On the evidence of a single coin toss
Posted by Tom Moertel Tue, 07 Dec 2010 07:22:00 GMT
Say that you bump into me at the coffee shop. After saying hello, I hand you a coin and claim that it’s special: when flipped it always comes up heads. You inspect it and say that it looks like a normal coin.
“Seriously, it always comes up heads,” I say. “Flip it.”
You flip it. It comes up heads.
Now, here’s my question to you: After seeing the outcome of this single coin toss, how much more should you believe my claim that the coin always comes up heads, compared to what you believed before the coin toss? In other words, how much should this single toss sway your prior beliefs about my claim? Should it sway you at all?
Post your thoughts in the comments.
I’ll provide the answer later.
Update: For my answer, see part two: More on the evidence of a single coin toss
P(H|E) = (P(E|H) * P(H)) / P(E)
so our conditional probability P(E|H) is 1, given that the claim is we always see heads. P(H) is our original estimate of the hypothesis’s probability.
P(E) is the a priori probability of seeing the event, which is the coin coming up heads. Probably reasonable to say 0.5 here – you might just as well have a coin that always comes up tails, and of course the vast majority of fair coins will give 50/50 results.
so: P(H|E) = (1 * P(H)) / 0.5 = 2 * P(H)
so, you should be twice as likely to believe as you were before. Your prior is up to you… :)
Using Bayesian reasoning, the prior is conditioned by the observation.
Let P(A) = probability claim is correct (always heads)
Let P(O) = probability heads occurred (prior—provides a measure of the observer’s initial degree of belief)
P(O) = 1 (1 observation = heads)
So now P(A|O) = (P(O|A) P(A))/P(O) = P(O|A) P(A)
Hmm.. Does it depend on how special a special coin is?
Assuming I don’t believe in special coins, it should not sway me. This would happen to normal coins half the time anyways. P <= 0.5 does not make a convincing argument.
If I do believe in special coins, and estimate the probability that you procured one to be x, then Bayes says the probability would be 2x/(x+1). I.e. if there’s a 10% a priory chance of you having a special coin, then this increases my confidence to 18%.
Does the coin have a tails side.
If it does:
Look at T-tests versus Bayesian priors.
The t-tests will tell us that you don’t have enough information.
The Bayesian priors mean that I doubt your coin is so biased it is always heads, why? Because I generally expect the coin to produce both heads and tails. I can use previous knowledge from previous situations to evaluate claims that seem wild.
I am guessing your answer has something to do with the Anthropic Principle. So no, it should not sway me at all. But then I’m pretty sure I know your punchline :)
I’ll assume you are either a lucky liar with a normal coin, or you tell the truth and have a perfectly biased or two-headed coin. Let’s call these two possibilities FiftyFifty and AlwaysHeads, respectively, and outcomes heads (H) and tails (T)
L(H|AlwaysHeads) = 1
L(H|FiftyFifty) = 0.5
My prior belief in AlwaysHeads (p(AlwaysHeads)) is q, and thus p(FiftyFifty) = 1-q
By Bayes’ Theorem, P(AlwaysHeads|H) = L(H|AlwaysHeads) * p(AlwaysHeads) / (sum of L*p for both possibilities)
= 1 * q/ (q+0.5(1-q) = 2q/(1+q)
which is what we wanted.
For a few seconds i would look at you astonished, after that i will inspect the coin to see what is it that it’s special (like where is the trick ;) ). Of course i will toss it again a few times. Basically just telling me is special won’t be enough to make me believe :)
If you never toss the coin again the statement is 100% accurate, so I’d pocket it and trust him absolutely, vowing to never toss it ;)
A throw of the dice will never abolish chance.
http://www.poetryintranslation.com/PITBR/French/MallarmeUnCoupdeDes.htm
obligatory xkcd link
http://xkcd.com/221/
int getRandomNumber() { return 4; // chosen by fair dice roll. // guaranteed to be random }
You can never say whether the coin truly is special or not, you can only say the likely-hood that the coin is fair based on outcomes over time verses expected outcomes.
Starting from what tanielson says above, I’d say that’s only halfway there because we don’t know the value of q (ie how much you believe your friend).
Summing over a range of q from 0 to 1 gives a mean value of P(AlwaysHeads|H) as ~0.61.
This I think matches what I’d expect, namely that you should be more inclined to believe your friend following one successful toss. (By extension, following 100 successful tosses you’d believe your friend for sure; fewer tosses reduces this down until you have no tosses = no information.)
I have a feeling this is relevant:
http://en.wikipedia.org/wiki/Sunrise_problem
You basically assume the coin has a p chance of being heads and then you can determine the probability of it being heads again as a function of p. You don’t get a conclusive answer, though you could integrate from 0 to 1 on p and maybe get a result…
I agree with tanielson.
