Interpretation of coin toss: odds of getting a tail in two coin flips

Question

Let's say someone asks me: What are the odds of getting a tail when a coin is flipped twice?

Assuming the coin is fair and tosses are independent, I see two ways of solving this problem:

1. If we tabulate the results, we know the possible outcomes are: TT, HT, TH and HH. Therefore, the probability of getting at least one tail is 0.75.

2. Using a more intuitive approach, when I flip a coin, the probability of getting a tail is 0.5, so I can expect to get a tail every 1/0.5 = 2 flips. So, if I flip 2 times, I expect 1 tail, so 100%.

I know right answer is 75%, but I can't find a way to properly explain why solution 2 is incorrect. Maybe there is a difference in interpretation?

Answer 1

These are answers to different questions.

$75\%$ or $0.75$ is the answer to the question "What is the probability of at least one Tail when tossing a fair coin twice". You might calculate it from your equally likely TT, HT, TH and HH outcomes using an indicator variable as $$\frac14 \times 1 + \frac14 \times 1 + \frac14 \times 1 + \frac14 \times 0 =\frac34.$$

$1$ is the answer to the question "What is the expected total number of Tails when tossing a fair coin twice". You might calculate it from your equally likely TT, HT, TH and HH outcomes using a counting variable as $$\frac14 \times 2 + \frac14 \times 1 + \frac14 \times 1 + \frac14 \times 0 =1.$$

Answer 2

Let $T_1$ and $T_2$ denote the respective events that each of the tosses come up tails. The second statement is effectively asserting that:

$$\mathbb{P}(T_1 \cup T_2) = \mathbb{P}(T_1) + \mathbb{P}(T_2) \quad \quad \quad \quad \quad \text{(Erroneous)}$$

or that:

$$\mathbb{P}(T_1 \cup T_2) = \frac{\mathbb{E}(T_1 + T_2)}{2} \quad \quad \quad \quad \quad \quad \text{(Erroneous)}$$

The correct rule is the addition rule of probability:

$$\mathbb{P}(T_1 \cup T_2) = \mathbb{P}(T_1) + \mathbb{P}(T_2) - \mathbb{P}(T_1 \cap T_2).$$

Answer 3

The term "expectation" in statistics means the probability-weighted average over the whole possibility space. It does not mean "the result that will happen with 100% probability". For instance, if you roll a die, the expected value is 3.5, even though the probability of a single die having a value of 3.5 is zero. Going from "the expected number of tails is 1" to "The probability of getting a tail is 100%" is a complete non sequitur.

Answer 4

In addition to the good answers already given, there is in this situation a simple trick you can use to compute directly the probability.

Solve the inverse problem - what is the probability of not getting any tails? This is $$P_1 = 0.5$$ for the first throw and $$P_2 = 0.5$$ for the second throw, therefore this event has the probability of $$P = P_1*P_2 = 0.5*0.5 = 0.25$$

Since $P$ is the probability of not getting tails, the probability of getting at least one tail is $$Q = 1-P = 0.75$$ as in your first case.