European Roulette Payout

[retrochoir]

On a European Roulette Table (36 numbers plus a no win zero) what should be the average payout in percentage terms?

[Peter Pedant]

If you are betting on numbers alone

it is 36/37 X 100 whatever that is about 97% I think

The average is less than 100 because you LOSE in the long term at Roulette

[helpmetoo]

the in built house advantage in single zero roulette is 2.7%
playing live and on some net casinos zero pays 50% of stake return on evens bets only
the advantage for that bet is 1.35%
to understand more
and overcome the house advantage
serious players only
look up mean18.co.uk

[retrochoir]

Thanks, I thought it was something like that.

I am different roulette betting systems i.e. covering your previous bets.

These systems do give you a better chance in the short term but the casino will always win in the end.

I was playing on roulette the other night with pretend money. I started with �600 and betting on 1 number only I eventually got up to about �40,000!

Knowing my luck, this would never happen if I was betting real money!

[helpmetoo]

numbers do reappear
if you placed a �1 accumulator for the chance that in 37 spins every number would appear once (in any order)
how much would the payout be?
i know
i bet youre not within 10% with your guess
mean18.co.uk

[helpmetoo]

it aint going to happen!!!!!!
play with the knowledge of roulette physics
mean18.co.uk

[retrochoir]

I dont know what you mean. Are you talking about putting a �1 bet 37 times on the same number.

[helpmetoo]

no, in lay terms you could expect in 37 spins each number should appear once?
put a �1 bet on the second spin being different from the first and then that bet including true odds return would be �1.02 that amount is bet on the expectation the third number to appear would be different to the first two...and so on in .fractions the first spin would be 37/37=1 ie guaranteed a new number, the second spin would be 36/37 and so on.
i'll give you a hint....its big big big numbers

[retrochoir]

Yes but the law of probability states that even though 1 number may have already appeared once, it still has the exact same chance of appearing again i.e. 1 in 37.

Even if the number 10 appeared 3 times in a row, it would still have the same chance of appearing on the next spin as any of the other 36 numbers.

[retrochoir]

I understand what you are saying now, but can you place this sort of bet on a roulette table?

[retrochoir]

My guess for you example would be a payout of 1,900% or �19 from a �1 stake.

[retrochoir]

I've changed my mind. The payout on an initial bet of �1 would be �2,276,930.617

That's my final attempt.

[helpmetoo]

not bad but light years away
695,000,000,000,000.
it escalates a bit
the accumulated bet would pay 50850 for the first 25 numbers all being different
the next bet would be your accumulated stake of 50850 at true return against a 12 in 37 chance of another new number appearing = x 3.083333
you would then carry over 156621 to the next stage which if successful x 3.363636 = 526245
youre now at the lively part of the bet

[retrochoir]

I'm still trying to get my head around that but it is really interesting.

It's a shame this kind of result can't be obtained in a real situation.

You have mentioned mean18 a number of times. I've discovered they have an e-book available. Are you aware of this publication? Are you a contributor? Is it any good?

I appreciate your interest in advance.

Thanks a lot

Jeff