You should be swayed from q, where q is your prior, exactly 2q / (1 + q).
This means that if you had 50% confidence in your friend (from past experience you know he lies half the time), after one heads coin toss you should believe him:
2(.5) / 1 + .5 = 1 / 1.5 = 2/3
This makes sense: if you believe him 100% (very trustworthy), you should still believe him 100% (but no more):
2(1) / (1 + 1) = 2 / 2 = 1
The claim “this coin always comes up heads” is an arbitrary assertion with no reference to causality. WHY would it always come up heads? There is no reason, ergo you have no reason to believe the claim to be true. No amount of coin tosses will provide evidence for such a causal connection.
tanielsen: I don’t think you can assume he’s either definitely right or completely wrong. What if he just has a coin that’s more likely to flip heads? Say it is heads 75% of the time.
Tyler @14: “I hand you a coin and claim that it’s special: when flipped it always comes up heads”.
In terms of “how much more should you believe my claim”: taking tanielson @4 and myself @10, I think the answer is “11% more”.
This starts from the assertion that I know nothing about your trustworthiness, and have to distribute the chance that you’re lying evenly from 0 to 1 (#1). So before the toss I think there’s a 50% chance you’re telling the truth.
After the toss, tanielson’s formula allows us to calculate P(AlwaysHeads|H) for given values of q. q is unknown, so integrate from 0 to 1 (I did an approximate sum because my math is a little rusty (#2)) – this gives ~0.61 expected value for q. So now I think it’s 61% likely you’re telling the truth, an increase of 11%.
(#1) I don’t like this assertion as it ignores a lot of real-world stuff that would invalidate it. But I think in the context of the problem you can’t do any better. (#2) For “a little rusty” read “exposed to the sea air and unloved for several years rusty” :-)
I put a lengthy response to this on my blog, but the key observation is that it matters what I believe about Tom’s truthtelling; the summary was, if I suspect Tom might be stating the claim even if the claim were false, then my belief in the claim increases only negligibly upon seeing a single coin toss. Full post: http://www.blahedo.org/blog/archives/001081.html
We start with zero tosses and zero information. Adding one toss that does not falsify your claim does nothing to sway one towards affirmation.
However, our expectation that the coin is fair should change since we’ve excluded half the space.
I would believe you just as much as I did before. There’s still a 50/50 chance with only one toss, so with not further information, there’s no need to believe it really is a special coin.
I agree with tanielsen and Eric. Or rather, I assumed q – the probability of a random coin being heads-only to be very small, so the claim is twice as trustworthy as before. Which maybe isn’t so surprising, since you just added one bit (one coin toss) of information.
Let p be the prior probability that the coin will comes up heads. Then the answer is that your new probability is your old probability times (1/p).
Here’s my reasoning. First find the probability that the coin is always-heads given that it came up heads.
p(always heads | was heads) = (p(was heads | always heads) * p(always heads))/(p(was heads)) = p(always heads)/p(was heads)
You wanted to know “how much more”. So let’s find the ratio of the probability that you assign to always-heads before seeing the coin come up heads, to the probability afterwards.
p(always heads | was heads) / p(always heads) = (p(always heads)/p(was heads))/p(always heads) = p(always heads)/(p(was heads) * p(always heads)) = 1/p(was heads)
A little clarification. “your new probability” means “your new probability that the coin is an always-heads coin”. Similarly for “your old probability”.
Operating in the space of likelihood ratios makes this easier (for me).
We have two worlds: A) The coin always comes up heads. B) The coin is fair.
The probability of a heads in world A is 1. The probability of tails in world A is 0.
The probability of either heads or tails in world B is 0.5.
The likelihood ratios given heads or tails is therefore 1/0.5 = 2 and 0/0.5 = 0.
Each throw coming up heads multiplies your aggregate likelihood ratio by 2.
Each throw coming up tails multiplies your aggregate likelihood ratio by 0.
If I think you’re 100 times more likely to be lying than telling the truth, the initial likelihood ratio is 0.01. Throwing one heads would create an aggregate likelihood ratio of 0.01 * 2 = 0.02. I would need to see 7 heads in a row to bring the aggregate likelihood ratio above 1.0 (also known as 50% probability).
Assuming that I severely doubt the initial claim, the single coin toss would approximately double my belief in the claim.
I would yell “Black Magic”, “Voodoo” or something of the sort… and then have you arrested (or at least publicly flogged).
Really I would believe you no more, knowing that on the next flip there is a 50/50 chance that either side will come up. Every flip has this same chance. The economist would tell you that “PAST PERFORMANCE IS NOT INDICATIVE OF FUTURE RESULTS”