How many ways can you get 21 in blackjack?

3 Comments

1. mikey ^_^ says:

   February 10, 2012 at 7:04 pm

   2 card combo:
   Ace, King
   Ace, Queen
   Ace, Jack
   Ace, 10

   Total: 4

   3 card combo:
   Ace, King, Queen
   Ace, King, Jack
   Ace, King, 10
   Ace, Queen, Jack
   Ace, Queen, 10
   Ace, Jack, 10
   Ace, 10, 10
   Ace, 9, Ace
   Ace, 8, 2
   Ace, 7, 3
   Ace, 6, 4
   Ace, 5, 5
   King, 9, 2
   King, 8, 3
   King, 7, 4
   King, 6, 5
   Queen, 9, 2
   Queen, 8, 3
   Queen, 7, 4
   Queen, 6, 5
   Jack, 9, 2
   Jack, 8, 3
   Jack, 7, 4
   Jack, 6, 5
   10, 9, 2
   10, 8, 3
   10, 7, 4
   10, 6, 5
   9, 9, 3
   9, 8, 4
   9, 7, 5
   9, 6, 6
   8, 8, 5
   8, 7, 6
   7, 7, 7

   Total: 35

   You realize I can keep going, but you need to understand the depth of the question you’re asking lol There’s probably over a thousand ways to make 21 if there’s no card quantity limit.
2. Divide By Zero says:

   February 10, 2012 at 7:04 pm

   These are known as partitions, and I don’t know of a simple way to calculate them.

   I use this partition calculator — http://www.btinternet.com/~se16/js/partitions.htm

   I’ll disregard suits.

   EDIT — for some reason I was excluding 11-valued aces before except in blackjacks.

   If all 10-valued cards are counted the same:
   "Max number of terms" = 11
   "Each term no more than" = 11
   Result = 598 <—–

   But if 10′s, J’s, Q’s and K’s all count as different cards, more work is needed.

   In order to use the calculator I’ll have to compute the 21′s that use 10′s separately from the ones that don’t.

   For the 21′s without 10′s, "Max number of terms" is 11 and "Each term no more than" is 9.
   Result = 502
   Those results also don’t include 11′s. For those we say "Largest term exactly 11". The result is 42, but 1 of those results (blackjack) uses a 10, so we exclude it for now.
   # 21′s without 10′s = 502 + 41 = 543

   For the 21′s with 10′s, we must distinguish between ones with 1 10 and 2 10′s. We’ll call them X and Y respectively.

   To get X, first we tell the calculator "Largest term exactly" 10 .
   Result is 54, but now we subtract the possibilities that use two 10′s, of which there is only 1. So the new result is 53.
   Now since there are four different cards that equal 10, we multiply that by 4 and get 212.
   And finally, we include blackjacks, of which there are 4.
   X = 216

   Now for Y we already said there’s only 1 possibility (two tens and an ace), but we still have to account for the 4 types of tens. We’re choosing 2 of them, and the same type is allowed twice so instead of regular combinations we use multiset combinations:
   Y = 4 multichoose 2 = (4+2-1) C 2 = 10

   And so our total #ways to get a 21 = 543 + 216 + 10 = 769 <—-

   If you wanted us to account for suit or for how many of each card there is, that would be too time-consuming. Also, you didn’t specify how many decks are in the shoe.
3. John F says:

   February 10, 2012 at 7:04 pm

   To take Mike’s answer to the extreme:

   If you have a 6-deck shoe, there are 24 aces in the deck. You could theoretically use 21 aces to make 21. Or 19 aces and a deuce. Or 18 aces and a trey. …

   You get the idea.