A simple blackjack variant

[Yoruichi-san]

I remember hearing from college-mates that some grad student did their thesis on Blackjack and proved that there was a 51% chance of winning if the player counted cards.

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.

[Rob_Gandy]

I just might take the first seat. I'd have about a 15.5% chance of winning after getting two cards. Just a first thought. Still considering numbers.

[bonanova]

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?

[CaptainEd]

Must be Bonanova is right. Cards are dealt one at a time, and completion criterion is tested each time. As soon as the first person hits 20 or 21, or the second person busts, the hand is over.

[CaptainEd]

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

[CaptainEd]

"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?

[~andy~]

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

[TheChad08]

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+?

[Rob_Gandy]

I'm gonna stick with seat one. I'd have about a 15.5% chance to win after getting my second card. Then I have around a 55% chance to not bust on my third card. I did it all on paper so I can't really go much further without getting the computer into it. I may try to do this as I have time, but I'd have to do it on excel because that's all I have at work and I wouldn't know how to write a program to do it for me.

[Yoruichi-san]

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.

[TheChad08]

Oops, forgot to put my post in spoilers.

Sorry about that.

As for the calculations, I don't know how to write programs to do that for me as there would be thousands of possibilities.

[superprismatic]

My 1,000,000,000 game simulation gave the following probabilities of winning:

Seat 1: 0.342568700

Seat 2: 0.318651372

Seat 3: 0.338779928

[CaptainEd]

Would you say the winning seat benefited more from 20/21 or from the other two going bust?

[superprismatic]

Would you say the winning seat benefited more from 20/21 or from the other two going bust?

I don't know, but I can modify my program to count those things. I have an appointment soon.

When I return from that, I'll fix up the program. Perhaps knowing such things may make it easier

to analyse the problem analytically.

[Yoruichi-san]

Admittedly the ace duality is a little annoying, but setting that aside, treating it akin to rolling 13 sided dice (with some repeats) and summing the values on the dice might be helpful.

[superprismatic]

Would you say the winning seat benefited more from 20/21 or from the other two going bust?

Out of 1,000,000,000 simulated games:

Seat 1 won 342,546,409 times, 281,000,589 of which were because seat 1 got 20/21, and 61,545,820 of which were because Seats 2 and 3 both went over 21. In this simulation, Seat 2 won 318,668,459 times and Seat 3 won 338,785,132 times.

[CaptainEd]

Thank you, Superprismatic! (And thank you, Y-san. Such a simple question, with so little intuition available for guessing how it should go, and such a tiny, but repeatable effect. Great fun!)

I'm quite surprised how much more important the ability to hit 20/21 is than the ability to be the survivor. Over 80% of the wins due to 20/21!

And yet, even so, the first position is only very slightly stronger than the other two...

Fascinating!

[brifri238]

Similar to Superprismatic I too ran a simulation (over only a mere 100,000,000 hands) and got similar results with Hand1 winning and Hand 3 in second.(Hand1: 34.257% / Hand3: 33.854% / Hand2: 31.889%). I also ran the simulation for a 'REAL WORLD' execution using one deck of cards without replacement. (Other assumptions the same, and in addition: if a hand goes over 21 no further cards are dealt to that person during that round-[this assumption is irrelevant when using an infinite number of decks]). In this case Hand 3 is the most likely to win! - (Hand 3: 34.0045 %, Hand 1 in a close second with 33.9777 % and Hand 2 with 32.0178%)

Edited November 2, 2012 by brifri238

[brifri238]

I re-ran both my simulations: scenario 1: with an infinite number of card decks, and the more real world scenario 2: with one deck of cards where cards are dealt without replacement. Both simulations were run over 1000,000,000 hands and the previous results and rankings were confirmed:

Scenario 1: (infinite number of decks): Hand 1 (34.2521%), Hand3 (33.8808%) and Hand 2 (31.8671%)

Scenario 2: (one deck without replacement): Hand 3 (34.0031%), Hand 1(33.9674%) and Hand 2 (32.0294%)

Edited November 5, 2012 by brifri238

[Yoruichi-san]

Wow, that's pretty interesting, thanks brifri .

If anyone is still attempting the semi-analytical approach, I find that comparing the winning chances of each chair during a round is a good way to do it.

[Prime]

The analytical solution for the probability of win for each hand is as following:

1st hand = .342538484

2nd hand = .318676316

3rd hand = .338785200

A recursive program, which would traverse the entire game for 3 hands, would be simple enough to code. However, I did not chance it fearing unmanageable number of iterations. After all, given the infinite deck, a single hand in this game can be as large as 18 cards.

Instead of writing a program, I built a spreadsheet table with probabilities of win/bust/continue for a single player for each round. From that probabilities of winning a “sudden death game” for each seat among 3 players may be derived. It’s a tedious solution, but I had some similar tables from my own projects with BJ probabilities, so it did not take too much of my time.

I could not see any simple and elegant solution, which would avoid calculating actual probabilities.

Some more trivia from the table if you play this game without partners, by your lonesome:

Probability to bust = .621058

Probability to get 21 = .159379

Probability to get 20 = .219563

Probability to finish the game with a hand of 6 cards or less = .996315