# Thread: An insight into Mickey Crimms math background

1. I originally posted this here: https://vegascasinotalk.com/forum/sh...l=1#post113257

In 2015, Mickey Crimm announced to the world he didnt know what a geometric distribution was.

https://math.stackexchange.com/quest...-game-question

I have no idea what a geometric distribution is. LOL!  mickey crimm Jun 7 '15 at 13:50

This post gives insights on Mickey Crimms exploits from casino activities. This was 2015.

Food for thought.  Reply With Quote

2. Most people that I've met who have a deep understanding and comprehension of statistics have been failures at making money with AP. Math is cheap. You can buy all you want for about \$125 an hour. Yours I wouldn't pay 2¢ for.  Reply With Quote

3. I originally posted this here: https://vegascasinotalk.com/forum/sh...l=1#post113257

In 2015, Mickey Crimm announced to the world he didn’t know what a “geometric distribution” was.

https://math.stackexchange.com/quest...-game-question

“I have no idea what a geometric distribution is. LOL!” – mickey crimm Jun 7 '15 at 13:50

This post gives insights on Mickey Crimm’s exploits from casino activities. This was 2015.

Food for thought.
It's worse than that, nerftard-AP. I never even took Algebra. All I ever knew how to do is add, subtract, multiply and divide. Learned all that shit by the 8th grade. But contrary to what you are selling that's all that's needed in gambling. All that fancy smancy made up bullshit you push is wasted time and wasted effort. Gambling math is simple. Here....I'll show you.

It's video keno. You are looking at a 4-spot. There are 20 balls out of 80 drawn. The payscale is:

4 out of 4 pays 65 for 1
3 out of 4 pays 5 for 1
2 out of 4 pays 1 for 1

While you are playing the 4-spot you get a free 3-spot.

So you are playing one way of 4 and one way of 3.

1 out of 3 and 2 out of 3 pay nothing.

3 out of 3 gets you 12 free games on the 4-spot with a 2X multiplier and 24 draws from the 80 ball tank. So that:

4 out of 4 pays 130 Units
3 out of 4 pays 10 units
2 out of 4 pays 2 units

In your opinion, ex-ap, is this a positive play or a negative play?  Reply With Quote

4. Now, while we await your answer, nerftard, if you found me on stackexchange you are doing a pretty thorough investigation of everything I've ever put up online. Since you are reading so much of me it looks like you are my biggest fan. Thanks. But I think you should take a break. Your fascination with me is looking more and more like an obsession. We know that is not healthy. Perhaps you should seek psychiatric help. It wouldn't hurt, you know.

Anyway, good luck with the keno math. A person of your mathematical caliber should have no problem with it.

BTW, it takes only a couple of minutes for someone that knows the math to solve this keno problem I put up.  Reply With Quote

5. [QUOTE=mickeycrimm;113290]
It's video keno. You are looking at a 4-spot. There are 20 balls out of 80 drawn. The payscale is:

4 out of 4 pays 65 for 1
3 out of 4 pays 5 for 1
2 out of 4 pays 1 for 1

While you are playing the 4-spot you get a free 3-spot.

So you are playing one way of 4 and one way of 3.

1 out of 3 and 2 out of 3 pay nothing.

3 out of 3 gets you 12 free games on the 4-spot with a 2X multiplier and 24 draws from the 80 ball tank. So that:

4 out of 4 pays 130 Units
3 out of 4 pays 10 units
2 out of 4 pays 2 units

In your opinion, ex-ap, is this a positive play or a negative play?

Bump. It's been 9 hours and no answer to this simple problem from ex-ap. Strange because he has told us all more than once of his advanced ability in mathematics. Seems that he would want to impress us by explaining this very simple math problem that only takes a few minutes to solve.

