Probability: finding the minimum draws required. There is a basket filled with $24$ balls. Half of them are red and half are blue. person $X$ draws a ball from the basket wearing a blindfold. what is the MINIMUM number of balls $X$ needs to draw in order to be absolutely sure that he now has two of the same color?
 A: We can use the pigeon-hole principle to figure it out.
Rewording: MINIMUM number of balls X needs to draw in order to be absolutely sure that he now has two of the same color -> MAXIMUM number of balls without duplication happening +1
(Why +1?
Let balls inside bag be X(1),X(2)....,X(n-1),X(n)
After choosing n times, there must be X(1),X(2)....,X(n-1),X(n) balls if there is no duplicate.
Assume that the n+1 time we choose the balls is not duplicating.
According to the construction, there is X(1),X(2)....,X(n-1),X(n) in the bag.
The n+1 th ball must not be X(1), because we have X(1) and this is not a duplicate.
The n+1 th ball must not be X(2), because we have X(2) and this is not a duplicate.
The n+1 th ball must not be X(3), because we have X(3) and this is not a duplicate.
Consider the case for X(4) to X(n), and we have X(4) to X(n) and this is not a duplicate.
So the ball is not X(1) to X(n), but there is only X(1) to X(n) balls inside the bag.Contradiction.(i.e,n+1 time we choose the balls is a duplicating one))
1st ball: Blue or red, that doesn't matter
2nd ball: If the first ball chosen is blue, then it is red.Vice versa.
3th ball: We can't have the blue ball, since we already have the blue ball, and We can't have the red ball, since we already have the red ball.So we cannot get a ball without duplicating.
Thus MAXIMUM number of balls without duplication happening = 2
So MINIMUM number of balls X needs to draw in order to be absolutely sure that he now has two of the same color = MAXIMUM number of balls without duplication happening + 1 = 3
