How's My Math? Multi-Card Keno strategy to hit FIVE 5-spots for the odds of a 4-spot

Hi folks,

First time poster here, so I apologize if I violate any decorum. I wanted to share a Multi-Card Keno betting system I've been using that has made me quite a bit of money over the last several weekends, and I wanted to know if I've just been on a good luck streak or if I'm on to something here.

I'm also horrible at math (I'm an alumnus of Prof. Shackleford's university, but I studied English, go figure), so if I'm way off, I hope I can at least give some of you a chuckle.

Here is the generous pay table I am using for this system. This can be found at the Yorky's on Fort Apache and Hacienda:

Pay Table

Okay, let me lay down some information before we get to the thesis. Your odds of hitting a single 4-spot in keno are 1 in 327. So, if you place the same exact 4-spot pattern on all 20 cards, you can expect to hit it, on average, 1 out of every 327 spins for a total payout of 2,000 credits (according to my pay table which pays 100 credits per 4-spot if you are playing the minimum 20 credits).

Now, in my system, I place the same 4-spot pattern on all 20 cards. BUT, I cover two whole rows by putting a unique fifth spot on each card. Here's my setup:

Setup

It is my belief, and my observation based on the dozen times I've played this, that I will hit an average of FIVE 5-spots 1 out of every 327 spins for a payout of 4,385 credits. That's because each 5-spot pays 838 credits on this pay table. (838*5 = 4,190, plus an additional 195 credits for the 15 three-out-of-four spots).

Why do I believe that I will hit an average of five 5-spots?

There are eight rows on the keno table. With 20 balls being drawn, this averages out over a couple hundred spins to 2.5 balls landing on each row. Now, you can't have 1/2 a ball, so if we group the rows by two, that averages out to five balls on each double row. Here is an illustration of what I mean:

5 Hit Per Covered Double Row

Indeed, I tested this out by performing 200 simulated spins and counting how many hits landed on the double covered rows 41-60. Here are the results:

5 Hits Per Double Row

In other words, all you have to do is hit the 4-spot pattern (1 in 327 chance), and you can expect (on average) to hit 5 additional numbers in the covered rows. With the same odds it takes to make 2,000 credits, you can instead make 4,385.

Here is what it looks like when you hit it!

Five 5-spot hit
(The wins are so high because I was betting 80 credits instead of 20, but you get the idea).

Of course, of course, it's not going to hit exactly 5 numbers in the covered rows each time. In the times I've made this hit in real life (not on the simulator), the lowest I hit was 2 and the highest I hit was 9. The idea is that you will hit 5 on average.

I started using this strategy with $200 and have since turned that initial money into $3,900 over several weekend visits to the bar. The bartenders have been going nuts about this lol.

So, my questions are:

1. Is any of this even right? Have I just been getting lucky?
2. How do I calculate the RTP of this strategy?
3. If you are familiar with the Gambler’s Bonus system available at Yorky’s and many local Vegas bars, how do I factor points earned and won into the RTP?

I don't know why the hyperlinks I created don't seem to be working in the final post, but they were just fine in the preview? Whatever, hopefully you get what I'm saying.

There are only 16 balls being drawn to your extra spots. Four of the balls are already used up to make your fixed pattern. Any probability that you calculate for any other spot on the card needs to be conditional on the the four-spot pattern hitting.

Flip the problem around. Assume that none of your four-spot numbers hit. Now, there is a significantly better conditional probability that you hit your extra spot on any of the cards. You are choosing 20 balls to cover only 76 spots.

You cannot calculate the two probabilities in isolation from each other.

The RTP is the same as for any 5-spot pattern you could choose. You have not changed the RTP at all.

It is good that you are thinking outside of the box. Your math skills are probably better than 95% of adults. But if a result seems to good to be true, then you probably missed some important part of the calculation.

Quote: BrotherAron

I don't know why the hyperlinks I created don't seem to be working in the final post, but they were just fine in the preview? Whatever, hopefully you get what I'm saying.
link to original post

I suspect your links were nuked because you are new to the forum. The forum managers don't want newbies posting spam links. Stick around and make more posts. Eventually, you will be able to post images that stay visible.

Quote: Mental

Quote: BrotherAron

I don't know why the hyperlinks I created don't seem to be working in the final post, but they were just fine in the preview? Whatever, hopefully you get what I'm saying.
link to original post

I suspect your links were nuked because you are new to the forum. The forum managers don't want newbies posting spam links. Stick around and make more posts. Eventually, you will be able to post images that stay visible.
link to original post

Almost certainly. My PM's are open; I often review and post links for new members.

