Tricky question involving sets of cards

I asked this on math.stackexchange :

http://math.stackexchange.com/questions/320579/probability-question-involving-sets-of-cards

I have an infinite deck built out of sets of 10 cards (in other words 10*n cards). The sets are identical so one '2' is identical to another '2'.
A player draws 6 cards. If he draws:

• any '1' AND a '2', or
• any '3' AND a '4', or
• any '5' AND a '6', or
• any '7' AND a '8', or
• any '9' AND a '10',

he wins. In other words there are 5 pairs and if the player draws a complete pair he gets a point.
What is the probability he won't win any points at all?
To expand on the problem, if the player gets a point for every pair he completes in a hand, what is the probability he'll get 1, 2, or even 3 points? (3 points being 6 cards of 3 completed pairs)
From what I know of Newton's Binomial, there are : binomiall[FONT=MathJax_Size2]([FONT=MathJax_Main]10 [/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]210[/FONT] different hand combinations.[/FONT]
To expand even further, how do the probabilities change if the source deck ceases to be infinite? From trial and error I can see that if the deck has only 10 cards then the player has to draw at least 1 complete pair.
EDIT: For example, a hand of {1,1,3,5,5,9} will get no points. A hand of {1,1,2,3,4,5} will get 2.

EDIT2: Please don't forget the last question of how the size of the card deck changes those odds: e.g. the deck is of n*10 size. Or at least 40, or something...

antonadelson said:

I asked this on math.stackexchange :

http://math.stackexchange.com/questions/320579/probability-question-involving-sets-of-cards

I have an infinite deck built out of sets of 10 cards (in other words 10*n cards). The sets are identical so one '2' is identical to another '2'.
A player draws 6 cards. If he draws:

• any '1' AND a '2', or
• any '3' AND a '4', or
• any '5' AND a '6', or
• any '7' AND a '8', or
• any '9' AND a '10',

he wins. In other words there are 5 pairs and if the player draws a complete pair he gets a point.
What is the probability he won't win any points at all?
To expand on the problem, if the player gets a point for every pair he completes in a hand, what is the probability he'll get 1, 2, or even 3 points? (3 points being 6 cards of 3 completed pairs)
From what I know of Newton's Binomial, there are : binomiall[FONT=MathJax_Size2]([FONT=MathJax_Main]10 [/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]210[/FONT] different hand combinations.[/FONT]
To expand even further, how do the probabilities change if the source deck ceases to be infinite? From trial and error I can see that if the deck has only 10 cards then the player has to draw at least 1 complete pair.
EDIT: For example, a hand of {1,1,3,5,5,9} will get no points. A hand of {1,1,2,3,4,5} will get 2.

EDIT2: Please don't forget the last question of how the size of the card deck changes those odds: e.g. the deck is of n*10 size. Or at least 40, or something...

Click to expand...

Since you haven't shown any work, I have to assume you are having trouble setting up the problem.

Consider 6 draws from the deck. Whatever the first card is, there is exactly one value that gets a point. What is the probability the 2nd card is NOT that value?
If two have been drawn but no point yet, there are two possible values that give a point and 8 values that do not.
If three have been drawn and no points, then...
If four...
If five...

antonadelson said:

I asked this on math.stackexchange :

http://math.stackexchange.com/questions/320579/probability-question-involving-sets-of-cards

I have an infinite deck built out of sets of 10 cards (in other words 10*n cards). The sets are identical so one '2' is identical to another '2'.
A player draws 6 cards. If he draws:

• any '1' AND a '2', or
• any '3' AND a '4', or
• any '5' AND a '6', or
• any '7' AND a '8', or
• any '9' AND a '10',

he wins. In other words there are 5 pairs and if the player draws a complete pair he gets a point.
What is the probability he won't win any points at all?
To expand on the problem, if the player gets a point for every pair he completes in a hand, what is the probability he'll get 1, 2, or even 3 points? (3 points being 6 cards of 3 completed pairs)
From what I know of Newton's Binomial, there are : binomiall[FONT=MathJax_Size2]([FONT=MathJax_Main]10 [/FONT][FONT=MathJax_Main]6[/FONT][FONT=MathJax_Size2])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]210[/FONT] different hand combinations.[/FONT]
To expand even further, how do the probabilities change if the source deck ceases to be infinite? From trial and error I can see that if the deck has only 10 cards then the player has to draw at least 1 complete pair.
EDIT: For example, a hand of {1,1,3,5,5,9} will get no points. A hand of {1,1,2,3,4,5} will get 2.

EDIT2: Please don't forget the last question of how the size of the card deck changes those odds: e.g. the deck is of n*10 size. Or at least 40, or something...

Click to expand...

Since you haven't shown any work, I have to assume you are having trouble setting up the problem.

Consider 6 draws from the deck. Whatever the first card is, there is exactly one value that gets a point. What is the probability the 2nd card is NOT that value?
If two have been drawn but no point yet, there are two possible values that give a point and 8 values that do not.
If three have been drawn and no points, then...
If four...
If five...

