# π²Root Raffle

Last updated

Last updated

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

**Base Eligibility**: Each holder must lock at least 5,000 $BROOT for a minimum of two weeks to participate.**Increasing Chances**: Each additional 5,000 $BROOT locked increases the holder's chances by an additional "entry".**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: $π=βπ₯/5000β$. 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:

$π=minβ‘(βπ₯/5000β,πΆ)$

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

**Calculating Individual Odds**

$N = \sum_{i=1}^{\text{total players}} n(x_i)$

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

$P_i = 1 - \left(1 - \frac{n(x_i)}{N}\right)^5$

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