🎲Root Raffle

The beetroot raffle is a weekly lottery that distributes half of the reward pool to 5 participants. the main points are as follows:

  1. Base Eligibility: Each holder must lock at least 5,000 $BROOT for a minimum of two weeks to participate.

  2. Increasing Chances: Each additional 5,000 $BROOT locked increases the holder's chances by an additional "entry".

  3. Cap on Maximum Entries: To maintain fairness and avoid excessive dominance by high stakeholders, there will be a cap of 100 entries (500,000 $BROOT).

Step-by-Step Breakdown of the Root Raffle

Let x be the total amount of $BROOT locked by a participant.

Let n be the number of entries a participant receives in the lottery.

The formula for calculating the number of entries n based on the amount x locked is: 𝑛=⌊π‘₯/5000βŒ‹π‘›=⌊π‘₯/5000βŒ‹. This denotes the floor function, which rounds down to the nearest whole number. This ensures that only full 5,000 $BROOT increments count towards additional entries.

To prevent excessive influence by very large holders, there is a cap on the maximum number of entries. This cap is set at 500,000 $BROOT for a maximum of 100 entries.

The formula with a cap C (e.g., 100) would be:

𝑛=min⁑(⌊π‘₯/5000βŒ‹,𝐢)𝑛=min⁑(⌊π‘₯/5000βŒ‹,𝐢)

Practical Example

  • Participant A locks 5,000 $BROOT: They receive 1 entry (n = 1).

  • Participant B locks 27,000 $BROOT: Without a cap, they would receive 5 entries (n = 5). With a cap of 100, they still get only 5 entries.

  • Participant C locks 505,000 $BROOT: Without a cap, they would receive 101 entries. With a cap of 100, they receive only 100 entries (n = 100).

Calculating odds to win

Total Entries (N)

Sum the entries of all participants to get the total entries in the lottery. This can then be used to calculate individual odds

N=βˆ‘i=1totalΒ playersn(xi)N = \sum_{i=1}^{\text{total players}} n(x_i)

Calculating Individual Odds

Given there are 5 winners each week, the chance 𝑃𝑖𝑃𝑖 that a participant wins at least one prize is derived from the complement of the probability of not winning any of the 5 draws:

Pi=1βˆ’(1βˆ’n(xi)N)5P_i = 1 - \left(1 - \frac{n(x_i)}{N}\right)^5

This formula calculates the probability that a participant wins at least one of the five available prizes, given their number of entries 𝑛(π‘₯𝑖)𝑛(π‘₯𝑖) relative to the total 𝑁𝑁.

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