[Closed] The best way to play roulette?

For colour, you win $50 18x out of 37. For each number you win 180 1x out of 37.
Therefore you get 50*18/37 + 5*180/37 = $48.65 on average.

(And you always get $48.65 EV no matter how you bet on roulette. Every bet has 36/37 = 0.973 EV.)

Dip,

All bets there are -EV long term, it is all about the Variance np{1-p}

But I think everything as betting in the numbers (single 11 or split 10/11) has the same EV

I think this goes as well for the Red/Black, dozens and stuff

EV (1st dozen) = .324*$30 - (1-.324)*20$
EV = -$3.8

EV (Single Number) = (1/37)*35*10$ - (36/37)*$10{??}
EV = +$0.8~ but the VARIANCE IS SICK

I will look it up in a Math Exam since to refresh my mind ^^

you massive loser

Originally posted by Gerv
Dip,

All bets there are -EV long term, it is all about the Variance np{1-p}

But I think everything as betting in the numbers (single 11 or split 10/11) has the same EV

I think this goes as well for the Red/Black, dozens and stuff

EV (1st dozen) = .324*$30 - (1-.324)*20$
EV = -$3.8

EV (Single Number) = (1/37)*35*10$ - (36/37)*$10{??}
EV = +$0.8~ but the VARIANCE IS SICK

I will look it up in a Math Exam since to refresh my mind ^^

I never said it could be +EV, that's why I made the long intro

Interesting that everything has the same EV, didn't expect that... And what are you saying that a single number actually has a positive EV, but the variance is huge?! If that would be true, couldn't you just bet the same amount on all the numbers and win everytime without variance?

Originally posted by Dippy19
And what are you saying that a single number actually has a positive EV, but the variance is huge?!

No, it has the same negative EV like anything else (36/37 of the bet).

Originally posted by Dippy19
If that would be true, couldn't you just bet the same amount on all the numbers and win everytime without variance?

Yes, but it's not true. ♦

OP i suggest you read this

http://en.wikipedia.org/wiki/Roulette

Now here are some thing that contradict other posters:

1) an instance where the edge is not the same

pay odds

Five Number Bet 0, 00, 1, 2, 3 6 to 1 6.6 to 1 −$0.079

NB: obvs assumers american wheel

this is more lost (different) compared to the rest which are −$0.053

2) i think you mean losing the least! In which case: I'll refer you to this:

1st and 3rd column strategy - search the wiki for that

my understanding is that this strategy breaks even in the long run (or VERY close to)

Of course, I have not tested this myself and am not interested enough, but assuming these are correct it should be insightful for you as it has been for me.

Originally posted by shortfuse
1st and 3rd column strategy - search the wiki for that

Ok, you have 12 numbers in the 1st column and 12 numbers in the 3rd one.
You win 3 times your bet I believe.
Therefore you get 3*(24/37)/2 your bet (you win 3 times your bet, you win if you hit one of your 24 "outs" out of 37 numbers and you place 2 bets). Surprisingly the resulting EV is 36/37 of your bet (like always in roulette).

Originally posted by Meiffert

Originally posted by shortfuse
1st and 3rd column strategy - search the wiki for that

Ok, you have 12 numbers in the 1st column and 12 numbers in the 3rd one.
You win 3 times your bet I believe.
Therefore you get 3*(24/37)/2 your bet (you win 3 times your bet, you win if you hit one of your 24 "outs" out of 37 numbers and you place 2 bets). Surprisingly the resulting EV is 36/37 of your bet (like always in roulette).

did you even read the article? no

go read it again

Originally posted by Dippy19
The best way to play roulette?

With play money.

Originally posted by Gerv
EV (Single Number) = (1/37)*35*10$ - (36/37)*$10{??}
EV = +$0.8~ but the VARIANCE IS SICK
I will look it up in a Math Exam since to refresh my mind ^^

Please, roullete in never +EV ..
EV (Single Number) = (1/37)*35*10$ - (36/37)*$10 .. That is exactly 35/37 - 36/37 and that can never be > 0 ...

I already thought I made a mistake, thanks

[ So I calculated a bit and I found out that the way I placed the bets(shown above) I get 49.32$ for every 50$ invested.

You aint got 0 covered; so when it hits u lose $50

I've seen three/four 0's ina row Siiiiiiiiiick

i only ever bet on roulette sometimes; when i do i load onto 1-12, 12-24, or 24-36 then cover eacch other the other 12;s

e.g.
50p per number 1-12
then bet like £4 on 12-24 and 24-36

then pick a few numbers from the 'loaded' section 1-12 in this case and put more money on them i.e. £1.30 on 2, 5, 7, 10 or something and then spiiiiiiiiiiiiiiiiiiiiiiiiiiiin the wheel lol its not exactly a 'system' but its fun when u feel like having a punt

Sabo

Originally posted by shortfuse
did you even read the article? no

go read it again

Ok, I'm sorry. :-)
I've read it now.
It only breaks even in comparison with other possible bets on roulette.
In the article they explain it themselfs: "This "strategy" breaks even in the long run, of course, for the simple reason that any linear combination of zero-expectation bets is itself zero-expectation".
Obviously the "partial" bets (1st column, 3rd column and black) are not zero-EV, they are negative EV (-1/37). Therefore the sum of these (sum is a special case of linear combination) is also negative EV. You will loose 6/37 bets on average with this system (as you always bet 6 bets total).