Yes, you're right. I don't know exactly where to begin. So far I've only managed to calculate the probability of having a 'x' card in the hand. I'll show the calculation later.

In reply to your idea. I've reached a stumbling block.

(i'm using an infinite deck in this calculation)
After the first card is drawn, there's a 9/10 chance the second card is not going to be a matching pair.
Then the chances split. Because if the second card is identical to the first then there's still a 9/10 chance the third card is not going to be a matching pair.
But if the second card is not identical, then there's a 8/10 chance.

Example#1:
[1] <- first card | 9/10 chance the second card is not going to be a matching pair
[1] <- second catf | 9/10 chance the third card is not going to be a matching pair
or in total of 81/100 chance that three cards will not have a matching pair
Example#2:
[1] <- first card | 9/10 chance the second card is not going to be a matching pair
[3] <- second card | 8/10 chance the third card is not going to be a matching pair
or in total of 72/100 chance that three cards will not have a matching pair

how do I combine the two different probabilities?

P.S. To calculate the chance of drawing at least a single '1' in the entire hand I calculate the chance of not drawing it at all and then substract it from one. 1-(9/10)^6 ~ 46.8559% | which fits my computer analysis.

Wow. Thanks to your set up and this page, I've managed to calculate the chance of getting no pairs in an infinite deck. (the first question in the original question).

I'll try to describe the solution is good as I can without drawing and using only 4 cards in a hand.

Below are the possible outcomes without any matching pairs and the chance for each such outcome (with the first card being drawn as '1' by default):
('X' is any card different from the ones drawn AND not being a pair to any of the drawn cards)
('P' is any card already drawn so, for example, {1,3,P} can be {1,3,1} or {1,3,3}, and {1,P} is actually only {1,1})
DRAW SECOND CARD:
#1{1,1} 1/10
#2{1,X} 8/10 (because X is not '1' or its pair)

Now this is where I split it in 2. For example, if the first outcome actually happens, it has its own set of probabilities for drawing the third card which will not be a matching pair. If #2 happens, then the probabilities are going to be different. This creates sort of a probability tree. Later I'll have to multiply the probabilities in the same branch to get the total probability after all 6 cards are drawn.

THIRD CARD:
#1.1{1,1,1} 1/10
#1.2{1,1,X} 8/10
#2.1{1,3,P} 2/10 (because P is either '1' or '3')
#2.2{1,3,X} 6/10

FOURTH CARD:
#1.1.1{1,1,1,1} 1/10
#1.1.2{1,1,1,X} 8/10
#1.2.1{1,1,3,P} 2/10
#1.2.2{1,1,3,X} 6/10
#2.1.1{1,3,P,P} 2/10
#2.1.2{1,3,P,X} 6/10
#2.2.1{1,3,5,P} 3/10
#2.2.2{1,3,5,X} 4/10 (because X is neither 1,3,5 nor any of the matching pairs: 2,4,6)

Now to get the actual probabilities of drawing the 4th non-matching card I multiply the probabilities of each of the 8 possible results with the probabilities earlier in the "tree".
For example, to get the probability of the result {1,1,1,9} I multiply these probabilities: #1.1.2 * #1.1 * #1 = 8/10*1/10*1/10 = 8/1000
or another example, for the result of {1,3,5,3} I multiply: #2.2.1 * #2.2 * #2 = 3/10*6/10*8/10 = 144/1000

Then I add all the final probabilities and get the result: 53.7%, which matches the results of my js script.

Now to work some more on answering the other 2 questions *sigh* Any help?

Excellent!
I had not gone far enough into the solution to recognize the complication of having multiple cards with same value.

Suppose the deck has n sets. Original number of cards is 10*n.

DRAW SECOND CARD:
#1{1,1} 1/10
#2{1,X} 8/10 (because X is not '1' or its pair)

Click to expand...

becomes
#1{1,1} (n-1)/(10n-1)
#2{1,X} 8n/(10n-1)

THIRD CARD:
#1.1{1,1,1} 1/10
#1.2{1,1,X} 8/10
#2.1{1,3,P} 2/10 (because P is either '1' or '3')
#2.2{1,3,X} 6/10

Click to expand...

#1.1{1,1,1} (n-2)/(10n-2)
#1.2{1,1,X} 8n/(10n-2)
#2.1{1,3,P} 2(n-1)/(10n-2)
#2.2{1,3,X} 6n/(10n-2)
The largest affect seems to be that the probability of drawing the same number multiple times goes down. Probability of "different" numbers only increases because the total number of cards available decreases.

Are you allowed to to a Monte Carlo calculation? Or is that what you are doing in your js script?

Yeah, that's what I do with my script to check my theories.

By the way, using this: http://mathforum.org/library/drmath/view/64626.html I've found a way to prove that there are totally of 5005 different variations for a hand in an infinite deck. Although I have no idea at the moment on how to use it.

All in all, I'm quite proud that I found a problem which stumbles so many mathematicians =)

Before I forget, I'd like to use this opportunity to immensely thank you, Dr. Phil, for sharing some of your knowledge and experience with me!