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rado_kotorov · Jan 9, 2011, 04:04 PM

If two people are given two different card decks, what is the probability they to draw exactly the same pair of cards - let us say each draws Ace and King of hearts.

What will be also the probability if one draws two cards and the other draws 4 cards the two to have a matching pair?

ebaines · Jan 10, 2011, 08:18 AM

To adresss the first question: it doesn't matter what the first person draws, it only matters whether the second person draws the same cards as the first. So for that second person: on his first draw he has a 2/52 chance of matching one of the first person's two cards. If that is successful, then on his second draw he must hit the one remaining card out of the 51 remaining. To be successful both of tehse events must occur, so you multiply their individual probabilities:

2/52 x 1/51

For the second question he is successful if he has two matches and two misses. First suppose the two matches are on the first two draws- the probability of that heppening is:

2/52 x 1/51 x 50/50 x 49/49

Same answer as before. BUT, you must now take into account the fact that the two matches can happen on different draws - for example draw 1 and draw 3. So - you must determine how many different ways there can be 2 matches in 4 draws, and multiply that by the answer above.

rado_kotorov · Jan 10, 2011, 11:25 AM

This is helpful. Thanks. Let me now put it into the context I am interests - what is the probability of two people giving 4 or more identical responses on a set of 7 binary questions.Is it 4/7x3/6x2/5x1/4?

ebaines · Jan 10, 2011, 01:54 PM

Don't use the comment box to ask a follow up question - use the answer box instead.

No. The probability of getting 4 or more the same is equal to the probability of getting for the same, plus the probability of getting 5 the same, plus 6 the same, plus 7 the same. The probability of getting any N the same out of 7 is C(7,N)*(1/2)^7. Do you see why that is? So add up C(7,4)+C(7,5)+C(7,6)+C(7,7) and multiply by (0.5)^7.

Or you could do a shortcut . Realizing that C(7,4) = C(7,3), and C(7,5) = C(7,2), etc, this means that the probability of getting 4 or more the same is equal to the probability of getting 3 or fewer the same. Since the sum of Prob(4 or more) + Prob(3 or fewer) = 1, that means the each of these equals 1/2.

rado_kotorov · Jan 10, 2011, 02:59 PM

Thanks - this is helpful. So I understand adding C(7,4) etc. but why the multiplication by (0.5)^7?

Your approach seems to give me half of the value if I use the binomial distribution formula ∑i=4(7i)pi(1−p)7−i. Your (0.5)^7 is equivalent to the formula.

ebaines · Jan 10, 2011, 03:07 PM

Unfortunately you can't cut and paste formulas into the ansewer box. You have to either use Latex or write things out long hand.

As to why I myultiply ny (1/2)^7: that's because there are 2^7 = 128 ways that 7 binary questions can be answered. If we assume that this is like tossing a coin 7 times, the probability of two people getting exactly the same sequence of answers (or coin flips) is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2, or (1/2)^7.

rado_kotorov · Jan 10, 2011, 03:18 PM

The coin analogy is great. So if I translate this into a statement that means that the probability of ANY matching sequence is 1/128, but the probability of four or more heads in the matched sequence is 0.50x1/128

rado_kotorov · Jan 10, 2011, 03:20 PM

The coin analogy is great. So if I translate this into a statement that means that the probability of ANY matching sequence is 1/128, but the probability of four or more heads in the matched sequence is 0.50x1/128

ebaines · Jan 10, 2011, 03:26 PM

Not quite. Yes, the probabilty of an exactly matching sequence is 1/128. But the probability of getting 4 or more to match (in seqence) is:

[C(7,4) + C(7,5) + C(7,6) + C(7,7)]*(1/2)^7 = (35+21+7+1)*1/128 = 64/128 = 1/2.

rado_kotorov · Jan 10, 2011, 03:36 PM

Here is where it is confusing. There are indeed 64 sequences of 4 or more head flips/ Yes answers. But each user will independently select one. So you may think of it as all combinations being tossed in a hat and user selecting one.

The formula I sent you estimates the success of n trials out of k possibilities. p^n * (1 - p)^7-n. Which is analogous to coin flip.

ebaines · Jan 10, 2011, 03:46 PM

Sorry, I'm not following you. The formula you sent previously in post #3 was 4/7 x 3/6 x 2/5 x 1/4 which equals 1/35. That is not the probaility of 4 matches out of 7 binary choices. Instead, that's the probability of matching 4 cards drawn from a deck of 7 without replacement - a different kind of problem entirely. And in post #8 you suggested an answer of 1/2 x 1/128, or 1/256, which is not correct either.

rado_kotorov · Jan 10, 2011, 04:01 PM

Apologies for the confusion. You have been very patient. I understand very well your statement that the probability two people to draw exactly the same sequence is 1/128. I also understand that the probability of two people selecting 4 or more "Yes" answers is 50 percent. That is still 64 possible sequences. So it seems to me that each respondent may answer in 64 different possible ways. This is where I am confused.

[C(7,4) + C(7,5) + C(7,6) + C(7,7)]*(1/2)^7 = (35+21+7+1)*1/128 = 64/128 = 1/2.

Unknown008 · Jan 10, 2011, 11:19 PM

Yes,

35 ways to get 4 'correct' answers, 21 ways to get 5 'correct' answers, 7 ways to get 6 'correct' answers and 1 way to get all 'correct' answers.

Hence, 64 ways to get 4 or more 'correct' answers.

That is to say, if I understood your confusion well.