Pick 2 cards from deck. Then pick another as many cards, as is sum of 1st two cards.

AndreyMironov

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One pick two cards from a deck. Then pick another as many cards, as is the sum of first two card's score. What is the prob to get three aces in a picked cards?

Card's score: ace=1, two=2, ..., ten=10, jack=11, ...

Please, help!
 
One pick two cards from a deck. Then pick another as many cards, as is the sum of first two card's score. What is the prob to get three aces in a picked cards?

Card's score: ace=1, two=2, ..., ten=10, jack=11, ...

Please, help!
First, please read
https://www.freemathhelp.com/forum/threads/109845-Guidelines-Summary?p=422890&viewfull=1#post422890

Second, what are you studying? In secondary school? In college? What thoughts have you had so far?

Third, it appears that you have translated this problem from some other language into English.

I suspect that the problem should read

Given a normal deck of 52 cards, each rank of card is given a value: ace = 1, 2 = 2, ..., queen = 12, king = 13.

Choosing 2 cards at random, the sum of their values is n. Choose another n cards at random so that there are n + 2 cards in total. What is the probability that of the n + 2 cards, exactly 3 will be aces?

Is that a correct reading?
 
Exactly correct reading. This problem is translated from chinese. I have my best try as creating a table of possible outcomes for the amount of scores of first two cards (to use conditional probability), but it is too large and complicated. There should be more elegant way to solve problems like this, I suppose.
 
Exactly correct reading. This problem is translated from chinese. I have my best try as creating a table of possible outcomes for the amount of scores of first two cards (to use conditional probability), but it is too large and complicated. There should be more elegant way to solve problems like this, I suppose.
I personally do not see an elegant way to do it, and your suggested approach does seem very tedious if done by hand. Are you allowed to use a spreadsheet? Alternatively, can you give the procedure without calculating the numeric answer?
 
Spreadsheets and calculators are not allowed. This enforces me to calculate by hand using just paper and rational fractions. Old school.
 
Choosing 2 cards at random, the sum of their values is n.
Choose another n cards at random so that there are n + 2 cards in total.
What is the probability that of the n + 2 cards, exactly 3 will be aces?
Soooo...from the good to the ugly:
1st 2 cards = aces; pick another 2 : end with 3 aces and a non-ace
..........
1st 2 cards = 13's; pick aNUTter 26 : end with 3 aces and 23 non-aces

Boy oh boy, wonder who made that up :confused:
 
Not sure that in 26 cards you've got 3 aces. What's the probability? Where is all intermediate steps? And what's their probabilities?
 
What's the probability? Where is all intermediate steps? And what's their probabilities?
Are you serious? You supply all the intermediate steps along with their probability and we'll tell you if you are correct and if you are not correct, we will tell you where you are wrong.

Denis listed more that you see by including .... in his response. Just before ..., you should see 1st 2 cards = 13's; pick aNUTter 26 : end with 2 aces and 24 non-aces.
 
Not sure that in 26 cards you've got 3 aces. What's the probability? Where is all intermediate steps? And what's their probabilities?
I have stayed until now but here is some hints for you AndreyMironov. Below is an expand form giving us the number of ways to get the sum of the first two cards dealt.

ex13.png
Look at the term \(\displaystyle 12x^{13}\) that tells us that there are twelve ways to get a sum of thirteen.
That means there are twelve pairs of cards that add to thirteen. For example \(\displaystyle (6,7)~\&~(7,6)\).
Now the probability \(\displaystyle \mathscr{P}\{(6,7)\}=\dfrac{4}{52}\cdot\dfrac{4}{51}=\dfrac{4}{663}\).
You must ask your self "why the fours?" Also can you tell us why that is the probability of any pair \(\displaystyle (x,y)\) where \(\displaystyle x\ne y.\)
What if \(\displaystyle \mathscr{P}\{(7,7)\}=~?\). And WHY ?
I hope you will show some of your own work. If you don't then no more.
 
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