I don’t care how you got the numbers, I care that they are incompatible with each other. That’s the point here.
If you calculate the probability that I have blonde hair is 33%, and you calculate the probability that I don’t have blonde hair is 90%, then it doesn’t matter how you calculated those probabilities, one of them has to be wrong. They are incompatible probabilities.
You are not listening to what I am saying, so here it is once more:
There are 51 blue balls total.
98% (50 of 51) are in the blue bag
2% (1 of 51) are in the red bag
100% = 51 of 51
End of story! The sum is 100%
For you to claim that doesn’t add up because it adds up to more than 100% when adding 33 of 66 women are blonde is nonsense!
You’re not engaging with what I’m saying, you’re changing it to something else.
There’s a 98% chance that the bag contains 0 red balls, and a 49.5% chance that the bag has a red ball.
That’s your position. That’s the position I’m talking about. You cannot hold both at the same time without being in contradiction.
The 49.5% is from 49 of 99 balls being red, it has nothing to do with 50 of 51 blue balls.
25% of cars (250 out of 1,000) cars need an oil change.
33% of stores (300 of 900) are out of baby formula.
They don’t sum to 100%, so they are wrong, is that what you are trying to say?
Your example shows two unrelated statistics, questions about two unrelated facts . Unrelated facts do not have to sum to 100%. We both agree on that.
The claim that “the bag contains 0 red balls” and the claim “the bag contains red balls” are not unrelated.
So no, I think you’re still misunderstanding something.
The numbers are not related.
50 of 51 is 98%
1 of 51 is 2%
That is a completely different calculation than:
49 of 99 are red (49.5%)
50 of 99 are blue (50.5%)
Each set of numbers sums to 100% separately, because they are two different calculations of two different things.
You haven’t explicitly denied either of these positions:
There’s a 98% chance that the bag contains 0 red balls, and a 49.5% chance that the bag has a red ball.
As long as you hold both of these positions, your position is in contradiction.
That is not what 98% is calculating…
It is a FACT (not a probability) that 50 of the 51 blue balls are in the blue bag and 1 of 51 blue balls are in the red bag.
A quote from you:
So there is a 50 out of 51 probability (98%) that you picked the blue ball out of bag #1, and therefore a 98% chance that what’s left in that bag are all blue balls.
If there’s a 98% chance that what’s left in the bag are all blue balls, and “all blue balls” necessarily implies 0 red balls, then there must also be a 98% chance that the bag has 0 red balls.
Is there anything incorrect about that logic?
There are 2 bags, each bag started with a 50% chance of being the bag that the ball was pulled from. So it is not right to claim the bag has a 98% chance of containing 0 red balls.
I don’t think that 50 of 51 equates to only a 2% chance of having a red ball in that bag.
So you’re now saying, there’s a 98% chance that the bag contains only blue balls, but there’s NOT a 98% chance that the bag contains 0 red balls? How does that work?
Because the 98% is the fact that 50 of 51 balls are in 1 bag, which had a 50% chance of being the bag the ball was pulled from.
I’m not sure how that works. 
If there’s anything greater than a 2% chance that the bag contains red balls, then you’re stuck with this contradiction:
There’s a 98% chance it has only blue balls,
And a (3+)% chance it does not have only blue balls (and I assume your 3+ is 49.5 here most likely)
It’s the same contradiction, just worded differently.
You can shift the contradiction around, but you’re not going to get rid of it without changing your position.
I am sure, it doesn’t work. You cannot produce a data set with these statistical properties, because they are impossible to coexist. The probability of A, and the probability of not(A), cannot add to more than 100% (barring some weird word play or trick question, I suppose, which this is not)
I don’t know if that is true.
I know that 98% of the blue balls are in 1 bag.
I don’t know that the bag the ball was pulled from has a 98% chance of containing only blue balls.
So, just to be clear, you posted this:
“So there is a 50 out of 51 probability (98%) that you picked the blue ball out of bag #1, and therefore a 98% chance that what’s left in that bag are all blue balls.”
Are you backing away from that position? Do you now think otherwise?
I’m not denying it, I’m saying I don’t know that it is true. I’m not sure that 50 of 51 blue balls in 1 bag means that there is a 98% chance the ball was pulled from that bag, because there was a 50% chance of pulling the ball from either bag at the start. Each bag had a 50% chance of being the bag the first ball was pulled from, even though we knew at the start one of the bags contained 98% of the blue balls.
Okay, well at least the good news is you’re finally grappling with the contradiction, that’s what I’ve been hoping for for the past few pages of this conversation.
The bad news is, of your two contradictory positions, you’re backing away from the one that was right.
I can show that the 98% position is correct, with Bayes theorem, and I’m capable of producing software that you can easily run that will run the experiment, randomised, hundreds of thousands of times and tell you the results.
Last time I offered experimental evidence of this sort, you were uninterested. I don’t know if your attitude towards that has changed by now.
You are going to produce software using your methods and show me your method is correct? LOL
I don’t know what you mean by “my methods”. I’m going to write software that simulates the experiment randomly. Ideally, it’s not “my methods” or “your methods” or anybody’s method, but just a fair recreation of the scenario using software. You’d be free to read and review the code, it would be possibly readable even to a non software dev.