/sci/ - Science & Math » Thread #13670347

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Anonymous Wed 22 Sep 14:54:11 2021 No.13670347 View Reply Original Report

Quoted By: >>13670383 >>13670972 >>13670988 >>13671002 >>13671110 >>13672271 >>13673861

How can you get a fair result from a biased coin ?

Notice : You can't use the Von Neumann procedure.

I remember asking that question to my students as a bonus of an exam and one of them gave me a totally original answer and I was surprised at how simple it was and that it wasn't the Von Neumann procedure. It was totally original and I couldn't find it online but I don't remember it !

So let's try to find it again !

Anonymous

Anonymous Wed 22 Sep 2021 15:03:35 No.13670376 Report

Quoted By: >>13670390

You got a totally original answer to a question like that and didn't bother writing it down, taking a picture, or saving the exam paper? Nah, man.
You're a college student trying to outsource to /sci/ for bonus points on your exam. You would have had better luck being honest.

Anonymous

Anonymous Wed 22 Sep 2021 15:05:43 No.13670383 Report

Quoted By: >>13671064

>>13670347
Suppose it takes N tries to get heads on average. Bet on whether it will take less or more than N times for a 50-50 chance of being right.

Anonymous

Anonymous Wed 22 Sep 2021 15:08:37 No.13670390 Report

Quoted By:

>>13670376

I'm 100% serious, bro. I figured it was so simple either there was a flaw in her answer or it was not that original or that I could come up with it anytime I wanted.

Turns out I was wrong. As much as I can remember there was no flaw, I never encountered a similar argument again and I can't come up with it.

Anonymous

Anonymous Wed 22 Sep 2021 15:21:30 No.13670439 Report

Quoted By: >>13670455 >>13670468 >>13671145

As much as I can remember, her procedure was something like :

1) flip the coin X times, write down the number of heads H1

2) flip the coin X times again, write the number of heads H2

if H1>H2 then it's equivalent to a fair H

if H1<H2 then it's equivalent to a fair T

if H1=H2 repeat the procedure.

Is there any flaw ?

Anonymous

Anonymous Wed 22 Sep 2021 15:27:31 No.13670455 Report

Quoted By:

>>13670439
X=1.

Anonymous

Anonymous Wed 22 Sep 2021 15:31:02 No.13670468 Report

Quoted By: >>13670476

>>13670439
this would roughly be equivalent to sampling from the same biased coin. The probability of accepting a "fair H" is in expectation the same as the original chance the biased coin is heads. It might work if you know ahead of time how the coin is biased and then add the appropriate constants to the inequalities.

Anonymous

Anonymous Wed 22 Sep 2021 15:33:31 No.13670476 Report

Quoted By: >>13670563

>>13670468
>this would roughly be equivalent to sampling from the same biased coin
No, it wouldn't. Only problem is that you need a large X to be truly fair.

Anonymous

Anonymous Wed 22 Sep 2021 16:02:10 No.13670563 Report

Quoted By: >>13670654

>>13670476
The number of heads is essentially a binomial dist. If X is large enough the distribution is approximately normal with a mean of X(biased probs of head) clearly P(heads > X/2) does not equal 1/2 if p > 1/2.

Anonymous

Anonymous Wed 22 Sep 2021 16:36:03 No.13670654 Report

Quoted By: >>13670932

>>13670563
>implying the universe has a tendency to give you more heads on one run than on the other
Are you using your brain at all?

Anonymous

Anonymous Wed 22 Sep 2021 17:44:52 No.13670932 Report

Quoted By: >>13670948

>>13670654

OP here, I agree with you, the two experiences are strictly identical. There is as much chance for the first one to have more heads as the second one.

Then again, it's statistics, and satistics are counter-intuitive so maybe there's a flaw in her argument I can't see.

Anonymous

Anonymous Wed 22 Sep 2021 17:54:26 No.13670948 Report

Quoted By:

>>13670932
The expected number of heads is obviously the same both times. To say that the method is biased, is to imply that on average, one of the runs would yield more heads than the other, which is to imply the expected number of heads is different. That's a contradiction.

Anonymous

Anonymous Wed 22 Sep 2021 18:04:28 No.13670970 Report

Quoted By: >>13671094

How about this: flip the coin twice. If you get HT then consider this heads, TH -> tails and given any other outcome flip twice again. This does give equal probabilities but you're only almost sure to get a result in finite time.

Anonymous

Anonymous Wed 22 Sep 2021 18:04:50 No.13670972 Report

Quoted By: >>13670985 >>13671152 >>13672737

>>13670347
If you don't know which way it's biased, the first flip is still 50/50 for you, because there's an equal likelihood for you that it's biased in either direction.

Anonymous

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Anonymous Wed 22 Sep 2021 18:09:18 No.13670985 Report

Quoted By:

>>13670972
200 IQ. If neither party knows, someone's gonna be a sucker, but there's a 50% chance it's gonna be you.

Anonymous

Anonymous Wed 22 Sep 2021 18:10:40 No.13670988 Report

Quoted By:

>>13670347
>How can you get a fair result from a biased coin ?
Flip a fair coin to decide who gets assigned tails.

