making at least a double pair out of 52 cards

[durexlw]

Question:
A deck of 52 cards, 2 cards are dealt. We assume that the card we have
in our hand are not a pair.
For example we get: '3 of hearts' and '2 of clubs'.

What is the probability that if you draw 3 cards at random from the remaining 50, you'll get at least another 2 and at least another 3? In other words: having two 'not pair' cards, what is the probability of making at least two pairs (or full house) if you draw 3 cards from a deck of the remaining 50 cards?

Difficulty:
2 Cards, unpaired, are now in our hands. 50 cards remaining.
Currently 6 cards can make one pair. So when the first card hits the
table, the chance of 'not making' a pair is:
(50-6)/50

When the second card hits the table (that becomes the 4th visible card
in total), 49 cards are remaining, but only 3 of these can make a
pair. So when the second card is placed on the table, the chance of
not making a double pair is:
(49-3)/49

Now what confuses me is: I could multiply (50-6)/50 *
(49-3)/49 = 2024/2450
Then 1 - (2024/2450) would give me the chance on having (at least a
double pair): 426/2450... this would mean that once out of 5.75
times would result in a double pair... this looks pretty high and this
is only after 2 cards have been dealt... one is remaining, so I'm
wondering: What did I overlook?

What do I do with the third card? How do I include this in this
calculation?
Keep in mind: we need *at least* (in this example) a 3 and a 2 on the
table. If on the table would appear: 3-3-2, this should be included in
the probability.

 

[Denis]

We assume that the card we have in our hand are not a pair.

If one card is dealt, then 3/51 remaining cards can make a pair. 48/51 cards can not make a pair.
2 Cards are now in our hands. 50 cards remaining.

Why are you "looking" at the 1st 2 cards: you said it is assumed that they're not a pair?

Can't your problem be worded:
you have a 2 and a 3 turned up:
what is the probability that if you draw 3 cards at random from the remaining 50,
you'll get at least another 2 and at least another 3 ?

 

[durexlw]

Why are you "looking" at the 1st 2 cards: you said it is assumed that they're not a pair?

Can't your problem be worded:
you have a 2 and a 3 turned up:
what is the probability that if you draw 3 cards at random from the remaining 50,
you'll get at least another 2 and at least another 3 ?

You are absolutely right. I'll edit the first post. Thanks for pointing this out.

 

[pka]

Just calculate the number of ways to get each of {2,3,X}, {2,2,3}, {2,3,3} and add.

 

[durexlw]

Just calculate the number of ways to get each of {2,3,X}, {2,2,3}, {2,3,3} and add.

Sir,

I agree this is one way, but this would lead to a very enormous list... we need to remember there are at least 44 cards, in addition to the 6 cards that can make a double pair and each of these 44 cards can in on of three positions... Only distributing 44 cards would be a tremendous work. For example: 2-3-t, t-2-3, 2-t-3, ... I'm lead to believe that this is the whole idea behind math: to make such tedious work redundant.

My goal is not to just solve this problem, but to use the solution to this problem to solve many others... whereas here the list might be a possibility in the remaining problems I face, this is no option at all.

Let it be clear that I appreciate you suggestion, considering my goal however: I'm in need of formula's, list are no option.

 

[soroban]

Hello, durexlw!

A deck of 52 cards, twp cards are dealt.
Suppose we get: .\(\displaystyle 3\heartsuit,\:2\clubsuit\)

What is the probability that if you draw 3 cards at random from the remaining 50,
you'll get at least another 2 and at least another 3?
In other words: what is the probability of making Two Pairs or a Full House?

\(\displaystyle \text{There are: }\:{50\choose3} \,=\,19,\!600\text{ possible outcomes.}\)


Two Pairs

There are 3 choices for the second deuce.
There are 3 choices for the second trey.
There are 48 choices for the third card.
. . \(\displaystyle \text{There are: }\:3\times3\times48 \:=\:\boxed{432}\text{ ways to get Two Pairs.}\)


Full House

There are two ways to get a Full Houses.. We must draw:
. . (a) two more deuces and a trey
. . (b) a deuce and two more treys

\(\displaystyle \text{(a) There are: }\,{3\choose2} \,=\,3\text{ ways to get two more deuces.}\)
. . .\(\displaystyle \text{There are 3 choices for the trey.}\)
. . .\(\displaystyle \text{Hence, there are: }\:3\times 3 \:=\:9\text{ ways to get "deuces over treys."}\)

\(\displaystyle \text{(b) There are: }\,{3\choose2} \,=\,3\text{ ways to get two more treys.}\)
. . .\(\displaystyle \text{There are 3 choices for the deuce.}\)
. . .\(\displaystyle \text{Hence, there are:}\:3\times 3 \:=\:9\text{ ways to get "treys over deuces."}\)

\(\displaystyle \text{So there are: }9 + 9 \:=\:\boxed{18}\text{ ways to get a Full House.}\)



\(\displaystyle \text{Therefore, there are: }\:432 + 18 \:=\:450\text{ ways to get Two Pairs or a Full House,}\)

. . \(\displaystyle P(\text{Two Pairs or Full House}) \;=\;\frac{450}{19,\!600} \;=\;\frac{9}{392}\)

 

[durexlw]

Hello, durexlw!

