I am not an expert but I do have to present very clear data information regularly - so some suggestion -
It is a just communication thing.
Ah, presentation, sure, that was a quickie from me and I think anybody who wants to understand it can. It’s not optimal presentation, but I’m not being paid for this shit, I’m trying to pay someone else for it!
Motor, I’m trying to pay you
I really don’t want your money!
I already showed you how I calculate the scenarios.
You are not capable of producing a data set with the statistical properties you’re proposing, because the statistical properties you’re proposing cannot both be true Motor. They can’t both be true. It’s literally impossible.
Don’t make a big data set, make a small one, with 10 items. Show me 10 items where 9 of them are blue first, from the blue bag, and where 4 of them are blue first, but the second ball picked is red. It’s easy if you think it’s possible.
I’ve got another way of explaining why you just can’t have both of these at the same time.
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I believe that the bag I selected has a 98% probability of being the bag that contained 50 of the 51 blue balls.
I believe that there are 99 balls left to pick from, 50 are blue and 49 are red, so there is a 50.5% probability of picking a blue ball.
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If you believe there’s only 50.5% probability of picking a blue ball, and a 49.5% probability of picking a red ball, then consider this: there can only be a 49.5% probability of picking a red ball out of the bag in front of you if the bag in front of you has at least a 49.5% chance of having a red ball inside it. Right?
So, according to your position, the bag in front of you has a 49.5% chance (at least) of containing a red ball, and a 98% chance of containing only blue balls.
And if it has a 98% chance of containing only blue balls, then it has a 98% chance of containing no red balls, 0 red balls.
So, according to your position, the bag in front of you has a 49.5% chance (at least) of containing a red ball, and a 98% chance of containing no red balls.
When I lay it out like that, do you see any problem with your position?
Right. The 49.5% comes from the fact that it is still undetermined which bag is in front of me.
If the bag in front of me has red in it then it is ALL red balls in it. There are no blue balls in the bag so there is 100% chance the ball I pick out of that bag will be red.
If on the other hand the bag in front of me has blue in it then there are ALL blue balls in it, so there is 100% chance the ball I pick out of that bag will be blue.
I get what you are saying, I just don’t think it’s correct, because the red balls are not accurately represented in the 98% number.
It could very well be that the first blue ball was picked from the red bag, and there is now only red left in that bag, so if it did happen like that, then it is 100% that the next ball picked is red.
“get what you are saying, I just don’t think it’s correct, because the red balls are not accurately represented in the 98% number.”
Why? We all agreed that it was a 98% chance that we choose the blue ball bag, why is that not accurate now?
Because it is either 100% that it’s all blue balls, or it’s 100% that it’s all red balls.
Sure, there was a 98% chance the first blue ball came from the all blue bag, but that puts all the eggs in one basket when in fact it could have came from the red bag, and if so there is NO CHANCE that the next ball picked will be blue. A 0% chance of the next ball picked being blue.
How does that calculate into your calculations?
If it’s a 98% chance it came from the bag of blue balls, that leaves a 2% chance that it came from the bag of red balls, naturally. So there’s a 2% chance that the next ball will be red.
In other words, 98% of the time there’s a 100% chance the next ball is blue, and 2% of the time there’s a 0% chance that the next ball is blue.
There is no 2% for the red balls, though.
The 98% came from 50 of the 51 total blue balls that are in one bag. The 2% is the 1 blue ball that is in the other bag.
So the 2% is not the chance the next ball will be red, the 2% is the 1 blue ball that is not in the all blue bag.
Since the 2% is the percent of blue balls not in the blue bag, then it’s 98% of that bag that are red, which is 49 of the 50 balls in that bag are red.
I feel like you’re so close, you’re so very close.
There’s a 98% chance, once we’ve seen the first blue ball, that the bag we chose is full of blue balls. You agree with that.
There’s a 2% chance, then, that the bag we chose is full of red balls. If that is the case, we’ve taken the only blue ball out of that bag and the rest are all red.
It’s really as simple as that. That’s all there is to it, and it doesn’t produce any contradictions like the ones you’re experiencing.
But I’m not sure you’ve really seen the contradiction in your view yet. Have you? Or should I go into more detail?
Do you agree that the 98% came from dividing 50 by 51?
Yes
That’s one way to calculate it anyway
The Bayes formula gives the same output, so I can get the 98.0392 by other means as well (Bayes gives me 0.5 / 0.51, so quite similar really)
Good. We are making progress. 
So you then must agree that the 1 of 51 is the 2%, which is the 1 blue ball in the red bag, along with the other 49 red balls.
So in that bag there are 2% blue balls and 98% red balls. Agree?
Yeah sure
Great!
So now we have 98% of blue balls are in the blue bag, and 2% of the blue balls are in the red bag.
Agree?
Yeah