Wolverine: probability question

[Don Schlesinger]

> Thanks for taking this further. We are very interested
> in the answer. For a moment, let us consider that
> perhaps this event (4 naturals) is very, very rare
> (like we said, 1 in 194,481 rare) and 3 naturals don't
> happen all that often (roughly 1 in 10,000). Should we
> be considering the thought that the casino is cheating
> somehow?

Don't get confused. If I say to you, "Deal four hands and try to get four naturals," that's 1 in 191,603 (more precise figure for double deck). That's not the same as dealing 5,000 hands and trying to observe one streak of four naturals. So, be careful; that event isn't nearly as rare, of course.

> Looking forward to the new answer.

Me too. But, I'm also looking forward to the formula and methodology for finding the answer. ("Give a man a fish, you feed him for a day; teach a man to fish, you feed him for a lifetime!" :-))

> Also, has anyone else following this thread ever
> seen, received, or been the victim of 4 naturals in a
> row (or even 3)?

Three for sure. Personally, I do not remember four in a row on either side of the table.

Don

[Zenfighter]

My bet is that this complicated problem can?t be solved through straight math calculations. The main reason is the nature of our game (dependence trials probability) plus practical dealing procedures. As an example of a DD dealt out 75% with a solo player on the table we can see that in reality what we get is:

104 * .75 = 78 cards

78/ 5.4 = 14.44 rounds

5000/14.44 => 346 double decks necessary to complete the required 5000 rounds. So our dealer could get four naturals in a row with quite a few different probabilities, all in all, very difficult IMHO to integrate them into one formula. E.g.

Dealer gets four naturals in a row:

1) Before reaching the cut card

2) One natural on the last round and three others on the first three rounds of the next deal.

3) Two naturals on the last two rounds and another two on the first two rounds of the next one.

4) Three naturals on the last three rounds plus another one on the first one of the next deal.

What we have here is, obviously, a mix of different dependent probability trials, interacting during the process.
As is usual the case, computer simulations seems the way to go to solve these types of questions, once the math goes over our heads. :-)

Sincerely

Zenfighter

[Don Schlesinger]

> My bet is that this complicated problem can?t be
> solved through straight math calculations.

Yes, actually, they can, if we limit the dealing to simply two cards. I have seen the formula before, and it's quite messy, but doable nonetheless.

> The main
> reason is the nature of our game (dependence trials
> probability) plus practical dealing procedures. As an
> example of a DD dealt out 75% with a solo player on
> the table we can see that in reality what we get is:

> 104 * .75 = 78 cards

> 78/ 5.4 = 14.44 rounds

> 5000/14.44 => 346 double decks necessary to
> complete the required 5000 rounds. So our dealer could
> get four naturals in a row with quite a few different
> probabilities, all in all, very difficult IMHO to
> integrate them into one formula. E.g.

> Dealer gets four naturals in a row:

> 1) Before reaching the cut card

> 2) One natural on the last round and three others on
> the first three rounds of the next deal.

> 3) Two naturals on the last two rounds and another two
> on the first two rounds of the next one.

> 4) Three naturals on the last three rounds plus
> another one on the first one of the next deal.

> What we have here is, obviously, a mix of different
> dependent probability trials, interacting during the
> process.
> As is usual the case, computer simulations seems the
> way to go to solve these types of questions, once the
> math goes over our heads. :-)

Yes, of course, actually playing the game can alter, somewhat, the probabilities, but not enough to really matter, as to the order of magnitude of the answer. I had stated that, for our purposes, we'll just use two cards dealt off the top of a double deck. You're right that, in actual play, the answer will be a little different, but not enough to really matter, in my view.

Thanks for your thoughts.

Don

[Don Schlesinger]

[Stewart]: The exact probability of at least one run of r consecutive successes in n independent trials, each with success probability p, is given by de Moivre's formula, which I quote from Uspensky's (1937) Intro to Mathematical
Probability:

In TeX symbolism,

$$
P=1-\sum_{k=0}^{[n/(r+1)]}(-1)^k{n-kr\choose k}(qp^r)^k
+p^r\sum_{k=0}^{[(n-r)/(r+1)]}(-1)^k{n-r-kr\choose k}(qp^r)^k.
$$

[Don]: Wow! I like my way better! :-)

[Stewart]: In your case, p=2*8*32/(104*103), r=4, and n=5000, so it suffices to consider the first few terms in each sum (the remaining terms are negligible). I get P=0.0245 (rounded). So your approximation was not too
bad.

[Don]: Surprisingly close. Doesn't make my way right, but didn't expect to be within one-tenth of the correct answer.

[Stewart]: In practice, the dealer doesn't necessarily shuffle between hands,

[Don]: This was foolish of me. Don't know what I was thinking. See below, for what I hope is a much better approximation to the true correct answer.

[Stewart]: so the true probability will be smaller than 2.45 percent.

