1. Consider the "coin flipping game", and let X record the �first time when the Head" player has $1. What is the probability that X � 7?
2. Continue with the "coin flipping game". Let Y record the fi�rst point in time when the "Head" player has $2. What is the probability that Y � 7? What is the probability that Y is even? What is the probability that Y is odd(Do these latter two probabilities sum to 1?)
for #1 What is the probability that X<=7
For #2 What is the probability that Y <= 7.
Typo
To answer these questions, we need to understand the rules of the coin flipping game.
The coin flipping game involves flipping a fair coin repeatedly. Each time the coin is flipped, if it lands on heads, the "Head" player gains $1. The goal is to track the number of flips it takes until the player reaches a certain amount of money.
1. Probability that X ≤ 7:
To calculate this probability, we need to determine the probabilities of the "Head" player reaching $1 in 1, 2, 3, 4, 5, 6, or 7 flips and add them up.
- In the 1st flip: The probability of landing heads is 1/2, so the player reaches $1 in one flip with a probability of 1/2.
- In the 2nd flip: The first flip must be tails and the second flip must be heads. The probability of this sequence is (1/2) * (1/2) = 1/4.
- In the 3rd flip: The first two flips must be tails, and the third must be heads. The probability is (1/2) * (1/2) * (1/2) = 1/8.
- Continuing this pattern, the probability for each subsequent flip decreases. So, we have:
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 = 127/128.
Therefore, the probability that X ≤ 7 is 127/128.
2. Probability that Y ≤ 7:
To calculate this probability, we need to determine the probabilities of the "Head" player reaching $2 in 1, 2, 3, 4, 5, 6, or 7 flips and add them up.
- In the 1st flip: The probability of landing heads is 1/2, so the player reaches $2 in one flip with a probability of 1/2.
- In the 2nd flip: The first flip must be tails and the second flip must be heads, resulting in $1. The probability of this sequence is (1/2) * (1/2) = 1/4.
- In the 3rd flip: The first two flips must be tails, and the third must be heads, resulting in $2. The probability is (1/2) * (1/2) * (1/2) = 1/8.
- Continuing this pattern, the probability for each subsequent flip decreases. So, we have:
1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 = 127/256.
Therefore, the probability that Y ≤ 7 is 127/256.
3. Probability that Y is even:
To calculate this probability, we need to determine the probabilities of the "Head" player reaching $2 in an even number of flips.
- The Base Case: The probability of reaching $2 in 0 flips (starting with $0) is 0.
- In the 2nd flip: The probability of reaching $2 in 2 flips is (1/2) * (1/2) = 1/4.
- In the 4th flip: The probability of reaching $2 in 4 flips is (1/2) * (1/2) * (1/2) * (1/2) = 1/16.
- Continuing this pattern, the probability for each subsequent even flip decreases. So, we have:
1/4 + 1/16 + 1/64 + 1/256 + ... = 1/3.
Therefore, the probability that Y is even is 1/3.
4. Probability that Y is odd:
To calculate this probability, we need to determine the probabilities of the "Head" player reaching $2 in an odd number of flips.
- In the 1st flip: The probability of reaching $2 in 1 flip is 0 because $2 cannot be reached in that short time.
- In the 3rd flip: The probability of reaching $2 in 3 flips is (1/2) * (1/2) * (1/2) = 1/8.
- In the 5th flip: The probability of reaching $2 in 5 flips is (1/2) * (1/2) * (1/2) * (1/2) * (1/2) = 1/32.
- Continuing this pattern, the probability for each subsequent odd flip decreases. So, we have:
1/8 + 1/32 + 1/128 + 1/512 + ... = 1/5.
Therefore, the probability that Y is odd is 1/5.
The probabilities that Y is even and Y is odd do not sum to 1 since there's still a small probability of reaching $2 after more than 7 flips. However, the sum of the probabilities of reaching Y is even and Y is odd is indeed 1/3 + 1/5 = 8/15, not equal to 1.