Turnabout

[plainglazed]

You and I each have a coin, select its orientation (heads or tails), and reveal to each other simultaneously.  If both are heads, you win a dollar.  If both are tails, you win a quarter.  But if both are different, I win fifty cents.  Fair enough?
 

[bonanova]

On 5/9/2018 at 7:10 AM, plainglazed said:

You and I each have a coin, select its orientation (heads or tails), and reveal to each other simultaneously.  If both are heads, you win a dollar.  If both are tails, you win a quarter.  But if both are different, I win fifty cents.  Fair enough?

Yes.

Spoiler

A selection strategy exists that gives us equal expected winnings, namely that we select H with probabilities p and q respectively, where

100pq + 25(1-p)(1-q) {my winnings} = 50[p(1-q) + q(1-p)] {your winnings}

This happens when p = q = 1/3 (Wolfram solution / contour plot); both sides of the equation evaluate to 200/9.

In fact, if either of us shows H with probability 1/3, the equation holds, regardless of opponent's strategy:
If p = 1/3 both sides evaluate to (50/3)(1 + q), and vv, due to symmetry.

Knowing this, both of us can avoid a net expected loss by presenting H 1/3 of the time.

Additionally, no advantage can be gained by either of us by slightly increasing or decreasing that value. If we both increase or both decrease, I come out slightly ahead. If one increases while the other decreases, you come out ahead by the same amount: in the case of 1/3 +/- 0.1, that amount is 2.25.

And as already stated, if only one of us increases or decreases, no advantage or disadvantage is gained if the other stays at 1/3.

.

 

 

[Molly Mae]

3 hours ago, plainglazed said:

You and I each have a coin, select its orientation (heads or tails), and reveal to each other simultaneously.  If both are heads, you win a dollar.  If both are tails, you win a quarter.  But if both are different, I win fifty cents.  Fair enough?
 

If we were choosing randomly with equal probability, I would take this bet and laugh my way to the bank.

Since we're not choosing randomly, I won't be taking this bet.  Optimal play from you actually has you win significantly over time.  This is tricky, though, and I haven't found the optimal play for you yet, but it's somewhere around choosing with probability 1/3 heads, 2/3 tails.

[CaptainEd]

Mixed strategies for you and for me

 

If you choose H with probability p and I choose H with probability q, then my expected outcome E is

E = pq - q(1-p)/2 - p(1-q)/2 + (1-p)(1-q)/4

dE/dp = 9q/4 - 3/4,  setting to zero shows my best q is 1/3

similarly your best p is 1/3

(partial derivatives, really, but I couldn’t find the curly d on my phone)

Edited May 9, 2018 by CaptainEd

[plainglazed]

No fooling you three, y'all were all over this one.  MM recognizing the slip fall right off, the good Captain doing the heavy lifting, and bn wrapping up the actual question which may have been apparent to you all just vaguely asked for in the OP.  Cheers

[CaptainEd]

Plainglazed, you asked perfectly clearly. I wish I had done such a clear job of answering as bonanova did. Thanks for the fun puzzle!