Let's discuss Mathematics

Discussion in 'Science & Technology' started by ParadigmShifter, Mar 16, 2009.

1. peter grimes...Retired Moderator

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Probably not the answer you're looking for, but rather than choose which side of the coin determines the winner, they could choose whether the face (or tail) is 'up or down'.

In other words, it would be determined by how far from an arbitrary 0 the face as rotated.

This could work with dice as well - are the rows closer to horizontal or vertical? Or is the number even or odd?

2. Miseisle of lucy

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I don't see how those problems aren't also present in a normal coin toss!

"Ok whatever I'll have tails."

Except replace "heads" and "tails" with "less than last time" and "more than last time".

My way works as long as P(X>E(X)) = P(X<E(X)).

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One person picks "heads or tails". Then have a shoot off, until one person has more wins than the other.

We both toss once - dutchfire ejaculates again - if we both get heads or tails we have more rounds, until one of us gets a head and the other doesn't.

Seems fair to me.

4. Miseisle of lucy

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When you think about it that's basically the same way as what I said, except much quicker.

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Why do people wrongfully insist that .999... is not equal to 1?

6. AtticusDeityRetired Moderator

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Yes, but the 100 tosses don't necessarily reveal what E(X) is. If the coin was fair, it would be irrelevant, what the other one chooses.

I think PS's method is fair, but here's another one, which I hoped you'd come up with:

They do double tosses, HT means ParadigmShifter wins, TH means DutchFire wins, and HH and TT means they'll toss again. If the coin lands H with the probability p, the probabilities of HT and TH are the same, namely p(1-p).

I suppose ParadigmShifter's method is quicker, but what's remarkable with this method is that you can make any random thing a perfect coin! Let's say you have a wheel divided into three sections, A, B and C. They look like equally big, but they don't even have to be! Now you choose that Ax means win for other party and xA for the other. Here x is B or C, and they do double spins, xx means they'll spin again.

Notice, that this assumes only that there is a constant p, which is the probability for one of the options to turn out, so this could -at least in theory- be used to find out whether a coin really has a constant probability to end up H or T up. Or to be more precise that how probable it is that there is such probability. I have read that there have been empirical investigations into the coins and dice, and the result has been that the used ones were unlikely fair. I'd be happy to know whether this thing has been investigated empirically.

EDIT: Shoot off does the same trick too, though.

And 0.999.....=1-thing, there has been debates about it here, and the last one of them was so depressing that I didn't visit this place for months.

7. Miseisle of lucy

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My method and PS's method are basically the same thing; I think you're reading it wrong if you think that knowing what E(X) is matters! You could rephrase PS's method as such:
Instead of PS and dutchfire flipping the coin, get two independent people, A and B, to flip the coin. Now, we all know that it doesn't matter whether A and B alternate their flips, i.e. A flips, then B flips, then A flips, then B flips, and so on -- or if A flips 100 times, then B flips 100 times, then you look at the results to decide on the winner using PS's method (sudden death with consequetive heads). Right? So basically, A is flipping the coin 100 times, then afterwards B flips 100 times. Then you look at the results and decide whether PS or dutchfire is the winner. But wait, who gets to decide who is A and who is B? Well, isn't that just like asking whether the first 100 flips is the winner, or the second 100 flips is the winner? You're just placing a bet on which side is winning, the probability of which is 50:50.

That's exactly the same as my method - the only difference is that the win condition is a bit different. Instead of it being about consequetive heads, it's about total number of heads. Both of those things depend on p, but don't require you to know what p is to make an informed decision about whether A or B is the better choice. They both reduce it to a 50:50 choice between which of two sets of random trial results better meets the win criteria.

Both of those methods can be applied to dice, wheel of fortune wheels, bus arrivals, or anything else just as easily, because you're just picking which set of random numbers better meets the arbitrary win criteria. It's just saying, "I got this one set of random numbers, and you've got that other set of random numbers, and now we're just going to see whether the string '927682893' appears first in your set or my set."

