Italia Keno 10e 20/90

Hello world! I'm was trying to obtain odds 2 on these 5 minutes lotteries and the only way 2/5 system at Italia Keno 10e 20/90 because there are 10 combinations and the minimum payout is x2, wich means I only hit 2 numbers from the 5 choosen and the maximum payout is x20 if I hit all the 5 numbers choosen. I don't think my math is right because I've made some simulations with my C++ program and got a very different result and I can't understand why, so let's start:

$10 TOTAL BET and let's say I choose numbers 1, 2, 3, 4, 5 for simplicity, we have these 10 combinations:

COMBINATION 1: 1$ bet, Number 1 2 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 2: 1$ bet, Number 1 ,3 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 3: 1$ bet, Number 1, 4 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 4: 1$ bet, Number 1, 5 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 5: 1$ bet, Number 2, 3 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 6: 1$ bet, Number 2, 4 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 7: 1$ bet, Number 2 ,5 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 8: 1$ bet, Number 3, 4 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 9: 1$ bet, Number 3, 5 hit, payout x2, chances to hit these numbers 4.938%
COMBINATION 10: 1$ bet, Number 4, 5 hit, payout x2, chances to hit these numbers 4.938%

And I asumed that I have an overall chance of 4.938 x 10 = 49.38% to hit atleast 2 numbers from the 5 choosen but my simulation with 10.000 tickets showed something else:

0 Numbers Hit: 3151 (31.51%)
1 Numbers Hit: 3913 (39.13%)
2 Numbers Hit: 2048 (20.48%)
3 Numbers Hit: 695 (6.95%)
4 Numbers Hit: 171 (1.71%)
5 Numbers Hit: 22 (0.22%)

Why? What's wrong with my math?

At first glance, the main problem appears to be, when you count 1,2, 1,3, and 2,3 separately, you are counting all of the combinations that include 1, 2, 3, three times.

Look at the problem this way:
Color the balls 1-5 red, and the balls 6-90 white. If you draw 20 balls, you want to know how many of them are white.
There are C(90,20) ways to draw the 20 balls.
There are C(5,0) x C(85,20) ways to draw 0 red and 20 white balls; this is about 27.54% of the total.
There are C(5,1) x C(85,19) ways to draw 1 red and 19 white balls; this is about 41.73%.
There are C(5,2) x C(85,18) ways to draw 2 red and 18 white balls; this is about 23.67%
There are C(5,3) x C(85,17) ways to draw 3 red and 17 white balls; this is about 6.26%, or 1 in 16.
There are C(5,4) x C(85,16) ways to draw 4 red and 16 white balls; this is about 0.77%, or 1 in 130.
There are C(5,5) x C(85,15) ways to draw 5 red and 15 white balls; this is about 0.03%, or 1 in 2835.

For those of you (what? Oh, okay...for those of us) interested in exact answers:
0 hits: 183,379/665,898 (about 27.5386%)
1 hits: 4,584,475/10,987,317 (about 41.7252%)
2 hits: 2,600,150/10,987,317 (about 23.665%)
3 hits: 76,475/1,220,813 (about 1 / 15.9636)
4 hits: 56,525/7,324,878 (about 1 / 129.5865)
5 hits: 1,292/3,662,439 (about 1 / 2834.7051)

Okay, if you want an overall return for this game, I need to know what the 3/5 and 4/5 would pay in addition to the 2/5 and 5/5 pays that you have provided.

As far as the probabilities go, that's just combinatorial math, so I can easily teach you how to do it.

Numbers Selected: 5
Numbers Not Selected: 85
Total Numbers: 90

So, you can use this:

https://web2.0calc.com/

Zero Numbers Hit: nCr(5,0)*nCr(85,20)/nCr(90,20) = 0.2753860200811536

One Number Hit: nCr(5,1)*nCr(85,19)/nCr(90,20) = 0.4172515455775054

Two Numbers Hit: nCr(5,2)*nCr(85,18)/nCr(90,20) = 0.2366501303275404

Three Numbers Hit: nCr(5,3)*nCr(85,17)/nCr(90,20) = 0.0626426815572901

Four Numbers Hit: nCr(5,4)*nCr(85,16)/nCr(90,20) = 0.0077168520758981

Five Numbers Hit: nCr(5,5)*nCr(85,15)/nCr(90,20) = 0.0003527703806125

Add Together: 0.2753860200811536+0.4172515455775054+0.2366501303275404+0.0626426815572901+0.0077168520758981+0.0003527703806125 = 1

Therefore, your overall probability of losing (zero or one hit) is 0.69263756565 or 69.263756565%.

The probability of any win is .30736243435 or 30.736243435%.

