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Revisiting the Crooked Bishop. Revisiting the Crooked Bishop.[All Comments] [Add Comment or Rating]

Peter Hatch wrote on Fri, May 10, 2002 09:05 PM UTC:

> When you're really stuck with probabilities, you can use the
> laborious case-by-case analysis.

Indeed.  Instead of checking each case for two-path bonus, it's easier to
just check each case to see whether it is blocked, and just add up the
probabilites.  With 5 squares that can be blocked (e3, d2, d4, f2, and f4)
there will be 2^5 = 32 total possibilities.  1 possibilty has no squares
blocked, 5 have one square blocked, 10 have two squares blocked, 10 have
three squares blocked, 5 have four squares blocked, and 1 has all five
squares blocked.  The probability of each possibility is 0.7^(number of
unblocked square) * 0.3^(number of blocked squares).

The no squares blocked possibility lets us reach the destination square, as
do 4 of the one square blocked possibilities (all but e3), and 2 of the two
squares blocked possibilities (d2 and d4, f2 and f4).  None of the rest do.
 So the total probability is 0.7^5 + 4 * 0.7^4 * 0.3 + 2 * 0.7^3 * 0.3^2 =
0.51793, which agrees with my formula.  So I'm confident I'm right again.
:)