Anonymous

Anonymous Wed 22 Sep 2021 18:15:08 No.13671002 Report

Quoted By:

>>13670347
she was probably wrong and you didn't know why

Anonymous

Anonymous Wed 22 Sep 2021 18:32:52 No.13671064 Report

Quoted By: >>13671079

>>13670383
Based. I started typing out a really long and semi-formal explanation, but you said it much more clearly and concisely.

Anonymous

Anonymous Wed 22 Sep 2021 18:35:53 No.13671079 Report

Quoted By: >>13671116

>>13671064
Okay, but it's completely wrong and I was just trying to get a rise out of autists.

Anonymous

Anonymous Wed 22 Sep 2021 18:39:09 No.13671094 Report

Quoted By:

>>13670970
You just invented Von Neumann procedure independently. Not bad

Anonymous

Anonymous Wed 22 Sep 2021 18:40:38 No.13671101 Report

Quoted By:

Probability is not an exact science. There are many answers possibles.

Anonymous

Anonymous Wed 22 Sep 2021 18:43:01 No.13671110 Report

Quoted By:

>>13670347
pick a side of the coin, heads or tails, let's say tails, ok, now person 1 flips the coin, if they get tails then they repeat and they try to see how many tails they can get in a row, person 2 flips the coin and does the exact same thing, if person 1 gets more tails in a row, person 1 wins, if person 2 gets more tails in a row, person 2 wins, if they get the same amount, repeat

Anonymous

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Anonymous Wed 22 Sep 2021 18:46:30 No.13671116 Report

Quoted By: >>13671167

>>13671079
Really? I don't know anything about probability, but if you know it takes an average of n trials to observe a particular value, say Heads, then half of the time it takes more than N trials, and the other 50% of the time it would take fewer than that many observations.

Alternatively, you could perform a sequence of observations of arbitrary finite length n, calculate the expected number of heads in a sequence of length n, and if the number of heads exceeds the expected number of heads, then count the sequence simulates an observation of "heads" for a fair coin, whereas if the number of heads is fewer than the expected value, then the sequence simulates an observation of "tails" for a fair coin.

This doesn't work?

Anonymous

Anonymous Wed 22 Sep 2021 18:58:49 No.13671145 Report

Quoted By: >>13672262

>>13670439

OK, let's try this.

We need a probafag to compute P(H1>H2) and P(H1<H2) and a pythonfag to verify it.

I conjecture that since 1) and 2) are results from strictly identical procedures, therefore P(H1>H2)=P(H1<H2)=0.5.

Anonymous

Anonymous Wed 22 Sep 2021 19:00:34 No.13671152 Report

Quoted By: >>13672281

>>13670972
You're assuming that coins are equally as likely to be biased heads as biased tails. There's no justification for that assumption. We don't have enough information about how biased coins are made in that world.

Anonymous

Anonymous Wed 22 Sep 2021 19:04:26 No.13671167 Report

Quoted By:

>>13671116
>half of the time it takes more than N trials
That would only be true if the distribution was symmetric, but that can't be the case: the minimum number of trials to get heads is bounded, while the maximum number of trials is unbounded, so you have an infinite tail on one side but not on the other.

Anonymous

Anonymous Thu 23 Sep 2021 01:04:52 No.13672262 Report

Quoted By: >>13673825

>>13671145
I'm too lazy to compute the actual function, but if we assume that X here is big enough that we can just ignore the chance that H1 == H2 then then any ordering between the two i.i.d random variables is equally probable, so at least in the limit Pr(H1 > H2) = Pr(H1 < H2) = 1/2. I coded up a script to test it and it seems to work, just fairly slow.

Anonymous

Anonymous Thu 23 Sep 2021 01:09:23 No.13672271 Report

Quoted By:

>>13670347
You grab the Jew who flipped it by the neck and tell him to do it right.

Anonymous

Anonymous Thu 23 Sep 2021 01:12:15 No.13672281 Report

Quoted By:

>>13671152
I have a coin which is either biased heads or biased tails, but I'm not telling you which. I'll offer you the following game: you pick either heads or tails, then flip the coin. If you call it right, you win $100. If you call it wrong, you pay me $99.99. Ignoring the marginal utility of money aspect, you should obviously play. QED: the probability is 50/50.

Anonymous

Anonymous Thu 23 Sep 2021 03:47:06 No.13672737 Report

Quoted By:

>>13670972
Why is this wrong?

Anonymous

Anonymous Thu 23 Sep 2021 14:17:36 No.13673825 Report

Quoted By: >>13673844 >>13673848

>>13672262

Post screencaps

Anonymous

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Anonymous Thu 23 Sep 2021 14:23:52 No.13673844 Report

Quoted By: >>13673891

>>13673825
code

Anonymous

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Anonymous Thu 23 Sep 2021 14:24:56 No.13673848 Report

Quoted By: >>13673891

>>13673825
output

Anonymous

Anonymous Thu 23 Sep 2021 14:30:14 No.13673861 Report

Quoted By:

>>13670347
Write down all possible sequences of N toss. Divide them to 2 groups, A and B, so that the total sum of times the coin face up is equal for both group. Toss the coin N times, and the sequence will belong to either A or B, 50/50

Anonymous

Anonymous Thu 23 Sep 2021 14:42:24 No.13673891 Report

Quoted By:

>>13673844
>>13673848

very nice thanks for the work

so I take it it does work and it is a more general procedure than Von Neumann's ?

Von Neumann looks like the particular case where X=2

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