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% 

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1. Some princesses

Recently I bought a Panini sticker album with Frozen. I know, but I like animated movies and it reminded me of the time when I was collecting and trading stickers myself 🙂

Some notes: In theory, you should not have any duplicate in a pack of 5 stickers.
Also, all the 192 stickers have the same representation in the envelopes. From my past experience, I know that’s not true. Even after swapping stickers with friends, some of the stickers were impossible to find.
Prices are pack of 5 stickers – 0.6€, 50 stickers 6€, 250 stickers – 30€.

2. The Birthday problem

So, I wanted to play a bit with the probabilities involved in this little collection game. The starting point was the Birthday problem, and the apparent paradox that in a class of 23 students, the probability of having 2 students with the same birthday is 50%.
The formula to find this probability is:

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