Royal Panda roulette

Author Topic: What to consider equally probable?  (Read 163 times)

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What to consider equally probable?
« on: November 12, 2018, 01:23:17 PM »
By matching same total of numbers being bet shall we expect the same overall probability?

For instance to bet 1 time 35 numbers VS 35 times 1 number, are they equally probable?
Another example, I'm betting 18 numbers for 10 times VS to bet 6 numbers for 30 times, are they equally probable regarding the expectation?

The expectation could be just 1 win with a progression, or net profit by flat bets.
1 number may or may not be part of the 35, 6 numbers may or may not be part of 18 numbers.

I believe the answer to the above questions is no, they don't have the same chance for a single win and for net profit.
So since they don't have the same chance then something has to be better, what is the optimal and why?

In general, more numbers for less times have higher chance to hit, this way is only better when you are using progression and aim to profit with a single win, thus only a negative progression could be used for such situation.
But if we would flat bet without progression involved then the more numbers we would bet the more negative would be the outcome after all bets have been placed.

Therefore we can conclude to a strategy for say betting 30 numbers with a 2 steps negative progression, 5 lines with 1 unit each, when we lose 5 units we'll raise the 5 selected lines to 6 units.
We go on as long as we are winning within these 2 steps, a bust would cost 5 from the first step plus 30 from the second, a total of 35 units.
Those 35 units are only a part of the BR, another part bets 35 numbers with 1 unit on each one of them, thus another 35 units at risk by on different way.
Those 2 parts are working independently from each other, what is happening on 1 does not affect the process of the other.
The third part has 3 steps, first it bets an EC, if loses bets for the second step 2 dozens or 2 columns with 2 units each, if this step lost too then proceeds to the final third which bets 5 lines with 6 units each.
For those 3 steps the total risk is 35 units to win 1, the same risk VS reward ratio have the previous 2 parts, but the difference between them is the steps according the amount of numbers being bet per step.

Part 1 } 35 numbers by 1 unit each for 1 step/time.

Part 2 } 30 numbers in the form of 5 6-lines by 1 unit each the first step and by 6 units each the second.

Part 3 } first step 18 numbers EC group by 1 unit, second step 24 numbers with 2 dozens or 2 columns by 2 units each, final third step 30 numbers in the form of 5 six lines by 6 units each.

Every part risks the same amount of units, 35 and wins 1 unit per coup.

Now I'm going to show you their 3 counterparts:

Part 4 } 1 number with 1 unit flat bet for 35 times (mirror of part 1)

Part 5 } 6 numbers in a form of six line with 1 unit, after every win parlay the 6 units for a second consecutive win. (mirror of part 2)

Part 6 } 18 numbers EC group by 1 unit, parlay every time to a dozen or column those 2 units, if the dozen/column won then you'd parlay for a final third time those 6 units to a 6 numbers in the form of a 6-line. (mirror of part 3)

Among parts 4,5 and 6 the risk versus reward ratio are the same, risking 1 unit for 35 times to win 35 units net.
Therefore the hedging effect is in place for 3 pairing parts, all of them cogs of the same betting mechanism.
When 1 part is being busted it'll be immediately discarded and continue only with what is performing positively on any given time.

Total fuel to last for our betting mechanism are 35 x 6 = 210 units, well if you had 200 it would not bother me!  8)
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