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Let's say we flip a fair coin $1$ time. The probability of obtaining at least one heads is $50\%$.

However let's say we flip the coin $2$ times. The probability of obtaining at least one heads becomes $75\%$.

I can't seem to wrap my head around why our probability of obtaining at least one heads increases as we increase the number of times we flip the coin since each flip of the coin is an independent event and each time we have a $50\%$ of getting heads so shouldn't the probability of obtaining at least one heads be $50\%$?

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    $\begingroup$ If you flip the coin one million times, it would be surprising if you never got a head. $\endgroup$ – Henry Aug 12 '15 at 20:09
  • $\begingroup$ Let's say a couple wants to have at least one boy. Are they better off having two children or one? $\endgroup$ – Jahan Claes Aug 12 '15 at 20:14
  • $\begingroup$ Wow I feel silly now. I understand where I went wrong. I kept isolating each event instead of listing all possibilities. $\endgroup$ – Daniel Aug 12 '15 at 20:18
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shouldn't the probability of obtaining at least one heads be 50%?

No it's not. The odds of an individual flip being heads is still 50%, but as you flip a coin more times, the odds increase that at least one of the flips will come up heads.

It's easier to grasp if you consider the inverse: If you flip a coin twice, what are the odds of getting two tails? The odds would be $0.5 \times 0.5$ or $0.25$. Thus the odds of getting at least one heads is $1-25\%$ or $75\%$.

If you genericize it to n flips, the odds of getting all tails would be $(\frac{1}{2})^n$ which gets smaller as n increases, so the odds of getting at least one head is $(1- \frac{1}{2}^n)$ which increases towards 1 as n increases.

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The probability of getting at least one heads is 50% for each throw. However when we throw a coin twice, we have four possible outcomes. For the first throw we can have heads or tails, and for the second throw we can have heads or tails. Let $(x,y)$ where $x$ is the first throw and $y$ the second. We'll write $H$ for heads and $T$ for tails. Then the possible outcomes are: $$(H,H)$$ $$(H,T)$$ $$(T,H)$$ $$(T,T)$$ As you can see 3 of these have at least one heads. It could be easy to understand as well if you think about throwin 10 times. The chance to get at least one heads is quite large. This is because the probability of getting less than one heads, (all 10 times tails) is $\frac{1}{2^{10}}$. Therefore the chance of getting at least one heads is $1-\frac{1}{2^{10}}$. I hope this is clarifies it for you.

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If you are going to flip the coin twice, you know that if do fail to obtain a head on the first try, there will still be another chance, so how can the probability of obtaining at least one head in two flips be any thing but greater than the probability of obtaining one head on the first flip?

$$\begin{align} & \; \mathsf P(\text{at least one head in two flips}) \\[1ex] = & \;\mathsf P(\text{head first try}) + \mathsf P(\text{tail first try})\,\mathsf P(\text{head second try}) \\[1ex] = & \quad \frac 1 2 + \frac 1 2\times \frac 1 2 \\[1ex] = & \qquad \frac 3 4\end{align}$$

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