|
From the past
It is not to difficult to calculate the probability of a losing streak when we start with a loss. Since we start with a loss, it is just the probability of a single loss raised to power of consecutive losses.
So, if q equals the probability of one loss, the probability of k losses simply equals q^k (q raised to the k power).
Example:
q (50% winners) = .5 and k =10 (10 consecutive losses)
q^k = .5^10 is approximately .001
However, let's ask a slightly different question. What is the probability that at least one losing streak of length k will occur somewhere among n independent wagers?
If k is less than n, this probability will be greater than q^k. Why would this be so? Because there now will be many opportunities to start a losing streak (each newly lost wager can be the start of a losing streak).
A good approximation for this probability is
P(>= 1 losing streak of length k in n wagers) is approximately equal to [1 + (n-k)p]q^k, where p is the probability of a win.
As expected this is greater than q^k.
Suppose you make 200 wagers in a month. With a win rate, p, of 50%, the probability of a loss q = 1-p, which is 50%. Most would calculate the probability of a losing streak of 10 losses as q^k = .5^10 = 0.001, or 1 out of 1000.
However, what is really needed is the probability of getting at least one losing streak of length 10 at some point during the 200 wagers. This probability is much higher. It is higher by the factor of [1 + (n-k)p].
[1 + (n-k)p] = [1 +(200 - 10)(.5)] = 96
So, the actual probability of at least one streak of length 10 over 200 wagers is 96 times as large as q^k or (96)(0.001) = .096 (nearly 1 out of 10).
That is quite a difference.
Mike (Dr Beav)
|
