[Math] Probability of drawing certain cards in Blackjack

probability

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?

Best Answer

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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[Math] Probability of winning blackjack dice game

You have not actually asked a question. I suppose you want to know what the player's expected loss is.

Nor have you said how the bank adjusts to the player's behaviour. For example, what happens if the player stays/sticks at 20 and the bank reaches 19?

Let's assume the bank is also required to follow the same rule as you give for the player, but if they both bust by going over 21 then the bank wins.

Then the probability the player busts is about $0.286205909$ (not that far away from $\frac27$, the limit of probability that the player if throwing indefinitely hits a particular large number). Similarly for the bank. So the probability they both bust is this squared, i.e. about $0.081913822$ (not far away from $\frac{4}{49}$). This is the only unfair part of the game so it is the player's expected loss if she has staked $1$.

To calculate the probability of a bust, try $p_0=1$, $p_n=\frac16 \sum_{i=0}^{i=n-1} p_i$ for $1 \le i \le 6$, $p_n=\frac16 \sum_{i=n-6}^{i=n-1} p_i$ for $6 \le i \le 19$, $p_n=\frac16 \sum_{i=n-6}^{i=18} p_i$ for $19 \le i \le 24$, and then add up $p_{22}+p_{23}+p_{24}$. The values of $p_n$ are about

n   p_n
==  ===========
0   1
1   0.166666667
2   0.194444444
3   0.226851852
4   0.264660494
5   0.308770576
6   0.360232339
7   0.253604395
8   0.268094017
9   0.280368945
10  0.289288461
11  0.293393122
12  0.290830213
13  0.279263192
14  0.283539659
15  0.286113932
16  0.287071430
17  0.286701925
18  0.285586725
19  0.284712810
20  0.238168945
21  0.190912335
22  0.143226680
23  0.095381442
24  0.047597788

[Math] Blackjack Probability

This is my second try, but I think is a MUCH better argument than what I had before. So I deleted the more complicated answer I tried to give earlier.

The total # of possible deals to the two players is: $$\binom{52}{2}\binom{50}{2}=1,624,350$$ (i.e., choose two cards for the first player and then two for the second.)

Then, the # of ways to deal a blackjack to both players would be: $$\binom{4}{1}\binom{16}{1}\binom{3}{1}\binom{15}{1}=2880$$ (i.e, choose which ace to give to the first player, then which 10,J,Q,K for the first player, then which ace for the second player, and then which 10,J,Q,K for the second player.)

Also, the # of ways that exactly one player gets a blackjack would be: $$\binom{2}{1}\binom{4}{1}\binom{16}{1}\times(\binom{3}{2}+\binom{3}{1}\binom{32}{1}+\binom{47}{2})=151,040$$ (i.e., choose which player to give the blackjack, then choose which ace to give them, and then which 10,J,Q,K to give them. Then for the other player we either give them two aces from the 3 that are left, give them one ace from the 3 that are left and 1 of the 32 2-9 cards, or give them two non-aces from the 47 non-aces that are left.)

So, the probability of at least one player getting a blackjack is: $$\frac{2880+151,040}{1,624,350}$$

Hence the probability of neither player getting a blackjack is: $$1- \frac{2880+151,040}{1,624,350} \approx 90.5\%$$

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