Here is a lookup table to check whether Henry Wan is useful or not (if no token ignore/cancel effect).
Strategy: I always reveal the same number of tokens. When I select the number, I choose for it to maximize the mean of success. Since I cannot choose to go ahead or stop when I fail, this strategy looks reasonable for me. But this does not consider the robustness of success.
Expectation Table: the table shows the required number of non-symbol(non-, , , , ) choas tokens to achieve the given success.
|
0 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
---|
0.50 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
7 |
8 |
9 |
0.75 |
1 |
3 |
4 |
6 |
7 |
9 |
11 |
12 |
14 |
15 |
1.00 |
1 |
4 |
6 |
8 |
11 |
13 |
15 |
17 |
20 |
22 |
1.25 |
2 |
5 |
7 |
10 |
13 |
16 |
19 |
22 |
25 |
28 |
1.50 |
2 |
5 |
9 |
13 |
16 |
20 |
24 |
27 |
31 |
34 |
1.75 |
2 |
7 |
11 |
15 |
19 |
24 |
28 |
33 |
37 |
41 |
2.00 |
2 |
8 |
13 |
18 |
22 |
27 |
32 |
37 |
42 |
47 |
2.50 |
3 |
10 |
16 |
22 |
29 |
35 |
41 |
48 |
54 |
- |
3.00 |
3 |
12 |
19 |
27 |
35 |
42 |
50 |
- |
- |
- |
4.00 |
4 |
16 |
26 |
37 |
47 |
57 |
- |
- |
- |
- |
5.00 |
5 |
20 |
33 |
46 |
59 |
- |
- |
- |
- |
- |
Row(0.5~5): expected success, Column(0~9): the number of tokens. (-: required number exceeds the total tokens in game (44+20)).
Usage: Find your chaos bag (close one), and then move left (by sealing tokens) and/or down (by adding bless/curse) until you reach your goal. For example, standard NotZ (5 symbols / 11 non-symbols) may exist between 0.75 ~ 1.00. For avg 1(), it is necessary to add 2 blesses/curses tokens or seal 1 token.
Revealed number selection: In my strategy, I select the revealed number based on chaos tokens for maximizing average. Here is my selection number (which is maximizing average).
The trial number is expressed as simple formula: (reveal) = (# of total + 1) / (# of symbols + 1) (round down). If no remainder, you may choose 1 less value.
For example, standard NotZ (5 symbols / 16 toal) case, I'll reveal 2 (17/6=2.xx) tokens for maximizing (avg: 0.92). If I add 1 tokens (5 symbols / 17 total), I'll revel 3 (18/6=3) or 2 tokens (avg: 0.97); success rate is 48% for 2 revealed, 32% for 3 revealed.