I have a question about probability, hope someone could help me.
Imagine we had a jar of 80 different color jelly beans (don't know if their are 80 different colors in the world, but it doesn't matter, this is all theoretical) and someone randomly chose 20 jelly beans from the jar.
I am told to choose 1 color & if my color matches any one of the twenty chosen jelly beans I win a jar of jelly beans.
The chance of me choosing the right color is 1 in 4.
Now, assuming I had insider information that the "purple" jelly bean had a chance of 1 in 3 of being chosen (don't ask me why, it doesn't really matter. Let us just assume this is the case. Maybe the purple / green jelly bean is 3 times the size of the others. But, honestly, the reason doesn't matter. Just assume that this is the case.)
Now, let's further assume that I had additional insider knowledge, that, again for some unknown reason, the "purple" jelly bean had a probability of 1 in 6, every 5 draws, instead of 1 in 3.
In other words, 4/5 draws, the purple jelly bean has a probability of showing up 1 in 3. But for 1/5 draws it has a probability of 1 in 6.
Assuming I could choose when to play, when the 5th draw comes up I would naturally avoid playing that game, since my overall chances of winning is 1/6. I would rather only play when I know the probability of choosing purple is 1/3.
My question is:
Assuming I played 4/5 games and avoided playing the one game with the chance of 1/6 - by how much did I enhance my overall chance of winning by not playing this game?
And what would be the formula to calculate this?
Thanks in advance
As I was writing it occurred to me that this would be the correct formula
1/3 times 4 = 4/12
1/6 times 2 = 2/12
So by avoiding playing that game I would increase my overall chances to 1/3 which is 25% increase?