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Author Topic: Generalizing the Law of the Third  (Read 469 times)

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Bayes

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Generalizing the Law of the Third
« on: January 15, 2017, 10:33:29 AM »
In this post I'll generalize the Law of the Third and show that it's just a special case of a simple general formula which finds the expected number of unique elements in a group in a given cycle of spins.

The data you need before applying the formula is the following:

   1. The length of the cycle, call this C.
   2. The probability of a single win for some element in a related group of outcomes (streets, dozens, etc). Call this P.
   3. The number of possible distinct members of the group. This just refers to the number of streets if the group consists of streets, the number of splits if the group consists of splits, etc. It's important that the group is mutually exclusive and exhaustive. I.e. The group must encompass the entire wheel and each member of the group must not overlap with any other member. In practice this just means the groups are restricted to:

single numbers (37)
Splits (18)
Streets (12)
Lines (6)
Dozens/Columns (3)
Even Chances (2)

I know these aren't quite exhaustive (single numbers excepted) because of the zero, but they're close enough. We'll call this number N.

   4. The probability of at least one hit in the cycle C for the outcome which has probability P. We'll call this A.
To find A you'll need a binomial distribution calculator. An easy to use one can be found here. You will will need to enter your                 value for C and the probability P.   

Having found A the formula is very simple:

Expected number of unique elements in the cycle = N x A

Example 1: (the law of the third)

The length of the cycle, C, is 37
The probability of a hit on a single number is 1/37 or 0.027, so P = 0.027
The number of members of the 'group' is 37, so N = 37.

Now calculate the probability of at least one win in the cycle. Enter P in the first box, C in the second box, and 1 in the third box. Important: The number which goes into the third box is always 1, no matter what P and C may be. Click 'calculate'. The probability A we need for the formula is the last value (hightlighted in red). See screenshot below:



Therefore, A = 0.64 (rounded up).

Now apply the formula:

Expected number of hits in the cycle = 37 x 0.64 = 24 (rounded up). This means that there are 37 - 24 = 13 numbers unhit after a cycle of 37 spins, which confirms the law of the third.

Note that having found A, we can apply the formula to successive cycles. e.g. For the second cycle of 37 spins we apply the formula to the unhit numbers (of which there are 13). So 13 x 0.64 = 8 (rounded) of the 13 numbers will have hit at the end of the 2nd cycle, leaving 5 numbers still unhit.

Example 2:

Here we'll find the expected number of unique streets in a cycle of 8 spins.

The length of the cycle, C, is 8
The probability of a hit on a single street is 3/37 or 0.0811, so P = 0.0811
The number of elements in the group is 12, so N = 12.

Using the binomial calculator, we find A = 0.49 (if you've been following my thread on 'High odds to EC' you'll recognize this probability).



So the number of unique streets which have hit after 8 spins is 12 x 0.49 = 6 (rounded). At the end of the next cycle of 8 spins the number of streets remaining will be 6 x 0.49 = 3.



 
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