Blackjack Probability Worksheet
Posted : admin On 7/19/2022
We first present the probabilities attached to card dealing and initial predictions. In making this calculus, circumstantial information such as fraudulent dealing is not taken into account (as in all situations corresponding to card games). All probabilities are calculated for cases using one or two decks of cards. Let us look at the probabilities for a favorable initial hand (the first two cards dealt) to be achieved. The total number of possible combinations for each of the two cards is C(52, 2) = 1326, for the 1-deck game and C(104, 2)=5356for the 2-deck game.
Probability of obtaining a natural blackjack isP= 8/663 = 1.20663% in the case of a 1-deck game andP = 16/1339 = 1.19492%in the case of a 2-deck game.
The probability of getting the face card or Ten as card 1 is 16/52 and then the probability of getting the Ace as card 2 is 4/51. Multiplying them, we get 64/2652 which simplifies to 16/663. Once you have determined the probability of your event, map it out on a probability line with 0 being 'impossible' and 10 being 'certain'. For kids wanting to learn more about how to present data in a visual way, check out our graphing and data worksheets.
Probability of obtaining a blackjack from the first two cards isP= 32/663 = 4.82654%in the case of a 1-deck game andP = 64/1339= 4.77968%in the case of a 2-deck game.
Similarly, we can calculate the following probabilities:
Probability of obtaining 20 points from the first two cards isP = 68/663 = 10.25641% in the case of a 1-deck game andP= 140/1339 = 10.45556%in the case of a 2-deck game.
Probability of obtaining 19 points from the first two cards isP = 40/663 = 6.03318%in the case of a 1-deck game andP = 80/1339 = 5.97460%in the case of a 2-deck game.
Find blackjack probability lesson plans and teaching resources. Quickly find that inspire student learning. Blackjack Probability Lesson Plans & Worksheets Reviewed by Teachers. Blackjack and Probability Chongwu Ruan Math 190S-Hubert Bray July 24, 2017 1 Introduction Blackjack is an usual game in gambling house and to beat the dealer and make money, people have done lots of research on it. They come up with several basic strategy which is consist of three tables corresponding to the different rules.
Probability of obtaining 18 points from the first two cards isP= 43/663 = 6.48567%in the case of a 1-deck game andP= 87/1339 = 6.4973% in the case of a 2-deck game.
Probability of getting 17 points from the first two cards isP= 16/221 = 7.23981%in the case of a 1-deck game andP= 96/1339 = 7.16952%in the case of a 2-deck game.
A good initial hand (which you can stay with) could be a blackjack or a hand of 20, 19 or 18 points. The probability of obtaining such a hand is calculated by totaling the corresponding probabilities calculated above: P = 32/663 + 68/663 + 40/663 + 43/663 = 183/663, in the case of a 1-deck game and P = 64/1339 + 140/1339 + 80/1339 + 87/1339 = 371/1339, in the case of a 2-deck game.
Probability of obtaining a good initial hand isP= 183/663 = 27.60180%in the case of a 1-deck game andP= 371/1339 = 27.70724%in the case of a 2-deck game.
The probabilities of events predicted during the game are calculated on the basis of the played cards (the cards showing) from a certain moment. This requires counting certain favorable cards showing for the dealer and for the other players, as well as in your own hand. Any blackjack strategy is based on counting the cards played. Unlike a baccarat game, where a maximum of three cards are played for each player, at blackjack many cards could be played at a certain moment, especially when many players are at the table. Thus, both following and memorizing certain cards require some ability and prior training on the player’s part. Card counting techniques cannot however be applied in online blackjack.
The formula of probability for obtaining a certain favorable value is similar to that for baccarat and depends on the number of decks of cards used. If we denote by x a favorable value, by nx the number of cards showing with the value x (from your hand, the hands of the other players and the face up card in the dealer’s hand) and by nv the total number of cards showing, then the probability of the next card from the deck (the one you receive if you ask for an additional card) having the value x is:
This formula holds for the case of a 1-deck game. In the case of a 2-deck game, the probability is:
Generally speaking, if playing with m decks, the probability of obtaining a card with the value x is:
Example of application of the formula: Assume play with one deck, you are the only player at table, you hold Q, 2, 4, A (total value 17) and the face up card of the dealer is a 4. Let us calculate the probability of achieving 21 points (receiving a 4).
We havenx = 2, nv = 5, so:
.
For the probability of achieving 20 points (receiving a 3), we havenx = 0, nv = 5, so:
.
For the probability of achieving 19 points (receiving a 2), we havenx = 1, nv = 5, so:
.
If we want to calculate the probability of achieving 19, 20 or 21 points, all we must do is total the three probabilities just calculated. We obtainP = 9/47 = 19.14893%.
Unlike in baccarat, where fewer cards are played, the number of players is constant (two), and the number of gaming situations is very limited, in blackjack, the number of possible playing configurations is in the thousands and, as a practical matter, cannot be entirely covered by tables of values.
