Alex, Bill, and Chen each in turn toss a balanced coin. The first one to throw a head wins.
a. What are their respective chances of winning if each tosses only one time?
b. What are their respective chances of winning if they continue, given a maximum of
two tosses each?
c. What are their respective chances of winning if they continue until someone wins?
Answers:
(a)P(A)=1/2, P(B)=1/4, P(C)=1/8
(b)P(A)=9/16, P(B)=9/32, P(C)=9/64
(c)P(A)=4/7, P(B)=2/7, P(C)=1/7
Help I don't understand how to arrive at these answers:(
To understand the chances of winning for each person in these scenarios, we need to consider the probability of getting a head on a single coin toss and then apply it to each situation.
a. For each person to win on their first toss, they need to get a head. Since the coin is balanced, the probability of getting a head on a single toss is 1/2. Therefore, the chances of each person winning on their first toss are:
P(A) = 1/2, P(B) = 1/2 * 1/2 = 1/4, P(C) = 1/2 * 1/2 * 1/2 = 1/8
b. In this case, each person has a maximum of two tosses. To calculate the chances of winning, we need to consider the combinations of outcomes that result in one person winning within these constraints. Let's break it down:
- Person A can win in two ways: (H), (TH)
- The probability of (H) is 1/2 * 1 = 1/2
- The probability of (TH) is 1/2 * 1/2 = 1/4
- The total probability of Person A winning is 1/2 + 1/4 = 3/4
- Person B can win in three ways: (T,H), (TH), (TT,H)
- The probability of (T,H) is 1/2 * 1 = 1/2
- The probability of (TH) is 1/2 * 1/2 = 1/4
- The probability of (TT,H) is 1/2 * 1/2 * 1 = 1/4
- The total probability of Person B winning is 1/2 + 1/4 + 1/4 = 9/16
- Person C can win in four ways: (T,T,H), (T,H), (TH), (TTT,H)
- The probability of (T,T,H) is 1/2 * 1/2 * 1 = 1/4
- The probability of (T,H) is 1/2 * 1 = 1/2
- The probability of (TH) is 1/2 * 1/2 = 1/4
- The probability of (TTT,H) is 1/2 * 1/2 * 1/2 = 1/8
- The total probability of Person C winning is 1/4 + 1/2 + 1/4 + 1/8 = 9/16
c. In this scenario, the game continues until someone wins. The chances of each person winning can be calculated using the concept of geometric series.
- The probability of Person A winning is equal to the probability of them winning on their first toss (1/2) plus the probability of them winning on their second toss (1/2 * 1/2) plus the probability of them winning on their third toss (1/2 * 1/2 * 1/2), and so on. This can be written as:
- P(A) = 1/2 + 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + ...
- Using the formula for the sum of an infinite geometric series, we can calculate P(A) = 1/2 / (1 - 1/2) = 1/2 / (1/2) = 1
- Similarly, the probability of Person B winning is:
- P(B) = 1/2 * 1 + 1/2 * 1/2 + 1/2 * 1/2 * 1/2 + ...
- Using the formula for the sum of an infinite geometric series, we can calculate P(B) = (1/2 * 1) / (1 - 1/2) = 1/2
- The probability of Person C winning is also:
- P(C) = 1/2 * 1/2 * 1 + 1/2 * 1/2 * 1/2 + ...
- Using the formula for the sum of an infinite geometric series, we can calculate P(C) = (1/2 * 1/2 * 1) / (1 - 1/2) = 1/4
Therefore, the chances of each person winning if they continue until someone wins are P(A) = 1, P(B) = 1/2, and P(C) = 1/4.