Food for thought.  Reply With Quote

6. Ex-AP, you're coming across as creepily obsessed with Mickeycrimm. :/  Reply With Quote

7. [QUOTE=mickeycrimm;113295]
It's video keno. You are looking at a 4-spot. There are 20 balls out of 80 drawn. The payscale is:

4 out of 4 pays 65 for 1
3 out of 4 pays 5 for 1
2 out of 4 pays 1 for 1

While you are playing the 4-spot you get a free 3-spot.

So you are playing one way of 4 and one way of 3.

1 out of 3 and 2 out of 3 pay nothing.

3 out of 3 gets you 12 free games on the 4-spot with a 2X multiplier and 24 draws from the 80 ball tank. So that:

4 out of 4 pays 130 Units
3 out of 4 pays 10 units
2 out of 4 pays 2 units

In your opinion, ex-ap, is this a positive play or a negative play?

Bump. It's been 9 hours and no answer to this simple problem from ex-ap. Strange because he has told us all more than once of his advanced ability in mathematics. Seems that he would want to impress us by explaining this very simple math problem that only takes a few minutes to solve.

Food for thought.
It's been 14 hours and still no response from ex-ap. I thought for sure he would want to impress us with his gambling math ability. He has sure done a lot of bragging about it. Here's a golden opportunity to show us something but he refuses to do it. What's up with that?

I'm beginning to think that ex-ap is all hat and no cattle.  Reply With Quote

8. Originally Posted by Mickey
It's video keno. You are looking at a 4-spot. There are 20 balls out of 80 drawn. The payscale is:

4 out of 4 pays 65 for 1
3 out of 4 pays 5 for 1
2 out of 4 pays 1 for 1

While you are playing the 4-spot you get a free 3-spot.

So you are playing one way of 4 and one way of 3.

1 out of 3 and 2 out of 3 pay nothing.

3 out of 3 gets you 12 free games on the 4-spot with a 2X multiplier and 24 draws from the 80 ball tank. So that:

4 out of 4 pays 130 Units
3 out of 4 pays 10 units
2 out of 4 pays 2 units

In your opinion, ex-ap, is this a positive play or a negative play?

Instinctually this looks really good. You're playing about a 1/3 loser 2/3 of the time but the other 1/3 of the time you're getting about 130% at no additional cost.

I'm looking forward to the math lesson. Get cracking Ex-AP. A rocket scientist could probably do it in his head. I can only imagine what a guy that does options type math on the fly before putting down a Scarab play can do.... LMAO  Reply With Quote

9. Checking back in......

"Crickets".....LOL

Oh well,,,,,Maybe Monday when the school math department opens back up......go Ex-AC...lol  Reply With Quote

10. It's been 34 hours. Ex-Ap is not going to answer on the keno math problem. He would solve the problem if he could just to rub my nose in it. But I'll give him 48 hours which will be tonight at 8 PM mountain time.  Reply With Quote

11. It's been 34 hours. Ex-Ap is not going to answer on the keno math problem. He would solve the problem if he could just to rub my nose in it. But I'll give him 48 hours which will be tonight at 8 PM mountain time.
Ex-AP is coming across as a Joke and a Clown. It's obvious he's being a Troll who is trying to look relevant by trying to nake YOU look bad when he is the only one who looks bad whereas YOU look good instead.  Reply With Quote

12. [QUOTE=mickeycrimm;113295][QUOTE=mickeycrimm;113290]
It's video keno. You are looking at a 4-spot. There are 20 balls out of 80 drawn. The payscale is:

4 out of 4 pays 65 for 1
3 out of 4 pays 5 for 1
2 out of 4 pays 1 for 1

While you are playing the 4-spot you get a free 3-spot.

So you are playing one way of 4 and one way of 3.

1 out of 3 and 2 out of 3 pay nothing.

3 out of 3 gets you 12 free games on the 4-spot with a 2X multiplier and 24 draws from the 80 ball tank. So that:

4 out of 4 pays 130 Units
3 out of 4 pays 10 units
2 out of 4 pays 2 units

In your opinion, ex-ap, is this a positive play or a negative play?