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Well my method also makes anything a perfect coin, as Atticus mentioned (as long as the probability of a head or a tail isn't 1). The closer the probability of landing on the winning choice gets to 0 or 1 though, the longer it will take (on average ) to resolve the winner.

9. AtticusDeityRetired Moderator

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Oh, I actually did read your first suggestion wrong, that will do also.

I just calculated, which method is quicker PS's or the one I told, and noticed they are exactly the same! Both will continue, if the coin lands HH or TT (both succeed or fail), and TH or HT will end it.

Now when I think of it, it's also the same as Mises proposition, with small adjustments: Forget the calibration, assume that 0<E(X)<1, which is of course necessary condition for this thing to make sense Then shorten the length of competition from 10 to 1 flips. So the one who first gets more than E(X) wins (that's the same, since you have to have tie break too).

Oh boy, hope my brain functions better next week...

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Nope, my method is quicker.

I'll use WL instead of HT. Remember only one person picks the winning flip (we could change pick and/or picker every round, no effect though)

WW vs. LL - player 1 wins EDIT: And it's a "no result" in Atticus' method
WL vs LW - player 1 wins 1 toss earlier than Atticus' method

EDIT: Whoops Said WW vs. LL was a win under Atticus' method.

Conclusion: my method is the superior in every way!

11. AtticusDeityRetired Moderator

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I think you got wrong what I was saying: each double flip in my method equals one flip from both parties in your method.

Picture it this way: You have come up with your method, and Dutchfire agrees it's fair. Now you toss the coin and Dutchfire is about to grab the coin, but you object: "I don't want to get my coin dirty, your fingers might be sticky. I'll do all the flipping. The even flips are yours and odd mine".

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Ah OK then. Yeah, they are identical. Great minds think alike.

Spoiler :
And fools seldom differ.

EDIT: I think dutchfire's fingers will be sticky considering the amount of ejaculating he seems to be doing in this scenario.

13. Miseisle of lucy

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I'm happy I got to an answer but sad that it was by far the least useful.

14. AtticusDeityRetired Moderator

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I didn't come up by myself with the method I proposed, it was an exercise in a book to prove that it's fair and to calculate what's the expected number of tosses before the conclusion is arrived at, so I knew the answer before the question. It would have been too easy to ask you that way.

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That makes me absolutely super then, I suppose, for coming up with the answer all on my own Saying you need an even number of tosses was a good clue though.

I think it is probably the quickest possible method.

EDIT: And reading Mise's earlier post, he said that "as long as p(X>E[X]) = p(X<E[X])"... isn't that always true? Furthermore, for a continuous pdf, isn't p(X>E[X]) = 0.5 as well?

EDIT2: Whoops. It's obviously not true for discrete values, e.g.

p(X=0) = 1/3
p(X=1) = 2/3

E[X] = 0 * 1/3 + 1 * 2/3 = 2/3

I think it is true for continuous pdf's though.

EDIT3: Ok, scratch that. The median is defined so that the area under the pdf from (-inf, median) is equal to 0.5, and that isn't true for non-symmetric distributions in general.

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OK, found what the difference between the mean and the median actually represents...

median splits area under the pdf into 2 halves, mean (expected value) is the value of X where the centre of gravity lies.

17. Miseisle of lucy

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I don't think what I said earlier was correct. Or maybe it's correct only because of the central limit theorem. (X is "number of heads" btw.) Anyway I think what I said about generating a random string makes more sense.

18. SouronThe Dark Lord

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All of these methods of re-tossing the coin have one problem: they may never terminate.

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Yeah, but the probability of that occurring is 0 though, so

EDIT: Unless one of the probabilities is 0 also.

EDIT2: The meaningful question that can be asked of course is - "what is the probability that this process will not terminate after a certain number of iterations?", and as long as the probabilities aren't 0 and 1, this tends to 0 as the number of iterations tends to infinity.