(Quote clipped, space)

Quote: BrotherAron

1. Is any of this even right? Have I just been getting lucky?
2. How do I calculate the RTP of this strategy?
3. If you are familiar with the Gambler’s Bonus system available at Yorky’s and many local Vegas bars, how do I factor points earned and won into the RTP?
link to original post

1.) You have just been, 'Running well.'

2.) You can use the Wizard of Odds Keno Calculator here:

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

Zero of Five: 0

One of Five: 0

Two of Five: 0

Three of Five: 3

Four of Five: 13

Five of Five: 838

(Per One Unit Bet)

Return to Player: 0.949452446287889

You are correct in stating that an RTP of basically 95% is extremely generous by Video Keno standards.

However, the return of each individual bet doesn't care what numbers you pick. You can pick the same five numbers twenty times, pick the same four numbers (with the fifth always being different) or pick the same three twenty times (and make the other two always different); it doesn't matter.

In this case, what is influencing your short-term results is the fact that all of your cards are highly correlated.

But, let me ask you this: How well would you be doing if, instead of having the fifth number always be different, the fifth number was always the same? Naturally, you would be paid 838-FOR-1 twenty times. Of course, you would also be betting twenty units.

Think of it this way: If you inserted twenty units worth and hit that on the first draw, then your bankroll would be 16,760 units, as you don't get the original bet back. In other words, you could completely whiff on the next 837 draws and would still have the units you originally inserted to the machine.

The Difference in Variance

Another thing to note is that your Keno game is simply a higher Variance game when it comes to the five-spot. Let's compare the two games using the WoO Calculator:

Four Spot Returns (For One):

Two of Four: 0.425270931600046

Three of Four: 0.216239456745786

Four of Four: 0.306339230389863

SUM OF VARIANCE: 31.667243286177953

The four spot also has a slightly lower RTP, but not so much lower as it would explain the difference in short-term results by itself.

Five Spot Returns (For One):

Three of Five: 0.251805156868448

Four of Five: 0.157200394542167

Five of Five: 0.540446894877275

SUM OF VARIANCE: 454.792058559047575

In simple terms, Variance is the degree to which a particular result varies from the average expected outcome.

The reason that the Five-Spot Variance is remarkably higher is twofold:

1.) Five of Five is significantly less likely than four of four.

2.) Five of Five, relative to probability, pays significantly more than Four of Four. This must be true because both events have a fixed probability, but hitting Five of Five represents a greater percentage of the game's return. In fact, Five of Five represents a greater return percentage than the most likely event on a Four of Four game, which is hitting two of four.

In short run terms, given that this is a negative expectation game, Variance is your friend. That's true of any negative expectation game if the goal is to be profitable, at some point.

For example, a Keno game with no variance where you Pick Four with a 95% RTP would look like this:

Zero of Four: PAYS .95
One of Four: PAYS .95
Two of Four: PAYS .95
Three of Four: PAYS .95
Four of Four: PAYS .95

The Variance is zero because the result is always the same. Regardless of what hits, the result does not deviate from the mean expectation whatsoever. You bet twenty and you lose one of twenty units, which is a 5% loss.

With that, let's jump over to the Beating Bonuses simulator and simulate 10,000 attempts of both games. We will start with a bankroll of 10,000 just so that busting out is impossible, unless there is zero return on every single draw, which won't happen. This simulator isn't perfect, but the percentage of those who finish these 10,000 trials, I am extremely confident, won't even be close comparing four spot with five spot.

https://www.beatingbonuses.com/simulator.htm

ENTRY: Blank (We have to do this so we can put a custom paytable)

FOUR SPOT SIMULATION

The first thing that we are going to do is just combine the probabilities of Zero + 1 because they don't pay, so it's functionally the same thing.

0.308321425410033 + 0.432731825136888 = 0.74105325054

That is the probability of a Return of -1. Again, it doesn't matter if we hit one ball, or zero.

Two of Four: 0.212635465800023

This will have a return of 1, which you can think of as +1.

(This game is on a FOR ONE basis, so we actually want all results to pay stated -1)

Three of Four: 0.043247891349157

This will have a return of four, rather than five.

Four of Four: 0.003063392303899

This will have a return of 99, rather than 100.

The Simulator calculates a 5.22% House Edge; that is good because that is correct.

