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 Chessboard (Posted on 2004-01-14)
Three different squares are chosen randomly on a chessboard.

What is the probability that they lie in the same diagonal?

 No Solution Yet Submitted by Ravi Raja Rating: 3.5000 (8 votes)

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 solution | Comment 1 of 20
For the purpose of counting available positions in the same diagonal, there are 10 possible types of position of the first chosen square. It could be on one of the two main diagonals, either in a corner or once, twice or three times removed from the corner (that's 4 so far). It could be on an edge, once, twice or three times removed from the corner (we're up to 7). It could be adjacent to an edge row and either near the center of that row or near the main diagonal, or it could be 2 rows removed from an edge row and adjacent to the center line.

It simplifies matters if we consider only one triangle of 10 squares and multiply each one's subsequent probabilities by the probability that the first square chosen would be of that type. The four types that are on a main diagonal have probability 4/64 each, as there are four corner squares, four squares that are on a main diagonal one removed from the corner, etc. The types that are not on a main diagonal have probability 8/64 each as there are duplicates, one on either side of the nearest main diagonal.

If we number the rows 1-8 (from top to bottom) and the columns 1-8 (left to right), let's consider the ten typical squares as those in which the column number is equal to or greater than the row number and less than or equal to 4. Those where the column number equals the row number are of course on a main diagonal, and the probability is 1/16 of our initially choosing a given such type of square. The others have a probability of 1/8 of being chosen as the initial square. We'll multiply the conditional probabilities given that type of initial square by these initial probabilities and add them up to get the overall probability.

For most squares, there are two possible diagonals to consider. The restricted triangle of squares we have chosen makes the formula simple for how many choices of other squares there are in each of these diagonals. The SW-NE (directionally speaking) diagonal will have c+r-2 squares besides the one at row r, column c (for example, the square at row 1, column 3, has 2 other squares in its SW-NE diagonal: row 2, column 2 and row 3, column 1). The NW-SE diagonal has 7+r-c other squares besides the one under consideration. (Again for example, the square at row 1, column 3, has 5 other squares sharing this diagonal, at r 2, c 4; r 3, c 5; r 4, c 6; r 5, c 7 and r 6, c 8.) So, for a given row and column, the probability of completing one diagonal is (c+r-2)/63 * (c+r-3)/62, and of completing the other diagonal is (7+r-c)/63 * (6+r-c)/62. These are mutually exclusive, so their probabilities can be added, before multiplying by the probability that the first one was of that type to begin with.

Taking the ten types by row and column, we get:
```
r 1 c 1 : 1/16 *  (0/63 *-1/62 + 7/63 * 6/62) = 1/1488

r 1 c 2 : 1/8  *  (1/63 * 0/62 + 6/63 * 5/62) = 5/5208

r 1 c 3 : 1/8  *  (2/63 * 1/62 + 5/63 * 4/62) = 11/15624

r 1 c 4 : 1/8  *  (3/63 * 2/62 + 4/63 * 3/62) = 1/1736

r 2 c 2 : 1/16 *  (2/63 * 1/62 + 7/63 * 6/62) = 11/15624

r 2 c 3 : 1/8  *  (3/63 * 2/62 + 6/63 * 5/62) = 1/868

r 2 c 4 : 1/8  *  (4/63 * 3/62 + 5/63 * 4/62) = 2/1953

r 3 c 3 : 1/16 *  (4/63 * 3/62 + 7/63 * 6/62) = 3/3472

r 3 c 4 : 1/8  *  (5/63 * 4/62 + 6/63 * 5/62) = 25/15624

r 4 c 4 : 1/16 *  (6/63 * 5/62 + 7/63 * 6/62) = 1/868

```

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Note that the cases where there are no choices or only one choice do have the corresponding probability nulled out, as the formula introduces a zero in each of these two cases.

The probabilities add up to 7/744, or 1/106.285714... (the 285714 repeats).

The above was calculated by
```
10   Tot=0

20   for Row=1 to 4

30    for Col=Row to 4

40      if Row=Col then P=4//64:else P=8//64

50      W1=Col+Row-2

60      W2=7-Col+Row

65      print Row;Col;":";P;tab(14);"* ";

70      P=P*(W1*(W1-1)+W2*(W2-1))//(63*62)

75      print str(W1);"/63 *";str(W1-1);"/62 +";str(W2);"/63 *";str(W2-1);"/62 =";P

80      Tot=Tot+P

90    next Col

100   next Row

110   print Tot,1/Tot

```

----
with its output cleaned up (putting the letters r and c, adding parentheses and eliminating the double / that UBASIC produces in its output of rational numbers).
Edited on January 14, 2004, 3:57 pm
 Posted by Charlie on 2004-01-14 15:43:24

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