Proof-of-stake

Round 1's leader: D — a validator with 10% selection probability.

Slide the round number and watch a different validator win each time. There's no computation to race — just a deterministic lottery where each validator's ticket size is exactly its stake. A's 40-stake ticket wins 4 times as often as D's 10-stake ticket, on average.

1. Lay every validator's stake out on a line

   Validators A, B, C, D hold stakes 40, 30, 20, 10 (total 100). Laid end to end on $[0, 100)$:

   • A occupies $[0,40)$
   • B occupies $[40,70)$
   • C occupies $[70,90)$
   • D occupies $[90,100)$
2. Hash the round number into a ticket on that line

   $\text{ticket}(\text{round}) = \text{toyHash}(\texttt{"epoch-}\text{round}\texttt{-x"}) \bmod 100$ — a pseudo-random position, uniform over $[0,100)$, standing in for a real chain's verifiable random function (VRF).

3. Whoever's range the ticket lands in proposes the block

   Because each validator's slice of the line is exactly proportional to its stake, the probability validator $v$ is selected is $P(v) = \text{stake}(v) / \sum_i \text{stake}(i)$ — no computation race, just weighted chance.

4. Nothing-at-stake: voting costs nothing without a penalty

   In proof-of-work, hash power spent mining one fork can't also be spent mining a competing fork — it's a physical resource. Stake isn't consumed by voting, so without a rule against it, a validator can support every competing fork at once for free. Slashing closes this gap: forfeit your entire stake if you're caught supporting more than one fork.

• stake(v) — validator $v$'s stake; determines both its slice of the line and its slashing exposure.
• ticket — the round's hashed lottery draw, uniform over $[0, \text{totalStake})$.

No rounds run yet.

Run rounds one at a time and watch the tally bars grow. Over many rounds, D (10% of stake) wins roughly a tenth as often as A (40% of stake) — the same proportional guarantee as the formula, just visible as frequency instead of probability.

Six rounds, validators A(40) B(30) C(20) D(10), stake ranges A:[0,40) B:[40,70) C:[70,90) D:[90,100).

1. Round 1

   $\text{toyHash}(\texttt{"epoch-1-x"}) = 896$, so $\text{ticket} = 896 \bmod 100 = 96$. That falls in D's range $[90,100)$ — D wins, despite holding only 10% of the stake.

2. Round 2

   $\text{toyHash}(\texttt{"epoch-2-x"}) = 857 \Rightarrow \text{ticket}=57$, inside B's $[40,70)$ — B wins.

3. Rounds 3–6

   • Round 3: $\text{toyHash}(\texttt{"epoch-3-x"})=818 \Rightarrow \text{ticket}=18$, falls in A's $[0,40)$ → A
   • Round 4: $\text{toyHash}(\texttt{"epoch-4-x"})=779 \Rightarrow \text{ticket}=79$, falls in C's $[70,90)$ → C
   • Round 5: $\text{toyHash}(\texttt{"epoch-5-x"})=740 \Rightarrow \text{ticket}=40$, falls exactly at B's lower bound → B
   • Round 6: $\text{toyHash}(\texttt{"epoch-6-x"})=701 \Rightarrow \text{ticket}=1$, falls in A's $[0,40)$ → A
4. Six rounds: D, B, A, C, B, A

   • A: 2 wins (stake 40%)
   • B: 2 wins (stake 30%)
   • C: 1 win (stake 20%)
   • D: 1 win (stake 10%)

   A small sample, but already leaning the way the stakes predict.