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 Chess Strategy (Posted on 2006-08-01)
A man has to win two games in a row in order to win a prize. In total, he has to play only three games. The opponents are weak or strong. He has to at least play one strong opponent, and he cannot play consecutively two weak opponents. What sequence should he choose to play?

 No Solution Yet Submitted by Salil Rating: 3.2857 (7 votes)

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 re: probablistic proof :-D | Comment 5 of 7 |
(In reply to probablistic proof :-D by Daniel)

<<wss
the odds of getting two in a row are

w*s+s*s=s*(w+s)>>

Actually the prob is s*(1-(1-w)*(1-s)) = s*(w+s-w*s)

The formula given by Daniel counts the event of wins against all three twice, once as wins against the first two and once as wins against the last two.

<<sws
odds of getting two in a row here are
s*w+s*w=2*s*w>>

Actually w*(1 - (1-s)^2) = 2*s*w - w * s^2

not mentioned by Daniel is wsw, where the prob is

s*(1 - (1-w)^2) = 2*w*s - s * w^2

Of
s*(w+s-w*s)
and
2*s*w - w * s^2,
from w>s it follows that

sw > s^2
2sw-ws^2 > sw - ws^2 + s^2
2sw-ws^2 > s(w+s-ws)

so sws beats wss or ssw.

Of
2*s*w - w * s^2
and
2*w*s - s * w^2

the former is larger as a smaller amount is being subtracted from the same first term.

So sws is best.

 Posted by Charlie on 2006-08-01 11:40:04

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