Odds of getting a blackjack?

Factual Questions
1

I play a bit of blackjack here and there, but something I read recently astounded me.

I was told that:

If I had a full deck of 52 cards that were randomly shuffled, and I flipped over the first 2 cards, the odds that I would have a blackjack would be ‘x’. (One of the cards being an Ace, the other being either a 10, J, Q, K).

Then, if I combine 2 full decks of cards, randomly shuffled, the odds of the first 2 cards off the deck(s) being a blackjack are ‘y’.

I read that ‘x’ odds are better then ‘y’ odds.

I also subsequently read that for each additional deck of cards you add, randomly shuffled, the odds of getting a blackjack decreases.

Is this true?

2

I’m sure a statistician will be along shortly to help you out, but in my experience playing blackjack recently, I got 15 a hell of a lot more often than I got 21.

This was while playing with about four or five decks of cards, so you might be on to something…

3

yes, but not by a significant amount.

With 52 cards there are 1326 two card combinations (note - NOT permutations, combinations are unordered). 64 of these are blackjacks (one of the 4 aces combined with one of the 16 10’s or face cards, again this is unordered).

With 104 cards, there are 5356 two card combinations. 256 of these are blackjacks (one of the eight aces combined with one of the 32 10’s or face cards - and yes, I’m counting the two copies of the same card as distinct because I did when I calculated the number of combinations).

64/1326 = .048265, 256/5356 = .047797 (to 6 places).

In general, with N decks you will get (52N)(52N - 1)/2 combinations. (4N)(16N) will combine an ace with a face card or ten. The ratio will be:

64N[sup]2[/sup]/(1352N[sup]2[/sup] - 26N)

which will approach 64/1352 = .047337 as N approaches infinity. Note that this limit is the same as drawing with replacement, which it should be:

(1/13)*(4/13)*2 = 8/169 = 64/1352.

(probability of drawing an ace times probablity of drawing a ten or face card times two for the ways to order the choices)

4

The odds of getting a blackjack with the first two cards does decrease with the number of decks. The probability is the number of aces in the deck times the number of cards worth 10, times 2 (because the order doesn’t matter) divided by the number of cards in the deck before you pick the first card, divided by one less than that (the number of cards in the deck before you pick the second card). For n decks, the probability is is P=2416n/[(52n52(n-1)]


number of decks        probability     percentage
---------------        -----------     ----------
     1               2*4*16/(52*51)        4.83%
     2               2*8*32/(104*103)      4.78%
     3               2*16*64/(208*207)     4.76%
     4               2*32*128/(416*415)    4.75%
     5               2*64*256/(832*831)    4.74%
5

I’m just not getting this, here is the way I think of it:

Assume I have all the cards for one suit, let’s say clubs.

The odds I am going to draw an Ace on my first card is 1 in 13. The odds I will then draw a 10-value card next is 4 in 12, or 1/3.

Add another suit.

The odds I am going to draw an Ace on my first card is 2 in 26, or, 1 in 13.

The odds I am going to draw a 10-value card next is 8 in 25, or… hang on, that’s just OVER 1/3.

I am beginning to see it now. The more decks you add, the more cards you are adding which lessen your chances of a blackjack. Namely the Ace?

6

exactly… the “extra” ace is the key.

assuming you pull an ace the first card, your chances of pulling a 10 are 16/51, or .3137.

with two decks, your odds of pulling a 10 are 32/103, or .3010.

There is an “extra” ace in the deck which would not be present had there only been one deck.

This is a simplified example of why single-deck games are more advantageous to the player than multi-deck games (even discounting card-counting). The more decks you add, since the player plays his hand first, the more ways there are for the dealer not to bust. i.e., in a multi-deck game, the dealer can draw 8 twos in a row, something which is not possible in a single-deck game. but i digress…

7

Just go through either my explanation or bibliophage’s CAREFULLY. We arrived at the same numbers, and the only difference in our reasoning was that he chose to calculate the ratio of ORDERED pairs which represented blackjacks to all ORDERED pairs, while I chose to calculate the ratio of UNORDERED pairs representing blackjacks to all UNORDERED pairs. In the end, he wound up multiplying by two in his numerator, while I wound up dividing by two in the denominator.

One thing I pulled out of the air because I just remember it and use it all the time was the calculation for the number of combinations of N objects taken K at a time:

N! / (K! * (N-K)!)

For K=2 this boils down to N*(N-1)/2.