Jack Jack

One of the local casinos recently added a Blackjack variant. The game has rules common for Blackjack in Midwest casinos (Dealer hits S17; no surrender; etc) and uses 8 decks.

The variant is this:
- Blackjack with two blacks cards pays 2-1
- Blackjack with two red cards pays 3-2
- Blackjack with red & black cards pays 1-1.

I have no doubt that like most variants this favors the House more than the player, but if some could calculate the House Edge I would appreciate it.

Based on on eight decks, I show an increase in house edge of 0.58%, compared to 3-2 blackjack. Here is the effect for other number of decks:

1 deck: 0.26%
2 decks: 0.45%
4 decks: 0.54%
6 decks: 0.57%
8 decks: 0.58%

You can use my house edge calculator to determine the 3-2 paying house edge. You didn't mention whether double after split is allowed or re-splitting aces.

https://www.jackentertainment.com/cincinnati/wp-content/uploads/sites/8/2017/05/JackJackBlackjac4x111.pdf

Edit: my work below contradicts that of the Wizard, so I must've misunderstood.

Just looking at the average payout for blackjack:

both black - 25%
both red - 25%
one of each - 50%

Average payout = 2*1/4 + 3/2*1/4 + 1*1/2 = 1/2 + 3/8 + 1/2 = 1-3/8 = 11/8

The average bj pays 11/8 to your 8/8 of a bet, so payout odds of 11-8. Not as good as 3-2, and not as bad as 6-5.

To see the difference between 11-8 bj and 3-2 bj, note that 3-2 is equivalent to 12-8, so with the new pay scheme, you are getting 11/12, or 91.667%, of the 3-2 bj payout. If you know the contribution to the return of bj's in 3-2, multiply that by 91.667% for 11-8.

I get slightly different answers but perhaps my assumptions are wrong.

If you're dealt an Ace then your second card is one of the pictures. The Ace could be Red or Black - and the chance is 50%. There are the same number of pictures left having been dealt an Ace, so their colour chances are also 50%.

It would seem fair to me that there would be no change to the average payout if they paid 2/1 if the colours were the same and 1/1 if the colours were mixed. Half the time you'd get paid 2 and half the time paid 1, for an average payout of 1.5.

The system quoted only pays 3/2 where both cards are Red. Thus 25% of the time they underpay by 1/2, thus whenever you have a winning Blackjack you lose 1/8th. (You can see this being argued earlier by comparing 11/8 to 12/8).

The probability of winning a Blackjack is the chances you get one multiplied by (1-chances Dealer then gets one).

For one deck
P(Player gets BJ) = 4 (aces) * 16 (pictures) * 2 (either order) / 52 / 51 = 4.826 546%
P(Dealer now gets BJ) = 3 (aces) * 15 (pictures) * 2 (either order) / 50 / 49 = 3.673 469%
Chances player wins = 0.04826 * (1-0.0367) = 4.649 244%
So reduction in House Edge is 0.04649 / 8 = 0.581 156%

For more decks your chances of getting a BJ goes down a little bit, and the dealer's chances goes up. All this means your chances of winning with a BJ gradually goes down. Thus the change in House Edge gets closer to 0.5637%.

Number of Decks	P (Player gets BJ)	P (D=BJ given P=BJ)	P (Players wins with BJ)	Decrease in HE
1	.048 265	.036 735	.046 492	0.581 156%
2	.047 797	.042 128	.045 783	0.572 291%
4	.047 566	.044 755	.045 437	0.567 964%
6	.047 489	.045 621	.045 323	0.566 537%
8	.047 451	.046 052	.045 266	0.565 827%
Lots	.047 337	.047 337	.045 096	0.563 706%

I get the same as charliepatrick. I just calculated with a blackjack pay of 1.375. (25% * 2 + 25% * 1.5 + 50% * 1).

Double after spit is allowed; no-re-spitting of Aces. Thanks for the response!

I suppose if even money is available. Its not clear to me if it is in this game. You would take even money for the 1:1 blackjack. That might make a small difference in your calculations.