blackjack test

getting snowed in in New England on a late saturday night after a beating at the tables friday (the player next to me played ALMOST perfect basic strategy but managed to lose 16 hands STRAIGHT in a shoe that headed south and just stayed there!, I faired better but should have wonged out)

1. Question #1 - what is the most number of cards a DEALER can get without busting and in what order. What is the most number of cards a player can get without busting and in what order.

(I did figure this out but its not what you might think right off the bat)

2. Question #2 - what is the most number of cards a dealer can get with no two card values the same without busting, and whats the value of his/her highest possible card, what is the answer for a player.

3. Ok thse are fairly easy, here is one thats almost impossible and I will leave it as a challenge to anyone with real computing power, Question #3, What is the total number of possible different hands a dealer can get disregarding suit and assuming all 10-king are just 10 value cards, group together each set of hands where the order doesnt matter and count it as one (like a 10/7 versus a 7/10, thats one). For insane extra credit whats the total number of all possible combinations, what is it for a player. (no I have no idea what the answer might be but an educated guess with some rationale counts for something)

I am just curious to know if its in the hundreds, thousands, tens of thousands or 100k+ range.

#3, possible dealer hand combinations, came up on bjmath a short time ago. My answer

#4) Just for fun, what are the number of distinct combinations possible for an entire deck (1,2,4,6, and 8 decks have different answers) where there is no distinction between equivalent ranks (any ace is equivalent to any other ace, any two is equivalent to any other two, etc.?)

kaycee

2. Question #2 - what is the most number of cards a dealer can get with no two card values the same without busting, and whats the value of his/her highest possible card, what is the answer for a player.

Six.

2,3,4,5,A,6

There are other solutions that permute the cards listed.

For the Dealer:

If it is 3+ decks and H17: A (up card), A down, Hit cards order of: A, A, A, A, A, 5, A, A, A, A, 5 = 21...for a total of 13 cards.

If is it 3+ decks and S17: A (up card), A down, Hit cards order of: A, A, A, A, 6, A, A, A, A, 5 = 21...for a total of 12 cards.

For the player it would be differnt for the number of decks too. And I don't wanna take the time to do all of it.

do A, A, A, A, A, 5 and then hit? that's a soft 20 at that point.

Same problem in the second example (S17 game). A, A, A, A, 6 is 20.

Maybe you meant A, A, A, A, A, A (soft 16) followed by 6 (hard 12),
followed by A, A, A, A, A to get to hard 17. So 12 cards rather than 13?

Or did I mis-read something??

Read too quickly. Thought the A, A, .. stuff included the first two...