4 ms·

Ok, so what the fund manager did, is sell his BTC holdings to lock in profits, and buy back severely out-of-the-money (i.e. worthless) options, just in case he

by tomp 8y ago

Ok, so what the fund manager did, is sell his BTC holdings to lock in profits, and buy back severely out-of-the-money (i.e. worthless) options, just in case he was really wrong (i.e. if BTC didn't just go to 25k, but up to and over 50k). Presumably, his thought process was something like "if I sell now, I make $Xm. If I wait until BTC goes to $36k, I make $2Xm... but I think that's quite unlikely. However, it's possible, and my clients would be really mad at me if BTC skyrockets, goes to $60k, and I chickened out at only $18k. So, what I can do, is buy this option for cheap, forgo the profits between $18k and $50k, but if I'm really wrong, at least I'll cash in if it goes above $50k." So, hopefully he paid much less ($1m) than the total profit he made ($Y0m)?

patio11 8y ago
Assuming he sold as many Bitcoin as he bought calls for, the options cost more than 20% of Bitcoin's highest price ever. (Black-Scholes is a harsh mistress.) c.f. http://data.ledgerx.com/top100 http://data.ledgerx.com/top100, Ctrl-F "2017-12-20 8:39 am"
- tomp 8y ago
  Maybe, but that literally makes no sense - there's no BTC price at which the (sell BTC + buy options) trade makes more money than the (keep BTC) trade (unless I miscalculated something, which is also an option).
  - patio11 8y ago
    Sell at ~$18k, pay $3.6k for the call, outperforms hodling in worlds where Bitcoin ends the year below ~$14.4k.
  - pathseeker 8y ago
    Do you mean the sell + buy options never makes more money, or do you just mean in this current case? Are you forgetting that you can leverage up with more options to offset that gap? Sell securities at $X, buy calls for 2x the number of securities you had with strike $X+$Y for $Y/10, and have the product go up to $X+($Y2). Profit = ($Y22) - 0.1$Y = $Y3.9. Keep the securities, have the security go up to $X+($Y2). Profit is $Y*2.
    - tomp 8y ago
      in this specific case, if the option is more expensive than what you're selling in the first place.

by tomp 8y ago

patio11 8y ago

Assuming he sold as many Bitcoin as he bought calls for, the options cost more than 20% of Bitcoin's highest price ever. (Black-Scholes is a harsh mistress.) c.f. http://data.ledgerx.com/top100 http://data.ledgerx.com/top100, Ctrl-F "2017-12-20 8:39 am"

tomp 8y ago
Maybe, but that literally makes no sense - there's no BTC price at which the (sell BTC + buy options) trade makes more money than the (keep BTC) trade (unless I miscalculated something, which is also an option).
- patio11 8y ago
  Sell at ~$18k, pay $3.6k for the call, outperforms hodling in worlds where Bitcoin ends the year below ~$14.4k.
- pathseeker 8y ago
  Do you mean the sell + buy options never makes more money, or do you just mean in this current case? Are you forgetting that you can leverage up with more options to offset that gap? Sell securities at $X, buy calls for 2x the number of securities you had with strike $X+$Y for $Y/10, and have the product go up to $X+($Y2). Profit = ($Y22) - 0.1$Y = $Y3.9. Keep the securities, have the security go up to $X+($Y2). Profit is $Y*2.
  - tomp 8y ago
    in this specific case, if the option is more expensive than what you're selling in the first place.