Pokhran II wrote:
Thanks all for trying!
The question is from Manhattan CAT.
The OE and OA:
The problem asks us to find the number of cards that must be turned over before the probability is greater than 1/2 that a pair shows up.
This is a difficult problem involving events in succession.
The probability of having a pair on the first card is 0, regardless of which card is selected.
For the second card, there are 51 cards left in the deck and 3 of them will form a pair with the card already on the table, so the probability that there is a pair after two cards is:
For the third card, there are 50 cards left in the deck and 6 of them would pair with one of the two cards already on the table (if a pair has not already been formed). As this involves an OR event, we add the probabilities. The probablity that there is pair after three cards is:
(3/51) + (6/50). This sum is less than 1/2.
For the fourth card, there are 49 cards left in the deck, and 9 of them will pair with one of the three cards already on the table (if a pair has not already been formed). The probability that there is a pair after four cards is:
(3/51) + (6/50) + (9/49). This sum is less than 1/2.
For the fifth card, there are 48 cards left in the deck, and 12 of them will pair with one of the four cards already on the table (if a pair has not already been formed), so the probability that there is a pair after five cards is:
(3/51) + (6/50) + (9/49) + (12/48). This sum is greater than 1/2, as can be quickly estimated given that the numerators add up to 30, which is clearly more than half of the average denominator of approximately 50.
When you are dealt five cards at random, the probability is greater than 1/2 that you will have a pair after five cards have been dealt.
The correct answer is B.
i'd argue that this OE is not correct.... i'll explain on the first wrong statement in my view:
the probability of having a pair of same value in 3 cards is not 3/51+6/50.
true... for 2 cards probability of a pair is 3/51
but for three cards you have:
the reason for the 6/50 to be added it must be, at the same time that the first two didn't yield a pair (which is 48/51).
to help visualize this correction suppose there are only 3 possible values A,B,C each with two cards... so we have A1,A2,B1,B2,C1,C2 cards...
what is the probability to have 2 of the same letter in a three card drawing...
according to the reasoning of the OE it would be:
1/5 (for the second card)+2/4 (for the third card) = 7/10
there are 120 possibilities for 3 card drawings (6*5*4)
lets count all triplets that contain a pair
A1,A2 (4 out of 120)
A2,A1 (4 out of 120)
similarly (B1 B2), (B2 B1), (C1 C2) and (C2 C1) add 4 each.
total 24 triplets whose first 2 are pair (no surprise... we already calculated it would be 1/5 of 120)
lets count those triplets whose first and third form a pair
similarly for (B1 * B2) (B2 * B1) (C1 * C2) (C2 * C1)
total of 24.
and last, we count those triplets whose second and third form a pair - and in a similar manner we have 24...
total - 72 out of 120. that represent a probability of 0.6 not 0.7 !!!!!
hence you can see that the reasoning of the OE is wrong.
i still think that the answer to the original question is 6 (C).
for 2 cards it would be p(2)=3/51
for 3 cards p(3) = p(2)+(1-p(2))*6/50 = 3/51+48/51*6/50
for 4 cards it is p(4)=p(3) + (1-p(3))*9/49
and so forth.... a bit more complicated formula...
my claim was (and still is) that my approach provide a simpler view of the "opposite" of the required formula. hence help to solve the question correctly.