2
$\begingroup$

Here is the situation: I have a two-sided, fair coin that I'm going to flip twice in a row. Before starting the experiment, the probability of any possible result of 2 flips (head-head, head-tail, tail-head, tail-tail) is 25% because there is a 50% chance of any given result per flip, twice in a row or 1/2 * 1/2 = 1/4 = 25%

My question is what happens if I evaluate the probability of the second flip, after the first flip has already occurred. In my math class, the teacher answered that the probability of a result on the second flip is still 25%, but this doesn't make sense to me. If the coin is memoryless, why would it not be 50% that say, after having gotten a heads on the first flip, the probability of tails coming up on the second flip would be 50% since past results can't influence future outcomes?

$\endgroup$
  • 2
    $\begingroup$ Your argument is right, there may have been miscommunication. The probability of a head on the second flip is $0.5$, both before and after the first flip. $\endgroup$ – André Nicolas Dec 19 '13 at 23:34
0
$\begingroup$

Yes, your argument is right (your teacher may have misunderstood you). Flip 1 ($F_1$) and Flip 2 ($F_2$) are independent. The outcomes of one do not affect the other. We can also look at it in terms of conditional probability:

$p(F_2 = H \mid F_1 = T) = p(F_2 = H)$, because the outcome of the first flip does not impact the outcome of the second flip. Note that this is different from $p(F_2 = H \land F_1 = T) = p(F_2 =H) \cdot p(F_1 = T)$, which is the multiplicative rule for independent events.

(If you are not familiar, $\land$ means and.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.