Without looking it up, how would you answer this brain teaser:

You have two ropes. Each takes exactly 60 minutes to burn. They are made of different material so even though they take the same amount of time to burn, they burn at separate rates. In addition, each rope burns inconsistently. How do you measure out exactly 45 minutes?

Great question, I loved this. I think my response is correct but am not sure. SKIP OVER MY RESPONSE IF YOU WANT TO DO THIS UNAIDED:

The intuitive answer is “welp, just wait until the ropes are 3/4 burnt and you should be at 45 mins.” However, this assumes a steady burn rate, which Chad has says is not the case.

So, we’re left with knowing that once either of these [different length] ropes is completely burned, starting from one end and running to the other (as in lighting a fuse) should be 60 minutes. However, that also means that if we light both ends of the rope, we should be able to burn through the full length in 30 minutes. If we cut the rope in half and light each end of the two ends of the now two segments, when both segments are consumed, 15 minutes should have elapsed.

So, my solution:

Take one rope and light both ends. When this rope finishes burning, a half hour has elapsed. When this rope finishes burning, you take the other rope and cut it in half. Light both ends of each of the two segments. When all segments have been burned, 15 minutes has elapsed. All ropes are gone, and you should be exactly at 45 minutes.

Edit: I’m loving the brain teasers, Chad. Keep 'em coming!

Cutting rope in half doesn’t work. As stated in the problem, each rope burns inconsistently.

Burn both ropes. One at both ends (call it rope A), the other at only 1 end (rope B).

When rope A burns out completely, that means 30 minutes have elapsed. At this instant, you know rope B has 30 minutes of life left if left burning only on 1 side.

Light the other end of rope B. When rope B burns out, that’ll measure 45 minuts.

This one I’ll take a stab at. Take the best of two tosses, but on toss number two, reverse the menaing of Heads and tails, so that HT and TH = heads, HH and TT = tails (or alternately HH and TT = heads and HT and TH = tails).

np is expected value of heads, expected value is at 50% of binomial distribution, (it is not skewed), from there whatever to the right is ‘head’ to the left ‘tail’

bchad, you are nearing the right track. The trick is to consider the possible outcomes of multiple coin tosses. I think you can get the answer if you write down the probabilities of each combination.

Yours is not the correct approach. Since you do not know P(Head) or P(Tails), you cannot make any assumption about the expected outcome or the distribution of outcomes. You also do not need to toss the coin many times, although more than one toss is necessary.

At first I thought yours was the more graceful solution but I sat with pen and paper for a sec and tried an example - you are correct and the second half of my solution is wrong! I can’t think of an alternate to the solution you’ve proposed. Nice work!

Maybe you toss it N times, and then you toss it N times again, and record whether you have more or fewer heads than you had the first time. If you get the exact same number, then you repeat (perhaps incrementing N by 1 for each set of tosses). The more unfair the coin, the higher the number N needs to be in order for this to work.

So now you’re testing differences in the outcome rather than the raw outcome. Almost like trying to get rid of a unit root.

This is similar to CSK’s solution. You can probalby make an argument about how multiple tosses are effectively turning a binomial process into a normal distribution via the central limit theorem, and then taking advantage of the symmetry of the normal curve to get a fair toss.