My “Day I Left Pennsylvania” led me to some archived website posts (before blogs were invented) I had written many years ago. I’m reposting them now. Bear in mind that most of the content in this series is over 5 years old. I have left the content more or less intact. I have removed some links and added some others — but that’s it. Enjoy!
You are presented with the following puzzle:
In a box there are 20 balls: 10 white and 10 black. With a blindfold on, what is the least number you must draw out in order to get a pair that matches?
Some people get this one right off the bat, other people don’t. If you answered 3 then you are one of the people that DO get it. If you answered 11, then you are in the latter category. Why is the answer three? The best way to explain it is by analyzing the worstcasescenarios of draws, which will give us the minimum number of draws necessary. (W = White, B = Black)
Drawing #

Current Draw

Total Draws

1

W

W

2

B

WB

3

W

WBW

The problem most people run into is that they forget that the problem asks for ANY TWO balls of the same color. If the problem said “how many must you pick until you can guarantee two white balls” then the answer would be 12.
How about this one:
In a box there are 13 balls; 6 white, 4 black, and 3 with a star design. With a blindfold on, what is the least number you must draw out in order to get three matching balls?
Again, the best approach to determine the minimum is to consider the worst case scenario. Making a table like the one above, we have: (S = Star, W and B are as above)
Drawing #

Current Draw

Total Draws

1

W

W

2

B

WB

3

S

WBS

4

W

WBSW

5

B

WBSWB

6

S

WBSWBS

7

W

WBSWBSW

So here it takes seven draws. While it is certainly possible to draw three of the same color/type in a row on the first three draws, we are only concerned with a guarantee of getting three of a kind.
Also note that the amount of starting balls is irrelevant provided there are enough to meet our goals. (the fact that there are differing amounts of White, black, and starred balls doesn’t matter, as long as there is at least three of each, since three is our target number) The formula for calculating the solution to this puzzle is: (x = number of different types of balls, y = required number to draw) [ ( x * ( y – 1 ) ) + 1 ]
Applying this to the examples above we get:
Problem 1 (2 balls, one pair)  Problem 2 (3 balls, one trio) 
[ ( 2 * ( 2 – 1 ) ) + 1 ] [ ( 2 * ( 1 ) ) + 1 ] [ ( 2 ) + 1 ] [ 3 ] 
[ ( 3 * ( 3 – 1 ) ) + 1 ] [ ( 3 * ( 2 ) ) + 1 ] [ ( 6 ) + 1 ] [ 7 ] 
Confused by the formula? Think of it this way, using the second problem as an example:
types of balls


(1) W

(2) B

(3) S

number of draws

(4) W

(5) B

(6) S


(7) W

I’m sure the answer probably doesn’t just jump out at you right there, so I’ll explain further.
Columns represents the number of different types of balls. In this case the number is three, and we have White, Black, and Starred.
Rows represents the number of balls needed for the problem, in this case also three. Now the fact that I picked “white” to be the one to get three of is completely incidental. In all probability, you wouldn’t draw them in a repeating sequence as in the example, but we’re not concerned with realism here, only with the least efficient scenario that still meets requirements.
Each cell above is numbered (the number in parentheses). Read them across from left to right, and top to bottom, just like you are reading the text on this page.
In the first and second rows, we are assuming the least efficient exhaustive scenario, meaning we go through all possibilities before repeating. The third row only has one cell because once we reach our goal, there is no reason to keep going. (Once you find what you’re looking for, do you keep looking for it afterwards?)
So the columns represent the number of steps in each iteration, and the rows represent the number of iterations needed.
However we will never need a full iteration the last time, so we subtract one from the number of iterations, but add one to the final result, since we still have to get that very last ball. Do you understand? (While we don’t need to go through all the motions the last time around, we do still need to account for that last balldraw).