Incredible Blackjack Hand Last Saturday night I played at Bally's in Atlantic City and got a hand I could not believe. Dealer had 9 and I was dealt 2 8s. I split the 8s and was given a third card. It was an 8 so I split them again. The next card I was dealt was a fourth 8. This has happened to me three other times in my life, so no big deal. The fifth card was again an 8 and the sixth consecutive 8 followed. No one at the table or the dealer or even the pit boss had ever seen that before. I do not even know how to start calculating what the odds are in getting 6 straight cards of the same denomination from an 8 deck shoe, which holds 416 cards. Can you help me?
 A: In 8 decks there are 32 8s.  To pick up six of them there are $\binom{32}{6}=906,192$ possible ways for that to happen.  There are, similarly, $\binom{416}{6}=6,942,219,827,088$ ways to get just any six cards.  Dividing these, that's about a 1 in 7.6 million chance for this to happen.  This is about 1/12 as likely as drawing a royal flush on five cards in a single deck.
Of course, you asked about six of any card, which is considerably easier (13 times more common than above), somewhat more likely than a royal flush.
A: $8$ decks gives a total of $52 \times 8 = 416$ cards.
$8$ decks with 4 cards each as an Eight gives a total of $32$ possible Eights to draw.
So a probability of drawing each Eight in sequence is:
$$\underbrace{\frac{32}{416} \times \frac{31}{415} \times \dots \frac{28}{412} \times \frac{27}{411}}_{\text{6 draws}}$$
If you take into account that the dealer doesn't draw an Eight, then you have $415$ cards to choose from, so a more accurate probability is:
$$\frac{32}{415} \times \frac{31}{414} \times \dots \frac{28}{411} \times \frac{27}{410}$$
Since there are 13 possible cards that can be drawn in sequence, the answer to 

what the odds are in getting 6 straight cards of the same denomination from an 8 deck shoe

is 
$$13 \times \frac{32}{415} \times \frac{31}{414} \times \dots \frac{28}{411} \times \frac{27}{410}$$
which is about $1$ in every $580,798$ attempts.  Keep in mind that if you split all the Tens out of the deck, any other competent player will be angry and leave the table.
A: We can assume that the dealer's hole card is irrelevant. So we are looking for the probability that the first six cards are $8$ out of the $415$ cards left in the deck (excluding the dealer's up-card $9$). Among the $415$ cards, there are $32$ $8$'s. There are $\binom{32}{6}$ ways of choosing six $8$'s, and there are a total of $\binom{415}{6}$ ways of choosing the first six cards. Hence, the probability is
$$\frac{\binom{32}{6}}{\binom{415}{6}} \approx 1.32\times 10^{-7},$$
which is very very small.
A: Clarification: I have interpreted the phrase "what are the odds in getting 6 straight cards of the same denomination from an 8 deck shoe"
 as asking for the chance that there are  6 straight cards of the same denomination from 
an 8 deck shoe. My answer addresses this interpretation only. 

Ignoring the details of the game, I'll just consider the 
chance that  a  well-shuffled 8-deck shoe has  6 or more  eights in a row
somewhere.  I solved a similar problem here.
For your problem,  put $b=384$ and $w=32$ in my answer above to 
arrive at 
\begin{eqnarray*}
\mathbb{P}(\mbox{at least 6 eights in a row})
&=&{378917534435104330038751954618647 
\over 7539892080833060495675366062952229323}\\[5pt]
&=&0.000050255,\end{eqnarray*}
 or about 1 in 20,000. 
If you ask for the probability of at least 6 in 
a row of any of the 13 possible values, an approximate answer is to 
multiply the above by 13, giving $P\approx .0006533$
or about 1 in 1350. Not a common occurrence, but 
 not that rare! 
A: Calculations from symmetricuser were all perfect when trying to calculate the probability of taking, for example, 6 8's in a row. Even though you ask for 6 cards in a row, of any denomination, so you are not forcing the number. Therefore, you have to multiply that for all the 13 possible denominations.
$$13\times \frac{\binom{32}{6}}{\binom{415}{6}} \approx 1.72\times 10^{-6},$$
A: I was in s black jack game at Harrah's Cherokee in February
I received a pair of 5s against a dealer's 6. I split and got 2 more 5s. Now ihad 4 hands and a pair of 5s on each. One hit on each hand resulted in 4 more 5s. I had doubled down on all 4 so i could only get 1 card. The dealer turned up a hard 16 and the next card out of the shoe was another 5 for s 21. I lost only 400.00 but, i think the electronic card shuffling can align cards to the casino advantage as the odds of 13 5s consecutively have to be astronomical. A true story witnessed by 4 players the dealer and the pit boss. Can someone calculate the odds?
