SB's BJ Aptitude Quiz Question #2

The game is 6D and about 156 cards are in the discard tray. The running count happens to be 21 and the true count is what it is. As the remaining playable cards are used, is it more likely that the true count will:

1A. increase to a higher number
1B. remain
1C. decrease toward zero
1D. stay approximately 3 times lower than the running count

Is it more likely that the running count will:

2A. remain
2B. increase to a higher number
2C. decrease toward zero

In the above scenario, the true count is at:

3A. 11
3B. 21
3C. 7
3D. 63
3E. 0
3F. none of the above

The question was

As the remaining playable cards are used, is it more likely that the true count will:

1A. increase to a higher number
1B. remain
1C. decrease toward zero
1D. stay approximately 3 times lower than the running count

The TC is 7.0, and the expected value for the TC remains 7. However, the TC is more likely to increase. These are not inconsistent.
If Peter Griffin wrote the Exam, I would assume that he was looking for the correct answer, 1A.

However, if know that a Free Page poster made the exam, I would guess that they mathematically ignorant, and that the answer they were looking for is 1B. (I did not answer the meaning of choice 1D, �3 times lower than the running count�).

Finally, the big issue, What the Heck does this have to do with AP aptitude . Frankly, if I were ever involved in a Team, and if I were interviewing potential team members, these are not among the list of questions that I would ask.

You say "the TC is more likely to increase".

Is this because of remaining decks estimation to a fraction of a deck, causing it to drop suddenly at the next "boundary"?

I supposed that has to be the reason, since the RC is supposed to drop linearly over the remaining cards.

If that is the case, what about the perfectionist that estimates to the exact number of remaining cards? Would the TC still tend to climb there???

Not trying to cause a long discussion, but I was definitely intrigued by that comment. Not that I particularly like the concept of an AP IQ test. :)

I was looking at the Exact True Count, and not an approximation.

Consider this: Suppose you were looking at a random variable T.

Suppose that there are two possibilities:

T can down up 1 with a probability of 75%

T can go down 3 with a probability of 25%.

T is more likely to go down than up.
The expected change in T is 0.

This is roughly what is happening with True Count, although obviously that random variable is a lot more complicated.

Here is another illustration. If you play a martingale at Roulette, your bankroll is very likely to increase. However, the expected change in you bankroll is negative.

>I was looking at the Exact True Count, and not an approximation.

>Consider this: Suppose you were looking at a random variable T.

>Suppose that there are two possibilities:

>T can down up 1 with a probability of 75%

>T can go down 3 with a probability of 25%.

>T is more likely to go down than up.
>The expected change in T is 0.

OK. I follow that easily enough...

>This is roughly what is happening with True Count, although obviously >that random variable is a lot more complicated.

What I didn't follow was your statement "The TC is more likely to increase".

Say we have exactly one deck left. And we have a RC of +4, which gives a TC of +4. This means that in 52 cards, ignoring the 7/8/9 since I'm a hi-lo'er, there must be 24 small cards and 28 big cards.

Now, on the next hand, and again I am assuming your statement above about exact true count is to be taken at face value, the curreent TC is clearly 4, with 24 small and 28 big cards left, and the TC is exactly 4.0...

To keep my head from exploding, I am going to play heads up. The dealer deals four cards, two to himself, one of which I can see, and two to me, both of which I see. It would seem to me that the probability of the first card dealt ... [interrupt...] wait a minute. I think I see your point.

In this case, your statement doesn't hold because there are no neutral cards in the deck. So lets change the above to 19 low cards, 23 high cards, and 10 neutral cards. Now I see where your statement is coming from. The TC will fall only if 10's are dealt. And that will happen with a probability of 23/52. But if a low card falls, the TC will increase, but also if a neutral card falls the same happens. So there is a 29/52 chance of a non-high card falling, and the TC will rise.

So OK. I see your point. Took a bit of thinking. And, in fact, I discovered this for myself in another discussion here a long while back. But apparently 57 years of age is enough to see some things go forgotten. I neglected the fact that when a neutral card falls, the denominator in the TC calculation gets smaller making the TC climb even though no low card fell.

Thanks for the hints... as counter-intuitive as it seems it makes perfect sense assuming I did not have a total mental failure above.

-SSR

The game is 6D and about 156 cards are in the discard tray.

Stealth Bomber,

Perhaps you should change this to say instead: The game is 6D and you have seen about 156 cards go into the discard tray.

The way you wrote the problem, a wonger just approaching the table would still say the RC=TC=0, since he hasn't seen any of the cards. Also, the answer will be different for a play-all player if the casino burns a large number of cards: I've seen some that burn a half deck at the start of each shoe.

Dog Hand

I went to one casino on Fremont street when it was not crowded, and approached an empty DD table. Sat down, dealer un-fanned the cards shuffled, and immediately took about 1/2 deck off the top and stuck 'em in the discard tray. Then put the shuffle/cut card at about the 50% point of what was left. I played one hand, got a "phone call" and bailed... never went back...

I think I also saw a "bunch of burn cards" a couple of years back at Caesar's although I am not certain that is the right casino, it was Summer '04... this was a shoe game $10 min.

Only good thing there was that the dealer didn't cut off multiple decks off the back-side as well... so it was playable.

You example where there were no neutrals got me thinking more about this. Another issue is: what happens when multiple counts are drawn.
I put together a Sim where I sampled the TC Change, whenever the pre-deal TC was >+ 3. This was 6 decks, dealt to 1`.5 decks. First Sim had 4 players:

 

Pct Up     =     47.75%    Mean  =   1.1254 of SE  0.00126 

Pct Down =     52.11%    Mean  =  -1.0300 of SE  0.00103 

My head just exploded, sorry. Thought I had this figured out, but now it seems not. :)

I assume that extra down turn is due to varying number of neutral cards left in the shoe? And the case to worry about is any case where low + neutral < high (or something along those lines), which makes it more likely for the TC to drop? Also I notice you only sampled for TC >= +3, while in negative counts the same question arises to affect the overall average??

It is so much more fun just to count, bet and play, and not try to understand "why". :)

With a TC of 7, you put out a big bet expecting high cards which will lower the TC and hopefully make you money. If as you suggest it is more likely that the TC will go higher, that means that it is likely that more low cards will come out, thus this is a negative betting situation and you should wong out or bet one unit.

OR

the next round will be neutral, further increasing the TC, however this is statistically impossible in the long run of +21 shoes. Again if the next round is neutra, its a bad betting situation so best to wong out o bet one unit.

In reality with a TC of +7 we put up the money hoping that the TC goes down.

> With a TC of 7, you put out a big bet expecting high cards which >will lower the TC and hopefully make you money. If as you suggest it >is more likely that the TC will go higher, that means that it is >likely that more low cards will come out, thus this is a negative >betting situation and you should wong out or bet one unit.
>
> OR
>
> the next round will be neutral, further increasing the TC, >however this is statistically impossible in the long run of +21 >shoes. Again if the next round is neutra, its a bad betting situation >so best to wong out o bet one unit.
>
> In reality with a TC of +7 we put up the money hoping that the TC >goes down.

I don't expect the "TC" to come down, just the RC (running count). (see the true count theorem references scattered around for the reasoning). This was (I believe) MathProf's point. I _expect_ the TC to remain the same. But if it is going to move, which way is it more likely to move? (see the above discussion for why it might well move up more frequently even though his sim said it would drop slightly more frequently).

He made the point about expectation vs probability. We expect the TC to remain the same, as the running count comes down but the number of remaining cards drops as well.

Remember that the TC rises, if the RC is unchanged, because the remaining decks continually drops as cards are dealt. That's the peculiar nature of counting low and high cards, and ignoring the 7/8/9 cards for hi-lo, since when ignored cards fall, the TC goes up since the RC is unaltered while the cards left drops.

Good idea to wrap duct tape around your head before thinking about this too much, of course, or your head will explode.

"Remember that the TC rises, if the RC is unchanged, because the remaining decks continually drops as cards are dealt. That's the peculiar nature of counting low and high cards, and ignoring the 7/8/9 cards for hi-lo, since when ignored cards fall, the TC goes up since the RC is unaltered while the cards left drops."

This is the most easily understood description of a complex concept that I have seen on this board in a long time.

I thought you were going to say something like "hey, you used too many danged "sinces" to be able to parse/understand the sentence".

But MathProf had great influence on my understanding this particular point, of course. Always fun to pick his brain...