The first part of the problem is the regular game pays. Its easy enough to just go to the keno analyzer at Wizard of Odds, punch in the payscale and get the answer, 62.8%. Of course the Wiz's keno calculator hasn't always been around and I had to do this stuff from scratch.

With 80 balls in the tank and 20 balls drawn how many 4-spots combinations are there in the tank?

80*79*78*77*/4*3*2*1 = 1,581,580 total 4-spot combinations.

How many 4-spot combinations will be in the 20 balls drawn?

20*19*18*17/4*3*2*1 = 4,845

What are my chances of hitting the 4-spot?

1,581,580/4845 = 326.4355

What are my chances of catching 3 out of 4 numbers? The three numbers I catch must be part of the 20 drawn balls and the 4th ball must be one of the 60 balls not drawn.

20*19*18/3*2*1 = 1140
The 4th ball will be 1 of 60
1140*60 = 68,400
1,581,580/68,400 = 23.1225

What are my chances of catching 2 out of 4? Two numbers must be part of the 20 drawn balls and 2 numbers must be part of the 60 undrawn balls.

20*19/2*1 = 190
60*59/2*1 = 1770
190*1770 = 336,300
1,581,580/336,300 = 4.7029

So now we have our frequencies. So how much is each pay worth?

4/4 pays 65 units
65/326.4355 = 19.9121%

3/4 pays 5 units
5/23.1255 = 21.6211%

2/4 pay 1 unit
1/4.7029 = 21.2636%

Add 'em up and it comes to 62.7968%

The slight difference from Wiz's 62.8% is due to rounding.

The above is the first part of the problem. We'll move on in the next post.  Reply With Quote

13. The next part of the problem is the 3-spot. Hitting zero out of 3, 1 out of 3, and 2 out of 3 pays nothing. The only action is on hitting 3 out of 3.

How many total 3 spots combinations are there when 20 balls are drawn out of 80?

80*79*78/3*2*1 = 82,160

How many 3-spots are there in the 20 balls drawn?

20*19*18/3*2*1 = 1140

So the freq. of the 3 spot is:

82,160/1140 = 72.0702  Reply With Quote

14. So now we need the frequencies for hitting 4/4,3/4,2/4 when 24 balls are drawn from 80.

4/4
24*23*22*21/4*3*2*1 = 10,626
1,581,580/10,626 = 148.8406

3/4
24*23*22/3*2*1 = 2024
The 4th ball must be 1 of 56 undrawn balls.
2024*60 = 121,440
1,581,580/121,440 = 13.0235

2/4
24*23/2*1 = 276
Two balls must be in the 56 undrawn balls.
56*55/2*1 = 1540
276*1540 = 425,040
1,581,580/425,040 = 3.7210

4/4 pays 130 units
130/148.8406 = .873418

3/4 pays 10 units
10/13.0235 = .767843

2/4 pay 2 units
2/3.7210
2/3.7210 = .537490

Add em up and its 2.1788 units per free game.

There are 12 free games
12*2.1788 = 26.1456 units returned per 12 free games

The frequency of the 3-spot is 72.0702. The 26.1456 units returned in the free games represents

26.1456/72.0702 = 36.2780% of the payback.

The regular games payback is 62.7968%.

62.7968 + 36.2780 = 99.0748%

So the game is not quite positive. I would need some kind of promotion that would put the game over 100% to play it.  Reply With Quote

15. That's why interpreting the rules is so important. When I looked at that problem I was thinking 3 out of 4 numbers which would have had the person getting the bonus about 3x as much. Which at a glance looked GOAT.

