Blackjack Probability Worksheet
- 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.
- A game of Blackjack is played. Suppose you are the first player dealt with a new pack of cards. Quiz & Worksheet Goals. This quiz will test your ability to. Find probability of getting two.
- Blackjack Probability Math
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- Blackjack Probability Calculator
- Blackjack Probability Worksheet
- Blackjack Probability Theory
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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.
The Wizard's spreadsheet file consists of several worksheet tabs. Clicking a tab displays the sheet for that tab. Each sheet covers one aspect of the blackjack strategy analysis, summarized as follows: dealer: This sheet shows the probability of each possible final outcome (17 through 21 or bust), given the dealer's current total. Twenty problems in probability This section is a selection of famous probability puzzles, job interview questions (most high-tech companies ask their applicants math questions) and math competition problems. Some problems are easy, some are very hard, but each is.
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.Blackjack Probability Math
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.
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