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Say, in the game of Pokemon Go, the probability of catching a shiny Pokemon is $\frac{1}{450} \approx 0.22222\%$.

Some players consider that "lucky". And

  1. by the Law of Large Numbers, we can say, if we catch 450 Pokemon per day, then over the long term, we get about 1 shiny Pokemon per day (assuming all Pokemon has a chance of being shiny, for simplicity, because in reality not all Pokemon has its shiny form released yet).

  2. by probability, we can also say that, the probability of getting one or more shiny Pokemon per day is

$$ 1 - \left(\frac{449}{450}\right) ^ {450} \approx 0.6325 = 63.25 \% $$

So how do we explain the difference between the "long term" $100\%$ vs the short term $63.25\%$? How do we explain this "lucky everyday" vs "not as lucky everyday at $63.25\%$? Where did that remaining $36.75\%$ go?

P.S. I think I know the answer and how the math relates to the philosophy of everyday life... but I will put the answer here 3 - 7 days later.

nonopolarity
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2 Answers2

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Over say $100$ days, we win at least one Pokemon on $63$ days, and also another $37$ Pokemon on those days, to make up for the days when we win nothing.

JMP
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1

You are confusing two different things.

The probability of getting $k$ shiny Pokemon in one attempt can be modelled using a Poisson distribution. In this case, the rate is $\lambda = \frac{1}{450}$, so the probability is $P(x = k) = \frac{\lambda^k \cdot e^{-\lambda}}{k!} = \frac{(1/450)^k \cdot e^{-1/450}}{k!}$ where $k$ must be an integer for it to make sense.

When you have $450$ attempts, the probability distribution is $450 P(x=k)$, and $k = 1$ gives an expected value of $1$ as you said. But this tells us nothing about $2$ Pokemon, or $3$ Pokemon and so on, which need different values of $k$. You could sum up the values of $k$ from $1$ to infinity, but much easier is to subtract $1$ from $k = 0$.

Toby Mak
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