1sir, i m not clear with the statement"M1 and M3 preserve parity,M2 changes it"
] On an in
finite chessboard (whose squares are labeled by (x; y), where x and y range over all integers),
a king is placed at (0; 0). On each turn, it has probability of 0.1 of moving to each of the four edge-
neighboring squares, and a probability of 0.05 of moving to each of the four diagonally-neighboring
squares, and a probability of 0.4 of not moving. After 2008 turns, determine the probability that the
king is on a square with both coordinates even. An exact answer is required.
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UP 0 DOWN 0 0 6
6 Answers
62Hint..
suppose x right moves, y left, m up, n down
also, remaining constants..
we have to find the probability that x-y=m-n
which is same as x+n=m+y
now can you try this..
there is still some more work left..
341An exploratory:
Let
M1: a diagonal move
M2: a lateral move
M3: stay in the same square
represent the three possible moves.
Notice that while M1 and M3 preserve parity, M2 changes it.
Hence we only need to ensure that we have an even number of M2 moves.
Thus the required probability works out to
2009 (0.4)^{2008} + \binom{2008}{2} \times (0.2)^2 \times 2007 \times (0.4)^{2006}+....+ (0.2)^{2008}
= (0.4)^{2008} \left(2009 + 2007 \times \binom{2008}{2} \frac{1}{2^2} +2005 \times \binom{2008}{4} \frac{1}{2^4}....+ \frac{1}{2^{2008}}\right)
which can be summed up with known summation techniques.
1341Parity refers to odd and even. In a diagonal move you go from an odd square to an odd square, or even square to an even square. Staying put, i.e. M3 does the same thing.
But a lateral move changes the parity.
You can now see that you need an odd number of M2 moves. From then on its binomial probability distribution (i am a bit rusty on this part, but i trust you guys know it).
1consider a situation where the king moves (1, 0) with probability 0.2, moves (0, 1) with probability 0.2,
moves (1, 1) with probability 0.2, and stays put with probability 0:4. This can be analyzed using the
generating function
f(x; y) = (0.4 + 2 * 0.1x + 2* 0.1y + 4 * 0.05xy)
2008
=
((2 + x + y + xy)^2008)/5^2008
:
We wish to fi
nd the sum of the coeffcients of the terms (x^a)(y^b)
, where both a and b are even. This
is simply equal to
(f(1, 1) + f(1,1) + f(1, 1) + f(1,1))/4. We have f(1, 1) = 1 and f(1,-1) =
f(-1, 1) = f(-1,-1) = 1/[5^2008]
Therefore, the answer is .
0.25(f(1; 1) + f(1;1) + f(1; 1) + f(1;1)) =
0.25(1+3/[5^2008])