Weekly draw

The commitment

Published at window open

1aa21ed8eb21505a83b574b84629428506137673df59521ff60dbaf76b435bd9

Server seed

e2012b3205553441d463579654308bcaf7af38d7f9b5d8d945e14096f01ca0a0

Pool id

35

Total tickets at draw

0

Prize pool

0 credits

Result

One ticket is removed per placement, not the whole holding, so somebody holding many tickets can legitimately take more than one place.

Recompute it yourself

Three steps. You need nothing but a shell and Python, and none of it depends on us.

1. Check the seed matches the commitment

The hash below was published before any ticket in this window existed.

echo -n "e2012b3205553441d463579654308bcaf7af38d7f9b5d8d945e14096f01ca0a0" | shasum -a 256

That must print 1aa21ed8eb21505a83b574b84629428506137673df59521ff60dbaf76b435bd9

2. Derive each winning index

Placement n is HMAC_SHA256(key = seed, message = "weekly|35|n"), read as a big endian integer and reduced modulo the tickets still in the pool.

import hmac, hashlib
seed = "e2012b3205553441d463579654308bcaf7af38d7f9b5d8d945e14096f01ca0a0"
for n, remaining in []:
    msg = f"weekly|35|{n}"
    digest = hmac.new(seed.encode(), msg.encode(), hashlib.sha256).hexdigest()
    print(n, int(digest, 16) % remaining)

That must print the winning index column above: .

3. Walk the ticket snapshot

Tickets are laid end to end in ascending user id order. The winner of a placement is whoever owns the ticket at that index. Then one ticket is removed from that holder and the next placement is drawn from what is left.

The snapshot is stored at draw time and never edited. If we had changed a single ticket afterwards, the recomputation in step 2 would still produce the same indices but they would point at different holders, and the result above would no longer match.