a and b look good, but for c, why wouldn't a bid of $15,000 maximize your chances?
Suppose you are interested in bidding on a piece of land and we know one other bidder is interested. The seller announced that the highest bid in excess of $10,000 well be accepted.. Assume the competitor's bid x is a random variable that is uniformly distributed between $10,000 and $15,000.
I have the following:
f(x) = (1/(15,000-10,000))/0 (elsewhere) = 1/5000
a. What is the probablility that a $12,000 bid will be accepted?
P(10,000 < x < 12,000) = 2000(1/5000) = .40
Is this right?
b. What is the probablility that a $14,000 bid will be accepted?
P(10,000 < x < 14,000) = 4000(1/5000) = .80
Is this right?
c. What amount should you bid to maximize the probability that you get the property?
$14,000 is my answer. Should it be different?
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a and b look good, but for c, why wouldn't a bid of $15,000 maximize your chances?
I agree $15,000 would give me a P of 1. I am happy with a P of .80 since that gives me a strong possibility of winning the property and saving $1000.
I am a little greedy and cheap.
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That's fine, but the question only asks you to maximize the probability of winning the property, so I would go with $15k
just to be a smart alek:
i would go with 15,001 to ensure that the infinitely unlikely event of two bids of exactly 15,000 did not cause me to lose out.
cheers
jerry
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