Loss expectation in 3.000 bets on 2 dozens.

[RouletteExplorer]

I would like to know how much is mathematical the loss when we are playing 2 dozens at the same time in 3.000 betted spins.

Thank you.

[RouletteExplorer]

Is mt calculation correct?

3000 / 100 = 30 ... 30 X 2.7= 81 lost chips

[MadMax]

Hi!
I´m really no mathematician, but I would calculate this way:

(3000-expectet wins)*2

(3000-(3000*((1/37)*24)))*2 = 2108,11 lost chips

I think, thats right, because in simpler words (and without zero): you can expect to lose 1/3 of the bets with 2 units beted on the losing dozens. That means, 1000 lost bets á 2 units = -2000 units.
Ah, now I see, you just calculatet the 2,7% zero loss.

Or am I wrong?

[RouletteExplorer]

Mad Max my friend There is no posibility that in 3000 spins by placing 2 chips on the tabe to lose 2108 chips 

I think my calculation is correct of the 81 chips but I forgot to multiply it by 2 because I had forgoten that the betted chips on every spin are 2 and not 1

so 81 X 2= 162 chips....I think that now its correct.

[Fripper]

If you play one unit on a dozen then you should lose 81.08 units to the zero.
3000/37=81,08



In your case, bet on two dozens should lose 162,16 units to the zero.

3000/37=81,08*2=162,16


Hope I'm right.

[RouletteExplorer]

thanks  Fripper me too 

[MadMax]

Ah, yes, you both are right.
I have only calculatet the losses without the wins!
The expectable wins would be 1945,95, the losses would be 2108,11. And the difference: -162,16.
I should stop thinking for this day, nothing usefull comes out here today anymore, I guess!

[RouletteExplorer]

You just made a wrong calculation man....no big deal

[Chrisbis]

It didn't look right to me either guys.

Its already in here somewhere, as its been discussed before.

[RouletteExplorer]

Si if in 3000 bets we lose 90 chips instead on 162 does this mean that we have found a bet selection that has a less disadvantage than the -2,7? 

Or the 3.000 is a very little sample ?

[warrior]

Si if in 3000 bets we lose 90 chips instead on 162 does this mean that we have found a bet selection that has a less disadvantage than the -2,7? 

Or the 3.000 is a very little sample ?

WE should test like this bet selection = betting opposite 111111 meaning bet 23 now if you get 6 WS, SWITCH bet opposite 222222 meaning bet 13 every time you get 6 WS YOU SWITCH.If within in 100 spins you win 66% i think its enough to profit, naturally anything below this is a loss just a thought.

[nitrix]

Or the real math Gain formula:
(odds of winning * units won) -
(odds of losing * units lost) =
overall profit

(2/3 * 1) -
(1/3 * 2) =
0

Every bet combinations on the Roulette board have a neutral mathematical expectation.
Because of the ZERO though, the game is now a negative expectation game.

Now if you only want to know for average % of wins or losses, just take the row of the formula that relates to you, but I just wanted to say simply looking at parts of it will give you a false idea of what's going on.

In 3000 bets, betting 2 dozens, you'll lose exactly 1/3 of your bets and win 2/3 of them.

So 2000 times you'll win 1 unit due to (2/3) odds (Balance +2000).
Then are 1000 times you'll lose 2 units due to (1/3) odds (Balance -2000).

We're back to a balance of +0,
But also in mean time the Zero on the wheel should hit a few times... that's why you lose units.

Hope its clear enough.

[warrior]

IS THIS TOPIC DEAD ALREADY?

[warrior]

Or the real math Gain formula:
(odds of winning * units won) -
(odds of losing * units lost) =
overall profit

(2/3 * 1) -
(1/3 * 2) =
0

Every bet combinations on the Roulette board have a neutral mathematical expectation.
Because of the ZERO though, the game is now a negative expectation game.

Now if you only want to know for average % of wins or losses, just take the row of the formula that relates to you, but I just wanted to say simply looking at parts of it will give you a false idea of what's going on.

In 3000 bets, betting 2 dozens, you'll lose exactly 1/3 of your bets and win 2/3 of them.

So 2000 times you'll win 1 unit due to (2/3) odds (Balance +2000).
Then are 1000 times you'll lose 2 units due to (1/3) odds (Balance -2000).

We're back to a balance of +0,
But also in mean time the Zero on the wheel should hit a few times... that's why you lose units.

Hope its clear enough.
[/quote So were losing 1/3 of our bets must mean were winners if bettig 12 #.

[nitrix]

I'm not sure I understand you but the thing is, without a progression, usually a bet selection will have "that moment" where you're gonna lose everything you previously won..

and thats just the odds of the game. To defeat that, people make progressions of all kinds, wait for a trigger or anything else, yet nobody found a real solution but hey

'doesnt mean the Earth's flat because we believe so