New answers tagged probability
2
I don't think this is possible to exactly solve, but here is my attempt with probabilities, which leads to a complicated expression (which I have not quite worked out yet):
0
Literally thousands of answers.
Dish1 costs (price1) 1
Dish2 and Dish3 cost (price2, price3) a prime number greater than 3 where price2 != price3.
If either price is 3, the other cannot be 2 or 5.
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