0.25 Deck Blackjack

What would be the player advantage of a blackjack game where the deck contains only one of each of the 13 card values? You may assume the deck is shuffled after every hand and the player follows basic strategy (which only considers the player's total and dealer's upcard).

Would it still be a good bet if Blackjack pays 6:5?

For 0.25 deck, S17,BJ pays 3:2 I get a player advantage of 3.096%

A bigger advantage than I would've thought! Even blackjack paying 1 to 1 would make this game favorable for the player

Does anyone have the basic strategy chart for this game?

There probably can't be splits. What happens when the dealer runs out of cards?

Quote: AutomaticMonkey

There probably can't be splits. What happens when the dealer runs out of cards?
link to original post

Spits are possible because there are four T.
But really, the important question is what happens when the cards run out?

And so the program shows +3%

I think it would be impossible to run out of cards, because the total of all the cards (counting ace as 1) is 85, and if there's only one player, then the highest total anyone can get is 30 (from hitting a 20 and getting a 10)

Quote: SparklingStar

I think it would be impossible to run out of cards, because the total of all the cards (counting ace as 1) is 85, and if there's only one player, then the highest total anyone can get is 30 (from hitting a 20 and getting a 10)
link to original post

SparklingStar,

Dealt 10-J, split, get Q, split again, get K, split again. Now player has 4 hands, and the dealer has only 5 cards to deal, because six are on the table (counting the dealer's holecard), the dealer burned one, and will not deal the 13th card. Thus, if the player gets a second card on each X and then tries to hit each hand, the dealer will run out.

Granted, the above scenario is extremely unlikely, with a probability of the player getting all four X's in a row of 4!/13! = 3.854*10^-9.

Dog Hand

Quote: DogHand

Quote: SparklingStar

I think it would be impossible to run out of cards, because the total of all the cards (counting ace as 1) is 85, and if there's only one player, then the highest total anyone can get is 30 (from hitting a 20 and getting a 10)
link to original post

SparklingStar,

Dealt 10-J, split, get Q, split again, get K, split again. Now player has 4 hands, and the dealer has only 5 cards to deal, because six are on the table (counting the dealer's holecard), the dealer burned one, and will not deal the 13th card. Thus, if the player gets a second card on each X and then tries to hit each hand, the dealer will run out.

Granted, the above scenario is extremely unlikely, with a probability of the player getting all four X's in a row of 4!/13! = 3.854*10^-9.

Dog Hand
link to original post

I don't get that, I get 1 out of 715 of getting all 4 monkeys in a row. (4/13)*(3/12)*(2/11)*(1/10)

But that doesn't have to happen to split to 4 hands. As long as the first two cards are X, player can split, draw a 9 on the first half and another X on the second half. And then that can happen again. So only 4 out of the first 6 would have to be X. That will happen every night.

Now let's say the dealer has a 9 showing. First hand the player draws out 2,A,5. Second hand is 4,7. Third hand is 3,8. You now have deduced the holecard is a 6, and the dealer can't draw because there are no cards left and no discards either. So this game can't be dealt the way regular BJ is.

To simplify this 13-card mini game further, let’s fix the number of cards to be two each hand, for both the player and the dealer. The player has only one strategic decision to make, that is, the 2:1 insurance. A player’s blackjack is still paid 3:2. What is the house edge of this new game?

I think the game has to be single player, shuffle after every hand. Otherwise, with two players, no burn card, and no rule about not dealing the last card (although needing to deal the last card is mathematically impossible with 1-2 players except in cases where player is splitting Ten-valued cards.)