Too many 10's

Does anyone have any experience of a casion dealing themselves a consistently higher than average number of 10 face up cards, giving them a great advantage?

I always feel like these shoes and shufflers are rigged. 

What's your evidence for this?

Don

I seem to remember Peter Griffin, in one of his books, mentioning that he counted the frequency of tens as the dealer's up card, and that the frequency was higher than expected. I think Griffin mostly played in Reno in the days when single deck games were common.

I would think that a hand dealt single deck would be a lot easier to deal a ten upcard than would be possible from a shoe, not to mention an automatic shuffler.

I was at a Chinese casino. They played European style. They used an automatic shuffler, then dealt from a shue. The dealer got six 10 cards first up in a row. The odds of this are a 1,178-1. Most of the session the dealer were gettin higher than expected first up 10 cards, more than the players were getting. There were 6 players.

This number is not correct. In a six-deck shoe, there are 312 cards, of which 96 cards are Tens. In the 1st round of playing, the probability of a Ten as the dealer upcard is

96/312.

In the 2nd round, the probability of a Ten as the dealer upcard is

(95-5x95/311)/(312-6),

where 5x95/311 is the number of Tens that are used in the five other cards in that round. Here we assume there is only one player at the table and each round uses six cards on average.

In the 3rd round, if we set t= 95-5x95/311, then the probability becomes

[(t-1)-5x(t-1)/(311-6)]/(312-12).

In the 4th, 5th, and 6th rounds, the probabilities have the same recursive formula.

Therefore, the probability of six Ten-upcards in a row is the multiplication of all these six terms, that is 

0.000758, or 1 in 1320

*** (I've thought about this again. It looks like Don is correct, so I made revisions. No simulation is needed here.)

Sadly, you're both wrong. The OP didn't give his methodology, but the final answer is off a little. Aceside did offer his methodology and, as usual, it is grotesquely wrong. No simulation is necessary for this. It is a straightforward, simple probability calculation. And, as a matter of principle, other people at the table, and what cards they may or may not draw, have no bearing whatsoever on the dealer's probabilities.

The correct answer is simply: 96/312 x 95/311 x 94/310 x 93/309 x 92/308 x 91/307 = 0.000759 = 1 in 1,317.

Presence of other players or your or anyone else's hit cards or original holding(s) have no bearing whatsoever on the calculation.

Don 

The minor differences in outcomes from various calculation methods are insignificant. The reality is that the probability of the dealer receiving six consecutive 10-value cards in a row is greater than 1000 to 1. To me, this strongly suggests the game is rigged. I want to understand the methods they use to achieve this.

If there are 6 players at the same table, the probability becomes 0.000750, that is, 1/1332, slightly smaller than 1/1320 I got earlier on a single-player table.

Of course, these numbers are based on my earlier assumptions. In reality, players mostly play basic strategy, so the probability number should be different. Maybe a simulation is needed to find the probability?

No simulation needed. Don is correct.

Do you think that, if you repeat the same claptrap often enough, somehow it will become true? DON'T SAY IT AGAIN! I've explained this: 1) The other players don't matter, and 2) No simulation is required, but if you'd like to make a wager as to what such a simulation would produce, you know where to find me.

Don

I simulated it and I seem to be converging around one out of 1,313 or .0007616. It counts 10s or Faces up 6 times or more in a row. It counts these occurrences across shoe boundaries. 

The data is output in the simulator in the last 3 rows of largest table on the right as you scroll down. It is right above the first graph "Amount Up or Down Every Hand".

I think the original question has more to do with is the casino doing something nefarious 

Thanks,

Phil

I haven’t tried this part, but we should not count these events across two different shoes. Also, within one shoe itself, if there are 12 Ten-upcards in a row, they should be counted as two events. Another problem with this simulation is that the value of 0.00076 is too small to tolerate errors.  To be more reliable, we can pick a larger value to sim, such as a five Ten-upcards in a row.

Across a shoe is still in a row. If it is 7 in a row there are 2 6 cards in a row. 12 in a row would be 7 6 cards in a row. 

For the first part, I’m confident a string cannot cross shoe boundaries. For the second part, different ways of counting can be represented by different math formulas, so this part becomes more complicated.  For simplicity, we can just count any 6 or more-bit string as one single string. Furthermore, as soon as a 6-bit sting appears, we give up further counting into this shoe. 

A streak can continue regardless of new shoe or not. In simulating I have converged to 1 out of 1317 for this and probability of .00075914.

Thanks,

Phil

Maybe you are right, but the argument here is whether the number of players at the table affects the dealer streak probability or not. What would you say after this much effort?

The result calculated by Don was for a new 6 deck shoe. It is also accurate for any random point with a 6 deck shoe. It does however assume that the streak will be completed within the same shoe.

This part is still problematic for me. We consider this moment when the dealer jumps in right after the mid-shoe, so the most likely situation is a remaining deck of 48 Tens in 156 cards, so the probability becomes

48/156 x 47/155 x 46/154 x 45/153 x 44/152 x 43/151 = 0.000676 = 1 / 1480.

This result does not justify what you say above.

What does the dealer do immediately after shuffling? They cut the deck at a random point. Does that somehow change the expected frequency of the cards in the first round?

Your error is in considering "undealt" cards remaining when you should be considering "unseen" cards. FWIW novice counters often make this same error when jumping into a game mid-shoe.