Super Keno House edge -111?

How do i determine the house edge of this game . the base game has a house edge of 44.47
but the pay table with the last ball multiplier by four the edge comes out as -111.23
Whats my next step to figure this out?

Quote: BegoodJohnny44

How do i determine the house edge of this game . the base game has a house edge of 44.47
but the pay table with the last ball multiplier by four the edge comes out as -111.23
Whats my next step to figure this out?
link to original post

First of all, I assume the way it works is, if the last ball drawn is one of yours, then the payout is multiplied by 4.

It's going to depend on how many numbers you played. Since each of the 80 balls is equally likely to be the last one, the probability of triggering the bonus is N / 80.

My "gut answer" is, the overall house edge = N / 80 * (-111.23) + (80 - N) / 80 * 44.47 = 44.47 - N * 1.94625.
So if you play 10 numbers, the modified house edge is 25.
However, I am not entirely sure that is the way it is calculated, since if one of your numbers triggered the bonus, then you got at least one number, but the house edge includes the possibility of not getting any. I will look into this some more.

Quote: ThatDonGuy

Quote: BegoodJohnny44

How do i determine the house edge of this game . the base game has a house edge of 44.47
but the pay table with the last ball multiplier by four the edge comes out as -111.23
Whats my next step to figure this out?
link to original post

First of all, I assume the way it works is, if the last ball drawn is one of yours, then the payout is multiplied by 4.

It's going to depend on how many numbers you played. Since each of the 80 balls is equally likely to be the last one, the probability of triggering the bonus is N / 80.

My "gut answer" is, the overall house edge = N / 80 * (-111.23) + (80 - N) / 80 * 44.47 = 44.47 - N * 1.94625.
So if you play 10 numbers, the modified house edge is 25.
However, I am not entirely sure that is the way it is calculated, since if one of your numbers triggered the bonus, then you got at least one number, but the house edge includes the possibility of not getting any. I will look into this some more.
link to original post

i used a keno calculator and based on 9 numbers and yes last ball X4 and it states that my advantage is 111.23 but when i do only base pay the house has an edge of 44.47

Never mind - it turns out that method doesn't work.
It looks like you need to calculate it separately for each possible number of hits.

See The Wizard's page for this game for a better idea of how this works.

Since you are using a keno calculator, here's what I suggest you do:
Assume you are playing 9 numbers
For each possible number of hits H, you have an H/20 chance that one of your hits was the last number, which triggers the bonus, so the "effective" payout on a bet of 1 is P * (4 * H/20 + 1 * (1 - H/20)) = P * (1 + 3/20 H), where P is the payout for H hits.
Enter that number for that many hits.
For example, if 9 hits pays 2500, then instead of entering 2500, enter 2500 * (1 + 3/20 * 9) = 2500 * 47/20, or 5875.
If 8 pays 1200, then enter 1200 * (1 + 3/20 * 8) = 1200 * 44/20, or 2640.
Do this for the rest of the numbers, then calculate normally.

Quote: BegoodJohnny44

i used a keno calculator and based on 9 numbers and yes last ball X4 and it states that my advantage is 111.23 but when i do only base pay the house has an edge of 44.47
link to original post

What is the paytable?

Okay, the OP was kind enough to PM me the paytable and says that he must be out of posts for the time being.

Here is the apparent paytable for the nine-spot:

Numbers Selected 9 . In order to win the 2000 dollar jackpot you have to hit 9 out of 9 on a 0.04 cent bet.
SUPER KENO
4 out of 4 pays 0.08
5 out of 5 pays 0.20
6 out of 6 pays 0.60
7 out of 7 pays 2.40
8 out of 8 pays 8.00
9 out of 9 pays 2000.00

The first thing that we are going to do is simplify this by indexing it to credits won: credit bet, thus:

4/9: 2
5/9: 5
6/9: 15
7/9: 60
8/9: 200
9/9: 50000

Okay, with that out of the way, we have two different events that we need to calculate. The first event is every hit where the last ball is a match multiplies these winnings by four, so that would give you:

LAST BALL HIT:

4/9: 8
5/9: 20
6/9: 60
7/9: 240
8/9: 800
9/9: 200000

And...Wizard already has a calculator for this:

https://wizardofodds.com/games/keno/power/calculator/

The return comes out to: 0.985626587407409

However, we can verify that pretty easily. The first thing that we are going to do is look at the conditions for winning, which are simply to match four, or more, numbers. In the case of the nine-spot game, we would need to be starting with at least three of the first nineteen numbers for the 20th number to matter. How many numbers we catch in draws 1-19 will also impact how many possible 20th numbers there are, so with that, let's do some combinatorics:

