It’s been a LONG time since probability class, and I’m struggling with this this morning.
You have a bag with 1000 balls in it. Of those, 250 are red, 750 are black.
If you pick one ball, the probability is .25 that it will be red.
If you pick two balls, the probability is roughly .25 *.25 = .0625 that they BOTH will be red.
But, if you choose two balls, what is the probability that at least ONE of them will be red?
Part of me says it’s .25 +.25 = .5, but that doesn’t seem right because it implies that if I choose 4 balls, the odds of one being red is approaching 1.
Assuming the two teams are evenly matched, yes. But that’s not a valid assumption when it comes to sports.
While I understand the point your making, a better example is if you flip 3 heads in a row, the the probability of flipping a head on the 4th toss is still .5.
The point I am making is that, even if the teams are far from evenly matched there is no guarantee that the better team will finish on top over a short series of competitions.
Most people fail to see this and invoke nonsense theories.
Turek, ColdPhoenix’ analysis is of course correct.
What’s wrong with your .25 + .25 = .5 analysis is that you double-counted some outcomes : .25 that ball *one *is red already counts outcomes where ball *two *is red as well as black, and counting .25 that ball two is red counts *all *outcomes where ball two is red.
In other words, you double-counted the outcomes where both balls are red.
So your analysis requires subtracting that probability - which is 1/4 *1/4 = .0625.
It seems like there are 4 possibilities if you pick 4 balls. Both red, both black, first red second black, first black second red. In three of those cases you have at least one red ball. So 75%? But why do I have the feeling I am missing something?
Your calculation assumes the probabilities of picking red or black ball on any pull are the same, which they aren’t in the OP (three times as many black as red balls).
Are you putting the first ball back before you pick the second one? If you are, you’re right; otherwise, you have to account for the fact you have on fewer red ball and one fewer ball total.
I often would represent it to my students this way: say I’m in a boxing match with Mike Tyson. One of us will win, right? Therefore, I have a 50% chance of winning, right?
Most of them already know what when I say something, and then say, “Right?” I purposefully was wrong and I’m hoping they’ll figure out the mistake.
One thing not specified in the OP is whether the picks are with replacement or without replacement. That is, after you pick the first ball, do you put it back in and randomize the bag before picking out the second ball? Or do you leave the first ball out when you reach in to pick a second ball?
Also, one of the ways you can analyze the problem is to calculate the probability that both balls will be black, and then subtract that from 1 to calculate the probability that at least won’t be black; i.e., that one ball will be red.
E.g., in the first scenario you’ve got a 750/1000 chance each time to pick a black ball, and 1 - 0.75 * 0.75 = 0.4375, just as ColdPhoenix said. In the second scenario, the probability is 1 - 750/1000 * 749/999 = 0.437̅6̅8̅, like These are my own pants said.
I didn’t specify the replacement scheme in the OP, but I did specify a few posts later.
The actual, real life scenario is this: My wife and I are going to a reading by a famous author. For buying a ticket, you get to attend the reading and you get a copy of the book. 1000 tickets were sold. Randomly mixed into the 1000 books being distributed are 250 signed copies. I was trying to figure out the probability of us getting at least one signed copy.
