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## Checkerboard Chances 2

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #42307 Fun: (2.14) Difficulty: (2.81) Category: Probability Submitted By: billy314

Timothy has a four by four checkerboard. He uses scissors to cut out each square from the board. He then randomly arranges the pieces into four rows and four columns. What is the probability that this layout is in a checkerboard pattern?

The first black piece can go in 16 different places since the checkerboard has an equal number of red and black squares (8 red and 8 black). The next piece however has only 7 different places it can go since one space is taken and it must be in a checkerboard pattern. The next piece has 6 different places it can go, and so on. Therefore, the number of ways the black pieces can be correctly put into the checkerboard is 16*7*6*5*4*3*2*1=16*7! (7! is 7*6*5*...*1). The first red piece has 8 different places it can be put. The next piece has seven. The next piece has six, and so on. The number of ways the red pieces can correctly be arranged is 8*7*6*5*4*3*2*1=8! (this is read 'eight factorial'). The total number of ways all the squares can be arranged (correctly or not) is 16! (16*15*14*...*1). The number of arrangements that form a checkerboard is 16*7!*8!. The probability that a checkerboard will be formed is 16*7!*8!/16!. Simplifying this fraction by canceling (a lot!) you get 1/6435 or about 0.01554%.
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 marie7771 Jul 22, 2008 Complicated, but I got it! first comment woo hoo Herman Nov 24, 2008 I went about this a different way, actually. The prob that the first square was correct was 8/16. The prob that the next square selected was correct was 8/15. The next, 7/14, and so on. Multiply all of that out, then multiply by two, since the first square can be black or red (turn the board). lookagain Nov 10, 2010 There are (16 choose ways to assign one of the colors and then 1 way to assign the remaining color. There are two set-ups of the checkerboard (white left corner or black left corner). So the probability is 2/(16 choose . HFmanager Nov 19, 2010 I don't necessarily agree with this. It is assumed here that the 2 different red squares are uniquely identifiable, but shouldn't they be identical? So the number of ways the checkers can be arranged is 16!/(8!x8!), and there are only two arrangements that can have alternating red and black squares (i.e. RBRB...RB and BRBR...BR). So the answer should be 2x(8!x8!)/(16!). Am I right?