calculate probability for two draws in a row, with the first round indicating chances

Question

In order to win a game, Alice has to do two actions:

a) draw a number. She can draw any number from 1 to 100, each number having the same probability to be drawn. This first draw indicates her success chances. Let's call this number X.

b) she draws another number, the same from 1 to 100 with each number having the same probability to be drawn. Let's call this number Y.

In order to win, X must be greater or equal to Y.

How do I calculate her probability to win overall? First time she draws her probability level then she draws a number that must be less or equal to the first number. So, if she draws for X number 2, her probability is really really low to win, But if she draws for X number 98, her probability is really high.

How do I calculate the chances overall?

Answer 1

If Alice draws the same number twice, she wins. What is the chance of that? What is the chance she draws two different numbers? Given that she draws two different numbers, she wins half the time, because one order wins and one loses. Add the full chance of matching draws and half the chance of different draws to get the total winning probability.

Answer 2

Hint:

Think of this in a more simpler way, where you can only draw 1's and 2's.

Therefore, there can be 4 combinations.

$X=1, Y=1$

$X=1, Y=2$

$X=2, Y=1$

$X=2, Y=2$

In cases 1, 3, and 4, the player wins, so there is a 75% chance of winning.

Let's try it when you can only get 1's, 2's and 3's.

$X=1, Y=1$

$X=1, Y=2$

$X=1, Y=3$

$X=2, Y=1$

$X=2, Y=2$

$X=2, Y=3$

$X=3, Y=1$

$X=3, Y=2$

$X=3, Y=3$

The chances of winning here are 6/9 or $66.\overline6$ percent.

Try this when you get 1's, 2's 3's, and 4's.

$X=1, Y=1$

$X=1, Y=2$

$X=1, Y=3$

$X=1, Y=4$

$X=2, Y=1$

$X=2, Y=2$

$X=2, Y=3$

$X=2, Y=4$

$X=3, Y=1$

$X=3, Y=2$

$X=3, Y=3$

$X=3, Y=4$

$X=4, Y=1$

$X=4, Y=2$

$X=4, Y=3$

$X=4, Y=4$

Therefore, there is a 10/16 or $62.5$% chance of winning.

Do you notice the relationship we get?