A deck of 52 cards, twp cards are dealt.
Suppose we get: .\(\displaystyle 3\heartsuit,\:2\clubsuit\)

What is the probability that if you draw 3 cards at random from the remaining 50,
you'll get at least another 2 and at least another 3?
In other words: what is the probability of making Two Pairs or a Full House?

\(\displaystyle \text{There are: }\:{50\choose3} \,=\,19,\!600\text{ possible outcomes.}\)

Sir,

Thank you for taking the time.
I have seen this expression '50 choose 3' many times now. Would you be willing to explain what formula is behind this? I do not recollect seeing this formula when I got math and I'd like to know how to calculate the result of such expression.

Two Pairs

There are 3 choices for the second deuce.
There are 3 choices for the second trey.
There are 48 choices for the third card.
. . \(\displaystyle \text{There are: }\:3\times3\times48 \:=\:\boxed{432}\text{ ways to get Two Pairs.}\)

Correct me if I am wrong, but to me, this seems to overlook the fact that it could be that the first card is not a deuce. Also, we now calculated the 2-3-x but 3-2-x, x-2-3, ... are all possible ways of having a double pair.

Mainly this article here: http://mathforum.org/library/drmath/view/61824.html, has led me to believe that the above reasoning overlooks this fact. Am I missing something here?


Also, when I follow the above reasoning and extend this to the question: What is the probability that if you draw 5 cards at random from the remaining 50, you'll get at least another 2 and at least another 3?
According to the above, this should be: \(\displaystyle 3\times3\times48\times47\times46\:=\:\boxed{933984}\) ... which is more than the 19600 possible outcomes out of 50 cards.

 

[pka]

Two Pairs
There are 3 choices for the second deuce.
There are 3 choices for the second trey.
There are 48 choices for the third card.
. . \(\displaystyle \text{There are: }\:3\times3\times48 \:=\:\boxed{432}\text{ ways to get Two Pairs.}\)

Correct me if I am wrong, but to me, this seems to overlook the fact that it could be that the first card is not a deuce. Also, we now calculated the 2-3-x but 3-2-x, x-2-3, ... are all possible ways of having a double pair.

There is a slight mistake in Soroban’s numbers. There are only 44 ways to pick a third card.

But there is a huge mistake in your reasoning. ‘Hands’ in a card game are content driven not order driven. The reference on the MathForum page to straights has to do with the way the five cards can be arranged once they are dealt out. It has nothing to do with the order in which they are dealt. As to your question about changing from 3 to five new cards here is the way to count them:
\(\displaystyle \sum\limits_{a = 1}^3 {\sum\limits_{b = 1}^3 {\sum\limits_{c = 0}^3 {\text{if}(a + b + c = 5,{3 \choose a}{3 \choose b}{44 \choose c},0)} } } = {\text{136890}}\)

 

[Denis]

There is a slight mistake in Soroban’s numbers. There are only 44 ways to pick a third card.

Agree. Soroban has really included the "full house" possibilities twice.
As far as I'm concerned, there's no need to worry about full houses.

This simple coding finds 'em all (using 1 and 2 as cards to be paired):

Loop a from 3 to 50
u = a : if need be, reduce u to below 14
Loop b from a+1 to 51
v = b : if need be, reduce v to below 14
Loop c from b+1 to 52
w = c : if need be, reduce w to below 14

if u=1 or v=1 or w=1 and u=2 or v=2 or w=2 then t=t+1

Ends up with 19600 cases; t = 414 : includes 44*9=396 plus the 18 "full houses".

 

[pka]

t = 414 : includes 44*9=396 plus the 18 "full houses".

414 is correct: \(\displaystyle \sum\limits_{a = 1}^2 {\sum\limits_{b = 1}^2 {\sum\limits_{c = 0}^1 {\text{if}(a + b + c = 3,{3 \choose a}{3 \choose b}{44 \choose c},0)} } } = {\text{414}}\)