Ask the Wizard #286
I don't want to just tell you the answer without giving you the opportunity to solve it yourself.
First, here is a hint to help. If you don't already know this equation, you're unlikely to solve it.
Otherwise, I admit I'm bad. Just show me the solution.
For discussion about the equation in the hint, please visit my forum at Wizard of Vegas.
Johnny from Cambodia
Assuming eight decks, the probability of winning is 52*combin(8,2)/combin(52*8,2) = 1,456/86,320 = 1.69%. The house edge is 13.98%.
The probability of the dealer getting exactly a 9-card 21 under those rules is 1 in 32,178,035. Here is the probability for various numbers of decks and whether dealer hits or stands on soft 17.
Probability of Dealer 9-Card 21
|Decks||Stand Soft 17||Hit Soft 17|
|1||1 in 278,315,855||1 in 214,136,889|
|2||1 in 67,291,581||1 in 41,838,903|
|4||1 in 38,218,703||1 in 22,756,701|
|6||1 in 32,178,035||1 in 18,980,158|
|8||1 in 29,749,421||1 in 17,394,420|
Assuming six decks and the dealer stands on soft 17, here is the probability of the dealer getting a 21 (or a blackjack in the case of two cards), according to the total number of cards.
Probability of Dealer 21/BJ
by Number of Cards
|2||1 in 21|
|3||1 in 19|
|4||1 in 56|
|5||1 in 323|
|6||1 in 3,034|
|7||1 in 42,947|
|8||1 in 929,766|
|9||1 in 32,178,035|
|10||1 in 1,986,902,340|
|11||1 in 270,757,634,011|
|12||1 in 167,538,705,629,468|
Not that you asked, but the next table shows the probability of the dealer making any non-busted hand under the same rules by the number of cards.
Probability of Dealer 17-21/BJ
by Number of Cards
|2||1 in 3|
|3||1 in 4|
|4||1 in 12|
|5||1 in 67|
|6||1 in 622|
|7||1 in 8,835|
|8||1 in 193,508|
|9||1 in 6,782,912|
|10||1 in 424,460,108|
|11||1 in 58,597,858,717|
|12||1 in 36,553,902,750,535|
For more discussion about this question, please visit my forum at Wizard of Vegas.
I hope you're happy. To answer this question, I bought a big roll of tickets at the Office Depot. Then I put 500 of them in a paper bag, half folded in half, at about a 90-degree angle, and the other half unfolded. Then I had six volunteers each draw 40 to 60 of them one at a time, with replacement, as I recorded the results. Here are the results.
Drawing Ticket Experiment
So, 58.3% of the tickets drawn were folded!
If it's assumed that folding had no effect, then these results would be 2.89 standard deviations away from expectations. The probability of getting this many folded tickets, or more, assuming folding didn't affect the odds, is 0.19%, or 1 in 514.
I might add the subjects who drew tickets hastily were much more likely to draw folded ones. Those who carefully took their time with each draw were at or near a 50/50 split.
So, my conclusion is definitely to fold them.
For discussion about this question, please visit my forum at Wizard of Vegas.