https://blockchain.info will be considered as oracle.
1,000
3.00Are BTC options not well priced or something? I don't know much about them specifically but it seems like they're implying a much higher probability: https://www.deribit.com/options/BTC/BTC-29MAR24/BTC-29MAR24-75000-C
@dominic well, now BTC has crashed so it is closer to the "correct" price. I'm still curious about the options though if anyone knows more than me
@dominic Options on highly liquid lit markets are almost always very well priced. This thinly traded play-money market is the one with large inefficiencies.
@HarrisonNathan I agree in general, but I could imagine there being like large fees, or maybe these options aren’t the ones people actually trade or something
@dominic Is there a simple way to see the implied probability by these options, or do we need to calculate it by hand ?
And what probability did you get at the time ?
Also unfortunately I think Harrison is right, there are still large inefficiencies in Manifold, there aren’t enough people compared to the numbers of markets, and more time is needed for the mana to go in the hand of the good predictors on each subject and for people to become better at it.
For example here, I just bet enough to choose the probability of the market for some time, and I didn’t even know about these options.
@dionisos it’s the delta (second column in the linked chain)
edit: btw how were you picking a probability to set it to?
@dionisos It should be approximately 2x the delta, because what we’re looking for is the probability that it ever hits 75k not the probability that it is >75k at expiration
@dominic so, I think those deltas there are too frothy for my liking. But that just reflects BTC vol, which I tend to underestimate.
> edit: btw how were you picking a probability to set it to?
By feelings x)
And trying to vaguely adjust these feelings because I dislike bitcoin and because I bet against others.
Sorry this is bad x)
@dominic Is there a simple explanation of why it would be approximately 2X the delta (I assume you would have to convert the delta to an odd before here, otherwise it wouldn’t make much sense) ?
@dionisos lol nah it’s great! in the end the price (and all the technical metrics like volatility, deltas, greeks, etc.) is just the aggregation of everyone’s feelings.
@dionisos 2x is assuming it’s a random walk, which it likely isn’t as there’s bias (trend) and mean reversion factors.
@deagol It’s not exactly a random walk, but it should be pretty close for a highly traded financial instrument like BTC.
@dionisos the reason it’s 2x for a (symmetrical) random walk: imagine all the paths (P) hitting the strike or limit at any point before expiration, then from there until expiration, half of those paths (on average) will end up above the limit (p), and half below. Thus P=2p.
@deagol I am currently confused (and I know it is because I am somewhat dumb).
After reading it is assumed to be random walk, here is what I did :
P("meet 75k" and "end at 75k") = P("end at 75k") = P("end at 75k" | "meet 75k") X P("meet 75k")
= P("meet 75k") X 0.5 [assuming random walk]
But it seems like it can’t be right, because it would make the probability greater than 1
@dionisos i’m not sure from your formula, but it would only be >1 if P(end at >75k)>0.5, but in that case (for a symmetrical RW) the price must already be >75k so the “anytime before” condition has been met and P=1.
Note in the options chain, the >0.5 deltas are (usually) in the money (the price exceeds the strike).
@deagol
> but it would only be >1 if P(end at >75k)>0.5
It isn’t the case for 75k, but … oh wait, I assumed it was the case that this was >0.5 for some values, but after looking at it, it seems like it is <0.5 for any value above the current price.
So this makes sense.
Thank you for taking the time to explain it.