Give me your math problems

[theTruthshaper]

On 21/12/2020 at 11:12 AM, G2F4E6E7E8 said:

Let f be a real-valued function on the plane such that for every square ABCD in the plane, f(A) + f(B) + f(C) + f(D) = 0. Does it follow that f(P) = 0 for all points P in the plane?

Yes.
For every point P, we can take points W X Y Z, and points A B C D as such:

So, we have 4 squares, and adding the equations for each of those we get

4*f(P) + 2*( f(W) + f(X) + f(Y) + f(Z) ) + ( f(A) + f(B) + f(C) + f(D) ) = 0

Since the second term and the third term also represent squares, their sum is also 0.

Thus,

4*f(P) = 0  => f(P) = 0

Hence, proved.

 

On 7/2/2021 at 5:47 AM, AquaRegia said:

Card probability problem from real life: my wife and I are playing a variant of Pinochle.  4 players, 48 card deck, 9-10-J-Q-K-A only in 4 suits, so two of each card.  Each player holds 12 cards and bidding team pass 3 cards to each other before play begins.

If I need one specific card from my partner (say, an ace of clubs), what is the probability they will have it?

My thoughts: I already know the card I need is NOT in the 12 I have, so it must be in one of the other 3 hands.  P=1/3 that it is in my partner's hand, yes?  That means P=2/3 that it is in one of my opponents'.  But there are TWO aces of clubs, and I only need one.  So the only BAD outcome is if BOTH of them are in my opponents' hands.

Pbad = (2/3)(2/3) = 4/9 = 44.4%, therefore there is a 1 - 4/9 = 55.6% chance my partner has at least one of the aces I need.

Is this the correct solution?

And how would I extend this work to cover the case where I need TWO cards (say, K and Q of clubs)?  Is it just (5/9)(5/9) = 25/81?

This isn't exactly right, because the probabilities of the second ace are not independent of the first.

So,

P(bad) = (2/3)*(23/35) = (46/105) ~ 0.438.

Which implies that there is a 56.2% chance your partner has at least one of them.

Similarly, when you need two different cards,

P(bad) = 2*(2*23/3*35) - (2*23*22*21/3*35*34*33) ~ 0.696

Thus, there is a 30.4% chance that your partner has at least one of both of them.

Edited March 3, 2021 by theTruthshaper

[AquaRegia]

14 hours ago, theTruthshaper said:

Thus, there is a 30.4% chance that your partner has at least one of both of them.

Thank you!  Honestly I'm just happy that my initial estimates were both within 1%, as I've never taken a course in probability in my life.

[Silverblade5]

Lets try this one on for size

50 - 2sin(x) * (3 + 2cos(x)) = 0

Solve for x

[theTruthshaper]

8 hours ago, Silverblade5 said:

Lets try this one on for size

50 - 2sin(x) * (3 + 2cos(x)) = 0

Solve for x

No such x.

[Dunkum]

19 minutes ago, theTruthshaper said:

No such x.

definitely not in the reals.  maybe complex?  I don't really know what values sin and cos can take with complex arguments.  but yea for real x, sin and cos both max out at 1 so if you replace both of them with that then you get 50 -(2*(1)*(3+2(1))=50-(2*5)=40-10 which is obviously not 0.  and of course cos(x) and sin(x) can't both be 1 at the same time, but ultimately that doesn't matter.

[Silverblade5]

On 3/12/2021 at 7:18 AM, Dunkum said:

definitely not in the reals.  maybe complex?  I don't really know what values sin and cos can take with complex arguments.  but yea for real x, sin and cos both max out at 1 so if you replace both of them with that then you get 50 -(2*(1)*(3+2(1))=50-(2*5)=40-10 which is obviously not 0.  and of course cos(x) and sin(x) can't both be 1 at the same time, but ultimately that doesn't matter.

Fair enough lol. To make it possible, change 2sin(x) to 2tan(x), as tan(x) has an infinite real range. Alternatively,

Solve e^x = x for x.