Chess board probability problem. Three random squares are chosen from a regular chess board. Find the probability that they form the letter 'L'.
I cannot think about a general way to go about these type of questions. Need hints or solutions.
 A: Note: I'm assuming that an "L" is three squares that fall in a $2\times 2$ box. If you want the sides of the L to be able to have any size, please clarify.
Why don't you try counting the number of $L$s that are possible? We can count the number of "successes" (the number of times three squares form an $L$) and divide that by the number of ways to choose three squares.
If we fix a direction on the chessboard as "up", then there are four orientations our $L$ can take. By symmetry, there are the same number of ways to make each $L$, so let's count the ones that are oriented the way the letter $L$ usually is. If we think of an $L$ as being a $2\times 2$ box with one square removed, we are looking at the case where the upper right square is the one removed.
The location of the $L$ is uniquely determined by the location of the lower left square. This square can be placed anywhere except on the right-hand column or the top row, as that would cause the $L$ to not fit on the chessboard. Thus there are $7\cdot7=49$ ways to place the $L$ in the orientation we have chosen. There are four orientations, so the total number of $L$s is $4\cdot49=14^2=196$
You don't specify how the squares are chosen, but if we assume you want to choose three squares uniformly at random, there are ${64\choose 3}$ ways to do that, since there are $64$ squares on a chessboard. Since I'm assuming every choice is equally likely, the ratio of these two numbers is the answer: $\frac{7}{1488}\approx 0.0047$ or a $0.47\%$ chance.
A: Assuming you define an L to be any three squares such that two share a row and two share a column:
Pick any square to be the corner of the L ($64$ choices).  Pick any of the seven other squares in the same row as that square ($7$ choices).  Pick any of the seven other squares in the same column ($7$ choices).  There are thus $64 \times 7 \times 7 = 3136$ ways to produce an L.
There are $\binom{64}{3} = 41664$ ways to choose three squares in all.  So the probability is $\frac{3136}{41664} = \frac{7}{93} \doteq 0.0753$.
If you mean some other definition of L, please clarify.
A: A rather simple approach:
Suppose we pick the squares one by one, and the first one lies inside the inner $4 \times 4$ square. Then there are $63 \choose 2$ pairs of squares we can pick (assuming the three of them must be different) since we don't care about order. But only $8$ of those pairs will form a L with the one we chose previously.
You can proceed similarly with other regions of the board and use conditional probability:
$P(L) = P(L | region_1)P(region_1) + P(L | region_2)P(region_2) + ... $
assuming regions don't overlap. For example, $P(L | inner) = \frac{8}{63 \choose 2}$ and $P(inner) = \frac {16} {64}$.
Hope this helps.
A: I assume that the chessboard is 8x8, 3 squares picked are each of dimensions 1x1.
Now we want to pick out an L. If we visualize the chessboard in front of us, we can deduce that for every 2x2 square present on the board, we have 4 L -like shapes but only one with the correct orientation. Since now every 2x2 square has an unique L and each L has a unique 2x2 square so we have established a bijection. Now it is easy to count that the total number of 2x2 squares are (7*7)=49. Therefore number of L's are 49. Ways to randomly pick 3 squares=(64C3). So the required probability =(49/64C3)=0.00117(approx).
A: There is a 1.92 % chance of making an "L" without space in between. If you mean that they can form an "L" when they are far from each  other is a different story.
POSSIBLE Solution: 64/100=0.64    0.64x3=1.92,
And this was in percentage form. If you need the answer not in percentage and in something else i can`t help you right NOW.