Sources |
A big part of the gaming situations that require a decision, where the total value held is 15, 16, 17, 18, 19 or 20 points, is comprised in tables in the section titled Blackjack of the book PROBABILITY GUIDE TO GAMBLING: The Mathematics of Dice, Slots, Roulette, Baccarat, Blackjack, Poker, Lottery and Sport Bets.You will also find there other issues of probability-based blackjack strategy . See the Books section for details. |
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What are the chances of getting 4 blackjacks?
A single deck is used.
We get to play on 7 hands per game.
Blackjack Probability To Win
I'd encourage you to 'sound this out' and try to solve it yourself, even though I'm providing the info below. Problems like this sound really complicated, but really aren't when you 'sound it out' and take the probably one piece at a time.
In statistics when you need multiple events to happen (frequently when you use the word 'AND') the resulting probability is multiplicative... When any one of multiple events could happen (frequently when you use the word 'OR') the resulting probability is additive. Thus:
P(4 blackjacks) = P(1st blackjack) * P(2nd blackjack given 1st) * P(3rd blackjack given 1st 2) * P(4th blackjack given 1st 3)
P(1st blackjack) = P(1st ace BJ) OR P(1st 10 BJ) = P(1st ace 2nd 10) + P(1st 10 2nd ace) = [(4/52)*(16/51)] + [(16/52)*(4/51)] = (.0769*.3137) + (.3077*.0784) = .0241 + .0241 = .0482... or ~4.8% chance of getting dealt a blackjack (~1 in 21 hands, as shown previously by the Wiz).
P(2nd blackjack given 1st) = (same as above just remove one ace, 10, and 2 total cards from the deck)... [(3/50)*(15/49)]*2 = (.06*.3061)*2 = .0368, or ~3.7%
P(3rd blackjack given 1st 2) = (same as above with more removals)... [(2/48)*(14/47)]*2 = .0248, or ~2.5%
P(4th blackjack given 1st 3) = (same as above with more removals)... [(1/46)*(13/45)]*2 = .0126, or ~ 1.3%
Thus, P(4 blackjacks) = .0482 * .0368 * .0248 * .0126 = .0000005543, or ~ .00005543%... which is ~1 in 1.9 million.
I'd expect the payout to be a million bucks, and then they'd still be shorting you. So more than likely a terrible bet.
128/2652 * Number of hands = 128/2652 * 7 = 896/2652 ~ 1/3
Did you take into account that any of the 7 hands can make 4 blackjacks?
There are (7)C(4) = 35 groups of 4 players that can have the blackjacks
The first player can have any of 4 aces and any of 16 10-cards, or 64 possible hands
The second player can have any of the 3 remaining aces and any of the 15 remaining 10-cards, or 45 hands
The third player can have either of the 2 remaining aces and any of the 14 remaining 10-cards, or 28 hands
The fourth player can have the remaining ace and any of the 13 remaining 10-cards, or 13 hands
The first of the other three players can have any of the (44)C(2) remaining hands, the second any of the (42)C(2) remaining hands, and the third any of the (40)C(2) remaining hands.
Divide this product by (52)C(2) x (50)C(2) x (48)C(2) x (46)C(2) x (44)C(2) x (42)C(2) x (40)C(2), and you get about 1 / 51,685.
Simulation seems to confirm this calculation.
2 * (16/52 * 4/51) = 128/2652
128/2652 * Number of hands = 128/2652 * 7 = 896/2652 ~ 1/3
Did you take into account that any of the 7 hands can make 4 blackjacks?
I think he missed the part where it said they could play up to 7 hands.
If there are only 4 hands, the probability is about 1 / 1,808,900
If there are 5, about 1 / 361,800
If there are 6, about 1 / 120,600
6 x 2 x 15/50 x 3/49 = 540/2450
5 x 2 x 14/48 x 2/47 = 280/2256
4 x 2 x 13/46 x 1/45 = 104/2070
896/2652 x 540/2450 x 280/2256 x 104/2070 = 1/2153
I'd like to know if I am wrong please.
Thank you for your answers
...Did you take into account that any of the 7 hands can make 4 blackjacks?
No, I simply 4 hands in a row getting blackjack, without replacement.7 x 2 x 16/52 x 4/51 = 896/2652
6 x 2 x 15/50 x 3/49 = 540/2450
5 x 2 x 14/48 x 2/47 = 280/2256
4 x 2 x 13/46 x 1/45 = 104/2070
896/2652 x 540/2450 x 280/2256 x 104/2070 = 1/2153
I'd like to know if I am wrong please.
Thank you for your answers
You are counting every deal 24 times.
You appear to be saying, 'Any of the 7 players can have any of the four Aces, and for each one, any of the other 6 players can have any of the three remaining Aces,' but you are counting each hand where, for example, Player A has the Ace of Spades and Player B has the Ace of Hearts twice.
Any of the 7 players can have the Ace of Spades, but you should then be multiplying it by 1/52 instead of 4/52. Similarly with the Aces of Hearts, Clubs, and Diamonds.
Blackjack Probability Theory
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