First Nineteen Draws

Match 4:9 + Last Ball:

nCr(9,3)*nCr(71,16)/nCr(80,19) = 0.2320138702607924

Okay, so if this matches the fourth number, then it pays eight units:

0.2320138702607924 * 6/61 * 8 = 0.18256829135

Match 4:9 w/o Last Ball:

nCr(9,4)*nCr(71,15)/nCr(80,19) = 0.0994345158260539

0.0994345158260539 * 2 * 56/61 = 0.1825682913527547016

Match 5/9 with Last Ball:

0.0994345158260539 * 20 * 5/61 = 0.1630074029935309836

Match 5/9 w/o Last Ball:

nCr(9,5)*nCr(71,14)/nCr(80,19) = 0.0261669778489615

0.0261669778489615 * 5 * 57/61 = 0.1222555522451479918

Match 6/9 with Last Ball:

0.0261669778489615 * 60 * 4/61 = 0.10295204399

Match 6/9 w/o Last Ball:

nCr(9,6)*nCr(71,13)/nCr(80,19) = 0.0042107780446605

0.0042107780446605 * 15 * 58/61 = 0.0600553589976169672

Match 7/9 with Last Ball:

0.0042107780446605 * 240 * 3/61 = 0.04970098675

Match 7/9 w/o Last Ball:

nCr(9,7)*nCr(71,12)/nCr(80,19) = 0.0003976279509486

0.0003976279509486 * 60 * 59/61 = 0.02307545813

Match 8/9 with Last Ball:

0.0003976279509486 * 800 * 2/61 = 0.0104295855986518033

Match 8/9 w/o Last Ball:

nCr(9,8)*nCr(71,11)/nCr(80,19) = 0.0000198813975474

0.0000198813975474 * 200 * 60/61 = 0.00391109459

Match 9/9 with Last Ball:

0.0000198813975474 * 200000 * 1/61 = 0.06518490999

Match 9/9 w/o Last Ball:

nCr(9,9)*nCr(71,10)/nCr(80,19) = 0.0000003983522277

0.0000003983522277 * 50000 = 0.019917611385

Okay, so now we will take all of our returns in bold:

Match 4 w/Last: 0.18256829135
Match 4 w/o Last: 0.1825682913527547016
Match 5 w/Last: 0.1630074029935309836
Match 5 w/o Last: 0.1222555522451479918
Match 6 w/Last: 0.10295204399
Match 6 w/o Last: 0.0600553589976169672
Match 7 w/ Last: 0.04970098675
Match 7 w/o Last: 0.02307545813
Match 8 w/Last: 0.0104295855986518033
Match 8 w/o Last: 0.00391109459
Match 9 w/Last: 0.06518490999
Match 9 w/o Last: 0.019917611385

And, we add these all together:

0.019917611385+0.06518490999+0.00391109459+0.0104295855986518033+0.02307545813+0.04970098675+0.0600553589976169672+0.10295204399+0.1222555522451479918+0.1630074029935309836+0.1825682913527547016+0.18256829135 = 0.9856265873727024475

As we can see, this agrees with Wizard's calculator with differences due to rounding, so you can just use the calculator for any other paytables on this game. However, you may have somewhat similar Keno problems to analyze one day, so hopefully this method will be helpful, if so.

One thing that the OP will want to look into is that I know, for games like these, sometimes the top number hit result (9/9 in this case) is the most that it possibly pays, so it doesn't matter whether or not the last number is a hit on 9/9 results as it would not multiply anyway. I would encourage the OP to look into that as it will knock a few percentage points off of the return if $2,000.00 is the highest possible pay and cannot be multiplied.

Basically:

Quote:

Match 9/9 with Last Ball:

0.0000198813975474 * 200000 * 1/61 = 0.06518490999

Becomes:

Match 9/9 with Last Ball:

0.0000198813975474 * 50000 * 1/61 = 0.01629622749

Which then changes the return percentage by:

0.06518490999 - 0.01629622749 = 0.0488886825

Thereby making the return:

0.9856265873727024475 - 0.0488886825 = 0.93673790487

Which seems much more likely to me. I'd suggest that the OP read the Game Rules to determine if that is the case.

ADDED: The OP has informed me that 9/9 does not multiply, therefore, the Return to Player is roughly: 0.93673790487