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Sorry for my English, I translated everything with the google translator.

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

depends on the value of the cards and how many cards are in each deck

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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

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.

Quote:CarlosVillarI 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.

I considered it , but had NO CLUE (did NOT care, just wanted to show one could easily use combinations instead of some using permutations)Quote:Ajaxx7craps did an awesome job of showing his work but only considered the "hard 20" (a hand composed of two ten-point cards).

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

Quote:7crapsThe 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) = 284/2691gp > (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.