Tehilat Olam

🛡️ Fairness

Solving the Trust Issue

The underlying concept of provably fair is that players can prove and verify that every result is fair and unmanipulated. This is achieved through a commitment scheme combined with cryptographic hashing.

The commitment scheme ensures that you have influence on all results. Cryptographic hashing ensures that the server also remains honest. Both combined create a trust-less environment.

This is simplified in the following representation:

fair result = server seed (hashed) + client seed

How It Works

1

Before you play — The server commits to a secret seed by showing you its SHA-256 hash. It cannot change the seed after this point.

2

You play — The deck is shuffled using HMAC-SHA256(server_seed, client_seed:nonce). Your client seed is something you control, so the server cannot predict or rig the outcome.

3

After the game — The real server seed is revealed. You can hash it yourself and confirm it matches the commitment shown before you played.

4

Verify — Plug the three values (server seed, client seed, nonce) into the Verify tab. It re-derives the full deck in your browser — no server contact.

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To win, you must guess correctly across 5 stages in a row. Each stage narrows the deck further.

StageYour GuessBest Odds

1 Red or Black? 50%

2 Higher / Lower / Between? 72.4%

3 Higher / Lower / Between? 54.9%

4 Pick a Suit (♣♦♥♠) 26.5%

5 Exact Card ~2%

Overall chance to win

0.1099% · 1 in 909.83

💡 Tip: Always pick the option covering the most remaining cards — that's how these "best odds" are calculated.

Let P_i be the success probability at stage i, assuming optimal play. Here is the clean derivation.

1 Stage 1 — Red or Black?

P₁ = 2652 = 1/2

26 red cards and 26 black cards in the deck.

2 Stage 2 — Higher / Lower / Between?

P₂ = 113 ∑_r=1..13 4⋅max(r−1, 13−r)51

= 4(12+11+10+9+8+7+6+7+8+9+10+11+12)13⋅51 = 4⋅120663 = 160/221

For rank r, pick the larger of "below r" or "above r" cards (4 each). Rank is uniform over 1–13.

3 Stage 3 — Higher / Lower / Between?

best gap = max(min(r₁,r₂)−1, |r₁−r₂|−1, 13−max(r₁,r₂))

P₃ = 4⋅824120⋅50 = 206/375

120 valid rank-pairs after a correct stage 2; their best-gap values sum to 824.

4 Stage 4 — Pick a Suit

P₄ = 1349 ≈ 26.5%

After 3 draws, at least one suit is untouched → that suit still has all 13 cards, so always guess it.

5 Stage 5 — Exact Card

P₅ = 148 ≈ 2.08%

One specific card out of 48 remaining.

Final Product

P(win) = 12 × 160221 × 206375 × 1349 × 148

= 206187,425 ≈ 0.1099%

≈ 1 in 909.83

Statistics

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