Base Game Math¶
Model¶
- Each reel stops at one of 100 stop positions. Each stop position has the same probability.
- The reels are independent.
- The window of a reel at stop
sis the strip positionsstos+3.
For each reel r and each symbol X, the calculation uses the count c(r, X).
This is the number of symbols X in the window. The calculation finds c(r, X) for each of the 100 stops.
Ways pays¶
For a pay symbol S, let n(r) be the number of S symbols plus the number of WILD symbols in the window of reel r.
The symbol S gives exactly k of a kind when n(1) to n(k) are more than zero and n(k+1) is zero.
The number of ways is then n(1) x n(2) x ... x n(k).
The reels are independent. Thus the expected number of ways is a product:
E[ways for exactly k OAK] = M(1) x M(2) x ... x M(k) x Z(k+1) for k = 1 to 4
E[ways for exactly 5 OAK] = M(1) x M(2) x M(3) x M(4) x M(5)
M(r) = mean of n(r) for the 100 stops of reel r
Z(r) = fraction of the 100 stops of reel r where n(r) = 0
EV(S) = sum for k = 1 to 5 of pay(S, k) x E[ways for exactly k OAK]
EV ways = sum of EV(S) for all pay symbols
The product is correct because the term n(1) x ... x n(k) is zero when one of the counts is zero.
Condition
The formula is correct only if reel 1 has no WILD symbol. With a WILD symbol on reel 1, one way can pay for more than one symbol. The engine and the workbook reject such a strip.
Example: PIC1 on the base reel strips¶
| Reel | Mean count (symbol or WILD) | P(no symbol and no WILD) |
|---|---|---|
| 1 | 0.2000 | 0.8000 |
| 2 | 0.3200 | 0.7600 |
| 3 | 0.3200 | 0.7000 |
| 4 | 0.3200 | 0.7500 |
| 5 | 0.3200 | 0.7100 |
Results: base reel strips¶
| Symbol | Expected ways 3 OAK | Expected ways 4 OAK | Expected ways 5 OAK | EV (credits) | RTP |
|---|---|---|---|---|---|
| PIC1 | 0.015360 | 0.004653 | 0.002097 | 0.251453 | 2.5145% |
| PIC2 | 0.022084 | 0.008286 | 0.004031 | 0.292129 | 2.9213% |
| PIC3 | 0.030016 | 0.011648 | 0.007168 | 0.326592 | 3.2659% |
| PIC4 | 0.031808 | 0.012365 | 0.007168 | 0.277133 | 2.7713% |
| PIC5 | 0.044605 | 0.018536 | 0.011994 | 0.325818 | 3.2582% |
| AUS1 | 0.095040 | 0.064282 | 0.062208 | 0.534643 | 5.3464% |
| AUS2 | 0.085747 | 0.054996 | 0.049674 | 0.444111 | 4.4411% |
| AUS3 | 0.102359 | 0.048049 | 0.047206 | 0.387278 | 3.8728% |
| AUS4 | 0.100854 | 0.057321 | 0.047206 | 0.404318 | 4.0432% |
| AUS5 | 0.088310 | 0.048681 | 0.046363 | 0.371122 | 3.7112% |
| Total | 3.614598 | 36.1460% |
KEY and FISH collect¶
Let K be the number of KEY symbols in the window. Let N(f) be the number of FISH symbols of type f in the window.
Let V(f) be the mean value of FISH type f in the current mode.
The value of each FISH symbol is independent of the stop positions. Thus:
EV collect = sum for all FISH types f of V(f) x E[K x N(f)]
K and N(f) are sums of the reel counts. Counts on different reels are independent. Counts on the same reel are not independent.
E[K x N(f)] = (sum of E[k(r)]) x (sum of E[n(r, f)])
- sum for each reel r of E[k(r)] x E[n(r, f)]
+ sum for each reel r of E[k(r) x n(r, f)]
k(r) = number of KEY symbols in the window of reel r
n(r, f) = number of FISH type f symbols in the window of reel r
The last term uses the 100 windows of the reel directly. This formula is correct for KEY symbols on all reels.
Results: base game¶
| FISH symbol | Mean value V(f) | E[K x N(f)] | EV (credits) | RTP |
|---|---|---|---|---|
| FISH1 | 2.0775 | 0.064000 | 0.132960 | 1.3296% |
| FISH2 | 13.4500 | 0.019200 | 0.258240 | 2.5824% |
| FISH3 | 67.5000 | 0.004800 | 0.324000 | 3.2400% |
| FISH4 | 666.6667 | 0.004800 | 3.200000 | 32.0000% |
| Total | 3.915200 | 39.1520% |
The probability of one or more KEY symbols in the window is 11.53% (1 in 8.7 spins).
The FISH4 symbol gives the largest part of the collect RTP. A KEY symbol and a FISH4 symbol land together 1 in 217 spins.
Base game total¶
| Component | EV (credits) | RTP |
|---|---|---|
| Ways pays | 3.614598 | 36.1460% |
| KEY and FISH collect | 3.915200 | 39.1520% |
| Base game | 7.529798 | 75.2980% |