Asklemmy

All participants select their own random whole number and publish it to the group. All participants add all the numbers together. The result is either odd or even (heads/tails) and everyone arrives at the same result independently.

1. Decide on a random N and what tails (even) and heads (uneven) mean.

2. Each party generates a random number

3. Combine the numbers with a conmutative operation of some sort, the harder the operation the better.

4. Take the hash N times. (Can be done independently by each participant)

(4.5) optional: for extra robustness, do some hard-to-calculate transformations to the result of 4. (Can be done independently by each party)

5. The final result is either uneven or even === coin toss. (0 will be treathed as even*.*)

This is not infalibe, one party could get all the numbers a precalculate a answer to get a specific result but they will need to randomly try numbers. adding some timing constrains, using big numbers and hard operations would make that sort of attack not really practicable.

Nice question, had fun thinking about it!

Everyone throws a number at the same time, the result is the checksum/sum of the throws. Server throws first publicly, keeps the device Numbers secret until the last throw. It's not perfect, but it'll do.

Coin flipping

Suppose Alice and Bob want to resolve some dispute via coin flipping. If they are physically in the same place, a typical procedure might be:

Alice "calls" the coin flip,
Bob flips the coin,
If Alice's call is correct, she wins, otherwise Bob wins.

If Alice and Bob are not in the same place a problem arises. Once Alice has "called" the coin flip, Bob can stipulate the flip "results" to be whatever is most desirable for him. Similarly, if Alice doesn't announce her "call" to Bob, after Bob flips the coin and announces the result, Alice can report that she called whatever result is most desirable for her. Alice and Bob can use commitments in a procedure that will allow both to trust the outcome:

Alice "calls" the coin flip but only tells Bob a commitment to her call,
Bob flips the coin and reports the result,
Alice reveals what she committed to,
Bob verifies that Alice's call matches her commitment,
If Alice's revelation matches the coin result Bob reported, Alice wins.
For Bob to be able to skew the results to his favor, he must be able to understand the call hidden in Alice's commitment. If the commitment scheme is a good one, Bob cannot skew the results. Similarly, Alice cannot affect the result if she cannot change the value she commits to.

1. Everyone tosses three coins, and posts it in the chat
   • If a player tosses three of the same, they have to toss again.
2. Everyone chooses the mode coin from their neighbour, and adds it to their stack
3. Each player, with 3+N coins, picks the mode coin in their own collection.
   • Ideally: the player's own bias, is outweighed by the other player's biases.
4. The final coin is the mode of all players coins.

from numpy import median
from pprint import pprint

players = {"p1" : [1,0,1],  ## playing fair
           "p2" : [0,0,1],  ## cheating
           "p3" : [1,1,0],  ## cheating
           "p4" : [1,1,0],  ## cheating
           "p5" : [0,0,1]}   ## playing fair
print("Initial rolls:")
pprint(players)

get_mode_coin = lambda x: int(median(x))
get_all_mode_coins = lambda x: [get_mode_coin(y) for y in x]

for play in players: ## Players add the mode coin from their neigbours
    players[play] = players[play] + get_all_mode_coins(players.values())
print("First picks:")
pprint(players)

for play in players: ## Players collapse their collections to mode
    players[play] = [get_mode_coin(players[play])]
print("Last modes:", players)

print("Final choice:", get_mode_coin([x for x in players.values()]))

Which as you can see, is no better than simply picking the median coin from the initial rolls. I thank you for wasting your time.