Question

What is the probability of getting the sequence hhhttt in six tosses of a coin? If you know the first three are heads, what is the probability that the last three are tails? If you don't know anything about the first threc, what is the probability that the last three are tails?

Short answer

The probability of getting hhhttt is \( \frac{1}{64} \). If the first three are heads, the probability of the last three being tails is \( \frac{1}{8} \). Without knowing the first three, the probability of the last three being tails is \( \frac{1}{8} \).

Step-by-step solution

- Understand Event Probability

Each toss of a fair coin has a probability of landing heads or tails, denoted as P(H) = 1/2 and P(T) = 1/2.

- Calculate Probability of hhhttt

To find the probability of getting the exact sequence hhhttt in six tosses of a coin, multiply the probability of each independent event: \[ P(\text{hhhttt}) = P(H) \times P(H) \times P(H) \times P(T) \times P(T) \times P(T) = \left( \frac{1}{2} \right)^6 = \frac{1}{64} \]

- Conditional Probability For Tails

Given that the first three tosses are heads, the sequence is now reduced to three remaining tosses that must all be tails. The probability of three tails in three tosses is: \[ P(\text{ttt}) = P(T) \times P(T) \times P(T) = \left( \frac{1}{2} \right)^3 = \frac{1}{8}\]

- Probability of Last Three Tails

Without any information about the first three tosses, the probability of the last three being tails is still determined independently: \[ P(\text{ttt}) = P(T) \times P(T) \times P(T) = \left( \frac{1}{2} \right)^3 = \frac{1}{8}\]

Key concepts

Event Probability

Probability is a way to measure the likelihood that an event will happen. For a coin toss, this is straightforward: you have a fair coin, so the probability of landing heads (H) is 1/2 and the probability of landing tails (T) is also 1/2. These probabilities are because there are only two possible outcomes, and they are equally likely.
When we want to find the probability of a sequence of events, like hhhttt in six tosses of a coin, we multiply the probabilities of each individual event. Since each toss is independent (the outcome of one toss doesn't affect the next), we have:
\( P(\text{hhhttt}) = P(H) \times P(H) \times P(H) \times P(T) \times P(T) \times P(T) = \left( \frac{1}{2} \right)^6 = \frac{1}{64} \)
This means you have a 1 in 64 chance of getting the sequence hhhttt in six coin tosses.

Conditional Probability

Conditional probability is the probability of an event occurring given that another event has already occurred. In our case, you are given that the first three tosses are heads (H). So, to find the probability that the last three are tails (T), you're only focusing on a smaller sequence—the last three coin tosses—because the condition (first three heads) has already happened.
With the first three heads fixed, you then calculate the probability of the remaining sequence being ttt (three tails):
\[ P(\text{ttt}) = P(T) \times P(T) \times P(T) = \left( \frac{1}{2} \right)^3 = \frac{1}{8} \]
This result shows that the probability of getting three tails in a row, starting fresh from where the condition (first three heads) left off, is 1/8.

Independent Events

Independent events are events where the outcome of one event does not affect the outcome of another. Coin tosses are a classic example of independent events. The outcome of the second toss is not influenced by the first toss. Each toss has an equal chance of being heads or tails—always 1/2.
When calculating the probability of the sequence hhhttt without any special conditions, the last three tosses being tails (ttt) are treated independently of the first three tosses. Thus, whether or not you know the results of the first three tosses, the probability calculation for the last three tosses is independent:
\[ P(\text{ttt}) = P(T) \times P(T) \times P(T) = \left( \frac{1}{2} \right)^3 = \frac{1}{8} \]
Here, you get the same probability of the last three tosses being tails, confirming their independence from earlier outcomes.