Numbers of double-headed, double-tailed, and normal coins based on one toss per coin

#1
Suppose I have a large jar of evenly weighted coins. I know that some of the coins are double-headed, some of them are double-tailed, and some of them are normal, but I've no idea how many. I draw 100 coins at random and toss each coin once, throwing 90 heads and 10 tails. What can be said about the probable numbers of double-headed, double-tailed, and normal coins in my sample?

My untutored intuition is that there are probably more double-headed than double-tailed coins, but that's as far as I can take it. I'm not even sure how to calculate that probability.

(To my mind, this is related to the question I asked yesterday about survey responses, but I'm unsure about this, and I mention it only as evidence that I'm not fishing for an answer to a homework question.)
 
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fed2

Active Member
#2
i think p(head) in your experiment = proportion of pop that is double head + 0.5 proportion of pop fair, ?

Looking at this, my 'non-estimability' alarms are going off. It is a like spider-sense for stats. Whatever the outcome of your experiment, there will be many equally likely ways to achieve it. That is, many possible populations that would be equally likely to generate the observation.

you could observe 90% heads cuz you got a population with 90% double headed, or 85% double head and 0.5*proportion fair = 5% -> prop fair = 10%.

or something like that......
 

katxt

Active Member
#3
If you take the maximum likelihood approach, then the distribution which makes 90 heads and 10 tails most likely is 90 double heads and 10 double tails. Then the probability is 1. The likelihood of 90 heads and 10 tails if there are any normal coins is less than 1. A strangely unsatisfying answer.
 
#4
If you take the maximum likelihood approach, then the distribution which makes 90 heads and 10 tails most likely is 90 double heads and 10 double tails. Then the probability is 1. The likelihood of 90 heads and 10 tails if there are any normal coins is less than 1. A strangely unsatisfying answer.
Ha! That's the explanation missing from a reply to the same question on another forum. But does it have things the wrong way round? My question is about the probability of having 90 double-headed coins and 10 double-tailed coins given that I've thrown 90 heads and 10 tails. Your answer looks to me to be about the probability of throwing 90 heads and 10 tails given that I have 90 double-headed coins and 10 double-tailed coins. What am I missing?
 
#5
Whatever the outcome of your experiment, there will be many equally likely ways to achieve it.
Are the following all equally likely (see katxt's post above)?
  • 90 double-headed, 10 double-tailed
  • 89 double-headed, 9 double-tailed, 2 normal
  • 88 double-headed, 8 double-tailed, 4 normal
  • 87 double-headed, 7 double-tailed, 6 normal
  • ...
  • 80 double-headed, 0 double-tailed, 20 normal
 

katxt

Active Member
#6
No.
The idea of maximum likelihood is that you have some data and a range of possible explanations for that data. For each explanation you work out the likelihood for the data given the explanation, and choose the explanation which has the greatest likelihood.
So it would be possible to get 90 h and 10 t from 100 normal coins but the likelihood of getting that is so microscopically small that it is almost impossible to accept 100 normal coins as being the real situation.
90 h 10 t is probably not the right answer. In fact there is only a small possibility that it is correct. However, it is the most likely one even if that likelihood is small. No other answer is more likely.
kat
 
#7
I get that it's a form of abductive inference (inference to the best explanation). I misunderstood what you were giving the value 1, namely the data given the explanation, rather than the explanation itself. Very much my mistake!

In fact there is only a small possibility that it is correct.
Is there a formula that a stupid person can understand for calculating that?

Edit: Someone has pointed out to me that "probability" and "likelihood" have different meanings in a technical context. I didn't know that, but I'm reading about it now.
 
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fed2

Active Member
#8
katxt is computing the likelihood conditional on the observed sample, rather than in terms of the popultation. To me it seems that the problem describes sampling from an infinite population of coins, so thats how it is. the estimate for your 100 coins would 100 * estimated population proportion.
 

katxt

Active Member
#9
Is there a formula for calculating that?
=BINOMDIST(10,100,0.5,0) in Excel or about 1 chance in a billion billion
fed2 is right in that the maximum likelihood expressed above doesn't allow for the a priori probabilities of each type of coin. The fact that we see 10 tails / 100 must give some clue as to the population proportions.
However, I don't think that the maximum likelihood method is much help in practice.
 

fed2

Active Member
#10
anyway: math is sort of beside the point here. You got more of a basic science problem. You got one piece of info (90%) and you want to draw 2 independant conclusions (#double headed, # fair ). Thats just more of an 'epistemic' issue i think is what smart people call it. I'd be leary of any statistical framework that didn't give a meaningful accounting for this problem.
 

katxt

Active Member
#11
I know we both have other things to do, but if someone was to offer you a prize of $1000000 if you could guess correctly how many of each type of coin in Remster's original problem, what would be your choice? There are many possibilities but not all are equally likely given the data.
 

fed2

Active Member
#12
i don't even think the estimator you propose is mean square error consistent? if the population in fact contains any fair coins, your estimator has asymptotic bias. i could let you flip each coin a million times, and you would still tell me the wrong answer?

no $1000000 for you ;)

i don't want to get drawn into math-stats debate, i wouldn't be the one to ask, but I know a duck when i see one.

I know we both have other things to do
i doubt 'highly productive people' is really a theme of talk-stats.
 

fed2

Active Member
#16
grudge much? its like taking my cat to the vet...a week of stink eye. i would appreciate if you could resolve the apparent contradiction between the population and finite sample models. i think max liklihood is sometimes no good in finite samples.
 

katxt

Active Member
#17
You're right. Maximum likelihood sometimes gives odd results.
You are in a unfamiliar town where the busses are numbered 1 to n. You see a bus numbered 40. What is the best guess for n, the total number of busses? Maximum likelihood says n = 40 but that seems too low.
 

fed2

Active Member
#18
Just to clarify you were saying that the maximum likelihood solution was always just the
# double headed coins == number of observed heads
# double tailed coins == number of observed tails
?

I think that's 'right', but oh so wrong. Wish i could put my finger on why but i don't have enuf brain cells left.
 

Dason

Ambassador to the humans
#19
I mean... In this situation yes that is the MLE. Really though we are in a situation with one sample. We need more data to get a better estimate.

Side note: I don't really hold grudges - I was just giving you **** because you decided to give me **** previously :p - nothing personal I just thought it'd be funny
 

fed2

Active Member
#20
you had me going! well guess we're even now.

Yes, i guess more samples, in the sense that you know 90/100 on round one, 70/100 on round 2 etc would resolve it. Intuitively it seems the mle works out to minimum # heads = # double headed in this case?

On the other hand just adding more samples in the sense of 900/1000, ie all known is total heads probably just leaves it same as before.

I think the (conjectured) inability to construct an unbiased estimate at n=1 may have some important meaning here.................................................................................................................................