We insert the following:

DEPOSIT: 10,000
BONUS: 0
WAGERING: 10000
BET SIZE: 1
SIMULATIONS: 10,000

RESULTS:

Average Return: (523.13)---(Close to House Edge, so this is good)
Minimum Return: ($2,622)
Maximum Return: $1,992
Chance of Gain: 18%
Chance of Bust: 0% (By design)
Standard Deviation: 568.37

FIVE SPOT SIMULATION

You'll have to take my word that I haven't already done this, but I am confident that this is a principle that will hold. The Maximum Return from Five-Spot Play will be greater, the Standard Deviation will be greater AND a greater percentage of players will finish ahead. Let's watch the magic!

The first thing that we will do is, again, combine all loss probabilities because they are functionally the same.

0.227184208196866 + 0.405686086065833 + 0.270457390710555 = 0.90332768497

Naturally, those will have a return of -1. Of course, the first thing that we notice is that we are significantly more likely to lose on an individual trial in this game. The hit rate is much lower. The paytable, by design (which is what makes Keno a good game for players who understand it---you can get the experience you want out of it), will absolutely destroy you if you don't hit your five spots.

THREE OF FIVE: 0.083935052289483

Again, this functionally pays 2 for the purpose of the simulator.

FOUR OF FIVE: 0.012092338041705

This pays 12 for the purposes.

FIVE OF FIVE: 0.000644924695558

This pays 837.

The simulator gets a House Edge of 5.05%, which is good, because that's what it is. You always want to verify that because..sometimes it's just flat out wrong, but usually, you may have put something in incorrectly.

We insert the following:

DEPOSIT: 10,000
BONUS: 0
WAGERING: 10000
BET SIZE: 1
SIMULATIONS: 10,000

Results:

Average Return: ($494)
Minimum Return: ($6,143)
Maximum Return $8,299
Chance of Bust: 0% (By design)
Chance of Gain: 38.18%
Standard Deviation: 2,126.6

CONCLUSION

You got lucky. Strictly, I don't believe in luck. If enough people played the exact way you did, then some of those people WOULD experience results that are at least similar.

Of course, having a highly correlate (four of five of the picks are the same) multicard reduces your variance slightly; the way to get the most variance, naturally, would only to be betting one card...or to bet on twenty that are all the same.

As we see in our results, despite the sub-5% House Edge, our worst performer in the 5 of 5 simulation lost 61.43% of all monies bet. BRUTAL!!!. How brutal? Well, if they hadn't hit a single Five of Five, but ran exactly as expected on 4/5 and 3/5, they would have done better.

In the meantime, our Four-Spot players get a smoother ride, but fewer than half of them (compared to Five-Spot players) are still ahead after 10,000 spins.

Why? Vairance---which is our friend in a negative expectation game...unless you just want to play as long as possible and reduce your risk of ruin within x (where x is a relatively small number) of draws.

When we look at the Variance, what we notice for the Four-Spot is that the SD is equal to almost the average amount lost.

What are the implications of that?

Well, you look at the 68-95-99.7 rule, which says that 68% of players (in this example) will finish within one Standard Deviation. In the case of the Four-Spot, the majority of the players who finish within one SD to the right side of the mean still lost some amount of money.

In comparison, a tremendous percentage of the players to finish within 1SD to the right of the mean will be ahead after 10,000 attempts. The reason why is because 1SD is substantially larger than the mean average result is. That also explains why 38.18% of players finished ahead in that simulation.

When we look at the rule, we see that the distribution, roughly, would look like this:

34%---1SD Left
34%---1SD Right
13.5%---2SD Left
13.5%---2SD Right
2.5%---3SD+ Left
2.5%---3SD+ Right

With that, we can get rid of 16% of our winners as being outside of the first Standard Deviation. That still leaves us with 22.18% of players in this simulation who finished ahead. We know that approximately 34% of players will finish one SD to the right, therefore, just over 65% of our players who finished 1SD to the right also profited. I didn't calculate for the Four of Four, but the old eyeball test says that only 2% of players within 1SD made a profit after 10,000 attempts, as 16%/18% would have been 2SD+ to the right results.

Anyway, that is your success both explained and demystified. I'm extremely happy for you as I happen to enjoy Keno as a game, but this is absolutely expected to happen for some players and you did not break mathematical reality; you simply got the favorable end of that reality. May your streak continue should you choose to continue to play!

