Is this strong evidence a casino is cheating?

I don’t play blackjack but I heard a story about someone who got kicked out of a casino for accusing them of cheating. It seemed to me that what he experienced was probably not unlikely, statistically, and the conversation turned into a debate I’d like to settle.

Apparently on three separate times with new decks at a 6-deck table he only saw around 11 aces come out every time and only a small amount of cards left to deal before they shuffled (or however it works). His contention is that it’s unlikely the other half of the aces were in the small pile they didn’t use, all three times, and that the card factory is putting less than 24 aces in there before it gets to the casino.

Regardless of how they might manage to keep aces out of play, is seeing only 11 aces three times, or even five times in a night unlikely? I have no idea how it all works with shoes and cuts and discards and whatever but my hunch is that it’s not that uncommon.

Hi Nathan!

Hi, I’m not Nathan. Did you have a similar debate with someone named Nathan?

I imagine it’s pretty common for people who start losing at any casino game to start thinking it’s rigged. People probably get kicked out often for causing a scene over some unlikely but statistically possible occurrence.

Does your friend go to the restroom every single time the cards are changed on the table? They fan each new deck out face up as a part of that process & you can verify that the aces are all there...

Maybe your friend would have better luck with the pick 3 lotto.

Quote: rsactuary

Hi Nathan!

It's kentry to you sir.

Quote: rsactuary

Hi Nathan!

Probably very unlikely.

Are we talking about 18 yo's in a row here? I've lost track.

Quote: rsactuary

Hi Nathan!

My DNA testing says there’s only a 12% chance this is Nathan.

I see where you’re coming from. The Nathan-ish type writing style. However , the post is actually a valid question and makes sense overall , so it can’t be her

Quote: michael99000

My DNA testing says there’s only a 12% chance this is Nathan.

I see where you’re coming from. The Nathan-ish type writing style. However , the post is actually a valid question and makes sense overall , so it can’t be her

I dunno... BJ pro gets banned, and BJ cheat shows up? Nathan-ish style writing, designed to mislead but doesn't quite get there? The give away for me is the "I heard that". Nathan always talked about what he heard, never from experience.

(the "he" was intentional, because I think the "she" thing was always BS)

Quote: Bjcheat

I don’t play blackjack but I heard a story about someone who got kicked out of a casino for accusing them of cheating. It seemed to me that what he experienced was probably not unlikely, statistically, and the conversation turned into a debate I’d like to settle. ...

Regardless of how they might manage to keep aces out of play, is seeing only 11 aces three times, or even five times in a night unlikely? I have no idea how it all works with shoes and cuts and discards and whatever but my hunch is that it’s not that uncommon.

Hi BJCheat. Welcome to the forum.

In answer to you question... You present it as 'I heard a story about someone'. From that alone, YOU and WE have ZERO evidence that any cheating occured and zero reason to try to establish whether cheating occurred. Has as been said, the cards are normally presented face up at least once. Also, what 'someone' sees during the course of a shoe is seldom what happens. Such perceptions are usually flawed. Who counts aces as they come out, anyway? Rhetorical. There is no value in speculating further.

Meanwhile, if you are puzzled as to why you are greeted by all sorts of cryptic comments about being Nathan, Kentry or others, then please ignore them. Those comments come about because we only recently banned some members who had a writing style similar to yours.

To all members
This thread is not to become a discussion of the identity of BJCheat or anyone else.
If it does, it will be closed and members might get penalised.

Quote: Bjcheat

I don’t play blackjack but I heard a story about someone who got kicked out of a casino for accusing them of cheating. It seemed to me that what he experienced was probably not unlikely, statistically, and the conversation turned into a debate I’d like to settle.

Apparently on three separate times with new decks at a 6-deck table he only saw around 11 aces come out every time and only a small amount of cards left to deal before they shuffled (or however it works). His contention is that it’s unlikely the other half of the aces were in the small pile they didn’t use, all three times, and that the card factory is putting less than 24 aces in there before it gets to the casino.

Regardless of how they might manage to keep aces out of play, is seeing only 11 aces three times, or even five times in a night unlikely? I have no idea how it all works with shoes and cuts and discards and whatever but my hunch is that it’s not that uncommon.

How many cards are dealt before the cut card (shuffle point)? EG: Do they deal 5 out of 6 decks? 4 out of 6?

RS, I’m not sure and am not familiar with how dealing works, but as it was told to me, there weren’t many cards left when they shuffled, which is why he thought there should have been more aces out. How low can the stack get before it gets shuffled again?


OnceDear, thanks for the reply. I can’t and don’t care to prove that cheating occurred. I’m only trying to settle a debate about whether or not it’s not an unreasonable scenario. Apparently he decided to start counting aces because he suspected their decks had aces removed from them.

Dot-dot...

Quote: Bjcheat

RS, I’m not sure and am not familiar with how dealing works, but as it was told to me, there weren’t many cards left when they shuffled, which is why he thought there should have been more aces out. How low can the stack get before it gets shuffled again?

It's pretty common to leave between 1 to 2 decks un-played. Rare, but sometimes casinos will only leave 0.5 decks un-played.

Unfortunately I don’t know the probabilities for say only 12 aces coming out on a table that dealt 5 out of 6 decks. It does seem unlikely that the rest would be in the last 52 cards or so before shuffling but it still seems possible, even a couple times....

I wish I was better at math. Oh well

It is highly unlikely that their was any form of cheating going on here. Definitely a conspiracy of someone who may have lost some dough.

I heard a story about a dealer who brings his own Aces and adds them to the deck. I think he thinks more Aces equals more player wins, and more BJs,.hence more tips. Evidently, he changes jobs a lot because I always hear about him dealing at different casinos, in different cities even.
I haven't played with him yet, but look forward to it.

