The zero should only occur once in an average 37 spins

I can ignore the zero if I want to-my choice not yours .This does not mean I don't think it exists - just that it makes calculating easier.After calculating factor in a 1 in 37 loss to allow for the zero . Yes?.

"why do I want to reduce the house edge ? because maths geeks tell us it can't be done and I think it would give me a better chance to win .Don't you think so - assuming it can be done.

Quote:scepticusThere are only 8 sets of 2x2x2

The zero should only occur once in an average 37 spins

I can ignore the zero if I want to-my choice not yours .This does not mean I don't think it exists - just that it makes calculating easier.After calculating factor in a 1 in 37 loss to allow for the zero . Yes?.

"why do I want to reduce the house edge ? because maths geeks tell us it can't be done and I think it would give me a better chance to win .Don't you think so - assuming it can be done.

But it can't.

ZCore13

Quote:Zcore13What happens if a green comes up twice in a row after the indicator spin?

ZCore13

Your first green is the second spin/the first bet so the bet has lost and if the second spin loses then the third spin is not bet.

So we let the third spin go without a bet and also the next spin which is the first of the next three spins.

We ( theoretically) bet in series of three.

Quote:Zcore13But it can't.

ZCore13

" It can't " is not a reason.

Using my illustration - explain please.

Quote:scepticusexplain please.

We are greatly indebted to Michael Shackleford for providing a website that answers your question, so now we don't have to do the work! Check it out!

http://wizardofodds.com/

Quote:scepticusignoring the zero.

Once again, I may not have made myself clear.

ignoring the zero.

I did take some liberties with the quote, but you're ignoring the zero.

They're ALL good games, if you remove the house edge.

The problem is, even on a single-zero wheel, that little white ball doesn't ignore the zero 2.7% of the time. The fact that you only lose half of your even money bet is still a house edge of 1.35%.

If you want to place insurance on 0 ($1 per $72 on red/black, if I remember correctly...), you're creating a predictable loss rate, but you're not removing the house edge.

As far as I know, the ONLY way to remove the house edge in roulette is to find a biased wheel.

Quote:scepticus" It can't " is not a reason.

Using my illustration - explain please.

Maybe we are all not understanding each other. If you are saying you can reduce the house edge by placing certain bets over other bets, or by hedging bets, of course you can. If you are saying you can

You can reduce the house edge by playing a corner bet (5.26% HE) over the first five numbers (7.89% house edge).

If you are trying to say you can reduce any house edge by any type of money management or betting system, you can't. Nothing changes the house edge of a bet or combination of bets. You can not ignore certain outcomes no matter how infrequent they come up. They are part of the house edge.

Your illustration shows you can lose less money by better less often. That is reducing risk, not house edge.

ZCore13

Quote:scepticusWell ,O.K. TDG I didn't check my post properly .It is 1 in 4 but it is still a 3/1 chance - ignoring the zero.

Once again, I may not have made myself clear.

Consider I choose to bet BOTH Red AND Black over 3 spins-the 3 to be either all reds or all black.Betting both red and black on the first spin is not advisable because (a) we won't profit and (b) a zero would incur a loss.So we avoid the first spin of the three.

If a red wins the first spin we bet 1point on red and if it wins we put 2 points on red again on spin 3. Similarly with black.

If, on the second spin black wins when I'm betting red the bet is lost and we don't bet the third spin.

In theory we are betting BOTH red and Black to win 3 in a row- a 3/1 shot ignoring the zero.

The bet is over a series of 3 spins ,there is no fourth .

Clearer now ? The question is does it reduce the whole HE over the series of three or has it been reduced ?

Understood. If zero doesn't show up, the HE is zero. Of course, that's why the zero is there.

Going back to your original post concerning the zero:

Quote:scepticusIf it falls on the first spin we don't lose as we haven't bet.

Correct.

Quote:scepticusIf it falls on the second spin we lose half a point.

Correct.

Quote:scepticusIf it falls on the third spin then either,

a) we lose nothing if our second bet loses as our double has lost. or,

b) we lose the win of the second spin but not our original stake so we lose -or win - nothing.

So the chance of LOSING to the zero is reduced.

In other words, if you lose the second spin (because the first two spins were two different colors, or the second spin was zero), you don't lose anything if the third spin is zero because you didn't bet anything, and if you win the second spin, you lose nothing overall from the three spins because the 1 point that you lose (since your bet is 2) is the 1 point that you won on your second spin.

Here is what really happens:

Assume the first spin is not zero - if it is zero, you didn't bet anything, so wait for a non-zero spin.

1/37 of the time, the second spin is zero; you lose 1/2.

18/37 of the time, the second spin is a different color from the first spin; you lose 1.

18/37 x 18/37 of the time, the second and third spins are the same color as the first one; you win 3 overall.

18/37 x 18/37 of the time, the second spin is the same color as the first one, but the third one is different; you lose 1 overall.

18/37 x 1/37 of the time, the third spin is green; you break even overall.

The overall EV = (1/37 x -1/2) + (18/37 x -1) + (18/37 x ((18/37 x 3) + (18/37 x -1) + (1/37 x 0))

= (1/37 x -1/2) + (18/37 x -1) + (18/37 x 36/37)

= (1/37 x -1/2) + (18/37 x -1/37) = 1/37 x (-73/74).

It appears that your third bet has no edge on green because you're not comparing it to the alternative of stopping after the second spin, where you would win 1 overall. In effect, you are betting "one of yours and one of the house's" on the third spin, and if zero shows up, you lose nothing from your original 1 (which is why you claim that you lose nothing), but you lose 100% (not 50%) of the house's 1.