Examples
Examples
Try the starting example
- Total tickets
- 10
- Tickets held
- 2
- Prize draws
- 3
- Ticket replacement rule: without or with
- without
- Probability at least one winning draw
- 0.533333
- Expected winning draw count
- 0.6
- Winning-draw-count variance
- 0.373333
- Winning-draw-count distribution
- 0: 0.466666666667; 1: 0.466666666667; 2: 0.0666666666667; 3: 0
Try another scenario
- Total tickets
- 10
- Tickets held
- 2
- Prize draws
- 3
- Ticket replacement rule: without or with
- with
- Probability at least one winning draw
- 0.488
- Expected winning draw count
- 0.6
- Winning-draw-count variance
- 0.48
- Winning-draw-count distribution
- 0: 0.512; 1: 0.384; 2: 0.096; 3: 0.008
Calculation method
Calculation method
Probability at least one winning draw
One minus probability of zero held-ticket draws.
Expected winning draw count
Prize draws×held/total.
Winning-draw-count variance
Without replacement use finite-population correction; with replacement use binomial variance.
Winning-draw-count distribution
Hypergeometric counts without replacement; independent binomial draws with replacement. A held ticket can win multiple draws only in with mode.
Symbols and units
- a
- Total tickets
- b
- Tickets held
- c
- Prize draws
- d
- Ticket replacement rule: without or with
Use the units shown. A list keeps the order in which you enter its values. The symbol ^ means “raised to the power of”.
Assumptions & limits
Assumptions & limits
- Uniform ticket selection under the explicitly supplied replacement rule. This counts winning draws, not unique people, prize value or actual raffle eligibility; no tickets or draws are issued.
Are these inputs private?
These calculations run in your browser. A shared link contains the values you choose to share. Only questions submitted to the separate AI finder are sent to OpenAI.
Further reading
Background reading; this does not imply endorsement or independent review of SumFizz.