3:45-8:15
muppet said:
eh. ok so the events are not independent but is my math correct?
-- 10 hand losing streak within 30 hands:
there's 21 ways this can occur:
lose hands 1 through 10,
lose hands 2 through 11,
lose hands 3 through 12,
...
lose hands 20 through 29,
lose hands 21 through 30.
lose rate = 49.1%
let x = the chance of losing 10 hands in 10 hands: .491^10
Agree so far.
so what we want is 21x. which is .0171
Close. What are the chances of at least one 10 hand losing streak within a sample of 2000 hands? It isn't 1991x=1.62=162%. The calculation you did gives us the expected number of 10 hand losing streaks in 30 hands. The chance of having a 10 hand losing streak in 30 hands would be
1-(1-x)^21 = 0.0170
This is very close to your answer because for this particular example, your chances of one of those streaks within 30 hands is low, so your chances of multiple of those streaks is lower still.
-- lose 10 consecutive hands in 120 hands:
ok same deal but we want 111x. which is 0.0904
For the same reason, this becomes 1-(1-x)^111=0.0865
-- lose 9 consecutive hands in 240 hands:
let y = the chance of losing 9 hands in 9 hands: .491^9
so, 232y. which is 0.385
And this becomes 1-(1-y)^232=.320 (as you can see, over larger samples, the numbers are moving somewhat lower than what you calculated)
for the combined chance of these 3 events occuring we multiply them and get .000595 (or .0595%). convert that to a fraction and we get 1 in 1681.
With the corrected numbers above, this becomes 0.000471, or 1 in 2125.
Now...back to that independence thing...
If we take a huge sample (or even a small sample, actually), your calculation will correctly tell us the expected number of 10 hand losing streaks. For example, if we look at a 1,000,000 hand sample, we expect to see 999,991x=814 streaks. And so long as you evaluate each possible string of 10 hands as independent strings, that number is correct. What I mean by that is you must treat a 12 hand streak as a 10 hand streak, followed by a 10 hand streak (starting on the next card), followed by a third 10 hand streak. Following this procedure, you should see that your estimation of the expected number of streaks is good.
The independence thing rears its ugly head when we try to calculate the odds of seeing such a streak (not the expected number) within any given sample size. This is because your streaks will be "clumped" to some degree. If we know that a particular hand is the beginning of a 10 hand losing streak, then the very next hand has a 49.1% chance of being the beginning of another 10 hand losing streak (we know the first nine hands are losses, so we only care about the last hand). That's a lot more likely than 0.491^10.
On the other hand, if a particular hand is a win or a push, you have just eliminated 10 possible hands from contention for being be beginning of a 10 hand losing streak. This causes longer times between streaks than you would ordinarily expect. This longer time between streaks, followed by streaks that are clumped together, results in the correct number of streaks overall but they are not evenly distributed throughout our 1,000,000 hand sample.
I believe that is why the referenced website shows a lower probability of a particular losing streak within a given hand sample size. I don't know the math to calculate the correct odds given the dependence problem, so I would approach it with a simple simulation.
For another look at this topic, please watch
this video. The particular part related to this discussion is found from about 3:45 to 8:15, but if you've got 21 minutes, it's worth watching the entire video.