A CRAZY ODDS QUESTION

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

A six deck blackjack table. (Dealer hits on 16, standard blackjack.)

A player is dealt two aces (A1 & A2). She splits.
A1 is dealt a face card (blackjack #1, henceforth known as BJ1). A2 is dealt an ace (A3) - so she splits again.
(Review: there are now three aces in front of the player, one has been made into BJ1).
A2 is dealt a face card (blackjack #2, henceforth known as BJ2). A3 is dealt an ace (A4) - so she splits again.
(Review: there are now four aces in front of the player, two have been made into BJ1 & BJ2).
A3 is dealt a face card (blackjack #3, henceforth known as BJ3).
A4 is dealt a ten (blackjack #4, aka BJ4).
(Review: there are now four aces in front of the player, all four have been made into BJ1, BJ2, BJ3, and BJ4).

What are the odds of this happening?

Assuming by "Ten" for BJ4 and "Face Card" for BJ1, BJ2, and BJ3 you are referring to any ten-valued card and not specifically to JQK for the first 3 blackjacks and T for the 4-th blackjack, the answer is 4,034,212-to-1.

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).

WOW - thanks! No, I was not asking that general question, so thank you! How in the world does one go about figuring that out??

Quote: teliot

Assuming by "Ten" for BJ4 and "Face Card" for BJ1, BJ2, and BJ3 you are referring to any ten-valued card and not specifically to JQK for the first 3 blackjacks and T for the 4-th blackjack, the answer is 4,034,212-to-1.

Wow, that much? That's up there in Seven Card Straight Flush territory.

Quote: phoebeanne

WOW - thanks! No, I was not asking that general question, so thank you! How in the world does one go about figuring that out??

Chance of getting an ace as your first card is 24/312 (24 aces, 312 cards). Chance of an ace on your second card is 23/311 (23 aces remaining out of 311 cards). Then 64/310 for the face card. Then 22/309 for the next ace. And so on. Multiply all them numbers up and you get your answer.

(24*23*22*21*96*95*94*93) / (312*311*310*309*308*307*306*305)
:.
http://m.wolframalpha.com/input/?i=%28%2824*23*22*21*96*95*94*93%29+%2F+%28312*311*310*309*308*307*306*305%29%29&x=0&y=0
:.
http://m.wolframalpha.com/input/?i=1%2F%28%2824*23*22*21*96*95*94*93%29+%2F+%28312*311*310*309*308*307*306*305%29%29&x=0&y=0

= 1 in 4,034,212

Of course, the player had no blackjacks as a two card 21 after a split is simply a 21, not a blackjack.

Billryan,
Depends on the casino, definitions vary by casino. The place my friend was at considered her four amazing deals blackjacks (and paid appropriately).

Quote: phoebeanne

Billryan,
Depends on the casino, definitions vary by casino. The place my friend was at considered her four amazing deals blackjacks (and paid appropriately).

I think I may have heard of this rare rule, many moons ago.....where split A's with a T count as a 3:2 BJ.



I remember once playing blackjack with my brother. He got aces, split, got a T on one of them. When dealer paid him even money, he tried to get paid 3:2 on it (he thought it was considered a blackjack). At first I thought he was joking, but then realized he was serious. He tried getting the pit boss over and I was telling him the dealer was right, it pays even money. He thought we all missed the obvious.

From what I have read, that rule has been effect since casinos first adapted to Ed Thorpe in the early Sixties.
Anyone want to chime in with a casino that treats split 21s as a BJ?
When BJ pays 3-2, of course. I've seen video BJ machines that will say Blackjack when it happens but they pay even money.

The casino we were at is an Indian casino in Arizona. She was paid even money. I was wrong earlier when I said she got 3-2.

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?

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.

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.