But thanks for the math lesson Mickey. You actually laid out the solution in an elementary manner making it easy to understand. Seems like when you try to read some of these explanations about calculating combinations the writers always want to complicate it.  Reply With Quote

16. So now we need the frequencies for hitting 4/4,3/4,2/4 when 24 balls are drawn from 80.

4/4
24*23*22*21/4*3*2*1 = 10,626
1,581,580/10,626 = 148.8406

3/4
24*23*22/3*2*1 = 2024
The 4th ball must be 1 of 56 undrawn balls.
2024*60 = 121,440
1,581,580/121,440 = 13.0235

2/4
24*23/2*1 = 276
Two balls must be in the 56 undrawn balls.
56*55/2*1 = 1540
276*1540 = 425,040
1,581,580/425,040 = 3.7210

4/4 pays 130 units
130/148.8406 = .873418

3/4 pays 10 units
10/13.0235 = .767843

2/4 pay 2 units
2/3.7210
2/3.7210 = .537490

Add em up and its 2.1788 units per free game.

There are 12 free games
12*2.1788 = 26.1456 units returned per 12 free games

The frequency of the 3-spot is 72.0702. The 26.1456 units returned in the free games represents

26.1456/72.0702 = 36.2780% of the payback.

The regular games payback is 62.7968%.

62.7968 + 36.2780 = 99.0748%

So the game is not quite positive. I would need some kind of promotion that would put the game over 100% to play it.
I found a mistake here which is easy to do. One little mental slip throws the whole thing off. It's the 3/4. I multiplied 2024 by 60 when it was supposed to be multiplied by 56. So this throws things off a little. It should be

3/4
24*23*22/3*2*1 = 2024
The 4th ball needs to be 1 of the 56 undrawn balls.
2024*56 = 113,344
1,581,580/113,344 = 13.9538

10/13.9538 = .716651 This should be the decimal instead of .767843.

So a free game is worth

4/4 .873418
3/4 .716651
2/4 .537490

Total = 2.1276

12 free games = 25.5307 units.

25.5307/72.0702 = 35.4248

62.7968 + 35.4248 = 98.22%

So the game is 98.22%

It's easy to make a mental slip. I always have to go over it a few times to check for mistakes. Guys like Sklanky and Malmuth have mathematicians proof reading their math before they publish. I don't have that luxury.  Reply With Quote

17. The video keno games I play are either banking or progressive games. This example was neither. I just used it as an example of how to do keno math. The object of doing the math is to determine if the player has an advantage.

Gambling math is pretty simple. It's just addition, subtraction, multiplication and division. There's no "stochastics" or "heuristics" or "geometic distribution." It's just simple math.  Reply With Quote

18. That's why interpreting the rules is so important. When I looked at that problem I was thinking 3 out of 4 numbers which would have had the person getting the bonus about 3x as much. Which at a glance looked GOAT.

But thanks for the math lesson Mickey. You actually laid out the solution in an elementary manner making it easy to understand. Seems like when you try to read some of these explanations about calculating combinations the writers always want to complicate it.
Thanks. Steven Hawking wrote in the forward of his book, A Brief History of Time, that he was told by other astrophysicists that if he put just one equation in the book it would cut sales in half. LOL! Most people don't understand the fancy equations and avoid them. I learned keno math from John Scarne who wrote things out so even a 3rd grader could understand it. That's the way I try to do it.

I put up a slightly negative play because in the past I've put these problems up where the game is slightly positive and people guess it's slightly positive, not because they did the math, but because they think I wouldn't put a negative play up.

I threw in the extra draws in the free games so ex-ap couldn't just go and run it on the Wiz's keno caluclator. It only does 20 balls drawn from a tank of 80 balls.  Reply With Quote

19. The video keno games I play are either banking or progressive games. This example was neither. I just used it as an example of how to do keno math. The object of doing the math is to determine if the player has an advantage.

Gambling math is pretty simple. It's just addition, subtraction, multiplication and division. There's no "stochastics" or "heuristics" or "geometic distribution." It's just simple math.
Just for fun and if anyone reading this wants another way to solve it which (imo) isn't, "Too fancy," here's how I would do it:

BASE PAYS:

Like Mickey says, just use the WoO Keno Calculator:

2/4: 0.212635465800023

3/4: 0.216239456745786

4/4: 0.199120499753411

BASE GAME TOTAL: 0.627995422299220, which we can just call 62.8%.

