CarlosVillar Joined: Oct 20, 2019
• Posts: 2
October 20th, 2019 at 10:31:42 AM permalink
I would like to know what is the probability of getting a 20 in the first 2 cards in 4 and 6 decks?
Sorry for my English, I translated everything with the google translator.
dddkkk1 Joined: Oct 11, 2019
• Posts: 12
October 20th, 2019 at 1:54:05 PM permalink
10.55% for 4 decks and 10.58% for 6 decks.
7craps Joined: Jan 23, 2010
• Posts: 1977
October 20th, 2019 at 7:52:26 PM permalink
Quote: CarlosVillar

I would like to know what is the probability of getting a 20 in the first 2 cards in 4 and 6 decks?

good question
depends on the value of the cards and how many cards are in each deck
.
A standard deck of playing cards that use values from the game of Blackjack, for example,
a 20 can be a total of two cards with two-10 point value cards only

there are 16-10 value cards per deck
4 decks: 64-10 value cards, 208 total cards
6 decks: 96-10 value cards, 312 total cards

here is what I get
4 decks: 64/208 * 63/207 = 28/299 (0.093645485 about 9.36%)
using combinations: binomial(64,2)/binomial(208,2) = 28/299
6 decks: 96/312 * 95/311 = 380/4043 (0.093989612 about 9.4%)
using combinations: binomial(96,2)/binomial(312,2) = 380/4043
winsome johnny (not Win some johnny)
Ajaxx Joined: Sep 15, 2019
• Posts: 36
October 21st, 2019 at 1:21:19 AM permalink
To resolve the conflicting answers above: dddkkk1's numbers are correct. 7craps did an awesome job of showing his work but only considered the "hard 20" (a hand composed of two ten-point cards). You can also get a "soft 20", composed of an ace (counted as 11 points) and a nine. That hand can also be counted as a total value of 10, but it is considered a valid 20-point hand for the purposes of most side bets like Lucky Ladies that pay based on your first two cards. So the math is as follows:

4 decks:
• Hard 20 (drawn from among 64 ten-point cards): 64 × 63 = 4032 permutations
• Soft 20 (drawn from among 16 aces and 16 nines): 32 × 16 = 512 permutations
• Any hand (drawn from among 208 total cards): 208 × 207 = 43056 permutations
• Probability of any 20: (4032 + 512)/43056 = 284/2691 ≈ 10.5537%

6 decks:
• Hard 20 (drawn from among 96 ten-point cards): 96 × 95 = 9120 permutations
• Soft 20 (drawn from among 24 aces and 24 nines): 48 × 24 = 1152 permutations
• Any hand (drawn from among 312 total cards): 312 × 311 = 97032 permutations
• Probability of any 20: (9120 + 1152)/43056 = 428/4043 ≈ 10.5862%

If you'd like a more detailed breakdown of the numbers for 6 decks (including how often your first hand will be a suited or perfectly-matched 20, for example) you can find that in the Wizard of Odds article on Lucky Ladies.
Last edited by: Ajaxx on Oct 21, 2019
"Not only [does] God play dice... he sometimes confuses us by throwing them where they can't be seen." ~ Stephen Hawking
Ajaxx Joined: Sep 15, 2019
• Posts: 36
October 21st, 2019 at 1:25:33 AM permalink
Quote: CarlosVillar

I would like to know what is the probability of getting a 20 in the first 2 cards in 4 and 6 decks?
Sorry for my English, I translated everything with the google translator.

Carlos, mándeme un PM si prefiere usted una explicación en español.
"Not only [does] God play dice... he sometimes confuses us by throwing them where they can't be seen." ~ Stephen Hawking
7craps Joined: Jan 23, 2010
• Posts: 1977
October 21st, 2019 at 7:45:12 AM permalink
Quote: Ajaxx

7craps did an awesome job of showing his work but only considered the "hard 20" (a hand composed of two ten-point cards).

I considered it , but had NO CLUE (did NOT care, just wanted to show one could easily use combinations instead of some using permutations)
as to the complete meaning of the OP.
Great you figured it all out.
2.5 points to you.

The OP is very unclear on what is going on here, question-wise,
I DID not want to waste any more time on the OP as you did.

using combinations (as it can be easily applied to way more difficult type math questions)

4 decks only 20 total (10,10 or A,9 in any order)
(binomial(64,2) + (binomial(16,1)*binomial(16,1))) / binomial(208,2) = 284/2691
gp > (binomial(64,2) +  (binomial(16,1)*binomial(16,1))) / binomial(208,2)
%1 = 284/2691
winsome johnny (not Win some johnny)
Ajaxx Joined: Sep 15, 2019
• Posts: 36
October 21st, 2019 at 11:49:41 AM permalink
Quote: 7craps

The OP is very unclear on what is going on here, question-wise,
I DID not want to waste any more time on the OP as you did.

using combinations (as it can be easily applied to way more difficult type math questions)

4 decks only 20 total (10,10 or A,9 in any order)
(binomial(64,2) + (binomial(16,1)*binomial(16,1))) / binomial(208,2) = 284/2691

gp > (binomial(64,2) +  (binomial(16,1)*binomial(16,1))) / binomial(208,2)
%1 = 284/2691

You're right, the OP left it very vague, and I wasn't trying to suggest that your reasoning was flawed � I'm sorry it came off that way. It's just that dddkk1 didn't show his work, so at first glance yours looked like the more credible answer. I just wanted to clarify that his math is also valid but used a different starting assumption, which is why your numbers don't agree with dddkkk1's or what you'll find on professional sites like Wizard of Odds or Infarom's Probability Theory Guide and Applications if you google "What are the odds of getting 20 in blackjack?". Also, kudos for using binomial coefficients, that's an underrated approach for these kinds of problems.
"Not only [does] God play dice... he sometimes confuses us by throwing them where they can't be seen." ~ Stephen Hawking