here's a fun maths problem for you:
in Balatro, each hand gains a positive number of Chips and Mult, which are then multiplied together to get your final score. but on the Plasma Deck, the rules are altered such that Chips and Mult are "balanced", averaging the two before scoring.
does this "balancing" step always give you a better score than you would have gotten without the Plasma Deck? how would you prove this?
(I'll share my answer in a day or so)
my solution to balatro maths problem
since some folks have already given solutions, I figure I might as well give mine. it is not the best solution, but it is interesting
if we call Chips C and Mult M, then a regular score is CM and the Plasma Deck score is (½(C+M))²
expanding this out, we get ¼(C²+2CM+M²), which can be rearranged to ¼(C²+M²)+½CM
we can multiply the first term by "one" by multiplying by CM/CM. this is valid because both are nonzero. this lets us further rearrange it into ¼CM(C²+M²)/CM
now, this makes our score equal to CM(½+¼(C²+M²)/CM)
so, since both sides of our inequality have a factor of (positive) CM, we're really just comparing 1 and ½+¼(C²+M²)/CM, i.e. comparing 1−½=½ and ¼(C²+M²)/CM, i.e comparing 2 and (C²+M²)/CM
for C=M, we can see that they're the same. which is kind of obvious: averaging two equal values does nothing, so, if Chips and Mult are equal, the Plasma Deck does nothing
of course, the interesting thing is to determine whether we can actually go below 2. to do this, we need to take the gradient of this function using calculus
so, let f(C,M)=(C²+M²)/CM. the partial derivative ∂f/∂C is (2C(CM)−(C²+M²)M)/(CM)², which is (C²M−M³)/(CM)², which is 1/M−M/C². this means
∂f/∂M is 1/C−C/M² since they're symmetric
this is (0, 0) at C=M, which means that is a local minimum or maximum. we can kind of implicitly rule it out as a saddle point since the function is symmetric, but we can take the second partial derivative to check which it is
∂²f/∂C² is 2MC/C⁴=M/C³, and similarly, ∂²f/∂M² is 2C/M³. both of these are positive and never zero meaning that we have a global minimum
another way of wording (C²+M²)/CM is C/M+M/C, and this gives us the insight that the Plasma Deck is better by roughly the ratio of the two values, with a factor of a quarter (this comes from the factor of two combined with a sum of two values)
so, if the values are different by a factor of 100, the plasma deck will score around 25x more. so, 2×200 is 400, but 101×101 is 10201; we multiply by 100 but then divide by 4
we can see this also holds for 40×4000 = 160000, and 2020×2020=4080400
simple solution to balatro maths problem
this one comes from @wallhackio, so I can't take credit for it, but it is a substantially simpler solution
like before, the base deck score is CM and the Plasma Deck gives (½(C+M))²
subtract these two to get (½(C+M))²−CM, which expands to ¼(C²+2CM+M²)−CM, equal to ¼C²+½CM+¼M²−CM, equal to ¼C²−½CM+¼M², equal to ¼(C²−2CM+M²)
you then have to notice that this is (½(C−M))²; you can use the quadratic formula to explicitly factor it, but it's a common enough formula you can just know that's the solution
notice that this quantity is never negative: since we're squaring it, it doesn't matter whether Chips or Mult is bigger, since squaring it gets rid of the negative sign. it's only zero when C=M, which we also saw earlier
this also gives you a more explicit formula for the difference, and also the general rule that them being farther apart makes the Plasma Deck do even better than normal, which is similar to the intuition from the calculus solution
but I like it because you don't need calculus to get it
re: simple solution to balatro maths problem
@clarfonthey i am now curious about your calculus solution
re: simple solution to balatro maths problem
@clarfonthey i am the biggest silly billy
re: simple solution to balatro maths problem
@wallhackio I replied to it, so :p