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For such a question to make sense, don't we have to first define some distribution over how the coin might be biased in the first place?

ggm 2y ago
Interesting question. I would say yes. But, there are subtle biases: * coin always favours H or T. simple bias * coin has some component of behaviour which can be on, off or reset. For example a liquid mercury component, which can bias the H or T outcome but the right kind of "flip" resets it to a known-safe mode so the coin has less to no bias. * coin has bias which only manifests in skilled hands. a particular kind of flip. The point I'm making is that probably, the bias is always assumed to be H or T favouring, but doesn't admit more complex coin bias where it could be 2 or more actors and 2 or more capable of biassing, and a pigeon who can't (or a dummy, and a pigeon: a good con generally has more people involved than you think)
hervature 2y ago
This is the Bayesian vs. frequentist view point. What the OP is talking about is assigning a Beta(1,1) prior on the distribution and observing n heads in a row would yield a distribution of the bias of Beta(1, 1+n) and the mean of that distribution is 1/(n+2) which means the OP is off by one in the denominator but still good for memory. However, that is if you take the mean as your best estimate of the bias. If you take the mode, then the OP would be satisfied that even the Bayesian approach says that tails would be impossible. The frequentist view would say your best estimate is the average of the observations which would yield a completely unfair coin.
- yarg 2y ago
  It was twenty years ago, but I think it was n + 1. I'll redo the maths and check.
- yarg 2y ago
  With the simplified form of the binomial for n equal to k: ∫(0 -> x)(p^n)dp = ∫(x -> 1)(p^n)dp [p^(n+1)/(n+1)](0 -> x) = [p^(n+1)/(n+1)](x -> 1) x^(n+1) - 0^(n+1) = 1^(n+1) - x^(n+1) 2 * x^(n+1) = 1 x^(n+1) = 1/2 x = 1/(2^(1/(n+1))) Was I wrong? Have I been wrong for decades now?
  - hervature 2y ago
    Math is more than symbolic manipulation. You need to explain where the equations come from. The best I can decipher is that you are calculating the median of a Binomial distribution given k=0. Your "error" comes from the fact that you are positing that all biases are equally likely. Thus, you should not be surprised that you are getting a non-zero probability because you yourself are saying they are non-zero.
    - yarg 2y ago
      I'll write some code to test it tomorrow, but why the hell would it be zero? There are other probabilities that can lead to zero successes - every probability except for one. And yes, they diminish rapidly as the number of trials goes up (much in the same way as it does in my equation).
      - hervature 2y ago
        You have me confused a little because your original post seems to suggest you wanted it to equal 0: > I was uncomfortable with the use of biases to assign non-zero probabilities to events that fail to occur after some number of trials. Anyway, it entirely depends on your world view. When you say "There are other probabilities that can lead to zero successes" then that sounds like a Bayesian framework and that you have to pick your prior on the world. A natural choice is uniform (also Beta(1,1)) and update your prior as you collect data. You would then use the mean/median/mode of your posterior as your estimate for the bias. In your case, it appears you are operating in a Bayesian world but forcing your prior to be constantly uniform despite observing data. The frequentist perspective is that the probability is the relative frequency of observations. In this example, a frequentist would say that the probability of heads is 1 and tails is 0.
        yarg 2y ago
        > In your case, it appears you are operating in a Bayesian world but forcing your prior to be constantly uniform despite observing data. Got you now, thanks.
- LegionMammal978 2y ago
  Having a distribution over possible coins isn't related to frequentism vs. Bayesianism, I don't think. In frequentism, you can ask, "If I have a whole bunch of mystery coins with different biases, what is the distribution of their biases, and for how likely am I to flip k heads in a row for each possible bias?" And in Bayesianism, that distribution can just be the prior/posterior distribution for a single mystery coin.
yarg 2y ago
I just treated it as a binomial distribution with a fixed but unknown probability. Take the case of zero successes, that can happen for every probability except for one. The graph peaks at zero, but if you calculate 'x' such that the integral from zero to x of the binomial function is equal to the integral from x to one, you get a nice centre point.

