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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?

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    $\begingroup$ The missing 25% comes from the extra T in TT. $\endgroup$
    – Stef
    Commented Sep 14, 2023 at 11:53
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    $\begingroup$ Solution 2 is correct on its own, it just doesn't answer the question. Everything you said in your solution 2 is correct. It is just the answer to a different question. Note that your original question is slightly ambiguous and should probably be reworded as "at least a tail" to make it unambiguous and to make it clear why solution 2 is not an answer to this question. $\endgroup$
    – Stef
    Commented Sep 14, 2023 at 12:03
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    $\begingroup$ The incorrect part of Solution 2 is "so 100%". The expectation being one does not imply that the probability is 100%. $\endgroup$
    – usul
    Commented Sep 14, 2023 at 19:53

4 Answers 4

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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 $$\frac14 \times 2 + \frac14 \times 1 + \frac14 \times 1 + \frac14 \times 0 =1.$$

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    $\begingroup$ Upvoted for the first sentence, "These are answers to different questions." In probabilities it is extremely important to be careful and precise and specifics with words. That's important in all subfields of mathematics, but even more so in probabilities. $\endgroup$
    – Stef
    Commented Sep 14, 2023 at 11:58
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    $\begingroup$ @Stef The question used to include "Maybe there is a difference in interpretation or these are answers to different questions?" before it was edited $\endgroup$
    – Henry
    Commented Sep 14, 2023 at 12:18
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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:

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

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    $\begingroup$ Thus, with $\mathbb{P}(T_1) = 0.5$, $\mathbb{P}(T_2) = 0.5$, and $\mathbb{P}(T_1 \cap T_2)= 0.25$ we get $\mathbb{P}(T_1 \cup T_2) = 0.5+0.5-025 = 0.75$, same as in (1) $\endgroup$
    – pygri
    Commented Sep 15, 2023 at 7:22
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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.

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  • $\begingroup$ It would be sequitur in the particular case of a Bernoulli experiment, where expectation and probability of success are equal. $\endgroup$
    – Stef
    Commented Sep 14, 2023 at 15:28
  • $\begingroup$ @Stef Right. Unfortunately, the application in (2) is to a Geometric distribution. $\endgroup$
    – whuber
    Commented Sep 14, 2023 at 16:13
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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.

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