Sep 2014 Posts: 655

Not all wagers on the roulette table are independent.

A parlay is at least one bet.

Please keep amusing me.

Quote

01-27-2015 , 10:45 PM

#53

RickySteve

banned

Join Date: Sep 2014 Posts: 655

Quote:

Originally Posted by heehaww

Roulette players are so adorable. Especially when they try to play a game of skill or brains.

Remember when you thought that betting 2x Kelly leads to inevitable bankroll death? That was hilarious! Have you tried stand-up?

Quote

01-28-2015 , 12:50 AM

#54

NewOldGuy

Pooh-Bah

Join Date: Mar 2009 Posts: 5,935

Quote:

Originally Posted by Timlagor

It's worth noting that multiple -EV bets are only necessarily -EV together if they are independent

Quote:

Originally Posted by NewOldGuy

The above is false. If you know all the bets will be -EV, any relationship or dependence between them is irrelevant to their sum still being -EV

The EV of any series, whether the events are independent or not, is the sum of the individual EVs in the series.

Quote:

Originally Posted by RickySteve

You're wrong but that is amusing so please continue.

So are you ready to admit you were wrong in the exchange above? Or do you stil just want to be a snarky *******?

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01-28-2015 , 01:58 AM

#55

RickySteve

banned

Join Date: Sep 2014 Posts: 655

Quote:

Originally Posted by NewOldGuy

So are you ready to admit you were wrong in the exchange above? Or do you stil just want to be a snarky *******?

We can agree that one of us is wrong.

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01-28-2015 , 04:11 PM

#56

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Quote:

Originally Posted by RickySteve

Remember when you thought that betting 2x Kelly leads to inevitable bankroll death? That was hilarious!

Have you seen my reply to you in that thread which I posted yesterday? I stand by that statement.

Why don't you try simulating it and see what happens? Do you know how to write a simulation?

What do you predict will happen? Your bankroll will stay flat forever and you'll never experience a downswing that wipes you out? (Assuming there exists a minimum allowed wager, which was the context of that thread, not a hypothetical scenario where you can make infinitesimal wagers.)

Quote

01-28-2015 , 05:16 PM

#57

lkasigh

adept

Join Date: Jun 2013 Posts: 953

He is correct with respect to roulette that betting 2x Kelly will not lead to a loss of your bankroll.

Quote

01-28-2015 , 05:24 PM

#58

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Quote:

Originally Posted by lkasigh

He is correct with respect to roulette that betting 2x Kelly will not lead to a loss of your bankroll.

WRT roulette, betting anything will lead to a loss of your bankroll. A billionaire betting $1 per spin will lose his BR if he plays enough lifetimes.

Perhaps more importantly, betting anything in roulette is infinity times kelly.

Quote

01-28-2015 , 07:23 PM

#59

NewOldGuy

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Join Date: Mar 2009 Posts: 5,935

Quote:

Originally Posted by lkasigh

He is correct with respect to roulette that betting 2x Kelly will not lead to a loss of your bankroll.

LOL. Every possible betting system in roulette leads to a total loss of bankroll eventually. Not counting the change when you don't have enough for the minimum bet left.

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01-28-2015 , 08:26 PM

#60

lkasigh

adept

Join Date: Jun 2013 Posts: 953

That's what I meant, since the Kelly bet is $0, you don't bet at at all, hence you bankroll stays the same.

Quote

01-28-2015 , 08:33 PM

#61

NewOldGuy

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Join Date: Mar 2009 Posts: 5,935

Quote:

Originally Posted by lkasigh

That's what I meant, since the Kelly bet is $0, you don't bet at at all, hence you bankroll stays the same.

Right, after you are busted.

But I see your point now. Kelly is always 0 in a negative EV game.

Last edited by NewOldGuy; 01-28-2015 at 08:38 PM.

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01-28-2015 , 11:36 PM

#62

heehaww

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Join Date: Aug 2011 Posts: 5,081

Yeah I realized it after I left. You meant that 2x kelly still equals 0 because 2*0=0. Clever bastard ^^

So then correction: RickyTroll's assertion is true only for -EV games.

Quote

01-29-2015 , 11:18 AM

#63

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Back to the EV thing, I see that the debate about parlays arose from Timlagor saying the events have to be independent for expectations to sum together. That's wrong. In statistics there's the property known as "linearity of expectations". I'll state the relevant part.

With 2 random variables X and Y (independent or not),
E[X+Y] = E[X] + E[Y]

What Timlagor may have had in mind is multiplicativity. That does require independence.

E[X*Y] = E[X]*E[Y]
if and only if their covariance Cov[X,Y] = 0

I hope that puts this to rest. Anyone who still disputes it can break open any stats textbook and look up "linearity of expectations".

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01-29-2015 , 04:22 PM

#64

Lego05

Carpal \'Tunnel

Join Date: Mar 2006 Posts: 24,916

Quote:

Originally Posted by Lego05

If you bet a parlay, then you made 1 bet.

Quote:

Originally Posted by Lego05 as modified by RickySteve

If you bet a parlay, then you made at least 1 bet.

Quote:

Originally Posted by RickySteve

FYP.

YW.

I'll clarify:

One bet on one parlay is one bet.

Last edited by Lego05; 01-29-2015 at 04:28 PM.

Quote

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