You know, I've found about ten sites offering systems and aggregated results (which have zero impact on future drawings) and I can't find anything that just gives me the payouts. For the time being, I'll assume that everything just pays 2x, except 5/5, which you said pays 20x:

((0.2366501303275404+0.0626426815572901+0.0077168520758981)*2) + (0.0003527703806125*20) - 0.69263756565 = -0.07156283011 Which reflects a House Edge of 7.156283011%. As far as Keno games go, it could be worse.

If you intend to bet $10 every time, then you would expect to lose $0.7156283 of every bet you make.

***Wait a minute, does ONLY WAY 2/5 mean that 3/5 and 4/5 would lose?

No, with 3/5 I get x6, with 4/5 I get x12 and with 5/5 I get x20 because these payouts I knew how to calculate and I'm sure these are not wrong. Thanks anyway, I got the general idea with the whole thing :)) Do you have any Idea how can I get pure x2 odds without using systems? I guess it's not possible in any of these 5 minutes lotteries. I can get a x2.4 (x2.5 the real matematical odds) from 8/20 Hungary Putto by choosing 1 number, but It's not x2.00 wich I need to use with l'Abouchere betting system...

Quote: ThatDonGuy

At first glance, the main problem appears to be, when you count 1,2, 1,3, and 2,3 separately, you are counting all of the combinations that include 1, 2, 3, three times.

Look at the problem this way:
Color the balls 1-5 red, and the balls 6-90 white. If you draw 20 balls, you want to know how many of them are white.
There are C(90,20) ways to draw the 20 balls.
There are C(5,0) x C(85,20) ways to draw 0 red and 20 white balls; this is about 27.54% of the total.
There are C(5,1) x C(85,19) ways to draw 1 red and 19 white balls; this is about 41.73%.
There are C(5,2) x C(85,18) ways to draw 2 red and 18 white balls; this is about 23.67%
There are C(5,3) x C(85,17) ways to draw 3 red and 17 white balls; this is about 6.26%, or 1 in 16.
There are C(5,4) x C(85,16) ways to draw 4 red and 16 white balls; this is about 0.77%, or 1 in 130.
There are C(5,5) x C(85,15) ways to draw 5 red and 15 white balls; this is about 0.03%, or 1 in 2835.

For those of you (what? Oh, r those of us) interested in exact answers:
0 hits: 183,379/665,898 (about 27.5386%)
1 hits: 4,584,475/10,987,317 (about 41.7252%)
2 hits: 2,600,150/10,987,317 (about 23.665%)
3 hits: 76,475/1,220,813 (about 1 / 15.9636)
4 hits: 56,525/7,324,878 (about 1 / 129.5865)
5 hits: 1,292/3,662,439 (about 1 / 2834.7051)

As smart as last time and very well explained with that colors analogy, thank you very much again!

Quote: Mission146

If you intend to bet $10 every time, then you would expect to lose $0.7156283 of every bet you make.

***Wait a minute, does ONLY WAY 2/5 mean that 3/5 and 4/5 would lose?

Tell me if I'm right with this recalculated return for 2/5 system, please:

0.0003527703806125 x (20-1) + 0.0077168520758981 x (12-1) + 0.0626426815572901 * (6-1) + 0.2366501303275404 * (2-1)
- 0.4172515455775054 x 1 - 0.2753860200811536 x 1 =
= 0.0067026372316375 + 0.0848853728348791 + 0.3132134077864505 + 0.2366501303275404 - 0.4172515455775054 - 0.2753860200811536 = 0.6414515481805075 - 0.692637565658659 = -0.0511860174781515 wich is roughly 5.11% house advantage, wich is not that bad, I think, but It's bad in terms of loosing series because I can't use the L'abouchere betting system with 30.7% probability of any win, this system requesting a minimum of 33%...

Quote: Sacke86

Quote: Mission146

If you intend to bet $10 every time, then you would expect to lose $0.7156283 of every bet you make.

***Wait a minute, does ONLY WAY 2/5 mean that 3/5 and 4/5 would lose?

Tell me if I'm right with this recalculated return for 2/5 system, please:

0.0003527703806125 x (20-1) + 0.0077168520758981 x (12-1) + 0.0626426815572901 * (6-1) + 0.2366501303275404 * (2-1)
- 0.4172515455775054 x 1 - 0.2753860200811536 x 1 =
= 0.0067026372316375 + 0.0848853728348791 + 0.3132134077864505 + 0.2366501303275404 - 0.4172515455775054 - 0.2753860200811536 = 0.6414515481805075 - 0.692637565658659 = -0.0511860174781515 wich is roughly 5.11% house advantage, wich is not that bad, I think, but It's bad in terms of loosing series because I can't use the L'abouchere betting system with 30.7% probability of any win, this system requesting a minimum of 33%...

