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A castle and a bishop are placed on different squares of a chessboard. What is the probability that one piece threatens the other?

Guest Mar 10, 2015

Best Answer 

 #4
avatar+1037 
+10

A castle (usually known as a rook) moves laterally and a bishop moves diagonally. That means that they can never both threaten each other simultaneously. So the probability that one piece threatens the other is equal to the probability that the rook threatens the bishop, plus the probability that the bishop threatens the rook.

 

A rook placed anywhere on the board threatens 14 squares. The probability that the rook threatens the bishop is 14/63, or 2/9.

 

The bishop . . .

 

Threatens 7 squares when placed on one of the 28 edge squares.

Threatens 9 squares when placed on one of the 20 squares one square from the edge.

Threatens 11 squares when placed on one of the 12 squares two squares from the edge.

Threatens 13 squares when placed on one of the 4 center squares.

The average number of squares threatened by a bishop is

 

$$\dfrac{\left(7\times28 \right ) + \left(9\times20 \right ) + \left(11\times12 \right ) + \left (13\times4 \right )}{64}\; = \; \dfrac{35}4 \\\\

\small \text {The probability that the bishop threatens the rook} \;\dfrac{35}{4\times63} = \dfrac5{36}\\\\

\text {The probability that either threatens the other } \;
\dfrac{5}{36}+ \dfrac{2}{9} \;=\; \dfrac{13}{36}\\$$

Here you can see that getting hooked by a rook is 62.5% more likely than getting buggered by a bishop.

 

I’m not sure where the knight came in. Maybe Cphill is taking a break from his quest.

Nauseated  Mar 11, 2015
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7+0 Answers

 #1
avatar+17711 
+5

A bishop only threatens a piece of the same color; a knight only threatens a piece of the opposite color. So, the probability that they threaten each other is zero.

geno3141  Mar 10, 2015
 #2
avatar
0

Is it decided that the bishop will be placed on a black square or a white square, or is it that it is put down at random and that it can be either a black squared or a white squared bishop ?

Guest Mar 11, 2015
 #3
avatar+91436 
+5

Geno, I do not get your answer.  

do you mean that they both can't threaten each other at the same time?  That is true. 

 

Assuming that there are no other peices in the way. And assuming that one is white and the other is black. 

 

If a bishop is on any square it will be threatened if the castle is on the same row or the same column.

There are 7 squares left on the column and 7 on the same row and there are a total of 8*8-1=63 squares that the bishop could be on so 

P(castle threatens bishop) =  $${\frac{{\mathtt{14}}}{{\mathtt{63}}}} = {\frac{{\mathtt{2}}}{{\mathtt{9}}}} = {\mathtt{0.222\: \!222\: \!222\: \!222\: \!222\: \!2}}$$

 

NOW

The probability that the bishop threatens the castle is more complicated.   

I think I shall think about it later.  LOL     

Melody  Mar 11, 2015
 #4
avatar+1037 
+10
Best Answer

A castle (usually known as a rook) moves laterally and a bishop moves diagonally. That means that they can never both threaten each other simultaneously. So the probability that one piece threatens the other is equal to the probability that the rook threatens the bishop, plus the probability that the bishop threatens the rook.

 

A rook placed anywhere on the board threatens 14 squares. The probability that the rook threatens the bishop is 14/63, or 2/9.

 

The bishop . . .

 

Threatens 7 squares when placed on one of the 28 edge squares.

Threatens 9 squares when placed on one of the 20 squares one square from the edge.

Threatens 11 squares when placed on one of the 12 squares two squares from the edge.

Threatens 13 squares when placed on one of the 4 center squares.

The average number of squares threatened by a bishop is

 

$$\dfrac{\left(7\times28 \right ) + \left(9\times20 \right ) + \left(11\times12 \right ) + \left (13\times4 \right )}{64}\; = \; \dfrac{35}4 \\\\

\small \text {The probability that the bishop threatens the rook} \;\dfrac{35}{4\times63} = \dfrac5{36}\\\\

\text {The probability that either threatens the other } \;
\dfrac{5}{36}+ \dfrac{2}{9} \;=\; \dfrac{13}{36}\\$$

Here you can see that getting hooked by a rook is 62.5% more likely than getting buggered by a bishop.

 

I’m not sure where the knight came in. Maybe Cphill is taking a break from his quest.

Nauseated  Mar 11, 2015
 #5
avatar+91436 
+5

Thanks Nauseated,

I wasn't about to just take your word for it so I also worked it out.

I got exactly the same answer but my working was a little different.

I said that for 28 out of 64 squares the bishop would be a threat to 7 out of 63 squares

then  I said that for 20 out of 64 squares the bishop would be a threat to 9 out of 63 squares

then  I said that for 12 out of 64 squares the bishop would be a threat to 11 out of 63 squares

then  I said that for 4 out of 64 squares the bishop would be a threat to 13 out of 63 squares

So the probability that the bishop would be a threat to the rook is

 

$$\\\mbox{P(they are a threat to each other)}=0\\\\
\mbox{P(the rook is a threat to the bishop)}=\frac{2}{9}\\\\
\mbox{P(the bishop is a threat to the rook)}\\\\
=\frac{28}{64}\times\frac{7}{63}+\frac{20}{64}\times\frac{9}{63}+\frac{12}{64}\times\frac{11}{63}+\frac{4}{64}\times\frac{13}{63}\\\\
=\frac{(28*7)+(20*9)+(12*11)+(4*13)}{64*63}\\\\
=\frac{560}{4032}\\\\
=\frac{5}{36}\\\\
so\\
\mbox{P(that any threat is presented)}=\frac{2}{9}+\frac{5}{36}-0=\frac{13}{36}$$

 

Melody  Mar 12, 2015
 #6
avatar+91436 
0


I do not think that this is correct - I need to rethink.   

Melody  Mar 12, 2015
 #7
avatar+91436 
+5

What Nauseated and I both forgot is that bishops can only ever be on one colour.

This will not change the answer but I will still make the working more accurate.

 

So there is not 64 squares that they can be on there is only 32 possible squares!

I wasn't about to just take your word for it so I also worked it out.

I got exactly the same answer but my working was a little different.

I said that for 14 out of 32 squares the bishop would be a threat to 7 out of 63 squares

then  I said that for 10 out of 32 squares the bishop would be a threat to 9 out of 63 squares

then  I said that for 6 out of 32 squares the bishop would be a threat to 11 out of 63 squares

then  I said that for 2 out of 32 squares the bishop would be a threat to 13 out of 63 squares

So the probability that the bishop would be a threat to the rook is

 

$$\\\mbox{P(they are a threat to each other)}=0\\\\
\mbox{P(the rook is a threat to the bishop)}=\frac{2}{9}\\\\
\mbox{P(the bishop is a threat to the rook)}\\\\
=\frac{14}{32}\times\frac{7}{63}+\frac{10}{32}\times\frac{9}{63}+\frac{6}{32}\times\frac{11}{63}+\frac{2}{32}\times\frac{13}{63}\\\\
=\frac{(14*7)+(10*9)+(6*11)+(2*13)}{32*63}\\\\
=\frac{280}{2016}\\\\
=\frac{5}{36}\\\\
so\\
\mbox{P(that any threat is presented)}=\frac{2}{9}+\frac{5}{36}-0=\frac{13}{36}$$

Melody  Mar 12, 2015

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