Originally posted by Meiffert

Originally posted by shortfuse
did you even read the article? no

go read it again

Ok, I'm sorry. :-)
I've read it now.
It only breaks even in comparison with other possible bets on roulette.
In the article they explain it themselfs: "This "strategy" breaks even in the long run, of course, for the simple reason that any linear combination of zero-expectation bets is itself zero-expectation".
Obviously the "partial" bets (1st column, 3rd column and black) are not zero-EV, they are negative EV (-1/37). Therefore the sum of these (sum is a special case of linear combination) is also negative EV. You will loose 6/37 bets on average with this system (as you always bet 6 bets total).

no because in the instance numbers which are in 1st or 3rd AND are black come up, then you get a chance for a big payout.

NOTE: the best must be of equal value on 1st, 3rd, and black
(I'm unsure why you say 6 bets are placed?)

This is why it breaks even.

This is my answer to the thread; the best way to win = least way to lose

Originally posted by shortfuse
no because in the instance numbers which are in 1st or 3rd AND are black come up, then you get a chance for a big payout.

So? If you hit a red number in the 2nd column, you loose it all.
If you bet on a single number, the payout is much higher (like 35 times your bet).

Originally posted by shortfuse
NOTE: the best must be of equal value on 1st, 3rd, and black
(I'm unsure why you say 6 bets are placed?)

Yeah, you are right. On wiki they say you bet 2 bets on 1st, 3rd and black, 6 bets in total. You can obviously bet a different amount as long as all bets are equal.

Originally posted by shortfuse
This is why it breaks even.

It doesn't.

Lets do the math one last time.
You bet $2 on 1st column, $2 on 3rd column, $2 on black (that's the strategy from wiki).

If you hit a black number from 1st or 3rd column (there are 10 of them), you get 10 bets (4 for black and 6 for column).
If you hit a red number from 1st or 3rd column (there are 14), you get 6 bets (for column).
If you hit a black number from 2nd column (there are 8), you get 4 bets (for color).
If you hit a red number from 2nd column (there are 4), you get 0 bets.
If you hit zero (there is 1), you get 0.

That gives you grand total of (10*10 + 14*6 + 8*4 + 4*0 + 1*0)/37 (there are 37numbers), which is 216/37 = 5.8378 bets on average.
Because you always bet 6 bets total, your EV is 5.8378 - 6 = - 0.1622 (which is - 0.1622/6 = - 0.027 or - 1/37 EV on average on a single bet like always in roulette).

Originally posted by Meiffert

Originally posted by shortfuse
no because in the instance numbers which are in 1st or 3rd AND are black come up, then you get a chance for a big payout.

So? If you hit a red number in the 2nd column, you loose it all.
If you bet on a single number, the payout is much higher (like 35 times your bet).

Originally posted by shortfuse
NOTE: the best must be of equal value on 1st, 3rd, and black
(I'm unsure why you say 6 bets are placed?)

Yeah, you are right. On wiki they say you bet 2 bets on 1st, 3rd and black, 6 bets in total. You can obviously bet a different amount as long as all bets are equal.

Originally posted by shortfuse
This is why it breaks even.

It doesn't.

Lets do the math one last time.
You bet $2 on 1st column, $2 on 3rd column, $2 on black (that's the strategy from wiki).

If you hit a black number from 1st or 3rd column (there are 10 of them), you get 10 bets (4 for black and 6 for column).
If you hit a red number from 1st or 3rd column (there are 14), you get 6 bets (for column).
If you hit a black number from 2nd column (there are 8), you get 4 bets (for color).
If you hit a red number from 2nd column (there are 4), you get 0 bets.
If you hit zero (there is 1), you get 0.

That gives you grand total of (10*10 + 14*6 + 8*4 + 4*0 + 1*0)/37 (there are 37numbers), which is 216/37 = 5.8378 bets on average.
Because you always bet 6 bets total, your EV is 5.8378 - 6 = - 0.1622 (which is - 0.1622/6 = - 0.027 or - 1/37 EV on average on a single bet like always in roulette).

ok thanks for clearing that up, i was basing my facts on wiki- bad idea lol

Originally posted by sydney69

Originally posted by Dippy19
The best way to play roulette?

With play money.

ahhaha this imo!

I wrote a few notes on roulette into my blog, if anyone is interested
Augustus's Poker Simulations, Variance, Theory & NLSH Blog

hmm i heard the +ev to play roulette is to put let's say $100, if you dont hit then you put $200 on the same thing and so on and so on so that your next bet always has covered your previous losses and if you hit, you make a $100 profit - so if you have enough money to do this, you should be able to hit and once you do, you call it a day and come back tomorrow