(0.2366501303275404*1) + (0.0626426815572901*5) + (0.0077168520758981*11) + (0.0003527703806125*19) - .69263756565 = -0.05118601746

That's correct, it looks like a House Edge of 5.1186%. Please forgive my mistake in my first response to this thread. For some reason, I did it as if the original wager would be returned in addition to the winnings, but it appears that it's the winnings less the wager (wager presumed lost).

I agree that the House Edge is not that bad for a Keno game. That's better than most of the machines you'll get in land casinos, although, you wouldn't no0rmally have to bet $10 every play.

Betting systems will not change the house edge against you or give you a profitable expectation. All expectations are simply the negative of any amounts bet multiplied by the house edge expressed as a decimal, which I'm sure you already know.

With that, if you wanted to do a modified Labouchere simply for your own amusement, then you wouldn't necessarily need to win at a 33.33% clip. Simply modify the Labouchere and use a long Labouchere line...any of your wins on more than 2/5 hits cross out multiple numbers. Here's an example:

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

With an Even Money type game, completion of the Labouchere Line would lead to fifteen units won. You can do the same thing with this Keno game, if you really want to (I am NOT recommending it) in a different way. Suppose 3/5 hits on the first play, that's a profit of ten units on a two unit bet, so just do this with your line:

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

As you can see, having struck out your profit of ten units on that play, (you were betting two units) you're left with five "1's."

Strictly speaking, it's not a true Labouchere, but it basically accomplishes the same thing. On occasion, you'll win in excess of your remaining total of your Labouchere line...particularly when you're betting more than one unit and hit a 3/5, 4/5 or 5/5 result.

What's the maximum that you can bet, anyway? Starting with $10 bets and running a Labouchere, it seems that you might run into the maximum somewhat quickly. For example, if we look at this Labouchere line:

10, 10, 10, 10, 10

And you have a loss probability of .69263756565, then the probability of losing ten consecutive is:

.69263756565^10 = 0.02541326241 which is slightly more likely than one in forty attempts.

10, 10, 10, 10, 10, 20

10, 10, 10, 10, 10, 20, 30

10, 10, 10, 10, 10, 20, 30, 40

...and so on. The point is that, starting with $10 units, you're up to a bet of over $100 by the time this particular and relatively frequent losing streak ends.

So, your bet amounts can get out of control pretty quickly, which is just the nature of the Labouchere.

I'm sorry that I have to disclaim that betting systems do not work so many times, but I'm disclaiming it again. That said, what's wrong with just using a Martingale for this if you REALLY must play it with a system, or at all? The Martingale is usually applied to even money bets with the goal of winning one unit, but it doesn't have to be. Is it also because the bet amounts can quickly get out of control?

Another thing to remember is that past results have no bearing on future results. In other words, even if you have already lost ten (or any number) in a row, the probability of losing the next one remains nearly 70%.

That $10 bet was just for example, ofcourse I would not play this big bets because I'm aware that bets can go very big as it happend in the past when I played another games wich tehnically can be rigged, like blackjack where I've won 6 hands and lost 24 in 30 consecutive hands and I'm trying to find a game that is less likely to be rigged FOR ME because of a large number of players that shares the same result (same numbers in case of lotteries) and I think I would be less affected and only the players with big bets or big returns would be screwed. That's just my paranoia maybe (or not) but I want to be sure (even if it's not rigged) that the results can't affect me directly and affects all players, wich can be over 1000 let's say, wich means that results are less likely to e rigged JUST FOR ME particularly, especially when I play on small bets and small returns. When you get well under 30% winrates on almost even games (47%+ win probability) very often and endless loosing series you get quite aware and begin asking yourself questions like these and search for solutions / alternative games wich can't be rigged just to be sure they are not rigged.

Now I'll tell you how I want to play and how I played in the past and got screwed even if this strategy is pretty safe in my opinion:

1 unit is about $0.25 and my bankroll is $125 let's say (500 units) and this is where I start from:
I always start with just 2 numbers in the series like: $0.125, $0.125 and a starting bet of $0.25 and I keep the profit in a variable in a C++ program, wich could go negative over time or positive when I recover the losses and when my profit get's over 0 I'm starting the series again from $0.125, $0.125, bet $0.25.

It is not necessary to have 10 numbers in the series because if my profit is well under zero and I have let's say 5-6 numbers in the series and win a round with a big bet in the end and a return of x6 (3/5) I will recover all my lossses and even get a profit so then I will start again from 2 numbers in the series: $0.125, $0.125.

