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Suppose I flip a fair coin twice and ask the question,

"What is the probability of getting exactly one head (and tail) ?"

I was confused on whether I would treat this as a combination or permutation. My original thought was that it is a combination as we don't care about the order and just want the case of one head (or tail) giving the probability of 1/3.

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    $\begingroup$ How do you get $\frac 13$? The winning sequences are $HT, TH$. $\endgroup$ – lulu May 16 at 14:30
  • $\begingroup$ If we don't care about order then isn't HT TH treated as the same thing? So that leaves 3 outcomes? $\endgroup$ – jodeg May 16 at 14:30
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    $\begingroup$ Ok, but those three outcomes are not equi-probable. $HH, TT$ each have probability $\frac 14$ $\endgroup$ – lulu May 16 at 14:32
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    $\begingroup$ That's up to you. But if you treat them as the same case, then the cases don't have the same probabilities. Personally, I think the math is simpler if you stick to equiprobable cases, but you don't have to. $\endgroup$ – lulu May 16 at 14:33
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    $\begingroup$ Remember that the probability of an event can be calculated as the number of outcomes in the event divided by the total number of outcomes in the sample space only when the outcomes are equally likely to occur. There are two possibilities when playing the lottery, you either win or you lose, but you certainly don't win the lottery with probability $\frac{1}{2}$. $\endgroup$ – JMoravitz May 16 at 14:38
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I think it is worth stressing that you can think of the outcomes as ordered or not, as you wish. That does not change the problem, nor does it change the final answer, but it does alter the computation.

Ordered: There are $4$ possible outcomes $$HH, HT, TH, TT$$ each of which has probability $\frac 14$. By inspection, the winning outcomes are $HT, TH$ so the answer is $\frac 12$.

Unordered: Now there are $3$ possible outcomes $$\{H,H\}, \{H,T\}, \{T,T\}$$ but these three are not equi-probable. Indeed, there is only one way to get $\{H,H\}$, say, which makes the probability of that outcome $\frac 14$. Same for $\{T,T\}$. There are two ways to get $\{H, T\}$ so that has twice the probability, again yielding $\frac 12$.

This comes up for dice throws, for example. If you consider ordered outcomes, then each outcome has probability $\frac 1{36}$. If you consider them as unordered, then the probability of getting distinct values, like $\{1,2\}$ is $\frac 2{36}=\frac 1{18}$ while the probability of getting a double like $\{1,1\}$ is still $\frac 1{36}$. Personally, I try to stick to equi-probable cases wherever possible, as that simplifies the computations. But it's up to you, and in some situations you really can't avoid considering outcomes with differing probabilities.

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The possible outcomes are $$HH, HT, TH, TT$$ Two out of four qualify, so the probability is $1/2$.

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Probability of getting exactly one head, this can happen in two ways : $HT + TH$

Favorable cases = 2

Total Cases = 4

So, required probability is $\frac{1}{2}$

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