I toss two coins and look at the outcome.
I then tell you that at least one of the coins is showing up as "tails". What is the chance that the other one is showing "tails" as well?
(from techInterview.org)
(In reply to
re: I think the solution is WRONG by Lewis)
but it is precisely that the order does matter which creates the 1/3 answer.
Reading the problem carefully, consider the difference between the way it is stated "at least one" and the phrasing that would lead to your challenged answer of "the first one". In the problem, because the order is not stated does not mean it is not important, while counter-intuitively, while the alternate wording specifies the order, this makes it unimportant.
Go through this step by step. Flip two coins and look at the first one. If it is tails, then make the assertion that at least one of the coins is tails. If it is heads say nothing. At this point there is a 50% chance that you have made an assertion, and 50% chance that you've simply observed the first coin. Further break these two options by loooking at the second coin, resulting in 4 equally likely outcomes (therefore 25% each).
If, having already made an assertion, you see that the second coin is tails, then the problem conditions have been met with success - so far we're at 1/4 (success).
Next, if having already made an assertion, looking at the second coin you see a tails, we have a failure of 1/4.
Third, considering the case where no assertion was made after viewing the first coin, you see taht the second coin is tails, you make the assertion that at least one coin is tails, but we already know that both are not tails, so we have another failure (existing 1/4 plus current 1/4 makes 2/4)
Finally, if the second coin is a heads, then we cannot make an assertion at all, giving a not applicable result (of 1/4).
Now taking a close look at these outcomes, we see that the success rate is indeed 1/4 out of 3/4 (as the final 1/4 is impossible) simplifying to 1/3.
re(2): I think the solution is WRONG
