How many losing runs of length r playing x numbers in roulette in a sample of n bets?

Question

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.

Answer 1

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} $.