|
|
#1
05-20-2005, 05:03 AM
|
|||
|
|||
Blackjack problem
In blackjack the dealer gets two cards, one of which you can see and one of which you cannot. When the dealers visible card is an ace she offers you a chance to take out "insurance." You can bet $1 that the invisible card is a face card or a ten. If it is you win $2 otherwise you lose $1. What is the expected value of this bet if we assume that the dealers other card was chosen at randome from a second deck of 52? And if we use the information that the dealers ace and our two cards, which are six and eight, came from the same deck of 52.
SGS 
|
|
#2
05-20-2005, 06:24 AM
|
|||
|
|||
|
Re: Blackjack problem
[ QUOTE ]
In blackjack the dealer gets two cards, one of which you can see and one of which you cannot. When the dealers visible card is an ace she offers you a chance to take out "insurance." You can bet $1 that the invisible card is a face card or a ten. If it is you win $2 otherwise you lose $1. What is the expected value of this bet if we assume that the dealers other card was chosen at randome from a second deck of 52? And if we use the information that the dealers ace and our two cards, which are six and eight, came from the same deck of 52. SGS [/ QUOTE ] It's a straightforward calculation, but if the dealer is getting her hole card from a second deck of 52, I wouldn't play there.
|
|
#3
05-20-2005, 09:57 AM
|
|||
|
|||
|
Re: Blackjack problem
[ QUOTE ]
In blackjack the dealer gets two cards, one of which you can see and one of which you cannot. When the dealers visible card is an ace she offers you a chance to take out "insurance." You can bet $1 that the invisible card is a face card or a ten. If it is you win $2 otherwise you lose $1. What is the expected value of this bet if we assume that the dealers other card was chosen at randome from a second deck of 52? And if we use the information that the dealers ace and our two cards, which are six and eight, came from the same deck of 52. SGS [/ QUOTE ] I'll do this two ways - with or without the three cards you've seen. If you assume the mystery card comes from a brand new 52 card deck, you have 16 cards that win the bet for you, and 36 that lose: (16*2) - (36) = -4/52 EV = -0.077 If we know those 3 cards are out of a single deck: (16*2) - (33) = -1/49 EV = -0.02
|
|
#4
05-20-2005, 09:37 PM
|
|||
|
|||
|
Re: Blackjack problem
My fuzzy memory of reading blackjack strategy books as a kid is "don't take insurance unless the count is +2 or higher."
Given that the simplest count is "add one for each 3/4/5/6, subtract one for each ten-value card" (and ignore the other ranks) - that old bit of advice does indeed translate into "only take insurance if four non-tens have been removed from a single deck" as confirmed by the calculation in the last post.
|
 |
|
|
Linear Mode
