Quote from: ndnhc on October 17, 2015, 03:40:19 PM

Initial bet : 0.0001BTC
On lose multiply by : 1.0001010101x

Playing at 9900x

If you win your first bet you stop with a profit of 9899 * 0.0001 = 0.9899 BTC

The goal is to make a profit of 1 BTC.

I think you need to play with a higher multiplier or something.

Let's set the payout multiplier (m) and work everything out from there.

m = 10001, say

The probability (c) of a bet winning at that multiplier is:
  c = 0.99 / m

The starting bet to win 1 BTC at multiplier m is:
  a = 1.0 / (m - 1)

The amount we need to multiply the stake by when we lose to maintain the same net profit is:
  r = (1+a) / ((m-1)*a) = a+1 = m / (m - 1)

The stake on the nth bet will be:
  a * r ** (n-1)

The sum of stakes for the first n bets will be:
  a * (1 - r**n) / (1 - r)

The number of bets we can afford to make is:
  math.floor(math.log((a - 1 + r) / a, r)) = math.floor(math.log(2, r))

The probability of winning in the first n bets will be:
  (1 - (1 - c) ** n)

And so the probability of doubling up successfully is:
  (1 - (1 - c) ** math.floor(math.log(2, r)))

Using this we can quickly calculate the chance of doubling up for various payout multipliers:

Quote

>>> for m in range(2, 20):
  c = 0.99 / m; a = 1.0 / (m-1); r = a+1;
  print m, (1 - (1 - c) ** math.floor(math.log(2, r)))

2 0.495
3 0.33
4 0.43374375
5 0.484150392
6 0.417817125
7 0.456617399413
8 0.483416946706
9 0.4415940551
10 0.465006203908
11 0.483238980643
12 0.452678497552
13 0.469363352144
14 0.483175196779
15 0.494793694573
16 0.472032613906
17 0.483147625356
18 0.492797130255
19 0.473835743219

>>> for m in range(100, 1000, 100):
  c = 0.99 / m; a = 1.0 / (m-1); r = a+1;
  print m, (1 - (1 - c) ** math.floor(math.log(2, r)))

100 0.491634447195
200 0.495807062928
300 0.495521270088
400 0.495378549439
500 0.496292293111
600 0.496068806433
700 0.495909161472
800 0.49641338199
900 0.496251020552

1000 0.496121110593
2000 0.496285044722
3000 0.496505870593
4000 0.496491615824
5000 0.496483063523
6000 0.496477362221
7000 0.496473289977
8000 0.496470235856
9000 0.496467860466

10000 0.496515810046
20000 0.496507258615
30000 0.496521023585
40000 0.496515444608
50000 0.496522066276
60000 0.496518173205
70000 0.49651539243
80000 0.49651953749
90000 0.496517223088

100000 0.496520356059
200000 0.496521993186
300000 0.496520877415
400000 0.496521565628
500000 0.496521978562
600000 0.496521423111
700000 0.496521738439
800000 0.496521974923
900000 0.49652215884

1000000 0.496521807542
2000000 0.496522220475
3000000 0.496522191972
4000000 0.496522177798
5000000 0.496522169092
6000000 0.496522163469
7000000 0.496522159358
8000000 0.496522218726
9000000 0.49652220935

1000000 0.496521807542
2000000 0.496522220475
3000000 0.496522191972
4000000 0.496522177798
5000000 0.496522169092
6000000 0.496522163469
7000000 0.496522159358
8000000 0.496522218726
9000000 0.49652220935

10000000 0.496522201834
20000000 0.496522218205
30000000 0.496522223275
40000000 0.496522214705
50000000 0.496522217672
60000000 0.496522221587
70000000 0.496522222003
80000000 0.496522225028
90000000 0.496522226175

100000000 0.496522217899
200000000 0.496522220646
300000000 0.496522230602
400000000 0.496522220218
500000000 0.496522232255
600000000 0.496522238624
700000000 0.496522223637
800000000 0.496522212528
900000000 0.49652220586

This is consistent with my post back in February where I found:

Quote from: dooglus on February 21, 2015, 01:58:23 AM

I can make ever increasing longer sequences which do slightly better, but it appears never to beat 49.65222222222% chance of doubling.

What's the significance of that 49.65222222222% number?