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?
[Math] Incredible Blackjack Hand
combinatoricsprobability
Related Solutions
This is for the one player game and a initial hand of five cards. I will consider that you have a hand with $k$ cards if after throwing away all groups of four cards of the same value, you have $k$ cards in your hand. For instance, if your initial hand is $\{2,3,3,3,3\}$ I will consider it a one card hand, not a five card hand.
Let $P(k)$ be the probability that at some moment in the game you have a $k$ card hand. The following values are easy: $P(0)=1$ (at the end of the game you have no cards in your hand) and $P(40)=P(41)=\dots=P(52)=0$ (if you have $40$ cards, there must be a group of four cards of the same value). A little thought gives $$ P(39)=\frac{4^{13}}{\dbinom{52}{13}}=0.000105681 $$
For the rest of the values I have run a simulation in Mathematica of $10^7$ games. These are the results:
k H(k)
0 483
1 25839
2 596131
3 10000000
4 230004
5 10000000
6 10000000
7 10000000
8 10000000
9 10000000
10 10000000
11 10000000
12 10000000
13 10000000
14 10000000
14 9999996
16 9999945
17 9999720
18 9998600
19 9994328
20 9980514
21 9942513
22 9854106
23 9675123
24 9348731
25 8822787
26 8068444
27 7080395
28 5919525
29 4675434
30 3458156
31 2381202
32 1514558
33 876208
34 458693
35 213203
36 84852
37 28160
38 7216
39 1067
For each $k$, $H(k)$ is the number of games in which a hand of $k$ cards has been held before reaching the end of the deck. Observe that the value of $H(39)$ is in accordance with the exact value of $P(39)$. A graph of the results:
It is surprising (at least to me) that for certain values of $k$, like $k=5$, a hand of $k$ cards was held in all $10^7$ games, even if a deck like
1,1,1,1,2,2,2,2,3,3,3,3,...
will give only hands of $1$, $2$, $3$ and $4$ cards.
I don't know if this counts as an elegant solution in your book, but I think it's cute.
Let's say the "frequency state" of a deck is the number of cards of each face value remaining. A full deck, for example, has the frequency state "4 aces, 4 twos, 4 threes...," while an empty deck has the frequency state "0 aces, 0 twos, 0 threes...." There are $5^{13}$ possible frequency states. When you draw a card from a deck, the frequency state changes in a way that depends only on the face value of the card you drew.
You can turn the set of possible frequency states into a directed graph by drawing an arrow for each way the frequency state can change when you draw a card. I'll call this graph the "draw graph." Each vertex in the draw graph has at most 13 edges leaving it, one for each type of card you could draw.
You can turn the draw graph into a weighted directed graph by assuming that cards are drawn uniformly at random, and weighting each arrow by the probability of that kind of draw. The full-deck state, for example, has 13 arrows leaving it, each with weight $\tfrac{1}{13}$. If you draw a queen, you end up in a state that still has 13 arrows leaving it, but 12 of them have weight $\tfrac{4}{51}$, and one—the arrow for drawing another queen—has weight $\tfrac{3}{51}$. The weights of the arrows leaving each state add up to one, so the draw graph is a Markov chain.
Let's say the draw has been passed to the dealer. We know the dealer's hand and the frequency state of the deck. Here's a cool fact: from now on, we can figure out the dealer's hand just by looking at the frequency state of the deck. That's because, when the dealer starts hitting, all the cards she draws from the deck end up in her hand. Using this fact, we can translate properties of the dealer's hand, like whether it's bust, into properties of the frequency state.
Let's record this information on the draw graph by labeling its vetrtices. We'll label the states where the dealer stays as "stay states," the states where the dealer is bust as "bust states," and the states where the dealer keeps hitting as "hit states." When the dealer is hitting, she's walking randomly along the draw graph, with her direction from each state chosen using the arrow weights as probabilities. She keeps walking until she reaches a stay state or a bust state.
The dealer has to eventually stay or go bust, so the process we just described is an absorbing Markov chain. Like most things in graph theory, absorbing Markov chains have a very pretty linear algebraic description in terms of the adjacency map of the transition graph. If you know how this works, I can describe very quickly how to calculate the bust probability.
Cribbing from Wikipedia, let $Q$ be the map describing transitions from hit states to hit states, and let $R$ be the map describing transitions from hit states to stay and bust states. Let $\pi$ be the vector corresponding to the initial state, and let $\Pi$ be the projection map onto the subspace spanned by the bust states. The bust probability is the sum of the entries of the vector $$\Pi R (1 - Q)^{-1}\pi.$$ The $(1 - Q)^{-1}$ factor describes the part of the process where the dealer hits over and over, so I think it encapsulates the tricky "recursive bit" of the computation.
(Caution: my operators are the transposes of Wikipedia's, which is why the formula looks backwards.)
I think this question is related to a broader question that I ask myself all the time in research: what does it mean to have a "nice solution" to a problem? When I was young, I was taught that a "nice solution" is a formula for the thing you want to calculate, but that's not always true! Having a forumla for something often tells you very little about it, and other descriptions are often much more useful from a practical point of view.
I'm not sure whether the description of the bust probability given above is much use, but for this problem, I suspect that a linear algebraic description of this kind will be more useful than a formula.
Best Answer
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.