Quote: billryan

I heard a story about a dealer who brings his own Aces and adds them to the deck. I think he thinks more Aces equals more player wins, and more BJs,.hence more tips. Evidently, he changes jobs a lot because I always hear about him dealing at different casinos, in different cities even.
I haven't played with him yet, but look forward to it.

Doesn’t the shuffler machine recognize when there’s more than the correct number of cards and goes to a red light status?

I think we can handle this with just a little bit of combinatorics:

The first thing I am going to assume is that two decks were cut off, which would be pretty high, but not altogether unusual. Therefore, that leaves us with the following:

Cards: 312
Starting Aces: 24

Finishing Cards: 104
Finishing Aces: 13

Okay, so here is the combinatorial formula we can use to determine that probability:

First, we determine the total number of ways to select 208 of 312 cards:

nCr(312, 208) = 8.46496974×10^84

The next thing we have to determine is the probability of selecting eleven aces as well as 197 other cards.

nCr(24, 11)*nCr(288, 197) = 1.32766836×10^83

Now, to get the probability we have to take the total number of ways the eleven aces can happen and divide it by the total number of ways that anything can happen:

(1.32766836*10^83)/(8.46496974*10^84) = 0.01568426587

Which I did using this scientific calculator, though you have to put it in a little differently:

https://www.desmos.com/scientific

That means that the probability of EXACTLY eleven aces coming out given 208 cards of 312 drawn is about 1.568426587%

The probability of EXACTLY eleven aces coming out given three consecutive trials is:

.01568426587^3 = 0.00000385826

That's about 1 in 259,184, but by itself, it doesn't signify anything. Let's instead take a look at all of the other possibilities that could have happened involving eleven, or FEWER, aces and add them together:

nCr(24, 10)*nCr(288, 198) = (4.79435795×10^82)/(8.46496974×10^84) = 0.00566376265
nCr(24, 9)*nCr(288, 199) = (1.44553506×10^82)/(8.46496974×10^84) = 0.00170766713
nCr(24, 8)*nCr(288, 200) = (3.61835495×10^81)/(8.46496974×10^84) = 0.00042745042
nCr(24, 7)*nCr(288, 201) = (7.45484894×10^80)/(8.46496974×10^84) = 0.00008806704
nCr(24, 6)*nCr(288, 202) = (1.24862569×10^80)/(8.46496974×10^84) = 0.0000147505
nCr(24, 5)*nCr(288, 203) = (1.67044557×10^79)/(8.46496974×10^84) = 0.00000197336
nCr(24, 4)*nCr(288, 204) = (1.74004747×10^78)/(8.46496974×10^84) = 0.000000205558617
nCr(24, 3)*nCr(288, 205) = (1.35808583×10^77)/(8.46496974×10^84) = 0.0000000160435993
nCr(24, 2)*nCr(288, 206) = (7.46168074×10^75)/(8.46496974×10^84) = 0.00000000088147754
nCr(24, 1)*nCr(288, 207) = (2.5702912×10^74)/(8.46496974×10^84) = 0.000000000030363856
nCr(24, 0)*nCr(288, 208) = (4.17054462×10^72)/(8.46496974×10^84) = 0.00000000000049268276

You can actually pop the second part into Google and then, when it became necessary, I used this converter:

https://www.free-online-calculator-use.com/scientific-notation-converter.html#

Okay, so now we have to add all of these together to get the probability of eleven, OR FEWER, aces:

(0.01568426587)+(0.00566376265)+(0.00170766713)+(0.00042745042)+(0.00008806704)+(0.0000147505)+(0.00000197336)+(0.000000205558617)+(0.0000000160435993)+(0.00000000088147754)+(0.000000000030363856)+(0.00000000000049268276) = 0.02358815948

Okay, so the combined probability of seeing eleven, or fewer, aces under the circumstances described is 2.358815948%, so three times in a row:

0.02358815948^3 = 0.00001312448 or 1/0.00001312448 = 1 in 76,193.4949402

Well, it's MUCH more likely than a dealt royal, and one of those just happened at Downtown Grand recently.

OTHER OBSERVATIONS:

1.) This all assumes that two decks were cut off, when dealing with numbers like these, the number of decks cut off can substantially change the probabilities. It's also possible the casino cuts off more than exactly two decks, which would make this event more likely.

2.) It assumes that your friend was counting the aces accurately, which may or may not be the case.

3.) It assumes that the story is true.

CONCLUSION:

I would suggest that this, by itself, is not sufficient evidence of cheating. It's sufficient evidence that rare things occasionally happen, which any gambler already knows. If he's interested, then I would suggest he go to the table, watch, and then count the total number of cards dealt as well as the total number of aces dealt each time for maybe twenty shoes. Possible procedures to determine fairness could vary from there.

ADDED:

Going back to the OP, I now notice it says, "About eleven aces." What does about mean? Could it have been 12, 13, 14?

I also notice that I misread and assumed that this happened on three consecutive shoes, but now that I read the OP again it did not happen on three consecutive shoes. In my view, cheating would not be proven even if eleven (or fewer) aces had come out in three consecutive shoes assuming two decks cut off. For this to only happen three times in a night may well be unusual, but I don't even think ridiculously improbable.

You should ask your friend if he missed the shoe where 21 aces came out with two decks cut off:

nCr(24, 21)*nCr(288, 187) = (1.07772121×10^83)/(8.46496974×10^84) = 0.01273154238

I wonder if he would have accused the casino of cheating in the player's favor if that happened three times?