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A losing run is defined as a number of consecutive losing bets. I've written a program to simulate this but the results I'm getting are different from what the formula gives, so one or another is wrong (I suspect the formula).

Here is how I derived the formula :

The chance of winning the first bet betting $x$ numbers is $\frac{x}{37}$

So the chance of losing it is $\frac{37 - x}{37}$

By the Geometric distribution, the chance of a losing run of length r is

(1) $(\frac{37 - x}{37})^r * (\frac{x}{37})$

Now since I'm only concerned with the losing bets, the number of losing bets in the sample $n$ is

$\frac{37 - x}{37} * n$

And these losing bets consist of the sum of all the losing runs of length $1,2,3,\ldots$

So to find the number of losing runs of length r, I need to multiply the number of losing bets by (1), which is

$(\frac{37 - x}{37})^{r+1} * \frac{x}{37} * n$

So for example, if $x = 12$ (betting a dozen section on the layout), $n = 10,000$, and $r = 5$, I get

$(\frac{25}{37})^6 * \frac{12}{37} * 10000 \approx 309$ losing runs of length 5.

However, my simulation output is:

  Sum of Loss Streaks = 6751

  1   744
  2   456
  3   339
  4   188
  5   153
  6    98
  7    77
  8    34
  9    35
 10    24
 11    12
 12    11
 13     6
 14     2
 15     2
 16     2
 17     3
 18     2
 19     3
 20     0
 21     0
 22     0
 23     0
 24     0
 25     0
 26     0
 27     0
 28     0
 29     0
 30     0
 31     1

There are 153 loss streaks of length 5, so I seem to be out by a factor of about 2.

Thanks in advance for any help.

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  • $\begingroup$ You multiply the number of losing bets by the fraction of losing runs with length r, but the number of losing bets is not the same as the number of losing runs $\endgroup$ Dec 5 '19 at 11:09
  • $\begingroup$ I checked some more sim outputs using different parameters and they all matched the predictions of your revised formula. And more importantly, it makes sense - many thanks! $\endgroup$
    – Median Joe
    Dec 5 '19 at 12:10
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I think I see your problem. In this line:

So to find the number of losing runs of length r, I need to multiply the number of losing bets by (1), which is $ ({{37−x} \over 37})^{r+1} \cdot {x \over 37}∗n$

You mix the number of losing runs with the number of losing bets.

Let's reformulate a bit differently:

  • Let's re-define a "losing run" by having it include $r>=0$ consecutive losses but also the win at the end.
  • In this case, the number of "losing runs" is exactly the number of wins (i.e. $ nx \over 37 $), and the probability of a run with length $r$ ($r$ losses and 1 win) is indeed $ (1- {x\over 37})^r \cdot {x \over 37} $ .
  • So, the number of losing runs of length r is $ n \cdot ({x \over 37})^2 \cdot (1-{x \over 37})^r $

This differs from your derivation by a factor of $(x/37) \over (1-x/37)$... and you tested it with x=12 and was off by about a factor of 2, which fits $ {(12/37) \over (25/37)} = {12 \over 25} $.

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