# Math Help - Card Question

1. ## Card Question

You randomly draw 1 card from a standard deck of 52, and 1 card from another standard deck. What's the probability that the sum of the 2 cards is 7?

2. You can solve it by solving this question :

How many solution have to this linear equation:

x_1 + x_2 = 7 , when x_1, x_2 are from the group: {1,2,3,4,5,6}

?

3. Originally Posted by lifeunderwater
You randomly draw 1 card from a standard deck of 52, and 1 card from another standard deck. What's the probability that the sum of the 2 cards is 7?
First of all, what work have you done?

Second, are aces counted as having value 1 or not?

4. I thought of it that way >>> (not sure if I am correct though)
since there are 2 standard deck of cards and the sum of 2 numbers goes from 1 through 6
which means each card from 1-6 has 4 cards repeated. which makes 6*4= 24 and since
we have 2 deck of standard cards of 52 then >>
24/52+ 24/52 = 48/52= 12/13

But I don't think that is correct?!?!

5. Originally Posted by lifeunderwater
I thought of it that way >>> (not sure if I am correct though)
since there are 2 standard deck of cards and the sum of 2 numbers goes from 1 through 6
which means each card from 1-6 has 4 cards repeated. which makes 6*4= 24 and since
we have 2 deck of standard cards of 52 then >>
24/52+ 24/52 = 48/52= 12/13

But I don't think that is correct?!?!

Where you use the condition that the sum of the 2 cards is 7?

6. I don't understand what you asked above? could you please clarify. thank you!

7. Originally Posted by lifeunderwater
I thought of it that way >>> (not sure if I am correct though)
since there are 2 standard deck of cards and the sum of 2 numbers goes from 1 through 6
which means each card from 1-6 has 4 cards repeated. which makes 6*4= 24 and since
we have 2 deck of standard cards of 52 then >>
24/52+ 24/52 = 48/52= 12/13

But I don't think that is correct?!?!
Well you didn't answer my aces question so I'll just assume aces count as 1.

We have 52^2 ways to choose the cards when order is taken into account.

1 + 6 -> 4*4 = 16 ways

2 + 5 -> 16 ways

...

6 + 1 -> 16 ways

we have $\displaystyle P=\frac{6\cdot16}{52^2}$.

8. By the way it can be seen that this is the same as the probability of getting a sum of 7 if we only had two 13-card decks with just one of each card value in them. We get 6/169.

9. Quite different if 1 deck, and sum to be even; like:
"You randomly draw 2 cards from a standard deck of 52.
What's the probability that the sum of the 2 cards is 8?"

1st card: 28/52 = 7/13

2nd card: (6/7)(4/51) + (1/7)(3/51) = 9/119

Probability: (7/13)(9/119) = 9/221

Agree?

10. Originally Posted by Wilmer
Quite different if 1 deck, and sum to be even; like:
"You randomly draw 2 cards from a standard deck of 52.
What's the probability that the sum of the 2 cards is 8?"

1st card: 28/52 = 7/13

2nd card: (6/7)(4/51) + (1/7)(3/51) = 9/119

Probability: (7/13)(9/119) = 9/221

Agree?
Agreed; and just to be clear, that's drawing two cards from a single deck without replacement where aces count as 1.

Brute force check in PARI/GP

Code:
Input: s=0;for(i=0,50,for(j=i+1,51,if(i%13+1+j%13+1==8,s++)));s
Output: 54
Input: 54/binomial(52,2)
Output: 9/221

11. Thanks, Mr n/0