If 2 cards are drawn from a 52 card deck, what's the possibility that they add up to less than 10?

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Riley444

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Aces are equal to one, number cards are equal to what they show and face cards are equal to 10.

The greatest number you can choose is 8 because any card higher than that would exceed a total of 10. I went about finding the amount of options for the second card based on which of the 8 cards are chosen first:

Ace (1) = 31 cards 2 = 27 cards 3 = 23 cards 4 = 19 cards 5 = 16 cards 6 = 12 cards 7 = 8 cards 8 = 4 cards

If there's a mistake please let me know and I'm open to any help or new possible solutions!
 
Hi Riley444

You did not tell us if the cards are drawn with or without replacement!
 
Ace (1) = 31 cards 2 = 27 cards 3 = 23 cards 4 = 19 cards 5 = 16 cards 6 = 12 cards 7 = 8 cards 8 = 4 cards.
I assume that you meant Ace (1) = 31, cards 2 = 27, cards 3 = 23, cards 4 = 19, cards 5 = 16, cards 6 = 12, cards 7 = 8, cards 8 = 4 cards.
Sometimes you go down by 4 while another time you go down by 3. I have absolutely no idea what that line above means.

The greatest number you can choose is 8 because any card higher than that would exceed a total of 10. I agree that if you pick an 8 and then draw a larger card, then the sum would exceed 10. That is true if you pick a card 5 or higher. If you pick a 5 and then a larger card the sum will be larger than 10. Why are you even mentioning this? What if you pick a 8 and then an ace? That sum will be less than 10.

The only cards you can't pick 1st to have a chance on having a sum of less than 10 is a 9 or 10. Once you pick a 9 or a 10, the next card will cause the sum to equal 10 or exceed 10.

Either I am completely missing what you are saying or what you are not solving this problem correctly at all.
 
Jomo said:
Ace (1) = 31 cards 2 = 27 cards 3 = 23 cards 4 = 19 cards 5 = 16 cards 6 = 12 cards 7 = 8 cards 8 = 4 cards.
I assume that you meant Ace (1) = 31, cards 2 = 27, cards 3 = 23, cards 4 = 19, cards 5 = 16, cards 6 = 12, cards 7 = 8, cards 8 = 4 cards.
Sometimes you go down by 4 while another time you go down by 3. I have absolutely no idea what that line above means.

The greatest number you can choose is 8 because any card higher than that would exceed a total of 10. I agree that if you pick an 8 and then draw a larger card, then the sum would exceed 10. That is true if you pick a card 5 or higher. If you pick a 5 and then a larger card the sum will be larger than 10. Why are you even mentioning this? What if you pick a 8 and then an ace? That sum will be less than 10.

The only cards you can't pick 1st to have a chance on having a sum of less than 10 is a 9 or 10. Once you pick a 9 or a 10, the next card will cause the sum to equal 10 or exceed 10.

Either I am completely missing what you are saying or what you are not solving this problem correctly at all.
Click to expand...
I think I overcomplicated it for myself, if you think you know a solution would you mind explaining it? Thanks!
 
Riley444 said:
If 2 cards are drawn from a 52 card deck, what's the possibility [probability?] that they add up to less than 10?

Aces are equal to one, number cards are equal to what they show and face cards are equal to 10.

The greatest number you can choose [first] is 8 because any card higher than that would exceed a total of 10. I went about finding the amount of options for the second card based on which of the 8 cards are chosen first:

Ace (1) = 31 cards 2 = 27 cards 3 = 23 cards 4 = 19 cards 5 = 16 cards 6 = 12 cards 7 = 8 cards 8 = 4 cards

If there's a mistake please let me know and I'm open to any help or new possible solutions!
Click to expand...
As I understand this, you are saying that if your first pick is ... then the number of remaining cards that will yield a success are ...
  • Ace (1): 31 cards (namely 1 in another suit or 2, 3, 4, 5, 6, 7, 8 in any suit, which is 3 + 7*4 = 31)
  • 2: 27 cards (namely 2 in another suit or 1, 3, 4, 5, 6, 7 in any suit, which is 3 + 6*4 = 27)
  • 3: 23 cards (and so on)
  • 4: 19 cards
  • 5: 16 cards [why did this go down by 3 rather than 4? I see after thinking about it, but why not tell us?]
  • 6: 12 cards
  • 7: 8 cards
  • 8: 4 cards
  • 9 or 10: no cards
What you need to do is show your actual work to get these numbers, and then show us what you are going to do with them. But the basic idea is at least going in a right direction (though you may discover a more efficient way to get there as you work through this).
 
Riley444 said:
I think I overcomplicated it for myself, if you think you know a solution would you mind explaining it?
Click to expand...
A agree with you on that. If I were to work on this I would count the number of pairs in the space.
\(\dbinom{52}{2}=1326 \) that is 52 choose 2.
There are six pairs that are aces: \(\{A\diamondsuit,A\clubsuit\},\{A\diamondsuit,A\heartsuit\},\{A\diamondsuit,A\spadesuit\},\{A\heartsuit,A\clubsuit\},\{A\heartsuit,A\spadesuit\},\{A\clubsuit,A\spadesuit\}\)
Using that idea, there are twenty-four pairs of two-of-a-kind that work. What are they?
How many non-two of-a-kind fit the requirement of sum \(\le 9\)?
 
My suggestion is basically the same as pka’s, but maybe my explanation is easier for a beginner to grasp. (Pka’s suggestion is more computationally efficient.)

How many ways can the sum equal 2? Just one, namely 1 + 1.

How many ways can the sum equal 3? Just one, namely 1 + 2.

How many ways can the sum equal 4? Two, namely 1 + 3 or 2 + 2.

And so on all the way up to 9.

Now how many ways can you draw cards whose values will add up to 4.

[MATH]\dbinom{4}{1} * \dbinom{4}{1} + \dbinom{4}{2} = 4 * 4 + \dfrac{4!}{2! * (4 - 2)!} = 16 + 6 = 22.[/MATH]
Do you follow that?
 
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