Posted by Subhotosh Khan on September 19, 2002 at 08:23:14:
In Reply to: Re: quit gambling, Jim (n/t) posted by Soroban on September 18, 2002 at 13:33:52:
: : : You are given two coins and toss each one once.
: : : The probability of getting a head on the first
: : : coin is u and the probability of getting a head
: : : on the second coin is v where u<=v.
: : : The probability of getting 2 heads is 1/2.
: : : The probability of getting 2 tails is at a maximum
: : : subject to that condition.
: : : Calculate u,v each to 4 decimal places.
: : : HELP!!
: Good advice, Dennis!
: Jim,
: This is an interesting max/min problem.
: Probability of two heads is 1/2:
: P(HH) = P(H1)*P(H2) = u*v = 1/2, or v = 1/2u
: Probability of two tails is to be maximized:
: P(TT) = P(T1)*P(T2) = (1 - u)(1 - v)
: There is the function we are to maximize!
: P = (1 - u)(1 - v)
: Since v = 1/2u: P = (1 - u)(1 - 1/2u)
: You can finish it from here, right?
: (Differentiate, set equal to zero, solve, etc.)
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After maximizing, we get
u = v = 1/sqrt(2)
so now u + v = sqrt(2) > 1
Arn't you implicitly assuming u + v = 1 when you wrote P = (1-u)(1-v)!!!
Am I missing something (these probability problems always confuse me..)??