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Ice beam is a pokemon move that has a 10% chance to freeze the enemy pokemon each turn. What is the probability that I freeze the enemy pokemon if I have 3 turns?

On one hand 'logically' the answer should be 30%

On the other hand, if we add up all the cases we get 27.1%:

(1) Freeze on the first turn 10%

(2) Freeze on the second turn, but not the first: 90% * 10%

(3) Freeze on the third turn, but not the first/second: 90% * 90% * 10%

What is the correct way to go about this? Also, I observe that both methods are very close to each other. Is it generally true that the probability of an event firing within N turns is the same (or very close to) as the probability of the event firing * N?

Jason
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    Your (1),(2),(3) method is correct. If the probability is $p$ and the number of turns $N$, then $pN$ is a good approximation provided it is much less than 1. – almagest May 04 '16 at 07:33

1 Answers1

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Your second method is correct. The first one clearly is not, as probabilities don't add up like that. To convince yourself, what would happen if you had 11 turns? Probability of freezing would be 110%? Does not make any sense! :-)

In general if $p$ is the probability of freezing and you have $N$ turns, then what you're looking for is

$$1 - (1-p)^N$$

Now if $pN$ is "small", we get $(1-p)^N \sim 1 - pN$ hence

$$1 - (1-p)^N \sim pN$$

which explains your findings :-)

To understand why the approximation holds start with the taylor series of $(1+x)^\alpha$:

$$(1+x)^\alpha = 1 + \sum_{n=0}^\infty \frac{\alpha(\alpha-1)\dots(\alpha-n+1)}{n!} x^n = 1 + \alpha x + \frac{\alpha(\alpha-1)}2x^2 + \dots$$

(in our case $x=-p$ and $\alpha = N$). So if the $\alpha x$ is "small" you can ignore the higher powers and concentrate on the first term. The goodness of this approximation increases the more $\alpha x$ is small

Ant
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