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# A simple blackjack variant

### #1

Posted 25 October 2012 - 12:08 AM

Let's look at a simpler variant, with only one decision making node.

The game is thus: there are 3 players, and a first card is dealt to each, then a second card is dealt to each, then a third card, etc. The game ends when either one player has 20 or 21, in which case they win, or when 2 players have busted with 22 or above, in which case the remaining player wins.

Your decision: which seat do you sit in? For simplicity, assume the cards are dealt from an infinite stack of decks.

**Women are definitely stronger. We are [Fe]males, after all...**

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### #2

Posted 25 October 2012 - 01:41 AM

### #3

Posted 25 October 2012 - 02:46 AM

Are the groups of three cards dealt simultaneously?

Else how does seat position come into play?

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### #4

Posted 25 October 2012 - 04:57 AM

### #5

Posted 25 October 2012 - 06:25 AM

### #6

Posted 25 October 2012 - 03:05 PM

Is the purpose of this phrase to imply sampling WITH replacement?

### #7

Posted 25 October 2012 - 03:39 PM

If you are counting cards you need to be allowed to 'stick' many times

I think what is meant by 'infinite stack of decks' is that many decks get shuffled together (as in the proper casino game, usually six), not one deck after another

Therefore which seat you take does not make any difference whatsoever, card counting cannot come into it

### #8

Posted 25 October 2012 - 06:13 PM

You will have no advantage to counting cards if you have got to play to 20/21 or bust

If you are counting cards you need to be allowed to 'stick' many times

I think what is meant by 'infinite stack of decks' is that many decks get shuffled together (as in the proper casino game, usually six), not one deck after another

Therefore which seat you take does not make any difference whatsoever, card counting cannot come into it

The question is about probability of reach 20,21 or 22+

If the odds are higher than you'll reach 20 or 21, you want seat 1.

If the odds are higher that you'll bust, you want seat 3.

1/13 chance of an Ace

1/13 chance of a two

1/13 chance of a 3

etc.

1/13 chance of a 9

4/13 chance of a 10 card (10 or face).

Using those numbers, what are the odds of being dealt a 20, 21 or 22+?

### #9

Posted 25 October 2012 - 07:52 PM

### #10

Posted 25 October 2012 - 07:56 PM

Does the game end the moment the cards in one hand say so?

Are the groups of three cards dealt simultaneously?

Else how does seat position come into play?

Yes, cards are dealt one at a time.

Is an ace either 1 or 11 at the player's choice?

It will count as whichever will make the player win or not lose if there's a difference, otherwise it doesn't matter.

"infinite stack of decks"...One shuffled deck offers sampling WITHOUT replacement, which is what makes it possible for someone to benefit from card counting. If we stacked another shuffled deck on top, and another, etc. producing an "infinite stack of decks", there would still be sampling without replacement. (Indeed, with only three players, you're sure to exhaust the players before the first deck of cards.)

Is the purpose of this phrase to imply sampling WITH replacement?

Yes, it's to imply, say, the probability of getting an A on your next turn is always the same, regardless of what cards have already been dealt, for simplicity.

**Women are definitely stronger. We are [Fe]males, after all...**

*Some of what makes me me is real, some of what makes me me is imaginary...I guess I'm just complex. ;P*

<3 BBC's Sherlock, the series and the man. "Smart

*is*the new sexy."

Chromatic Witch links now on my 'About Me' page! Episode 3 is finally here!

When life hands me lemons, I make invisible ink.

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