Topic: NASCRASH - Risk free opportunity - page 6. (Read 988 times)

Your website can't be accessed, first I thought because it was blocked by the internet provider I tried to use a VPN but still couldn't access it, I hope you can fix it soon so members here can give their views on this game.

Introduction
I am pleased to announce the release of my new game, NASCRASH.  NASCRASH is a race simulation-style variant of the traditional crash gambling game.  Each round has 2 multipliers that are each represented by red and blue race cars.  We have 4 total bet options available per round of play (Crash bet Blue Car 1, Crash bet Red Car 2, 1st place Blue Car 1, and 1st place Red Car 2).

House Edge
Each of the 4 bet selections has a 1% house edge using provably fair technology.

Cashier
NASCRASH currently accepts Litecoin deposits and withdrawals.
1 LTC = 1,000,000 Nitros
BTC deposits coming soon

Bankroll Investing
We offer bankroll investing for all users.  Simply deposit LTC and transfer your Nitros into your bankroll balance.  Select what percent of your bankroll you would like to risk per round, per bet type with no dilution fee for early investors.
NASCRASH takes 50% of all profit that exceeds the previous all-time high profit, with the remaining 50% going to the investors.

NASCRASH will officially launch on Thursday, August 19, at 1 PM EST.

Provably fair seeding event
Seed
b8d4105eb74a537bbe869baf73e9392cef96e74990cd72e3dad0c0ce72997274 (Last hash from a chain of 10,000,000 SHA256 hashes, Game 1)
Salt Blue Car 1
Hash from Litecoin block 2,107,448
Salt Red Car 2
Hash from Litecoin block 2,107,449
The code used to convert the seed and salt to the result
multiplierFromHash(seed, salt) {
        const nBits = 52 // number of most significant bits to use

        // 1. HMAC_SHA256(key=salt, message=seed)

        const hmac = hmacSHA256(Hex.parse(seed), salt);
        seed = hmac.toString(Hex);

        // 2. r = 52 most significant bits
        seed = seed.slice(0, nBits / 4)
        const r = parseInt(seed, 16)

        // 3. X = r / 2^52
        let X = r / Math.pow(2, nBits) // uniformly distributed in [0; 1)

        // 4. X = 99 / (1-X)
        X = 99 / (1 - X)

        // 5. return max(trunc(X), 100)
        const result = Math.floor(X);
        return Math.max(1, result / 100);
    }