Blackjack: Probability of Suited Ace Jack

I recently started playing Blackjack in Zurich / Switzerland where there is a "Progressive Jackpot" that pays when the player and the dealer both have a suited Ace-Jack (different suits for dealer and player allowed). What is the probability of having a suited Ace-Jack? What is the probability that both dealer and player have a suited Ace-Jack?

My own calculations lead to something like 1:135000 for a single deck game. But I assume (not sure) that we actually play a six-deck game.

Quote: OnkaPonka

I recently started playing Blackjack in Zurich / Switzerland where there is a "Progressive Jackpot" that pays when the player and the dealer both have a suited Ace-Jack (different suits for dealer and player allowed). What is the probability of having a suited Ace-Jack? What is the probability that both dealer and player have a suited Ace-Jack?

My own calculations lead to something like 1:135000 for a single deck game. But I assume (not sure) that we actually play a six-deck game.

First of all, great question and welcome to WizardofVegas! I hope you stick around!

This is actually a fairly easy question, just separate the dealer and the player. Here is for six decks:

PLAYER SIDE:

Okay, so the player can get a Blackjack with the Ace-Jack or the Jack-Ace combo, there are 312 cards in six decks, 24 Aces, 24 Jacks:

(24/312 * 6/311) + (24/312 * 6/311) = 0.002968093

Okay, so above is the probability of the player getting a suited AJ Blackjack.

DEALER SIDE:

In order to win, the player must already have a suited AJ Blackjack, so that removes an Ace from one of the suits and a Jack from one of the suits. Now, we can look at the probability of the Dealer getting it in the same suit as the player:

(5/310 * 5/309) + (5/310 * 5/309) = 0.00052197515

And, now we will look at the probability of the dealer doing it in any other suit:

(18/310 * 6/309) + (18/310 * 6/309) = 0.00225493266

Okay, so now we add the ways the dealer can do it which already assumes the player has done it together:

0.00225493266 + 0.00052197515 = 0.00277690781

OVERALL PROBABILITY:

Now, all that remains is to multiply the probability of the player doing it to that of the dealer doing it assuming the player has already done it:

0.00277690781 * 0.002968093 = 0.00000824212

Okay, so there you go, it is 0.00000824212 or 1/0.00000824212 = Roughly 1 in 121,328 based on six decks.

The reason it is less likely with one deck is because the effect of removal is greater. Think of it this way, if the player gets a suited AJ in hearts on six decks, there are still five of those the dealer could get whereas if the player gets suited AJ hearts with one deck, then the dealer cannot get that.

ONE DECK PROBABILITY:

Player:

(4/52 * 1/51) + (4/52 * 1/51) = 0.00301659125

Dealer (Assuming Player has it)

(3/50 * 1/49) + (3/50 * 1/49) = 0.00244897959

Combined:

0.00244897959 * 0.00301659125 = 0.00000738757 or 1/0.00000738757 = Roughly 1 in 135,362.51

The greater the number of decks, the more likely it is due to the lower effect of removal.

Thanks for your quick and comprehensive response. I am happy to see my own ~ 1:135000 confirmed for a one deck game, so it seems I got the math right.

You’re welcome! I always do my best, and I’m glad my best was sufficient this time!