# Brain Teaser: Burning Rope

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?

Share this

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

notthe 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!

Nice solution. I would not have passed the test.

You want a quote? Haven’t I written enough already???

I knew the solution, but it’s because I’ve seen this question before. I like questions like this which have simple but unobvious solutions.

Here is another question, which is not really difficult, but has an interesting solution:

Q: You have an unfair coin, that is, P(Head) != P(Tail). How do you replicate a fair coin toss using this coin, and not knowing P(Head) or P(Tail)?

“Visit the Water Cooler forum on Analyst Forum. It is the best forum.”

- Everyone

Studying With

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

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

2) 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.

3) 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).

You want a quote? Haven’t I written enough already???

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’

trick is to toss multiple times

That is good

Bummer… mine didn’t quite work right (just checked it). :-(

Hmmm… but at least it was on the right path…

You want a quote? Haven’t I written enough already???

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.

“Visit the Water Cooler forum on Analyst Forum. It is the best forum.”

- Everyone

ohai, hows my solution?

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.

“Visit the Water Cooler forum on Analyst Forum. It is the best forum.”

- Everyone

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.

You want a quote? Haven’t I written enough already???

Ah, I have the solution to the first one:

1. When you run out of time at 45 minutes, you still have two burning ropes left.

2. So stuff them in the proctor/invigilator’s eye and keep filling ovals for another 15 minutes.

You want a quote? Haven’t I written enough already???

The true solution is much more simple (which is why it is a good question, in my opinion). You were much closer in your previous idea.

“Visit the Water Cooler forum on Analyst Forum. It is the best forum.”

- Everyone

I’ll try…

Fair coin H=T

So .5=.5

Lets say with the biased coin, probabiilty of heads = .8, so tails = .2

Possible outcomes

HH

HT

TH

TT

HT=TH since .8*.2=.2*.8

Ignore the TT and HH toss outcomes.

Tulips is correct. You win money - Chad, please send him $50.

“Visit the Water Cooler forum on Analyst Forum. It is the best forum.”

- Everyone

Oh i didnt realize you dont know P(Head) or P(Tail) - It was not stated in the problem.

I used to be able to solve problems without statistics or calculus but then i got arrow to the knee

You guys are good. I clearly don’t use my brain enough anymore and it is dying.

This space available.

Nice problem; nice solution! Thanks ohai.

(You’re right, the probabilities were sitting there in front of me, and I didn’t think to exclude the HH and TT parts - though if it’s highly unbalanced, it could take a LOOOONG time to get to an HT or TH)

You want a quote? Haven’t I written enough already???

The good news is that if you were able to do it in the past, you can learn to do it again. A lot of it is just practice.

Google has made us individually a bit weak in that category, even if it has made much of the rest of the world look stronger.

You want a quote? Haven’t I written enough already???

Speaking for myself, I don’t think the case is that I’m particularly good just that I’m a couple of years out of school and my solutions are still simple and naive :) Bchads answer went way over my head.

This is what popped into my head right away

+1. Have all my sweet rolls.

Yeah, but it isn’t that easy. You would have to sit there and flip coins and record the four outcomes. After many flips, the std error of estimate begins to converge to zero. For highly unbalanced odds, you would need a lot of flips. So let’s say we flip a thousand times. We get your probabilities.

HH ~64%

TT ~ 4%

You just take the sqr root of each and that is your probability for each side. Pretty slick, but since you can only solve this with brute force you need a ton of flips.

*Edit* ———–> You need to do thousands of flips and a bunch of trials for an unbalanced situation like that. 10,000 flips and 100 trials you still aren’t getting a “true” normal approximation. It’s close, but not perfect. I would post the graph if you could post pics.

"Essentially, all models are wrong, but some are useful..."

I think all you have to do is bet on whether the first unequal pair you flip will be HT or TH. Then flip in groups of two. Keep flipping until you get either HT or TH.

If you have a highly unbalanced coin, it could take a long time of waiting for something other than HH or TT, but you don’t actually need to approximate anything, because you don’t actually have to make any estimates with this system. All you know is that the chance of getting HT and TH in a group of two flips is equal, and so if you get HH or TT you just do two more flips.

You want a quote? Haven’t I written enough already???

No dude. Tulips has the right concept but the only way to solve this is through brute force. You need to get estimates of TT and HH and to get good estimates you need to look at the std error of the estimate for both TT and HH and make sure it is converging to zero. Reread my prior post.

"Essentially, all models are wrong, but some are useful..."

Damn kevin’s way is better than mine.

I would burn Rope A from both sides, and after that is done, burn rope B in three spots, both ends, and the middle which should cut the time to 15 minutes…..but the other way is better.

What is TT and HH?