# 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?

• I also wanted to add that having six $8$'s against a dealer's $9$ is not an envious position. :) – symmetricuser Oct 6 '14 at 21:39
• So how did the hand go? (Is it technically correct to split 8s against a dealer's 9?) – rogerl Oct 7 '14 at 3:23
• @rogerl Yup. Always split 8s! – symmetricuser Oct 7 '14 at 4:40
• In addition to any answer you might get, please understand that the probability of this happening for a given stretch of six cards is only spuriously interesting. That is, sure - the probability is very low, but what you really want to know is how likely this is to occur during some larger stretch of unknown size. You would have been just as surprised had this happened on the succeeding hand. – Christofer Ohlsson Oct 7 '14 at 8:35
• Given that about 1/3 of the deck has the value of "10" in blackjack, (bad) players splitting their hands 5 times isn't that uncommon. The pit boss was probably lying when he said he has never seen it before (it is their job to make you feel lucky). Sorry dude. But splitting a non-10 hand that many times may be as rare as the casino suggested. – DanielV Oct 9 '14 at 3:36

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.

• I think you mean $\frac{32\choose 6}{416\choose 6}\approx 1.3\times 10^{-7}$. If we allow other denominations besides 8, the number grows by a factor accordingly. Compare this with (the reciprocal of) the number of black jack games played / witnessed by the dealer and/or pit boss. – Hagen von Eitzen Oct 6 '14 at 21:26
• @HagenvonEitzen, we know one $9$ is exposed because of the dealer's up-card – symmetricuser Oct 6 '14 at 21:35
• I also said six cards, and then put $8$ in the binomial coefficients for some reason, so I edited those to $6$. – symmetricuser Oct 6 '14 at 21:37
• Great calculus, but have to take into account the 13 denominations. – Masclins May 12 '15 at 7:15

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.

• "6,942,219,827,088 ways to get just any six cards" -- only 6,842,091,656,505 because there's a card showing. "13 times more common than above" -- Actually 12.8125 (12 + 13/16) times, because there's a card showing. – Jim Balter Oct 7 '14 at 2:29

$8$ 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.

• 13 isn't quite right because it's harder to get 6 of the dealer's up-card than to get 6 of the other cards ... so it's about 1 in 589298. – Jim Balter Oct 7 '14 at 2:38
• Well if you are going to go that far, then you have to consider his downcard as well. You know that at least one of the 13 will be affected by his downcard as well. And we still haven't taken into account the rules of blackjack which influence the tendency of the dealer to reshuffle decks in certain situations. That is why "the" probability only exists in math classrooms, but for engineers and game players, there are many probabilities of progressively more accuracy. – DanielV Oct 7 '14 at 4:26
• No, you don't have to consider his downcard, because you don't know what it is, just like you don't know what the last card in the deck is. Probabilities are a function of known information. What you're trying to do is take it much further than what I wrote in an attempt to avoid its validity. The simple fact is that you took the up card into account to reduce 416 to 415, but not to reduce 13 to 12.8125, and that's simply a mistake. – Jim Balter Oct 7 '14 at 5:27

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!

• Your calculations address a different question than the one posted above. – Mico Oct 7 '14 at 0:04
• My calculations address this question, taken directly from the original post: "what the odds are in getting 6 straight cards of the same denomination from an 8 deck shoe, which holds 416 cards." – user940 Oct 7 '14 at 0:20
• This obviously doesn't answer the same question, because the OP is interested in the odds of drawing 6 of the same card, not the odds that there are 6 of the same card in a row somewhere in the shoe. – Jim Balter Oct 7 '14 at 2:42
• @JimBalter Are you saying that in blackjack you are always dealt the top cards off the shoe? I don't play blackjack, so I don't know. I thought that you sometimes be dealt cards from the middle of the shoe, so to speak. – user940 Oct 7 '14 at 2:58
• It doesn't matter where in the shoe the deal comes from, as long as you haven't seen any of the other cards ... probabilities depend on knowledge. In any case, a run elsewhere in the shoe could go to another player, or get split between players, or never get dealt ... it obviously isn't what the OP is asking about. The OP asked " odds are in getting 6 straight cards" -- "in getting" means being dealt 6 such cards in the manner that happened to the OP, not just having those cards in the shoe somewhere. – Jim Balter Oct 7 '14 at 3:03

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.

$$13\times \frac{\binom{32}{6}}{\binom{415}{6}} \approx 1.72\times 10^{-6},$$