Flip coin until you see heads: probability of seeing at least one tails I keep on throwing a coin until it lands heads. Now I want to know what the probability is of seeing at least one tail. I know that the probabilities for heads and tails are both 0.5 and that they are independent, but I don't know how I should start. The problem is that the amount of throws is only restricted by getting a heads or not. Can anyone help me?
 A: When you see the words "at least" it pays to look at the opposite (complementary) event.  In this case the opposite event is "seeing no tails". If you see no tails that means you only see heads > that means the first throw of the coin lands heads and the game is over. Hopefully you can finish the problem from there.
A: The sample space, i.e. the list of possible outcomes, is:
$H:\ $ Probability $\ = \frac12$
$TH:\ $ Probability $\ = \frac14$
$TTH:\ $ Probability $\ = \frac18$
$TTTH:\ $ Probability $\ = \frac{1}{16}$
etc.
We see that all the probabilities in this countably infinite probability space sum to $1,\ $ which is what we expect.
Now, the only way not get at least one tail is if the first toss lands on Heads. The probability of this happening is $\ \frac{1}{2}.$
Therefore, the probability of getting at least one tail
$=1-P(\text{not getting at least one tail)} = 1 - \frac12 = \frac12.$
A: The probability of getting at least one flip to land tails is equal to $1-$(the probabilitty that you get $0$ flips to land tails). This only happens when the first flip is a heads, which is $\%50$ of the time. Therefore, the chance of getting at least one tails is:

$(1-\dfrac12)$ or $\dfrac12$.

