Comments (13)

1. I’m too lazy to do the probability this morning. Does that assume an Ace always is/isn’t an 11? For an example, are you counting Ace/2 as 13 or 3?

   posted by Mean Joe on 10-14-2009 at 8:35 am

2. I would assume both, as thats what it would be in real blackjack rules.

   posted by Bill on 10-14-2009 at 9:14 am

3. Joe,
   I think that the Ace will always be 11 since with the first two cards you can’t go over 21. You would only consider Ace to be 1 if subsequent hits bring you over 10 for the rest of the cards.
   The solution is easy with that in mind. Just think about how you do the same thing with two six sided dice.

   posted by anomdebus on 10-14-2009 at 9:14 am

4. As mentioned in the preface to the puzzle, an Ace can be scored as either 1 or 11, so both scores work. We’re using typical blackjack rules here. Just remember that we’re dealing ONLY with the two initially-dealt cards, not any “hits” that might come afterward.

   posted by Sandy Wood on 10-14-2009 at 9:41 am

5. To put my claim a different way. The scoring is based on the maximum number that does not exceed 21. An A/K match is scored to be 21 and not 11. An A/2/K set is scored to be 13 and not 24, though in this case it was also scored to be 13 before the K was dealt.

   posted by anomdebus on 10-14-2009 at 10:41 am

6. Can someone help me figure out how there are 146 combinations for getting a 20? I keep getting 136.

   10 of clubs pairs with 15 other cards
   10 of diamonds with 14 other cards (10c-10d already paired)
   10 of hearts with 13 other cards
   10 of spades with 12 other cards
   jack of clubs with 11,
   etc.
   15+14+…+1 = 120

   4 Aces and 4 9′s make 16 more combos, so I get 120+16 = 136. What are the other 10 combos I’m missing?

   posted by Craig on 10-14-2009 at 11:04 am

7. Are we using two decks here?!:)

   posted by Kelly on 10-14-2009 at 11:28 am

8. I have the same question as Craig, and the second most has to be 11, as you have 64 Ace/Face combos, plus the rest of the combos (9/2,8/3,7/4,6/5) I come up with 128 combinations for 11. Just because Ace/Face can be 21, it can also be 11.

   posted by Rob on 10-14-2009 at 1:28 pm

9. According to my calculations 13 points has 128 two card combinations. 14 points has 115 two card combinataions. And 12 points has 118 two card combinations. This puts 12 points as more likely than 14 points. below is my computations. Where did I go wrong here?

   12 points
   A A = 3
   K 2 = 16
   Q 2 = 16
   J 2 = 16
   10 2 = 16
   9 3 = 16
   8 4 = 16
   7 5 = 16
   6 6 = 3
   total 118

   13 points
   A 2 = 16
   K 3 = 16
   Q 3 = 16
   J 3 = 16
   10 3 = 16
   9 4 = 16
   8 5 = 16
   7 6 = 16
   total 128

   14 points
   A 3 = 16
   K 4 = 16
   Q 4 = 16
   J 4 = 16
   10 4 = 16
   9 5 = 16
   8 6 = 16
   7 7 = 3
   total 115

   posted by Steve M on 10-14-2009 at 1:34 pm

10. I’m in near agreement with Craig and Steve M. I get 136 combos for 20, 124 combos for 12, 128 combos for 13, and 118 combos for 14. That would make 13 the second highest odds and 12 the third highest odds, both behind 20.

    My values are slightly different from Steve because he got the count of combinations of the same card wrong. It should be 6, not 3. For example, two aces to get 12 could be done as:
    club + diamond
    club + heart
    club + spade
    diamond + heart
    diamond + spade
    heart + spade

    So that would add 6 combos to the 12 count (for A-A and 6-6) and 3 combos to the 14 count (for 7-7) per Steve’s post. Then our numbers agree.

    posted by Jeff on 10-14-2009 at 1:58 pm

11. Fair enough with A/A, which would require that one of them be 1, though I think that is the only situation where an A==1.

    posted by anomdebus on 10-14-2009 at 2:02 pm

12. I agree with Steve and Jeff on the nubers they show but the wording leaves it open for another option. Since Sandy said “an Ace can be scored as either 1 or 11, so both scores work”, then A-9 would be included in the counts for 20 and 10, even though you would never count and A-9 as 10 in blackjack. Similarly, A-10 should be counted as both 21 and 11, even though you would never do that. If so, then 11 also has 128 combinations making the answer to the problem 11 and 13 (followed closely by 12).

    posted by Ric on 10-14-2009 at 4:56 pm

13. I would assume both, as thats what it would be in real blackjack rules.

    posted by BeyinOyunu on 10-19-2009 at 5:16 pm

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