[Don]: Shockingly smaller. Consider that, after three straight naturals, there are only five aces left, and the prob. of getting a snapper is dramatically decreased (by almost half, when one considers the reduced tens, as well).

[Stewart]: For example, if the dealer has had three straight naturals and is still dealing from the same deck, the fourth natural will be somewhat less likely, because three of the aces will be gone.

[Don]: MUCH less likely! Suppose I were to deal nothing but four straight hands of two cards each (but with no replacement), to simplify matters. Then the prob. of four straight naturals becomes:

[(2*8*32)/(104)(103)][(2*7*31)/102)(101)][(2*6*30)/(100)(99)][(2*5*29)/(98)(97)] = 1/1,442,515.

Compare this result to the 1/191,603 that we got with replacement! The latter is 7.53 times more likely than the former! So, if I now do it my way, I get a new probability of 1 - 0.99654 = 0.00346, or once in every 289 attempts, which, of course, is much rarer than once in 39. What do you get with Uspensky's formula, replacing p with the new 0.000000693?

Thanks very much Stewart; I appreciate your help.

Don

[bfbagain]

Don:And, maybe my memory is faulty, and I actually have seen such a streak.

Considering what I have personally been through of late, I would have to say it's a real stretch that this has not happened to you.

I can't say for sure if this has happened to me in the last month, but I did have a friend who witnessed my play at a heads up DD game, who was counting the number of dealer draws to 21's, and stopped counting at 39 in a two hour period. Yes, it's rare I would play for two hours, and even more rare at where I was playing, considering the stakes. That said, my losing, and the recent departure - I'm speculating here - of long time personnel probably contributed to my allowed play.

During this time, a most humbling experience I can tell you, there was no doubt that the dealer had at least three BJ's, and maybe multiple times. I was probably a little too jaded to concern myself with any apparent oddities, as the only oddity was me playing. But a heads up DD, at this place, with no heat, and a place I love to play, well....you don't leave good games, even when you're getting the tar beat out of you.

Sorry for the verbose answer, as I just got home and was catching up when I saw this thread, but yea Don, I believe you most definitely have witnessed this. :-)

Myself, if I never witness another 3 1/2 weeks like this, I'll be happy. :-)

cheers
bfb

[Don Schlesinger]

> Sorry for the verbose answer, as I just got home and
> was catching up when I saw this thread, but yea Don, I
> believe you most definitely have witnessed this. :-)

I now calculate that you could play 50 hours of DD a year for 60 years and still have a 72% probability of never seeing four consecutive dealer naturals.

And, for the record, I don't remember ever having seen it happen to me.

Don

[bfbagain]

> I now calculate that you could play 50 hours of DD a
> year for 60 years and still have a 72% probability of
> never seeing four consecutive dealer naturals.

> And, for the record, I don't remember ever having seen
> it happen to me.

I was referring to three naturals in a row. And for the record, three, four, or more snappers in a row all mean the same thing. You're losing.

It may be interesting to calculate probabilities, as an intellectual exercise, but dwelling on those things won't help players win.

The first thing that comes to mind when people ask about what may be perceived as blackjack anomalies, is the perception that there was cheating involved. There may be cheatng, but blackjacks are not the preferred method of cheating dealers, or so I'm told. The reason for the last comment is I probably couldn't tell you who was cheating or who wasn't if I watched with a microscope. However, when I do think I'm seeing extraordinary occurances, I make the dealer very aware that I'm "trying" to focus on their hands as they're dealing. It would take a pretty accomplished cheat to not react to "me" visually focusing, with extreme intensity, on their dealing the cards.

My conclusions are relatively simple, and that is, no cheating, just bad variance, in almost all cases.

Bad luck can come in many forms. What most people are not prepared for, are the incredible runs of bad cards that happen. The beans in a jar is always a good example of how streaky blackjack can become, and how small our edge really is.

The only real thing that pisses me off when I'm in one of these streaks, is the realization that I have to play so many more hours (and so many more hands), just to recover.

Anyway, I just thought I'd put some perspective to this thread.

cheers
bfb

[Karel Janecek]

> but didn't expect to be within one-tenth of the
> correct answer.

Actually, I was expecting that the numbers will be very close. A vague reason is that most trials are indpendent. Only the close neighbours are dependent, but not that much.

Regards,

Karel Janecek
(author of SBA)

[Wolverine]

Okay Don, do we have an answer to what the de Moivre's formula indicates without replacement? Your non-independent answer is 1.4 million to 1. In other words, very rare.

[Don Schlesinger]

> Okay Don, do we have an answer to what the de Moivre's
> formula indicates without replacement?

No, no further answer from Stewart.

> Your non-independent answer is 1.4 million to 1. In other
> words, very rare.

No, my answer was once in every 289 attempts of 5,000 hands dealt. "Rare"? Not so. The once in 1.4 million was if you try to get four naturals in a row, given one chance at dealing four hands. Here, we have 4,997 consecutive chances. You do understand the difference, right?

Don