Anthony Barnstable wrote:
Matthew Hayward wrote: Selling the random urs is complicated by the logistics:
A. How would the buyers be confident a swaperoo didn’t take place pulling out the more desirable ones?
B. If you get multiple orders your random Urs might not be delivered in a way that lets you say which ones are from what order. You might just get a bag with 19 URs if you place 2 8k orders for example. How would you handle that?
GTs rates are unknown - but I’d guess it at 1 per 2-3 8k preorders - so there auction would amount to an auction of the GT if one is found. This again intruduces a trust problem - if once the tokens arrive GTs are going for 1100 on the forums and a buyer won thee chance at one for 900 and the seller says “Too bad - didn’t get one.” What’s the buyer to make of that situation.
Posting a live video feed of opening the order would be one option to help with the trust issue, but I definitely see the concern.
For a bag of 19: flip a coin for each of the 19 and heads is lot 2 and tails is lot 2. The first lot to get to 10 causes all remaining tokens to go to the other lot. Once they have been sorted then let the picking begin.
Some such procedure could be generated that everyone could probably agree to.
As an interesting example of why I say this is logistically complicated, is that the procedure you've specified above is subject to manipulation by the seller who can pick the order that tokens are flipped for to direct the outcome.
The TL;DR of the below is that the auctioneer using the procedure Anthony provided above, by ordering the coins from least desirable to most desirable, can increase the odds of having a "really valuable" group and a "less valuable" group above what it should be.
Super nerdy shit follows:
To see this, consider the simpler example where I have 4 tokens worth: $1, $2, $3, and $4 - call them 1, 2, 3, and 4 respectively. I want am unscrupulous and for some reason want to divide these unfairly so that 3 and 4 end up together in a group more often than they should by random chance.
The correct random distribution of choosing two of the four items at random for the first group gives me with equal probability the first group having:
A. 1, 2
B. 1, 3
C. 1, 4
D. 2, 3
E. 2, 4
F. 3, 4
(and the second group having the leftovers).
Note that there are only 2 out of the 6 groupings A-F where a single grouping holds both items 3 and 4. This is grouping A (where 1 and 2 are together in one group, so 3 and 4 are in the other group), and grouping F.
The odds of 3 and 4 being together should be 2/6 = 1/3.
Any procedure of grouping these tokens together that does not give a 1/3 probability of tokens 3 and 4 being together is not a uniform at random selection procedure.
For the proposed coin flipping procedure, I could choose to flip on 1 first, 2 second, 3 third, and 4 fourth.
If I flip two heads, then 3 and 4 end up together (since I send 1 and 2 to the same group in this case).
If I flip two tails, then 3 and 4 end up together (since I send 1 and 2 to the same group in this case).
The odds of flipping either two heads, or two tails on two consecutive coin flips is 1/2.
So with the coin flipping procedure as described I can engineer a situation where tokens 3 and 4, the most valuable tokens, and put together in the same grouping 1/2 the time - in contrast with a truly random selection where they would be together 1/3 of the time.
P.S. If you wanted to fairly allocate the tokens into two groups, here's one way:
1. Assign each token a number from 1-N where N is the number of tokens.
2. Randomly generate a series of numbers between 1-N - perhaps by using some known RNG algorithm with a seed set by some mutually verifiable but unknown event, like the close of the S&P 500 stock index tomorrow.
3. Go down the RNG number list, assigning the Nth token to the first group until that group has half the tokens.
3a. If the same N is hit more than once, do nothing for all occurrences after the first.
There are more elegant ways I'm sure.
