Posted by Amelie on Saturday, February 23, 2008 at 7:14pm.
Yes, this looks like an expected value problem. The probability of one or the other getting a contract both times (2/3)^2 = 4/9. The probability of I or II getting one only is 2*1/3*2/3= 4/9. The probability of missing out both times is 1/9.
E.V. = (4/9)*180,000 + (4/9)*90,000) = ?
thank you drwls for your help I greatly appreciate it.
For firm 1
Probability of getting contract a only = probability of getting a - probability of getting both = 1/3 - 1/9 = 2/9
Probability of getting contract b only = 2/9
Probability of getting a and b both = 1/9
E($) = (4/9)90 k + (1/9) 180 k
= 40 + 20
= $60 K
I get 120K. With a total of 180K awarded, it seems to make sense that two of the three firms would, on the average, get 2/3 of it
for 2 firms together:
There are still 9 possible configurations
probability of 1 getting A and 2 getting B = probability of 1 getting B and 2 getting A = 1/9
so (2/9) * 180 k = 40k so far
Then 1 gets 0, 2 gets A 1/9
1 gets 0, 2 gets B 1/9
1 gets A, 2 gets 0 1/9
1 gets B, 2 gets 0 1/9
so 4/9 * 90k = 40k
Then 1 gets AB 1/9
2 gets AB 1/9
(2/9)*180 k = 40 k
so in the end
3 * 40 k = 120 k
Which is just what drwls said, but done in a much more difficult if simple way.
I did the one firm problem first.
I think you missed the first of 2 parts of the question.
Yes, I see that now. I just did the two firms together case. I should not skim through questions
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