Will the exponential mean of this market's probability be greater than ln(e-1)?
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Ṁ5524
resolved May 2
Resolved
NO

After the market closes, I will calculate the time-weighted exponential mean of its probability (using the exact probabilities, not the rounded ones in the display). Here, "exponential mean" means the natural log of the time-weighted average of e^P, where P is the market's probability. In other words, it is the generalized f-mean with f = exp.

I will resolve it to YES if the final mean is greater than ln(e-1) ≈ 0.541, which is the exponential mean of all values between 0 and 1 and therefore theoretically the fairest cutoff point.

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predictedNO

The final exp-mean was 25.57%.

predictedNO

The current average is 27.8%. Team NO is already guaranteed to win, and has been for a while.

@JosephNoonan can you recalculate what Id need to maintain to flip?

predictedNO

@MarcusAbramovitch 98.35%.

predictedNO

The exp-mean is now 35.78%. Team YES would need to bring it up to 81.36% for the rest of the time to win.

Currently calculating a mean of 39.12%.

The exp-mean over the remaining period would have to be above 69.95% for YES to win.

predictedNO

Currently calculating about 46.4%.

This one uses the exact probabilities like the others or the rounded ones?

@deagol Oh, yeah, I forgot to specify that. I'll update the description.

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