Brain Teaser: Burning Rope

[higgmond]

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

[bchad]

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)

[bchad]

@higgmond

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

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.

[Tulips]

@higgmond

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

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.

[comp_sci_kid]

@bchadwick

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.

This is what popped into my head right away

[Sweep_the_Leg]

@comp\_sci\_kid

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

+1. Have all my sweet rolls.

[system]

@Tulips

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.

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.

[bchad]

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.

[system]

@bchadwick

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.

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.

[Palantir]

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.

[Palantir]

What is TT and HH?

[bchad]

@Blake McCallister

@bchadwick

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.

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.

Dude, I did reread your post. It’s wrong.

You are correct if the goal is to figure out what the true P(heads) is, but that’s not what the puzzle is about. The puzzle just asks how to create a system that will generate one of two results with 50-50 probability using a coin that is possibly biased towards heads or tails.

TH and HT will appear with equal probability no matter what the true P(heads) is (though you do get a degenerate solution if P(H)=1 or 0.)

[black_label]

Burn rope A from both ends while burn rope B from only one end

After 30 mins rope A would be finished, just burn the rope B from the other end too

After 15 mins rope B would be finished

[black_label]

So there is A and B, you can ask A and B to choose either HT or TH

You toss the coin until HT or TH appears.

[black_label]

Actually you have to reset after every two tosses

[Turd_Fergeson]

haven’t read every response because i don’t want to be influenced by preivous guesses, but if this hasn’t been answered yet here’s my guess:

1. cut one rope in half two times to get four equal length ropes. you can do this by matching up the ends.

2. start burning all 4 cut ropes and the uncut rope at the same time. When the last of the 4 cut ropes is done burning, you know that 15 minutes have passed.

3. Begin the clock on the other rope and when it is done buring 45 minutes have passed.

//

[system]

@bchadwick

@Blake McCallister

@bchadwick

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.

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.

Dude, I did reread your post. It’s wrong.

You are correct if the goal is to figure out what the true P(heads) is, but that’s not what the puzzle is about. The puzzle just asks how to create a system that will generate one of two results with 50-50 probability using a coin that is possibly biased towards heads or tails.

TH and HT will appear with equal probability no matter what the true P(heads) is (though you do get a degenerate solution if P(H)=1 or 0.)

I missed the original question. I posted the methodology to find the exact probabilities for each side of the coin which is correct.

[Turd_Fergeson]

So what’s the answer to the burning ropes question? Other than it keeps Funyons in business…