# History dependency of probability

A coin flip has a 1/2 chance of being head or tail. It is said that the current flip has no relation to the previous flips and hence the chance remains at 1/2.

However, from what I was taught at school, the chance of getting 2 heads in a row is 1/2 * 1/2 = 1/4. If I go ahead and flip the coin for the third time, the probability of getting a head is now 1/4 * 1/2 = 1/8. Therefore the previous coin flip outcomes do have affect on the future ones.

Is the first claim about independency correct, or the math in the second claim correct? This is probably basic math but I cannot seem to find the answer anywhere or what to search for.

• The probability of getting two heads in a row is indeed $\frac{1}{4}$. But if you have flipped two heads, the probability of getting a third head is still $\frac{1}{2}$. Notice that $$\underbrace{P(3~\text{heads})}_{=\frac{1}{2^3}} \neq \underbrace{P(\text{heads} | \text{the two previous flips were heads})}_{=\frac{1}{2}}$$ – Matti P. May 23 '18 at 6:58
• Just one thing: $\frac14\times\frac12=\frac18$ (so not $\frac16$). I sincerely hope that is only a typo. – drhab May 23 '18 at 7:44

If you look at only one flip in front of you, the probability of guessing the outcome correctly is $\frac{1}{2}$. If you look at three flips that are in front of you, guessing all the three correctly is $$P(\text{3 correct guesses in a row}) = \left( \frac{1}{2} \right)^3$$ If you have flipped a coin twice and guessed correctly the outcome on both times, you have already entered into the "zone" of probability $\left(\frac{1}{2}\right)^2$. But looking forward in time, these do not have anything to do with the next flip, whose probability is still $\frac{1}{2}$. Does this make any sense to you?