So, in this game, there is totally 48 balls, and 35 balls will be drawn. Balls are organised like this: 1-8 blue, 9-16 green............. So, there is totally 6 colors - every color with 8 balls.

You can make bet whenever You want during round. There is different ways of betting. You can bet on:

- Next ball number
- Next two balls number
- Ball will be drawn in next 6 draws (You can select up to three balls to be drawn in next 6 draws)
- Next color
- Next ball will be odd / even
- In next 6 balls, there will be more odd / even / tie balls.

Betting on next 6 balls will be disabled after 30th ball was drawn.

Betting on next 2 balls will be disabled after 34th ball was drawn.

So, can I make advantage over house edge? Like, if we are on ball 20 and there is no blue balls drawn, we have a chance to next ball be blue? Or something like that? Do You need exactly odds for all of this types of bets?

Payouts are always changing according to a number of remaining balls.

For "next ball" bet it's always like this: (remaining number of balls - 1)

So for example, if 10 balls were already drawn and I bet for next ball, my odds will be x37 (48 balls total - 10 drawn = 38 remaining. 38-1=37), so If I bet $1 I will get $37.

When 0 balls were drawn, odd for next bet is 47.

But for everything else it's so complicated to explain, I can't explain how it's changed.

For next odd / even it's like this:

When 0 balls are drawn, odds are 1.80 / 1.80

For next colour it's like this:

When 0 balls are drawn, odds are 5.40 for every colour.

Some random situation for next colour bet:

bet on blue (6 balls remaining): x6.15

bet on green (8 balls remaining): x4.61

bet on violet (6 balls remaining): x6.15

bet on red (6 balls remaining): x6.15

bet on orange (7 balls remaining): x5.27

bet on grey (8 balls remaining): x4.61

I can't write everything or get a pattern for everything.

If the odds for each bet can change after each ball is drawn, then chances are you won't be able to beat it. If it is online, it is a simple matter for whatever app is running the same to adjust the odds on all of the bets so the house advantage is the same.

Is it possible to go through one complete game, and list the odds on all of the bets (a) at the start, and (b) after each ball is drawn (listing the number and color of each ball drawn, of course)?

If there are 20 remaining balls, odd for next ball will be 19 (for any ball). And in the situation like this: no blue balls were drawn - so, is it a bigger chance to blue be next ball? And if there are more odd balls remaining, will ODD BLUE balls have the biggest change to be drawn? And because we have same odds for any ODD BLUE ball and for any NON-BLUE EVEN ball, can we beat house edge in that way? Maybe we have a bigger chance and same odds?

Yes, it's possible to go through one complete game and list the odds. I will send You link to the game in PM.

Quote:RSOn pick #22 (for instance), will the green, blue, orange, etc. all have the same payout or do they differ from each other?

Betting on colour will not have the same payout, but betting on ball number will have the same payout. I was given an example:

Quote:mbackonja

Some random situation for next colour bet:

bet on blue (6 balls remaining): x6.15

bet on green (8 balls remaining): x4.61

bet on violet (6 balls remaining): x6.15

bet on red (6 balls remaining): x6.15

bet on orange (7 balls remaining): x5.27

bet on grey (8 balls remaining): x4.61

and also

Quote:mbackonja

For "next ball" bet it's always like this: (remaining number of balls - 1)

So for example, if 10 balls were already drawn and I bet for next ball, my odds will be x37 (48 balls total - 10 drawn = 38 remaining. 38-1=37), so If I bet $1 I will get $37.

This is balls colours:

1-8 blue

9-16 green

17-24 violet

25-32 red

33-39 orange

41-48 grey

So for example, if we are on pick #22 and:

4 blue balls were drawn,

4 green balls were drawn,

4 violet balls were drawn,

4 red balls were drawn,

5 orange balls were drawn

0 grey balls were drawn

so in total, 21 balls drawn, we have the same odd for betting on any number (regardless number ball's colour), we have 48-21-1=x26 payout for that bet. So if we bet on the ball "3 blue" we have same odds as betting on "45 grey".

But betting on next colour we will have different odds.

Can we combine two things like colour and parity to get most likely next number? For above example, let's assume that in 21 balls from the quote above ^ %70 of drawn balls was odd. Can we make next bet on every GREY EVEN ball? Will it increase chance? Or something like that.

Hope You understand me :/

Quote:mbackonjaBetting on colour will not have the same payout, but betting on ball number will have the same payout.

So for example, if we are on pick #22 and:

4 blue balls were drawn,

4 green balls were drawn,

4 violet balls were drawn,

4 red balls were drawn,

5 orange balls were drawn

0 grey balls were drawn

so in total, 21 balls drawn, we have the same odd for betting on any number (regardless number ball's colour), we have 48-21-1=x26 payout for that bet. So if we bet on the ball "3 blue" we have same odds as betting on "45 grey".

But betting on next colour we will have different odds.

Can we combine two things like colour and parity to get most likely next number? For above example, let's assume that in 21 balls from the quote above ^ %70 of drawn balls was odd. Can we make next bet on every GREY EVEN ball? Will it increase chance? Or something like that.

Not really, Every ball has the same chance of being drawn at any particular time. The color odds change because, after the first draw, there will be more balls of some colors than of others.

I am thinking that the odds for colors change after every ball drawn to take this into account. In your example, since there are still 8 grey balls left, the payout odds that the next ball will be grey will be lower than before, since the probability that the next ball will be grey is 8/27 instead of the 1/6 it was at the start of the game. I went to the web site you sent me in a PM, but it does not describe how the odds are set or changed after each ball; I will have to try to find a place that lets you play the game for free.