Hi Stephanie,
There is a long-winded derivation that follows this, but if we count the second total draw as 1 because it is the first draw for a match, then
the average number of cards drawn is 2, that is, two draws after the first card, or the third card drawn in total.
Of course, if you count total draws including the first card that you need to match, then the answer is 3 rather than 2.
The easiest way to answer this is with a tree graph, but I can't do that here, so I will talk through it. Instead of red and green, let whatever color is drawn first be called "X," and let the other color be called "not X." The expected number of cards drawn is sum of the probability that we draw a match at each draw times the probability that we reach that draw (i.e., because a match wasn't achieved earlier) times the number of the draw.
For the second card drawn, there are seven cards left, 3 of which are X and four of which are not X. Therefore the probability of drawing a match on the first draw after the initial draw is 3/7. Of course we reach the second draw with probability 1, so we have that
the first term in the sum is 3/7.
For the third card drawn, the only way we get to a third card drawn is if the second card was not X, which happens with probability 4/7 because 4 of the 7 cards at the first round were not X. If we get to the third draw, there will be 6 cards left, three each of X and not X, and the probability of drawing X will be 3/6 or 1/2. Therefore the term this adds to the sum is (probability we make a third draw) x (probability third draw is a match) x (2, because this is the second draw), or
the second term in the sum is 8/14 = 4/7
For the fourth card drawn, the only way we can get this far is if the second and third draws were both not X, which happens with 4/7 probability in the first draw and 3/6 probability at the second draw, the product of these, 2/7 = 4/7 x 3/6 is the probability that we need a third draw after the first draw to obtain a match. Also, if we get to the fourth draw, we know that not X has been drawn twice which would mean that there are 5 cards left and 3 are X and 2 are not X. Therefore the probability of drawing X on the fourth draw is 3/5. Putting all this together, (2/7) x (3/5) x 3, so that
the third term in the sum is 18/35
For the fifth card drawn, we only need a fifth draw if the first three draws after the first fail to produce a match, which happens at the first draw with probability 4/7, the second draw with probability 1/2, and the third draw with probability 2/5 - the probability that there will be a fifth card drawn is the product of these, which is 8/70, or 4/35. Also, we get to this draw only if we have drawn not X three times, which would mean that there are four cards left, three of which are X and one of which is not X, so that the probability of drawing X is 3/4. Therefore at the fifth draw, the contribution to the sum is (4/35) x (3/4) x 4 , or
the fourth term in the sum is 12/35
For the sixth card drawn, we can only get this far if we have drawn not X 4 times, which happens with the probabilities 4/7, 3/6, 2/5 and 1/4 - multiplying these together gives the probability of needing a sixth draw to be
(4/7) x (3/6) x (2/5) x (1/4) = 24/840 = 1/35. If we get to the sixth draw, then of course there are no not X cards left, so we draw a match at this point with probability 1. Therefore
the fifth term in the sum is 1/7
Add all of these terms together to obtain the result given earlier.
Kind regards,
Hugh
