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I'm interested in calculating the chance of receiving certain cards given the player's hand and the dealer's up-card after the dealer has "peeked" for Blackjack.

Typically, if the dealer shows an Ace or 10-value card (10,J,Q,K) the dealer will "peek" at their second card and check for Blackjack. If Blackjack is discovered all players (except those with Blackjack also) will lose immediately and the next round is dealt - this speeds up the game and thus speeds up casino profits.

If the dealer does not peek for Blackjack this calculation is easy:

For example, suppose the dealer shows a 4 and the player shows 2,3. Then the probability our next card will be a 5 is simply 4/49 (4 5's left in 49 remaining cards). The dealer's down-card is equally likely to be any of the 49 remaining cards and thus shouldn't affect the probability at all, we simply ignore it.

However, if the dealer shows a 10 and the player shows 2,3 and the dealer doesn't have Blackjack things are different. Intuitively, because we know the dealer doesn't have an Ace, we should be more likely to receive an Ace (or less likely to receive a non-Ace) - but I'm unsure how to calculate this exactly.

Any help?

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    $\begingroup$ I'm not quite sure what exactly is going on here, but it seems to be a problem that can be dealt with via conditional probabilities. $\endgroup$ – Fimpellizieri Jun 7 '17 at 2:35
  • $\begingroup$ BTW. In many casinos Backjack is played not with the standard 52-card deck but 6 such decks shuffled together, dealt from a holder called a shoe, because of its shape. $\endgroup$ – DanielWainfleet Jun 7 '17 at 2:43
  • $\begingroup$ By focusing so much on the face up cards, you are making the problem seem harder than it is. Ignore the irrelevant stuff. A good starting place with this kind of problem is "can you solve it if there are only 2 cards in the deck?" "If there are only 3 cards?" $\endgroup$ – DanielV Jun 7 '17 at 2:45
  • $\begingroup$ What exactly are you trying to calculate? You mention the probability of an Ace, the probability of a non-Ace, but don't say anything specific. Also, as noted many casinos use many decks for blackjack to make card counting and rich decks less common. What assumption would you like to make regarding the number of decks? $\endgroup$ – Ross Millikan Jun 7 '17 at 2:53
  • $\begingroup$ @DanielWainfleet that's true, i'm assuming a single deck shoe for this problem $\endgroup$ – Abovestand Jun 7 '17 at 3:07
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However, if the dealer shows a 10 and the player shows 2,3 and the dealer doesn't have Blackjack things are different. Intuitively, because we know the dealer doesn't have an Ace, we should be more likely to receive an Ace (or less likely to receive a non-Ace) - but I'm unsure how to calculate this exactly.

You know that four cards are not in the deck 10,2,3, and another non-ace, leaving 48 cards in the deck.

The four aces may be anyhere in the deck with equal probability, so the probability the next card drawn is an ace is $4/48$.

$$\mathsf P(X=A\mid P_1=2, P_2=3, D_1=10, D_2\neq A) = \tfrac 4{48}$$

For other cards we must also consider the probability that it is the dealer's second card.   One of the kings, say, will be the dealer's second card with a probability of $4/(52-3-4)$. On condition that one is, the probability that one from the other three kings is the next card is $3/48$, and on condition that the dealer's second card also ins't a king, the probability that the next card is one from the four kings is is: $4/48$.   So by total probability, for any from the nine such cards that are not showing (or an ace):

$$\mathsf P(X=K\mid P_1=2, P_2=3, D_1=10, D_2\neq A) ~{ = ~\tfrac 4{45}\tfrac 3{48}+\tfrac{41}{45}\tfrac 4{48} \\[1ex]=~ \dfrac{11}{135}}$$

Similarly, for cards that are showing, such as 2,3,10:

$$\mathsf P(X=2\mid P_1=2, P_2=3, D_1=10, D_2\neq A) ~{ = ~\tfrac 3{45}\tfrac 2{48}+\tfrac{42}{45}\tfrac 3{48} \\[1ex]=~ \dfrac{11}{180}}$$

You can now work out probabilities if the player and dealer are showing two or three cads from the same kind.

Reality Check: $3\cdot\tfrac{11}{180}+9\cdot\tfrac{11}{135}+\tfrac 1{12} = 1$ as it ought right should.

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For the question in the comment: dealer has $10$ plus non-ace, player has $2,3$ what is the chance the next card is $5$. The dealer's other card has $\frac 4{45}$ chance to be a $5$, so the average number of $5$s left in the deck is $\frac {176}{45}$ There are $48$ cards left, so the chance the next is a $5$ is $\frac {176}{45 \cdot 48}=\frac {11}{135}$

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