Quote: BrotherAron

Hi folks,

First time poster here, so I apologize if I violate any decorum. I wanted to share a Multi-Card Keno betting system I've been using that has made me quite a bit of money over the last several weekends, and I wanted to know if I've just been on a good luck streak or if I'm on to something here.

I'm also horrible at math (I'm an alumnus of Prof. Shackleford's university, but I studied English, go figure), so if I'm way off, I hope I can at least give some of you a chuckle.

Here is the generous pay table I am using for this system. This can be found at the Yorky's on Fort Apache and Hacienda:

Pay Table

Okay, let me lay down some information before we get to the thesis. Your odds of hitting a single 4-spot in keno are 1 in 327. So, if you place the same exact 4-spot pattern on all 20 cards, you can expect to hit it, on average, 1 out of every 327 spins for a total payout of 2,000 credits (according to my pay table which pays 100 credits per 4-spot if you are playing the minimum 20 credits).

Now, in my system, I place the same 4-spot pattern on all 20 cards. BUT, I cover two whole rows by putting a unique fifth spot on each card. Here's my setup:

Setup

It is my belief, and my observation based on the dozen times I've played this, that I will hit an average of FIVE 5-spots 1 out of every 327 spins for a payout of 4,385 credits. That's because each 5-spot pays 838 credits on this pay table. (838*5 = 4,190, plus an additional 195 credits for the 15 three-out-of-four spots).

Why do I believe that I will hit an average of five 5-spots?

There are eight rows on the keno table. With 20 balls being drawn, this averages out over a couple hundred spins to 2.5 balls landing on each row. Now, you can't have 1/2 a ball, so if we group the rows by two, that averages out to five balls on each double row. Here is an illustration of what I mean:

5 Hit Per Covered Double Row

Indeed, I tested this out by performing 200 simulated spins and counting how many hits landed on the double covered rows 41-60. Here are the results:

5 Hits Per Double Row

In other words, all you have to do is hit the 4-spot pattern (1 in 327 chance), and you can expect (on average) to hit 5 additional numbers in the covered rows. With the same odds it takes to make 2,000 credits, you can instead make 4,385.

Here is what it looks like when you hit it!

Five 5-spot hit
(The wins are so high because I was betting 80 credits instead of 20, but you get the idea).

Of course, of course, it's not going to hit exactly 5 numbers in the covered rows each time. In the times I've made this hit in real life (not on the simulator), the lowest I hit was 2 and the highest I hit was 9. The idea is that you will hit 5 on average.

I started using this strategy with $200 and have since turned that initial money into $3,900 over several weekend visits to the bar. The bartenders have been going nuts about this lol.

So, my questions are:

1. Is any of this even right? Have I just been getting lucky?
2. How do I calculate the RTP of this strategy?
3. If you are familiar with the Gambler’s Bonus system available at Yorky’s and many local Vegas bars, how do I factor points earned and won into the RTP?
link to original post

For anyone interested in also seeing the links, here they are:

Pay Table

Setup

5 Hit Per Covered Double Row

5 Hit Per Covered Double Row

Five 5-spot hit

Both BrotherAron's posts and the links are legitimate. The rule for new Members is so bots can't link spam us.

BrotherAron,

You know something? It occurs to me that you would probably be interested in knowing the probabilities of hitting particular numbers of Five/Five given that your shared Four/Four (between all Five-Spot Cards) has been hit.

Well, guess what, brother? Christmas came early for you this year, and as a present, I am going to do just that. I am going to do all of these as my gift to you, in order to spread joy, mirth and merriment.

But, you know how the old saying goes:

Quote:

If you give a man a fish, he still needs a pan.

I think so. Right?

Anyway, not only will I give you these fish, but I'm also going to share my cooking equipment (that doesn't actually belong to me) with you and teach you how to use it.

Let's have some fun with probabilities!

The first thing that we are going to need is a sexy, sexy, scientific calculator that makes combinatorics easy...but still a bit time-consuming.

Ladies and gentlemen, she even looks good in an elf get up, let's meet our scientific calculator:

https://web2.0calc.com/

She'll even let you push her buttons; just be classy about it.

Here we go:

FOUR OUT OF FOUR IN THE FIRST PLACE

nCr(4,4)*nCr(76,16)/nCr(80,20) = 0.0030633923038986

This is actually going to be easy since we're only looking at probabilities.

