I'm trying to calculate the probability that a certain scenario will be dealt in Blackjack (known as "21" to some), after a very random shuffle of a pack of 52 cards. Let me describe the "rules" that should be kept in mind for this question:
- There are 6 players and one dealer.
- The dealer deals each of the six players one card, and then themselves one card. A second card is then dealt to each player and the dealer.
- If the dealer has blackjack at this point (which is composed of a face card or a 10, paired with an ace), the cards are collected, and another hand is dealt the same way from the remainder of the 52 card pack.
- The cards showing for the players in this situation do not matter (there is no insurance, etc... Just the very basics as described above)
The specific question is "What is the probability that after a single completely random shuffle, the cards dealt will end with the dealer seeing blackjack three times in a row (as described above)?"
Now I'll describe how I worked out the odds myself. I may (likely) have made a mistake. Can someone confirm my thinking or correct my conclusion? I'm also interested hearing about other methods to determine the answer.
It seems to me that it's important that the A10 pairs be located in specific locations. Specifically, an ace and a 10 must be in positions 7 and 14 from the top of the shuffled deck, as well as a pair in positions 21/28, and 35/42. As long as this constraint holds true, the result will be as described above. In other words, the result (whether the "3 blackjacks" would occur) is determined immediately after the cards are shuffled. Once a shuffle is performed, these positions could be observed, and if we see three pairs of A/10, the dealer will win three times.
Therefore, I'll just pretend that instead of the pairs being at positions 7/14, 21/28, and 35/42, I could just change those positions to be 1/2, 3/4, and 5/6, without changing the odds.
Since each "pair" has two states (the ace first followed by the 10, or vice-versa), and there are three of them, there are 8 possible deck configurations that would satisfy the constraint (2^3). There are:
- ATATAT (ace-ten, ace-ten, ace-ten)
- ATATTA (ace-ten, ace-ten, ten-ace)
- etc.
For each of the 8 cases, I computed the probability that the particular case could appear.
For example, let's use the ATATAT case. The first card dealt must be an ace, and there are 4 of them in the 52 cards. The second card must be a 10, and there are 16 of these out of the remaining 51 cards in the pack. The following cards are 3/50 aces, 15/49 tens, 2/48 aces, and finally 14/47 for the final 10.
4/52 * 16/51 * 3/50 * 15/49 * 2/48 * 14/47 = 5.501382222e-6
The remaining 7 cases calculate out to the exact same value on my calculator (this was somewhat surprising to me).
I then multiplied that probability above by 8, to account for the 8 different cases that could cause the dealer to win three times. This value is 0.000033 Invert this and the number is 30295.
Therefore, I conclude that the probability that this will happen is once in every 30295 shuffles (approximately).
I agree with your analysis and your answer. An alternative way of computing the probability is:
Let $~E_1~$ denote the event that there are $~3~$ Aces and $~3~$ 10-King cards in the first 6 cards.
Let $~E_2~$ denote the event that (assuming that event $~E_1~$ has occurred), there is exactly one Ace in each of cards 1-2, 3-4, 5-6.
Then, the desired probability is $~p(E_1) \times p(E_2).~$
$$p(E_1) = \frac{\binom{4}{3} \times \binom{16}{3}}{\binom{52}{6}}.$$
To compute $~E_2~$ note that there are $~\displaystyle \binom{6}{3} = 20~$ ways of selecting the $~3~$ cards out of $~6~$ that will be the Aces. Further, of these $~20~$ ways, exactly $~2^3 = 8~$ of them are satisfactory (i.e. 1-3-5, 1-3-6, 1-4-5, 1-4-6, ...).
Therefore,
$$p(E_2) = \frac{8}{\binom{6}{3}} = \frac{8 \times 3! \times 3!}{6!}.$$
Then
$$p(E_1) \times p(E_2) = \frac{4! \times 16!}{3! \times 1! \times 13! \times 3!} \times \frac{46! \times 6!}{52!} \times \frac{8 \times 3! \times 3!}{6!}$$
$$= \frac{4 \times 3 \times 2 \times 16 \times 15 \times 14 \times 8}{52 \times 51 \times 50 \times 49 \times 48 \times 47}.$$