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I am wondering how knowing the initial physical conditions of a coin toss would affect the prior distribution. As far as I know, Bayesians think the parameter as a random variable, the values of which makes the prior distribution but I don't think the initial conditions do not make the prior distribution. It is confusing to picture the relationship between knowing the initial physical conditions of coin tosses and the prior distribution.

The following link is a youtube video saying that the probability of heads is the number of heads divided by possibilities for bayesians. Here, possibilities mean initial physical conditions of throwing the coin. I don't really understand the stuff when he talks about the bayesian way of probability.

https://www.youtube.com/watch?v=YsJ4W1k0hUg

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    $\begingroup$ It isn't enough to know the initial physical conditions, you also need to know the relationship between the initial physical conditions and the probability distribution of p(head), if you are going to go this route. A trivial example involves two alternatives: 1. Toss the coin very high and have it land on concrete -> p(head) is very tightly concentrated around 1/2, 2) Drop the coin from a height of 1 cm landing on its side -> p(head) is very tightly concentrated near 0 and 1 (as we don't know which side is up.) $\endgroup$
    – jbowman
    Commented Nov 9, 2020 at 22:13
  • $\begingroup$ Can you extend your answer with an example? $\endgroup$
    – xabzakabecd
    Commented Nov 9, 2020 at 22:16
  • $\begingroup$ Have done in the comment above. $\endgroup$
    – jbowman
    Commented Nov 9, 2020 at 22:17

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Physical experiments with mechanized tossing of coins have shown that it is very difficult to make an actual coin that is heavily biased. But it is easy to make unfair dice (by shaving edges or embedding lead weights), so I'll consider a die with 3 faces labeled H and three labeled T.

Suppose I've had a chance to look at a die and to know how it will be rolled, I see nothing suspicious, but I'm not absolutely sure the die is exactly fair. So I choose the distribution $\mathsf{Beta}(10,10)$ as my prior distribution. This distribution puts almost 95% of its probability in the interval $(.3,.7).$

qbeta(c(.025,.975), 10,10)
[1] 0.2886432 0.7113568

Also, suppose the die shows faces with H in $x = 1272$ rolls out of $n = 2500.$ Then using Bayes' Theorem to multiply the beta prior by the binomial likelihood, I get the beta posterior distribution $\mathsf{Beta}(10+1272=1282, 10+1228=1238),$ and a 95% Bayesian posterior probability interval for the Heads probability (credible interval) is $(0.489, 0.528).$

qbeta(c(.025,.975), 1282, 1238)
[1] 0.4892104 0.5282368

The die may not be precisely fair but I have faith that it will take a session with many more rolls of the die and at higher stakes than I'm ever likely to undertake before any slight bias will seriously impact my winnings or losses. (The Nevada commission that oversees casinos may require more evidence, but I am happy enough to use the die.)

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