Summer Problem Solving Marathon Solution #9

Let P_n be the probability of winning the game when it is played with n pairs of cards in the deck. Clearly P₁ = P₂ = 1, because the game cannot be lost with either 1 or 2 pairs in the deck.

Suppose we already know the value of P_n-1, and we are thinking about what happens in the game with n pairs. Here are four possibilities:

#1: The first two cards we draw form a pair. This pair is then discarded, and from this point on, the game proceeds as if it were a game with n-1 pairs, so the probability of victory from this point is P_n-1.

#2: The first two cards are different, but the third card matches the first. Then the pair formed by the first and third cards is discarded, and play continues as if it were a game with n-1 pairs (with the first card already drawn, but this does not affect the probabilities). So the probability of victory from this point is P_n-1.

#3: The first two cards are different, but the third card matches the second. Then the pair formed by the second and third cards is discarded, and play continues as if it were a game with n-1 pairs. Again, the probability of victory from this point is P_n-1.

#4: The first three cards are all different. The player then loses.

These four options are exclusive and exhaustive, so the sum of their probabilities of occurring must be 1. Consider the probability of the fourth option. Once the first card is drawn, 2n-1 cards remain. 2n-2 of these are different from the first card, so there is a ^2n-2⁄_2n-1 chance that the first two cards will be different.

After the second card has been drawn, 2n-2 cards remain. Of these, 2n-4 are different from both the first and the second card. So the probability that the third card will not match the first or the second is ^2n-4⁄_2n-2.

So the chance that the first three cards are all different is ^2n-2⁄_2n-1 x ^2n-4⁄_2n-2 = ^2n-4⁄_2n-1.

So the chance that one of the first three options occurs is 1 - ^2n-4⁄_2n-1 = ^{2n-1 - (2n-4)}⁄_2n-1 = ³⁄_2n-1.

Given that one of the first three options occurs, the probability of victory is P_n-1>. So P_n, the overall probability of victory when there are n pairs, is ^3P_n-1⁄_2n-1.

We already know that P₂ = 1. So using this formula, we have:

P₃ = ^3x1⁄₅ = ³⁄₅.
P₄ = ^{3 x 3⁄₅}⁄₇ = ⁹⁄₃₅.
P₅ = ^{3 x 9⁄₃₅}⁄₉ = ³⁄₃₅.
P₆ = ^{3 x 3⁄₃₅}⁄₁₁ = ⁹⁄₃₈₅.

So the probability of victory is ⁹⁄₃₈₅.

Summer Problem Solving Marathon Question #11

[Value = 3 points]

The negative of one of the solutions of x² - 3x + c = 0 is a solution of x² + 3x - c = 0.

What are the solutions of x² - 3x + c = 0?

Summer Problem Solving Marathon Solution #8

The equation x² + y² = 4 has as its graph a circle of radius 2 centered at the origin. The equation y = |x| then cuts off a 90 degree sector of that circle. The area of the whole circle is 4π, so the area of the 90 degree sector is ¹⁄₄ of that area, or π.

Summer Problem Solving Marathon Question #10

[Value = 9 points]

Let {a₁, a₂, ..., a_n} be a set of real numbers, such that a₁ < a₂ < ... < a_n. We define the power sum of {a₁, a₂, ..., a_n} to be a₁i¹ + a₂i² + ... + a_niⁿ.

Given any n, let S_n be the sum of the power sums of all non-empty subsets of the set {1,2,...,n}.

S₈ is -176 - 64i. What is S₉?

Summer Problem Solving Marathon Solution #7

The averages of X and Y are 14 apart. The average of X ∪ Y is 6 above that of Y and 8 below that of X. So the sizes of X and Y must be in the ratio of 6:8.

The averages of Y and Z are 18 apart, with the average of Y ∪ Z 10 above that of Y and 8 below that of Z. So the sizes of Y and Z must be in the ratio of 8:10.

Thus X, Y, and Z are in the ratio of 6:8:10. For convenience, we can take them to in fact be of sizes 6, 8, and 10.

Then X has 6*37 = 222 years of age in it. Y has 8*23 = 184 years of age in it. And Z has 10*41 = 410 years of age in it. That's a total of 222 + 184 + 410 = 816 years of age. There are a total of 24 people in X, Y, and Z combined, so the average age is 816/24 = 34.

Summer Problem Solving Marathon Question #9

[Value = 8 points]

A deck of cards contains twelve cards: two marked with the letter A, two with the letter B, two with the letter C, two with the letter D, two with the letter E, and two with the letter F.

A person plays a game with these cards in the following way. The deck is randomized. The player then draws cards one at a time from the deck. Any time the player has two matching cards in his hand, he discards them. If the player ever has three unmatched cards in his hand, he loses. If he draws every card from the deck without losing, he wins.

What is the probability that the player wins the game?

Summer Problem Solving Marathon Question #8

[Value = 3 points]

What is the area of the smallest region bounded by the graphs of y = |x| and x² + y² = 4?