March 20th, 2017 at 1:18:14 PM
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That math is intense! I'm very impressed.

The way I described that hand was really complicated. I think this is a better summary (certainly shorter and clearer):

I am seeking what the odds are of a very, very unusual blackjack hand!!! Ready for this scenario??

A six deck blackjack table. (standard blackjack.)

A player is dealt two aces (A1 & A2). She splits.

A1 is dealt an ace (A3) - so she splits again.

A1 is dealt an ace (A4) - so she splits again.

Review: there are now four aces in front of the player.

A1-A4 are all dealt a tens (3 face cards, 1 ten).

Review: there are now four blackjacks in front of the player.

What are the odds of this happening?

The way I described that hand was really complicated. I think this is a better summary (certainly shorter and clearer):

I am seeking what the odds are of a very, very unusual blackjack hand!!! Ready for this scenario??

A six deck blackjack table. (standard blackjack.)

A player is dealt two aces (A1 & A2). She splits.

A1 is dealt an ace (A3) - so she splits again.

A1 is dealt an ace (A4) - so she splits again.

Review: there are now four aces in front of the player.

A1-A4 are all dealt a tens (3 face cards, 1 ten).

Review: there are now four blackjacks in front of the player.

What are the odds of this happening?

March 20th, 2017 at 1:28:01 PM
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Quote:teliot...You may be asking more generally about the question, "how often do Aces split to 4 hands and make blackjack on each hand," but I have no idea if that's what you are really asking, so I will just answer the question you asked (with an assumption about Tens and Face Cards).

For teliot's other version of the question, there are five different patterns for the order in which the player's eight cards could come out of the shoe:

AAAATTTT

AAATATTT

AAATTATT

AATAATTT

AATATATT

So, the answer to this second question is 4,034,212/5-to-1 or 806,842-to-1.

March 20th, 2017 at 1:36:43 PM
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The math will be the same as before. I'm assuming you're interested in knowing what the odds of getting 4 aces, split, and getting 4 blackjacks? If so, your 3 face cards and 1 ten make it missleading.

If you do indeed mean misleading, then simply update RS's equation above so instead of searching for "any 10 valued card" it's literally any "10" left in the deck.

Simple Explanation of Simple Statistics

When you ask what are the odds of "this and this and this and this..." etc... the key word here is "and" which means the probability will be multiplicative. P(A)*P(B)*...etc.

When you ask what are the odds of "this or that or that or that..." etc... the key word here is "or" which means the probability will be additive. P(A) + P(B) +... etc.

Your Example

Thus, take them in order that they have to happen...

P(first 2 cards ace) = P(first card ace AND second card ace) = P(1st Ace) * P(2nd Ace) = (24/312) * (23/311)

AND

P(next card 3rd ace) = (22/310)

AND

P(next card 4th ace) = (21/309)

AND

P(next card 1st face) = (96/308)

AND

P(next card 2nd face) = (95/307)

AND

P(next card 3rd face) = (94/306)

AND

P(next card 4th face) = (93/305)

Remember: you said "AND THIS AND THIS AND THIS" so they're all multiplied together. That's where the

(24*23*22*21*96*95*94*93) / (312*311*310*309*308*307*306*305)

comes from... If you wanted to know 3 face values and the last one an actual "10" then you would change the 96, 95, 94, and 93 values to their respective values (how many KQJ's in 6 decks, then how many "10's" in 6 decks).

So if you actually multiply and divide those numbers out... you get: 20,332,308,648,960 / 82,024,863,079,573,814,400 = .00000025. If you just wanted to use ONLY the simple calculator on your computer you could do this, then just guess and check 1/x and see what comes up with 6 zeros and 25... hint: since they already solved it start with something like 4,000,000 ;-)... thus, a 1 in ~4,000,000 chance.

If you do indeed mean misleading, then simply update RS's equation above so instead of searching for "any 10 valued card" it's literally any "10" left in the deck.

Simple Explanation of Simple Statistics

When you ask what are the odds of "this and this and this and this..." etc... the key word here is "and" which means the probability will be multiplicative. P(A)*P(B)*...etc.

When you ask what are the odds of "this or that or that or that..." etc... the key word here is "or" which means the probability will be additive. P(A) + P(B) +... etc.

Your Example

Thus, take them in order that they have to happen...

P(first 2 cards ace) = P(first card ace AND second card ace) = P(1st Ace) * P(2nd Ace) = (24/312) * (23/311)

AND

P(next card 3rd ace) = (22/310)

AND

P(next card 4th ace) = (21/309)

AND

P(next card 1st face) = (96/308)

AND

P(next card 2nd face) = (95/307)

AND

P(next card 3rd face) = (94/306)

AND

P(next card 4th face) = (93/305)

Remember: you said "AND THIS AND THIS AND THIS" so they're all multiplied together. That's where the

(24*23*22*21*96*95*94*93) / (312*311*310*309*308*307*306*305)

comes from... If you wanted to know 3 face values and the last one an actual "10" then you would change the 96, 95, 94, and 93 values to their respective values (how many KQJ's in 6 decks, then how many "10's" in 6 decks).

So if you actually multiply and divide those numbers out... you get: 20,332,308,648,960 / 82,024,863,079,573,814,400 = .00000025. If you just wanted to use ONLY the simple calculator on your computer you could do this, then just guess and check 1/x and see what comes up with 6 zeros and 25... hint: since they already solved it start with something like 4,000,000 ;-)... thus, a 1 in ~4,000,000 chance.

Playing it correctly means you've already won.