There are 10 cards. 9 of them have goats and 1 of them have a car. I spread them on a table. I pick one and set it aside with out looking. Now there are two piles of them on the table. The one I picked in one pile and the rest of nine in another pile. Let's call the piles Pile_1 and Pile_9.
My hope is I picked the one with the car. I wonder now if I want to keep the card or do I want to pick again from the pile of 9 Pile_9 and return the one that I had picked earlier to the pile. Or do I want to stick with my original choice. Which one will be to my advantage? Should I change or stick with the original choice?
why would I change? what is my incentive? Because originally I picked 1 out of 10. If I were to change my mind, as the reason goes, then I will pick 1 out of 9 which seems better than 1 out of 10. Nevertheless there is a suspicion that the card may not be there in this pile Pile_9. How can I mathematically conclude that this picking from the pile of 9 again by setting one aside is same as 1/10?
9/10 times 1/9 = 1/10
Lets try putting 2 cards aside, making two pools of 2 and 8.
The probability of picking a car from the 2 pile is:
2/10 times 1/2 = 1/10
The probability of picking a car from the 8 pile is:
8/10 times 1/8 = 1/10
Pile_x, Pile_y, Pile_z so that x + y + z = 10
1. Probability that it is in pile_x
x/10
2. Probability that I pick my car from pile_x
1/x
2. The dependent probability that I got my car if I were to chose pile_x
x/10 times 1/x resulting in 1/10
The argument will apply to pile_y and pile_z as well. So it doesn't really matter in how many piles I break my cards, and it doesn't matter which pile I chose to pick my chances are the same.
Doesn't matter how I split the cards, as long I don't know what is in those piles, the probability of picking a car from any pile still stays at 1 out of 10. So it doesn't matter how many times you shuffle, your probability is not going to change.