You get the probability that the actual probability is on that region, but it's never 100%.
I have no idea what you're trying to say here. If you have a distribution for the probability of heads, you can calculate the probability of getting heads. For any symmetrical distribution, it will be 50%, reflecting the fact that you have no reason to favor heads over tails.
Think about it this way: Suppose that you have a much simpler distribution over p, the probability of heads, where it's 0.4 with probability 0.3, otherwise 0.7 Then by the law of total probability, the probability of heads is (probability of heads given p=0.4)*(probability p = 0.4) + (probability of heads given p = 0.7) * (probability p = 0.7) which is clearly 0.12 + 0.49 = 0.61. You might note this is also the expected value of p; in the continuous case, we would use the formula integral_0^1 xf(x) dx where f is the PDF. For your solution, Beta(9, 3), this is just 9/12 = 0.75. This is basically the same example as at the top of https://en.wikipedia.org/wiki/Law_of_total_expectation#Example
But the average value is not necessarily "the answer".
I never said it was? It was just another example where you can compute a specific property of the underlying random variable, given a distribution on one of its parameters.
If you have a distribution for the probability of heads, you can calculate the probability of getting heads.
Actually you can't. I don't think you quite understand the point. I can program a f() function that return heads p percent of the time. How many results do you need to accurately "calculate the probability of getting heads"?
Suppose that you have a much simpler distribution over p, the probability of heads, where it's 0.4 with probability 0.3, otherwise 0.7
OK.
You might note this is also the expected value of p
Yes, but the "expected value" is not "the answer".
I programmed your example of 0.3*0.4/0.7*0.7 as g(0.3), let's say that the threshold t in this case is 0.3, but I choose a different threshold for comparison and I run the function 10 times. Can you guess which results are which?
How many results do you need to accurately "calculate the probability of getting heads"?
What is "accurately"? The method I described will give the correct probability given all of the information available. As you gather more information, the probability changes. Are you getting confused between probability and statistics?
Actually you can't.
Yes, you can. I just gave you a complete worked example.
Yes, but the "expected value" is not "the answer".
In this case, it is. In fact, in any case where you have binary outcome and independent events, the expected number of successes is equal to the p*the number of trials. In the special case of n=1 coin flip, we have E(number of heads) = p. See https://en.wikipedia.org/wiki/Binomial_distribution
Can you guess which results are which?
is more likely to be 0.3 because the portion of 0.4 outcomes is 0.3, and in 2) it's 0.2, so the estimate for t is 0.2. Those are the choices which maximize the likelihood of the data. But I'm not going to do a bunch of math to figure out the exact probabilities (and there might not even be a closed form solution); what's the point of all this?
The method I described will give the correct probability given all of the information available.
It won't.
In this case, it is.
It's not.
is more likely to be 0.3
Yes, but it is not. You got it wrong.
so the estimate for t is 0.2
But it is not 0.2.
This is the whole point of the article: to raise doubt. But you are not even considering the possibility that you might be wrong, I bet even when I'm telling you the values of t in those examples are not the ones you guessed, you will still not consider the possibility that you are wrong, even when the answers are objectively incorrect.
It seems like you don't know enough probability to really have this discussion. "The expected number of heads from 1 flip is equal to the probability of heads" is a trivial calculation.
es, but it is not. You got it wrong.
That's not how probability works. I said it was more likely and that statement is correct. Just like how in your original coin flip example, it is more likely that p is 0.8 than 0.3, but if it really was 0.3 and you just got unlucky your answer would not have been "wrong" because that's not how probabilistic statements are judged.
But it is not 0.2.
Do you know not what an estimate is?
But you are not even considering the possibility that you might be wrong, I bet even when I'm telling you the values of t in those examples are not the ones you guessed, you will still not consider the possibility that you are wrong, even when the answers are objectively incorrect.
This blatant strawmanning isn't helping your case. The statements I made above about probability are correct, they are fairly basic mathematical ideas that get used all the time. If you think they're wrong, provide an actual argument. I have never said anything like "t is definitely 0.2" nor do I care what its value is, because it's an irrelevant exercise. Have you considered that you might be wrong?
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I have no idea what you're trying to say here. If you have a distribution for the probability of heads, you can calculate the probability of getting heads. For any symmetrical distribution, it will be 50%, reflecting the fact that you have no reason to favor heads over tails.
Think about it this way: Suppose that you have a much simpler distribution over p, the probability of heads, where it's 0.4 with probability 0.3, otherwise 0.7 Then by the law of total probability, the probability of heads is (probability of heads given p=0.4)*(probability p = 0.4) + (probability of heads given p = 0.7) * (probability p = 0.7) which is clearly 0.12 + 0.49 = 0.61. You might note this is also the expected value of p; in the continuous case, we would use the formula integral_0^1 xf(x) dx where f is the PDF. For your solution, Beta(9, 3), this is just 9/12 = 0.75. This is basically the same example as at the top of https://en.wikipedia.org/wiki/Law_of_total_expectation#Example
I never said it was? It was just another example where you can compute a specific property of the underlying random variable, given a distribution on one of its parameters.
Actually you can't. I don't think you quite understand the point. I can program a
f()function that return headsppercent of the time. How many results do you need to accurately "calculate the probability of getting heads"?OK.
Yes, but the "expected value" is not "the answer".
I programmed your example of
0.3*0.4/0.7*0.7asg(0.3), let's say that the thresholdtin this case is 0.3, but I choose a different threshold for comparison and I run the function 10 times. Can you guess which results are which?[0.7, 0.4, 0.7, 0.7, 0.7, 0.7, 0.4, 0.7, 0.7, 0.4][0.7, 0.7, 0.7, 0.7, 0.4, 0.7, 0.7, 0.4, 0.7, 0.7]Which is
g(0.3), which isg(t), and what do you guess is the value oftI choose?What is "accurately"? The method I described will give the correct probability given all of the information available. As you gather more information, the probability changes. Are you getting confused between probability and statistics?
Yes, you can. I just gave you a complete worked example.
In this case, it is. In fact, in any case where you have binary outcome and independent events, the expected number of successes is equal to the p*the number of trials. In the special case of n=1 coin flip, we have E(number of heads) = p. See https://en.wikipedia.org/wiki/Binomial_distribution
It won't.
It's not.
Yes, but it is not. You got it wrong.
But it is not 0.2.
This is the whole point of the article: to raise doubt. But you are not even considering the possibility that you might be wrong, I bet even when I'm telling you the values of
tin those examples are not the ones you guessed, you will still not consider the possibility that you are wrong, even when the answers are objectively incorrect.It will.
It is.
It seems like you don't know enough probability to really have this discussion. "The expected number of heads from 1 flip is equal to the probability of heads" is a trivial calculation.
That's not how probability works. I said it was more likely and that statement is correct. Just like how in your original coin flip example, it is more likely that p is 0.8 than 0.3, but if it really was 0.3 and you just got unlucky your answer would not have been "wrong" because that's not how probabilistic statements are judged.
Do you know not what an estimate is?
This blatant strawmanning isn't helping your case. The statements I made above about probability are correct, they are fairly basic mathematical ideas that get used all the time. If you think they're wrong, provide an actual argument. I have never said anything like "t is definitely 0.2" nor do I care what its value is, because it's an irrelevant exercise. Have you considered that you might be wrong?
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