This game stimulate Single dozen bet,
Where the *single dozen * hit 30%,
in every possible permutation.
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In a box,
There are 100 balls,
70 are white,
30 balls are black.
White=-1.
Black=+2.
You draw a ball, and NOT REPLACE BACK, THE BALL.(set aside)
Bet anyway you like,
How to bet, to win, at least 1unit, flatbet?
Think HARD!

Remember to account for the variable change. This game has a memory, just like counting cards in 21. That's how you beat it.

Thank Gizmotron, for your view.
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Gentlemen,
in this simple game,
when we draw out the first ball,
till the last ball,
there will be only ,
30-black ball [win=+2],

and only ,
70 white ball [ lose=-1],

in all possible PERMUTATION.

this stimulates a game,
that you will hit,
by 70 losses, of -1,
and gain 30 wins, of +2.

[same as bet single dozen, flatbet.
where you bet 1unit,
where win= +2, lose= -1]

You need to draw every ball, from 1st to 100th.
and cast them aside, not returning them into the box.

BUT,
you may choose to bet any way you deem feasible,

meaning, you may choose,
to bet or not to bet,
WHENEVER,
a ball drawn.

still no respond.....hmmmmm.

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There are 100 balls,
70 are white,
30 balls are black.
White=-1.
Black=+2.

thus, if we play all ball drawn, then,
30 x +2=+60.
70 x-1=-70,
thus the deficit of -10 or -10u losses..

30balls of black, could only win,
at less than 30/90spins.
where,
90-30=60,
the 30black x 2=+60,
and 60white=-60, =breakeven.

then , only 30black, in 89spins,
could produce profit,
or in other word,
we wait for -11 losses, before start to bet.