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Turnabout


plainglazed
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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?
 

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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.

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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 by CaptainEd
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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.

.

 

 

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