What a 5x playthrough costs on 100 SC
You hold 100 SC with a 5x wagering requirement, and you intend to play a game you assume returns 96%. What does meeting that requirement cost, on average?
Inputs
| Input | Value | Unit |
|---|---|---|
| Starting Sweeps Coins | 100 | SC |
| Playthrough multiplier | 5 | x |
| Game contribution | 100 | % |
| Assumed RTP | 96 | % |
Working
- The requirement is the balance times the multiplier: 100 x 5 = 500 SC of wagering.
- Because the game contributes 100%, every coin staked counts in full, so 500 SC must actually be staked.
- The house edge is 1 - 0.96 = 0.04. Over 500 SC of wagering the expected loss is 500 x 0.04 = 20 SC.
- That leaves an expected 100 - 20 = 80 SC. Note what this did not require: it does not matter whether the 500 SC is staked as 500 bets of 1 SC or 100 bets of 5 SC. With predetermined turnover T held fixed, expected value is linear in T, so bet size changes only how widely results spread, never the average.
- The break-even multiplier is 1 / 0.04 = 25x: the multiplier at which B − T×h = 0 under predetermined turnover T. That is the sign of expected remaining coins, not the chance of finishing the requirement.
Result
| Output | Value |
|---|---|
| Requirement (credited) | 500.00 SC |
| Must actually stake | 500.00 SC |
| Expected loss | 20.00 SC |
| Expected balance after | 80.00 SC |
| Break-even multiplier | 25x |
Takeaway
A 5x requirement at 96% RTP costs about a fifth of the balance on average. The multiplier and the house edge multiply together — neither number means much without the other.
The formula behind it
h = 1 − r
T = (B × m) ÷ c
E[loss] = T × h
E[balance] = B − (T × h)
m* = c ÷ h (the multiplier at which the expected balance reaches zero)
Assumptions
- Every coin staked is staked at the same assumed RTP r. Switching games changes r and therefore changes the result.
- With predetermined turnover T held fixed, expected value is linear in T. This follows from the linearity of expectation and holds regardless of how that wagering is split into individual bets — bet sizing does not change the expected total loss when T is unchanged, only how the results spread out around it.
- The full requirement is actually staked. The model does not stop when the balance runs out, which is why it is only valid while the expected balance stays above zero.
- No jackpot contribution, side bet, rake, or promotional credit sits outside the stated RTP.
Limitations
- This is an average over an extremely large number of repetitions, not a forecast of one session. Half of all real sessions finish below the average, and for a high-volatility game most finish well below it while a few finish far above.
- Because the model keeps staking after a real balance would be gone, B − T×h is usually a lower bound on the true expected balance rather than an exact figure. The published simulator can overshoot required turnover T by up to one stake on the last bet, so that bound is not a theorem under the simulator. Once the result reaches zero the model stops reporting a number, because a balance cannot go negative and a false precision there would be misleading.
- It cannot produce the probability of completing a requirement. That depends on the game’s full payout distribution and bet size, neither of which is an input here.
- Published RTP is a long-run property of a game’s design. It does not describe, predict, or guarantee any individual result.
This is a theoretical calculation, not a prediction. It describes an average over a very large number of repetitions and says nothing about what will happen in any particular session.
Run these numbers yourself · All worked examples · Methodology