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Tarcísio Gomes de Freitas - Figurinha repetida de novo? Calma! Vamos resolver isso daí. No Desaf...
50 statements · 1 politicians · June 21, 2026 · 4 min
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50 statementsFull transcript of the video. Statements the analysis classified carry topic and stance.
Folks, let's address a very important topic during World Cup time.
The World Cup sticker album.
My challenge today is a different challenge — the World Cup challenge.
And I want to know how much we need to spend to complete the World Cup album without trading any stickers.
And for that I initially need to know how many stickers our album has.
I'll call it n and n is equal to 980.
The World Cup album has 980 stickers.
Let's use probability and statistics to help us.
If I don't have any stickers and I'm going to take my first sticker, what's the probability that I get a sticker that I don't yet have?
The probability is exactly 1.
I don't have any stickers.
I'll certainly draw a sticker I don't have.
Now I'm getting to the very end of the album.
I already have 979 stickers.
What is the probability, assuming that the probability of drawing any sticker is the same, that I draw exactly the sticker I need?
And I'll call this P980.
This probability will be 1 over 980.
What is the probability that I draw, then, the sticker I need, knowing that I already have K stickers?
That probability will be... n minus k over n. And that is important information.
Because if I know that the probability of getting a sticker I don't yet have, given I have k stickers, is this, how many times will I have to repeat an event to draw exactly the sticker I want?
The inverse of the probability.
That is, 1 over p or n over n minus k. Knowing the number of events I need to perform to draw that sticker I need, it's possible to calculate the total number of stickers I'll need, which will be the sum of those events.
So Q — I'll call this Q — will be equal to
When K is zero, N over N, plus
In the next event, I'll already have one sticker.
How many times will I have to repeat this event to draw the sticker I need?
N over N minus 1, plus N over N minus 2, plus, and so on, up to N over 2, plus N over 1.
So here, notice that the n repeats.
I'll factor out the n, so my total will be the n that multiplies 1 over n, plus 1 over n minus 1, plus 1 over n minus 2, plus 1 over 2.
Plus 1.
And the sum of this harmonic series can be calculated as follows.
Approximately the natural log of n plus a constant.
This constant is very common in number theory.
It is called the Euler-Mascheroni constant.
And it can be defined as the limit of the difference between the natural logarithm and the harmonic series.
And it has an approximate value here.
of 0.58.
Knowing that, I can now calculate my number of stickers.
And my number of stickers will be N, which is 980, times the natural logarithm of 980 plus 0.58.
Natural logarithm of 980, more or less 6.88.
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