ggm 2y ago

Interesting question. I would say yes. But, there are subtle biases: * coin always favours H or T. simple bias * coin has some component of behaviour which can be on, off or reset. For example a liquid mercury component, which can bias the H or T outcome but the right kind of "flip" resets it to a known-safe mode so the coin has less to no bias. * coin has bias which only manifests in skilled hands. a particular kind of flip. The point I'm making is that probably, the bias is always assumed to be H or T favouring, but doesn't admit more complex coin bias where it could be 2 or more actors and 2 or more capable of biassing, and a pigeon who can't (or a dummy, and a pigeon: a good con generally has more people involved than you think)

hervature 2y ago

This is the Bayesian vs. frequentist view point. What the OP is talking about is assigning a Beta(1,1) prior on the distribution and observing n heads in a row would yield a distribution of the bias of Beta(1, 1+n) and the mean of that distribution is 1/(n+2) which means the OP is off by one in the denominator but still good for memory. However, that is if you take the mean as your best estimate of the bias. If you take the mode, then the OP would be satisfied that even the Bayesian approach says that tails would be impossible. The frequentist view would say your best estimate is the average of the observations which would yield a completely unfair coin.

yarg 2y ago
It was twenty years ago, but I think it was n + 1. I'll redo the maths and check.
yarg 2y ago
With the simplified form of the binomial for n equal to k: ∫(0 -> x)(p^n)dp = ∫(x -> 1)(p^n)dp [p^(n+1)/(n+1)](0 -> x) = [p^(n+1)/(n+1)](x -> 1) x^(n+1) - 0^(n+1) = 1^(n+1) - x^(n+1) 2 * x^(n+1) = 1 x^(n+1) = 1/2 x = 1/(2^(1/(n+1))) Was I wrong? Have I been wrong for decades now?
- hervature 2y ago
  Math is more than symbolic manipulation. You need to explain where the equations come from. The best I can decipher is that you are calculating the median of a Binomial distribution given k=0. Your "error" comes from the fact that you are positing that all biases are equally likely. Thus, you should not be surprised that you are getting a non-zero probability because you yourself are saying they are non-zero.
  - yarg 2y ago
    I'll write some code to test it tomorrow, but why the hell would it be zero? There are other probabilities that can lead to zero successes - every probability except for one. And yes, they diminish rapidly as the number of trials goes up (much in the same way as it does in my equation).
    - hervature 2y ago
      You have me confused a little because your original post seems to suggest you wanted it to equal 0: > I was uncomfortable with the use of biases to assign non-zero probabilities to events that fail to occur after some number of trials. Anyway, it entirely depends on your world view. When you say "There are other probabilities that can lead to zero successes" then that sounds like a Bayesian framework and that you have to pick your prior on the world. A natural choice is uniform (also Beta(1,1)) and update your prior as you collect data. You would then use the mean/median/mode of your posterior as your estimate for the bias. In your case, it appears you are operating in a Bayesian world but forcing your prior to be constantly uniform despite observing data. The frequentist perspective is that the probability is the relative frequency of observations. In this example, a frequentist would say that the probability of heads is 1 and tails is 0.
      - yarg 2y ago
        > In your case, it appears you are operating in a Bayesian world but forcing your prior to be constantly uniform despite observing data. Got you now, thanks.
LegionMammal978 2y ago
Having a distribution over possible coins isn't related to frequentism vs. Bayesianism, I don't think. In frequentism, you can ask, "If I have a whole bunch of mystery coins with different biases, what is the distribution of their biases, and for how likely am I to flip k heads in a row for each possible bias?" And in Bayesianism, that distribution can just be the prior/posterior distribution for a single mystery coin.

yarg 2y ago

I just treated it as a binomial distribution with a fixed but unknown probability. Take the case of zero successes, that can happen for every probability except for one. The graph peaks at zero, but if you calculate 'x' such that the integral from zero to x of the binomial function is equal to the integral from x to one, you get a nice centre point.