Quote from: dooglus on June 05, 2014, 09:29:24 AM

Quote from: organofcorti on June 05, 2014, 06:28:27 AM

EDIT: Can you explain this in more detail?

Quote

I'm told that the expected number of rolls before you see a 28 loss streak is around 514 million, but I don't know why.

No, not really, and apparently I'm misremembering anyway.

I was told "The expected value for 28 or more losses is 410,427,273" without explanation. I've asked for the reasoning.

... and was given it:

The expected number of rolls R(N) to get a streak of length N where each roll has probability p is:

R(0) = 0
R(N) = (R(N-1)+1)/p

I tested this via simulation and it does appear to be true.

That gives these expected numbers of rolls for these streak lengths of losing 49.5% rolls (p = 0.505):

Quote

1 1.98
2 5.90
3 13.67
4 29.04
5 59.49
6 119.78
7 239.17
8 475.58
9 943.72
10 1870.74
11 3706.41
12 7341.40
13 14539.40
14 28792.88
15 57017.57
16 112908.07
17 223582.32
18 442739.24
19 876713.34
20 1736068.01
21 3437760.41
22 6807448.33
23 13480097.68
24 26693264.72
25 52857951.91
26 104669213.69
27 207265771.66
28 410427272.59

Also, using simulation, I tried to figure out the number of rolls you need to make to have a 50% chance of seeing a streak of length N. It looks to me like the number is about sqrt(2) times smaller than the above numbers, and so we need to have seen 410M/sqrt(2) = 290M rolls to have a 50% chance of seeing a streak of length 28.

Since we've seen 357M rolls, we presumably have a greater than 50% chance of seeing such a streak.

This is pretty low quality research I did using simulation. The sqrt(2) thing in particular is very dodgy.