GETTING TO THE BONUS GAME:

The only possible mistake someone can make here is trying to come up with a bunch of ridiculous scenarios where a win on the base game impacts the probability of getting the 3/3...which in certain cases you WOULD need to know because there is at least one game where a win on the base game AND hitting the 3/3 combination is required. Fortunately, this is not one of them.

For this, I also use the WoO calculator (without attributing any pays---because just going to the bonus pays nothing) and lift the probability for 3/3, which is:

0.013875365141188

FREE GAMES RETURN:

The next step is to determine the overall return of the Free Games (as a whole) which we will then take and multiply by the probability of going to the Free Games, then having determined that, add to the base return.

Since it's a weird number of draws, we can't cheat with the WoO Calculator.

4/4 = 130
3/4 = 10
2/4 = 2

For this, I like to use an online scientific calculator you can find here:

https://web2.0calc.com/

(nCr(4,2)*nCr(76,22)/nCr(80,24)) * 2 = 0.5374878286270691

(nCr(4,3)*nCr(76,21)/nCr(80,24)) * 10 = 0.7166504381694255

(nCr(4,4)*nCr(76,20)/nCr(80,24)) * 130 = 0.8734177215189873

MickeyCrimm added them up and got 2.1276 units/game on Free Games, and:

0.5374878286270691+0.7166504381694255+0.8734177215 189873 = 2.1275559883154819

I agree with that, except I haven't rounded off yet just because I took the results the way the calculator gave them to me. Now, multiply by the number of Free Games:

2.1275559883154819*12 = 25.5306718597857828

Which is the expected return of all Free Games, so all that remains is to multiply that by the probability of Free Games:

25.5306718597857828*0.013875365141188 = 0.3542473943543810567001571019664

Which gives us our added expected return (per unit bet) that comes from Free Games, which we can call 35.4%, add to 62.8% and get 98.2%. Therefore, my results agree with those of MickeyCrimm with differences due to the way we rounded. It's not positive, anyway, so there would be very limited scenarios where I would care about this game.

Of course, if I had a multiplier day such that it came up to 2% return on the points multiplier and there was no reasonable video poker at this location (or any positive machines) this would be a good machine to know if I needed to earn points anyway...obviously I would not play this for well under a 0.5% edge alone.  Reply With Quote

20. COMBINATORIAL FUNCTION

I also want to mention that I don't use the Combinatorial function to look fancy, or anything like that. There are a few reasons why I like it, which include:

1.) I think it's really fast for all Keno-related math and lots of playing card probability math.

2.) You can easily check your work as you go: The first numbers in each parenthesis on the left side of the multiplication symbol (*) should ALWAYS add up to total whatever is on the left side of the parenthesis on the right side of the multiplication symbol. Same thing with the right side of it, let's break this down in simple terms to see what it means in plain English:

(nCr(4,2)*nCr(76,22)/nCr(80,24)) * 2 = 0.5374878286270691

There are 4 numbers that help me and 76 that do not for a total of eighty numbers. Of 24 total balls to be drawn, I want to know how likely it is that I get 2 numbers that help me which must mean I get 22 numbers that do not help me.

And...that's it. That's all you're asking. If you wanted to know how likely hitting 5/7 is in the same scenario, you'd just do this:

(nCr(7,5)*nCr(73,19)/nCr(80,24)

For this, the left side parenthesis numbers should still always add up to be the same as the spots on the right. No double-checking needed aside from that.

3.) Very copy-pastable. You can type the equation into the calculator and copy/paste over, or type it on your sheet and move it to the calculator. At that point, you can copy/paste the answer from the calculator back to your document or post.

4.) (ADDED) Also relatively easy to do from your phone if you need Keno math on the fly.  Reply With Quote