And that's it, if the game isn't rigged and could't be rigged chances to get endless loosing series and very often ar as math says... VERY RARE, so I have time to build a bigger bankroll meanwhile.

That's all I need, a game with a minimum of 45% winrate and x2 return, 50% winrate and x1.90 return or something like this wich isn't rigged, can't be rigged tehnically or it is rigged but could not affect me that much.

This 2/5 system isn't that great because I want to avoid getting big series of numbers and big loosng series and this is what happens with 30.7% win probability...

It sounds to me like you recognize the fact that the House Edge is immutable, but you want to play a system that lends itself to the possibility of prolonged small wins in exchange for the occasional devastating loss. That's all perfectly fine as long as you enjoy it and understand what your expected losses are going to be, which, as always, is just the total amount you will ever bet multiplied by House Edge.

While the probability of getting a prolonged losing series that would result in a devastating loss is pretty low, in your opinion, please keep in mind that devastating loss can come at anytime and it's not proof of a rigged anything. You basically acknowledge that some such loss is going to happen sooner or later and, speaking in terms of probability, it's just as likely to happen on your first attempt as it is on any other individual attempt.

Another thing to keep in mind when you utilize the Labouchere is that your effective winning percentage actually goes up as you incur losses. Think about it when it comes to an even money proposition: As you've stated, you need to win more than 33.33% of the time to complete the line.

That's fine, but once you've endured a losing streak, now you have to win MORE THAN 33.33% of the time the rest of the way out (to complete the line) in order to even get to 33.33% win rate overall. That's an example of one of those unfortunate, but virtually inevitable (if you play it long enough) situations where the Labouchere...as with almost any other negative progression betting system...gets completely out of control.

Anyway, you can gamble however you want and I hope you stay within your means, don't get any false hope of endless winning and have a lot of fun in the process. With that said, games don't have to be, "Rigged," in order for you to lose with this system...the fact that there is a house edge against you is all the, "Rigging," they need. For example, you could Labouchere at Baccarat in a land casino all day long and the casino will even provide a pencil and scorecard to make it even easier for you.

Quote: Mission146

It sounds to me like you recognize the fact that the House Edge is immutable, but you want to play a system that lends itself to the possibility of prolonged small wins in exchange for the occasional devastating loss. That's all perfectly fine as long as you enjoy it and understand what your expected losses are going to be, which, as always, is just the total amount you will ever bet multiplied by House Edge.

While the probability of getting a prolonged losing series that would result in a devastating loss is pretty low, in your opinion, please keep in mind that devastating loss can come at anytime and it's not proof of a rigged anything. You basically acknowledge that some such loss is going to happen sooner or later and, speaking in terms of probability, it's just as likely to happen on your first attempt as it is on any other individual attempt.

Another thing to keep in mind when you utilize the Labouchere is that your effective winning percentage actually goes up as you incur losses. Think about it when it comes to an even money proposition: As you've stated, you need to win more than 33.33% of the time to complete the line.

That's fine, but once you've endured a losing streak, now you have to win MORE THAN 33.33% of the time the rest of the way out (to complete the line) in order to even get to 33.33% win rate overall. That's an example of one of those unfortunate, but virtually inevitable (if you play it long enough) situations where the with almost any other negative progression betting ts completely out of control.

Anyway, you can gamble however you want and I hope you stay within your means, don't get any false hope of endless winning and have a lot of fun in the process. With that said, games don't have to be, "Rigged," in order for you to lose with this e fact that there is a house edge against you is all the, "Rigging," they need. For example, you could Labouchere at Baccarat in a land casino all day long and the casino will even provide a pencil and scorecard to make it even easier for you.

I still can't believe that some games are not rigged when you get LESS THAN EXPECTED WINRATES, BIG LOOSING SERIES and this happens VERY OFTEN, not rare as it should happen because it is just unnatural and obvious in my opinion, don't even want to think about roulette wich is sick, even the european one with just one zero, if you play on red/black odd/even 1-18/19-36. How the hell I get 2 reds, 2 blacks, 1 red, 1 black when I put 2 blacks, 2 reds, 1 black, 1 red? It feels like anyway I'm switching the colors with whatever pattern I'll get the oposite results and this is just unnatural. Do you think they are not aware of these players wich use pretty safe betting systems to screw them? I know how math works but this is how I feel...I lose more than I should lose and seems to be unnatural. I just want to find a game wich can't be tehnically rigged or is less likely to affect only me and is almost 100% random, that's all. If you have any idea please tell me what game should I choose? I don't bet lotteries can't be rigged because they can be but assuming that I share the same result with a lot of other players makes it less rigged FOR ME, not in general.