In this game you are competing against two other people to guess the number that I'm thinking of.
Each person hears the guesses stated by any preceding competitors, so you will be aware of the two prior guesses (provided to you as the vector guessesOfOpponents). Moreover, each guess must be unique.
If everyone guessed randomly, each person should have an equal chance of winning.
It might seem that you're at a disadvantage, having the last opportunity to guess. But actually you have the advantage of extra knowledge.
By guessing strategically, you should be able to achieve a success rate of 45% or more, in which
success rate = (wins + draws/2) / games
In some cases, it would be strategically beneficial to guess the same number as one of your opponents. For example, if they guess 7 and 9, for example, you would generally earn more points guessing 7 than any other number, as you would earn an extra half point from the draw if the number is 7 or 8 (which is a three-way draw) and not lose any of the full points if the number is 1-6.
Hi, James. That may be a valid point ...when playing with different rules. I deliberately set the constraint that all three guesses (from the three contestants) must be different. If my constraint were removed, then you would have freedom to implement that strategy; it would also mean you'd have to handle the situation of your opponents choosing the same number. However, I personally would not be scoring a three-way draw the same as a two-way draw. Consider if each person pays $1 into a 'prize pool' to enter: if you achieve a two-way draw, then you get back $1.50 (a 50¢ profit); but in a three-way draw nobody would get any profit, and hence I would be scoring it as if that 'game' never happened. —DIV
Understood, and I agree that a three-way draw (not possible with the current rule set) should definitely be scored differently than a two-way draw.
My code worked, although not quite as well as Tim's. I'll have to check the algorithms he used to generate his solution, and see how it differs from mine.
I figured out why Tim's solution scored higher than mine. My code gave me half credit for a few oddball cases where there was a tie between two of us but the third person was closer to the correct number. It didn't happen enough for the solution to fail the tests, but it was still sub-optimal.
This is a great problem. I was planning to eventually do one like this, but now I don't have to, as this one is very well implemented.
By the way, more problems like this one would be great, too.
Thanks. There are indeed some more that I have thought of like this....
An elegantly simple solution.
This one is a little shorter, but also longer.
Ah. Now I understand why your approach was guaranteed yield the strategically most effective guesses!
Well done! You actually devised a marginally more efficient strategy than my reference solution. :-) I've tweaked my reference solution to try to match your outputs, which makes me think that you could most likely reduce the size of your submission (if you were so inclined).
By my reading, this is close to a correct solution. The second half looks OK, but I think you may need to tweak some code in the first half.
2188 Solvers
217 Solvers
Back to basics 12 - Input Arguments
460 Solvers
60 Solvers
Given a window, how many subsets of a vector sum positive
659 Solvers