Written on 14 March 2019, 04:25pm

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You have 8 teams. They will be drawn one against each other, so 4 pairs in total.

Question 1: how many distinct pair sets are possible?

105. I got to this number after running a large number of simulations. Then I did a little bit of research and I also found the formula:

Question 2: if 4 of the 8 teams are from England, what is the probability that all 4 of them will be drawn together?

Again, after analyzing the 105 distinct pair sets, I found that only 9 of them have all-English pairs. The full probability set is:

• two English pairs: 9/105 or 8.57%
• exactly one English pair: 72/105 or 68.57%
• no English pair: 24/105 or 22.86%

## Random things #8

Written on 18 March 2015, 10:28pm

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## A/B vs Multivariate Testing

A/B testing: two versions (A and B) are compared, which are identical except for one variation that might affect a user’s behavior. Total number of variations: 2. More
Multivariate testing: multiple variables are modified for testing a hypothesis. The goal of multivariate testing is to determine which combination of variations performs the best out of all of the possible combinations. [Total # of Variations] = [# of Variations on Element A] X [# of Variations on Element B] ... More

## Permutations, Arrangements, Combinations

Given a set of n elements (ex – for n=3, the set is A, B, C)
Permutations: each ordered set of n elements P(n) = n!
In our example with n=3, P(3)=3!=6: АВС, АСВ, ВАС, ВСА, САВ, СВА
Arrangements: each ordered set of k elements A(n,k) = n! / (n-k)!
In our example with n=3, ordered pairs of 2, A(3,2)=3!/(3-2)!=6: AB, BA, AC, CA, BC, CB
Combinations: each unordered set of k elements C(n,k) = n! / k! (n-k)!
In our example with n=3, un-ordered pairs of 2, C(3,2)=3!/2!*1!=3: AB, AC, BC
And the relationship between P, A, C: C=A/P
Remember that for the permutations you don’t need a k! More

Learning isn’t done to you, it’s something you do. You need to take responsibility of your education. There will always be a new technology to learn, but this is not that important. Is the constant learning that counts.
Andy Hunt – Pragmatic Thinking and Learning

We all tend to learn best by doing and teaching. Active learning is a much more effective way to learn than any other way.
It seems a bit strange, but it should really be no surprise that play is a powerful mechanism for learning. […] This simple process that comes natural to us all, but somehow gets “taught” out of us, is the simplest and purest way to learn.
John Sonmez – Soft Skills

Written on 30 January 2014, 11:23pm

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The previous post about Panini stickers got into some mathematical formulas. However, the 2 main conclusions were referring to the duplicates probability and distinct probability. That was the mathematical approach to the problem.
Below – the geeky one 🙂

### 1. Duplicates probability

In a Panini pack of 17 stickers (out of 192 possible stickers), there are 50% chances to have a duplicate.

The geeky way:
– generate a random array of ‘n’ integers in the range [1,192]
– calculate how many duplicates has the array
– repeat this a number of times to get a reliable view.

Results (PHP code at the end of the post):

Number of stickers - Probability of duplicate
10 - 20.47%
11 - 25.8%
12 - 31.2%
13 - 37.13%
14 - 40.6%
15 - 45.47%
16 - 47%
17 - 53.4%
18 - 58.4%
19 - 63.27%
20 - 66.53%
21 - 69.87%
22 - 74.53%
23 - 76.53%
24 - 80.27%
25 - 82.33%
26 - 85.47%
27 - 86.27%
28 - 87.87%
29 - 89.93%
30 - 91.67%
31 - 93.73%
32 - 94.4%
33 - 94.87%
34 - 96.07%
35 - 96.53%
36 - 97.13%
37 - 97.47%
38 - 97.6%
39 - 98%
40 - 98.33%