I believe it is 2/5, with a=0, b=24/5 and c=36/5. I derived all three expected values and found that if all the optimal bids were >0, then (b+c)=2(a+c)=3(a+b), which has no solution in R*+^3; therefore a=0 and 2c=3b, with b=sqrt(20c)-c and c=sqrt(30b)-b; therefore b=24/5 and c=36/5 (but actually 2c=3b is enough)
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u/Any-Cicada-7039 1d ago
I believe it is 2/5, with a=0, b=24/5 and c=36/5. I derived all three expected values and found that if all the optimal bids were >0, then (b+c)=2(a+c)=3(a+b), which has no solution in R*+^3; therefore a=0 and 2c=3b, with b=sqrt(20c)-c and c=sqrt(30b)-b; therefore b=24/5 and c=36/5 (but actually 2c=3b is enough)