please I need help:
So, suppose I've a coin and I bet €x (that could be my Capital €100 or part of it) on throw H/C. If I win I double my €x (eg. if I bet €100, I'll gain €100*2) and if I loose I loose my €100.
Now:
1) If I bet 10% of my balance after 100 throws I'll have €4700
2) If I bet 25% of my balance after 100 throws I'll have €36100
3) If I bet 40% of my balance after 100 throws I'll have €4700
4) If I bet 51% of my balance after 100 throws I'll have €31
Please please help me to understand these results, I don't understand it at all. Which formula is used?
My idea (wrong): let me choose point 1) (so 10% of the balance every throw). Considering 50 throws (because on coin throw there's 50% of winnings)
Throw.....Bet...............Capital
0 .......... 0.00 € .......... 100.00 €
1 .......... 10.00 € .......... 120.00 €
2 .......... 12.00 € .......... 144.00 €
3 .......... 14.40 € .......... 172.80 €
4 .......... 17.28 € .......... 207.36 €
5 .......... 20.74 € .......... 248.83 €
6 .......... 24.88 € .......... 298.60 €
7 .......... 29.86 € .......... 358.32 €
8 .......... 35.83 € .......... 429.98 €
9 .......... 43.00 € .......... 515.98 €
10 .......... 51.60 € .......... 619.17 €
11 .......... 61.92 € .......... 743.01 €
12 .......... 74.30 € .......... 891.61 €
13 .......... 89.16 € .......... 1,069.93 €
14 .......... 106.99 € .......... 1,283.92 €
15 .......... 128.39 € .......... 1,540.70 €
16 .......... 154.07 € .......... 1,848.84 €
17 .......... 184.88 € .......... 2,218.61 €
18 .......... 221.86 € .......... 2,662.33 €
19 .......... 266.23 € .......... 3,194.80 €
20 .......... 319.48 € .......... 3,833.76 €
21 .......... 383.38 € .......... 4,600.51 €
22 .......... 460.05 € .......... 5,520.61 €
23 .......... 552.06 € .......... 6,624.74 €
24 .......... 662.47 € .......... 7,949.68 €
25 .......... 794.97 € .......... 9,539.62 €
26 .......... 953.96 € .......... 11,447.55 €
27 .......... 1,144.75 € .......... 13,737.06 €
28 .......... 1,373.71 € .......... 16,484.47 €
29 .......... 1,648.45 € .......... 19,781.36 €
30 .......... 1,978.14 € .......... 23,737.63 €
31 .......... 2,373.76 € .......... 28,485.16 €
32 .......... 2,848.52 € .......... 34,182.19 €
33 .......... 3,418.22 € .......... 41,018.63 €
34 .......... 4,101.86 € .......... 49,222.35 €
35 .......... 4,922.24 € .......... 59,066.82 €
36 .......... 5,906.68 € .......... 70,880.19 €
37 .......... 7,088.02 € .......... 85,056.22 €
38 .......... 8,505.62 € .......... 102,067.47 €
39 .......... 10,206.75 € .......... 122,480.96 €
40 .......... 12,248.10 € .......... 146,977.16 €
41 .......... 14,697.72 € .......... 176,372.59 €
42 .......... 17,637.26 € .......... 211,647.11 €
43 .......... 21,164.71 € .......... 253,976.53 €
44 .......... 25,397.65 € .......... 304,771.83 €
45 .......... 30,477.18 € .......... 365,726.20 €
46 .......... 36,572.62 € .......... 438,871.44 €
47 .......... 43,887.14 € .......... 526,645.73 €
48 .......... 52,664.57 € .......... 631,974.87 €
49 .......... 63,197.49 € .......... 758,369.85 €
50 .......... 75,836.98 € .......... 910,043.82 €
Unfortunately, as you can see, I can't reach €4700.
Also my idea is wrong because I'm supposing 50 throws continually winning.
I can't understand your four lines of data because you do not explain how many wins you are counting on. It could be instructive to look at what happens if you bet a fixed fraction of your bankroll, play only twice, and get one head and one tail. You always lose money and you lose more the higher fraction of your bankroll that you bet. It doesn't matter whether you win or lost the first toss. If you bet a fraction $f$ of the bankroll, when you lose the bankroll is multiplied by $1-f$ and when you win it is multiplied by $1+f$. After a win and a loss you have $(1-f)(1+f)=1-f^2$ of your bankroll left. The game is fair. This loss is compensated by the fact that if you win twice you have $1+2f+f^2$ and if you lose twice you have $1-2f+f^2$.
I managed to reproduce your first three values assuming that when you win you get your bet back plus twice the bet. Then a win and a loss gives $(1-f)(1+2f)=1+f-2f^2$. For $51\%$ bet I find you wind up with about $60$, not $31$. I suspect that is an error in calculation by whoever did it. Here is the Alpha result supporting $60$.