2x Dozens = 66% (2 Units) VS 1 Half = 50% + 1/5 unit Double Street (1.2 Units)

[mikeyboy]

Anybody thought of playing this way?

You're betting 100% extra (2 vs 1 unit) for just an extra 16% of the table (66% vs 50%)

Then to make up the remaining six numbers you can play 1/5 of the unit that is on the half (.20 if using 1 unit) and then if it lands on the half you win 0.8 units, if it lands on the six numbers you break even. If it lands on the remaining third you lose only 1.2 units instead of 2.

What does everybody think?

(before everyone says "but a half is less than 50% because of the zero", imagine no zero!  )

[Bazeegar]

Can you please explain it by giving example?

[JHM]

There has been a post concerning this.

It's the game, how less risk involved in the bet, how lower the pay out.

Bet a EC (even chances) vs. 1 dozen.

EC: bet 1 unit to cover 18 numbers and get paid 1 unit when won.
1 dozen: bet 1 unit to cover 12 numbers and get paid 2 units when won.

Is it really worth that to risk winning 1 unit in stead of 2 by only playing 6 more numbers?

[mistarlupo]

... and then if it lands on the half you win 0.8 units, if it lands on the six numbers you break even.

Okay, here's another comparison...

- bet 1.2u: win 0.8u / lose 1.2u (your bet)
- bet 1.0u: win 1.0u / lose 1.0u (random bet on even chances)

In both cases above you're expected to win 50% of the time (not break even!), right?
So I do not see any value in your bet compared to a simple EC bet... you bet more, lose more, win less.

Regards,
m

[See_Jerek]

Okay, here's another comparison...

- bet 1.2u: win 0.8u / lose 1.2u (your bet)
- bet 1.0u: win 1.0u / lose 1.0u (random bet on even chances)

In both cases above you're expected to win 50% of the time (not break even!), right?
So I do not see any value in your bet compared to a simple EC bet... you bet more, lose more, win less.

Regards,
m

Progressions for double dozen can kill,its worse then EC

[See_Jerek]

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