Instrument 4 · Spread
How far one session can land
The expected cost is an average. To see the spread around it, this instrument uses an invented game simple enough to work out exactly: on each spin, 1 chance in 20 of getting 18 × the bet back, otherwise nothing. Its return is 90%.
- Expected result
- −$60.00
- Low end of the band
- −$222.00
- High end of the band
- $102.00
- Sessions that finish inside the band
- 92.6%
- Chance of finishing ahead
- 25.1%
Worked by hand, with the defaults
- Each spin: $1.00 bet; 18 × $1.00 = $18.00 back with probability 0.05, otherwise nothing.
- Expected result: 600 × $1.00 × (0.05 × 18 − 1) = −$60.00.
- The number of paying spins out of 600 follows a binomial distribution. The band runs from its 5th to its 95th percentile: 21 to 39 paying spins.
- Net result for k paying spins = $1.00 × (18k − 600): −$222.00 at k = 21, $102.00 at k = 39.
- Because paying spins come in whole numbers, the band does not hold exactly 90 in 100 sessions: 3.2% finish below it, 4.2% above it, 92.6% inside.
- Chance of finishing ahead: the probability of more than 33 paying spins, 25.1%.
Why the band moves with the session length
The expected loss grows in step with the number of spins. The width of the band grows more slowly. So in this game, short sessions have a band that is wide compared with the expected loss, and finishing ahead is fairly common; over many spins the expected loss takes over and finishing ahead becomes rarer. Try 100 spins, then 3,000, and watch the share inside the band as well.
An invented game, on purpose
A game with many prize levels would have a different spread from this one. The invented game is here because its arithmetic can be shown in full; it is not a model of any real machine.