1/0.0030633923038986 = 1 in 326.4355005159993898556

Basically, we are going to take the four out of four for granted from now on. Besides that, if you want to figure out the probability for a specific number of five out of fives, then all you need to do is multiply any of the decimal probabilities (below) from the decimal probability of four out of four (above).

Okay, so assuming we have hit the four numbers we need, this is what remains:

NUMBERS COVERED: 20
DRAWS REMAINING: 16
COVERED NUMBERS THAT DO NOT HELP: 56

Basically, all of our combinatorial questions for the calculator are going to take this form:

(Numbers Covered, Numbers that Hit)*(DO NOT HELP NUMBERS, DO NOT HELP #'s HIT)/(NumbersCovered+DONOTHELPNUMBERS, TOTAL DRAWN)

The only variable inputs are going to be the number of each type of numbers to hit or not hit. Let's start with 16 out of 16 and work our way down:

REMEMBER THAT ALL PROBABILITIES BELOW TAKE HITTING THE FIRST FOUR FOR GRANTED!!! IN ORDER TO GET A COMBINED PROBABILITY YOU MUST ALSO MULTIPLY THIS BY THE FOUR OUT OF FOUR PROBABILITY!!! Also, all are rounded off.

nCr(20,16)*nCr(56,0)/nCr(76,16) = 0.0000000000004474 or 1 in 2,235,136,343,316.94***

nCr(20,15)*nCr(56,1)/nCr(76,16) = 0.000000000080168 or 1 in 12,473,805,009.48

nCr(20,14)*nCr(56,2)/nCr(76,16) = 0.0000000055115483 or 1 in 181,437,219.74

nCr(20,13)*nCr(56,3)/nCr(76,16) = 0.0000001984157381 or 1 in 5,039,922.79

nCr(20,12)*nCr(56,4)/nCr(76,16) = 0.0000042721388599 or 1 in 234,074.79

nCr(20,11)*nCr(56,5)/nCr(76,16) = 0.000059240325524 or 1 in 16,880.39+++

nCr(20,10)*nCr(56,6)/nCr(76,16) = 0.0005538970436498 or 1 in 1805.39

nCr(20,9)*nCr(56,7)/nCr(76,16) = 0.0035967340496738 or 1 in 278.03

nCr(20,8)*nCr(56,8)/nCr(76,16) = 0.0165224970406891 1 in 60.52

nCr(20,7)*nCr(56,9)/nCr(76,16) = 0.0542276825950821 or 1 in 18.44

nCr(20,6)*nCr(56,10)/nCr(76,16) = 0.127435054098443 or 1 in 7.84

nCr(20,5)*nCr(56,11)/nCr(76,16) = 0.213164090491941 or 1 in 4.69

nCr(20,4)*nCr(56,12)/nCr(76,16) = 0.2498016685452434 or 1 in 4

nCr(20,3)*nCr(56,13)/nCr(76,16) = 0.1989370754025467 or 1 in 5.03

nCr(20,2)*nCr(56,14)/nCr(76,16) = 0.1018368362179704 or 1 in 9.82

nCr(20,1)*nCr(56,15)/nCr(76,16) = 0.0300150675168755 or 1 in 33.32

nCr(20,0)*nCr(56,16)/nCr(76,16) = 0.0038456805255997 1 in 260.03

***By the way, this is an example of why the way you're doing it reduces the Variance slightly. In the case of all twenty cards being literally the same five numbers, so really, just one card twenty times...if you hit all five numbers, then you win 838 * 20 * (Bet per Card), right? In the case of the way you're playing the game, that's essentially never going to happen. You'd have to hit your four numbers AND all sixteen others.

With that, if we assume that you hit the four spot, then anything from 0-11 five spots hit wouldn't be anything too crazy.

The reason this is not a winning strategy is because you are playing the 5 spot paytable, not the 4 spot paytable.

Of course it's great when the 4 spot group does hit, but that's only 1 in 327. The other 326 in 327 games causes you to win less than if you were playing only the 4 spot.

If you get 3 of the 4 spots, for every one of the 5th unique spots you don't hit, you get the prize for 3 out of 5 (3 times bet instead of 3 of 4 which pays 5 times).

If you get 2 of the 4 spots, which is five times as likely as getting 3 of 4 spots, for every one of the 5th unique spots you don't hit, you get nothing, since 2 of 5 pays nothing, whereas 2 of 4 pays 2 times bet.

The few numbers in your 5th unique spots that hit in addition to 2 or 3 of the 4 spot group do win more than just a 4 spot bet, but they do not hit often enough to make up for it.