So after you pick the first ball, and it’s blue, and you can’t change bags, so you pick a second ball out of the same bag, what’s the probability that the second ball is blue, in your opinion?
If there are 50 blue balls and 49 red balls, then there are 99 balls total, and 50 are blue, so 50 of 99, or 50.5% that it is a blue ball and not a red ball.
Right, so this is what I’ve been building up to. Your approach has left you with two statements that, in my opinion, are not compatible with each other.
You said 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.”
And then, your answer to my previous question is this:
“50.5% that it is a blue ball and not a red ball.”
So let me just lay them out clearly, so you see what’s going on here:
You believe, after selecting a bag and a ball and seeing that the ball is blue, that the probability that every ball remaining in that bag is blue is 98%.
And you believe, after selecting a bag and a ball and seeing that the ball is blue, that the probability that the next ball you select from the same bag is blue is 50.5%.
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.
So, you have a bag in front of you, and you can say:
There’s a 98% chance that this bag is full of only blue balls, AND when I pick a ball out of it, there’s a 50.5% chance that it’s blue and a 49.5% chance that it’s not blue.
Is that right?
Until I know which bag I have to pick from I do not know if it was the bag full of blue balls or the bag that contained all but 1 red balls.
If the next ball I pick is blue, then there is a 100% that the bag has 48 blue balls left in it, at which time the NEXT choice will be 100% (48 out of 48) that it will be a blue ball.
If the next ball I pick is red, then there is 100% chance that the following ball will be red.
There is still not enough information after selecting only 1 blue ball to eliminate the red balls from the total. You still don’t know if the bag you have is all red or all blue at this point in the game.
1 blue ball has been selected. If it was from the 50 blue ball bag then that bag has only blue left in it. If it was from the 1 blue ball bag (49 red) then that bag has only red left in it.
The next selection will reveal 100% whether the bag is full of red or blue.
In my view, if I have a bag that has a 98% chance of being entirely full of only blue balls, then I would have to say that, if I pull a ball out, I’m expecting at least a 98% chance that it’s blue.
You know what I mean?
So in your calculations you calculated 98% that you will pick a blue ball next from what numbers, 50 of 51???
The exact same way you got the number.
I haven’t actually tried to calculate it myself, I haven’t tried the Bayesian approach but I’m confident it would give me the same value.
But the point I’m making isn’t about any specific number or value, it’s about the relationship of those probabilities.
If there’s a X% probability that the bag I’m holding is full of only blue balls, then the probability that the next ball I pick out of it is blue must be at least X%.
never mind
I got 98% from 50 of 51.
But there are only 49 balls remaining in the bag of blue if you think you picked from that bag. So you think there is a 49 out of 49 chance of picking a blue? That is 100% not 98%. So what numbers DID you use to calculate 98%, 50 of 51?
There can’t be 51 blue balls left because you already picked 1, and there was only 51 to start with. So there is only 50 blue balls left.
So in order for you to claim 98% chance of picking a blue, you must think there is 49 out of 50 blue.
You either picked the blue ball from the red bag, meaning the bag you are holding is all red remaining, or you picked the blue ball from the blue bag, which means there is 49 blue remaining, and the OTHER red bag still has 1 blue ball left in it.
So you either think there is 100% chance that all the balls in your bag are BLUE, or you think there is 100% chance that all the balls in your bag are red.
One bag has 49 balls and the other bag has 50 balls. You do not know which bag you have, 50 balls or 49 balls.
I don’t know which bag, no, but we both agree there’s a 98% chance that it’s full of blues.
But that does not eliminate the fact that there are 50 blue balls and 49 red balls remaining.
If I started a new scenario telling you 1 bag had 49 blue balls and the other bag had 1 blue ball and 49 red balls, what would you say the probability of picking a red ball would be?
So, I suppose what I’m seeing here that you’re not seeing is,
If there’s a 98% chance that the bag contains only blue balls,
and there’s a 49.5% chance that the next ball you pick is blue,
there’s significant overlap of those probabilities.
Let’s make it into a series of runs. Let’s assume we run the experiment 1000 times, and we get the perfect distribution every time.
Of those 1000 times, 980 of the times we had the bag that was full of blue balls.
Of those 1000 times, 495 of the times we selected a red ball.
That means that, at minium, 475 of the times we ran the experiment, the bag was full of Blue balls, and we somehow managed to pick a red ball out of it anyway.
Isn’t that strange?
Because they are two different calculations.
98% is the probability of the bag picked from contains all of that color, because there were 51 total blue in two bags, 50 of which were in 1 bag. A blue was selected hence the 98%.
If a red was selected then it would be 100% that the bag contained at the start 49 red and 1 blue.
If a red was selected then there is a 1 out of 49 chance the next ball selected from that bag would be blue, or a 2% chance of selecting a blue and 98% chance of selecting another red.
See how there was 100% that it contained 49 red and 1 blue at the start, but there is only a 98% chance of selecting a red from that bag?
At this point, I don’t feel like you fully engaged with what I said. I think you should.
If we assume your statistics are correct, and try to arrange a scenario where the experiment results match your expected probabilities, you will have situations where someone pulled a red ball out of a bag that had only blue balls in it.
That’s impossible! Every ball pulled would have a calculation done prior to pulling it, calculating the probability of the next ball selected being blue or red. If there is no red then there is 0% chance of it being red and 100% chance of being blue, and if there is no blue then there is 0% chance of it being blue and 100% of it being red.
There is NO CHANCE of predicting a color to be drawn when there isn’t any of that color. 17 blue and 0 red means 100% that a blue will be drawn. 456 red and 375 blue means a total of 831 balls, so there is a 456 out of 831 (54.87%) chance of drawing a red, and 375 out of 831 (45.13%) chance of drawing a blue. It does not matter if they are in 831 boxes or 3 boxes.
Then show me. Show me a way of distributing the experimental results that get even close to approximating your expected statistics.
Your expected statistics are
After pulling a blue ball first, 98% of the time you’ll have the bag full of blues
After pulling a blue ball first, 49.5% of the time the second ball you pull will be red
That’s the challenge. If you can do that, without pulling red balls of of blue bags, while staying inside the bounds of the game as it’s defined, I’ll Venmo you $100
How about you start by answering this: