Every day I enter a lottery in which I have a one in 400 chance of winning. If I enter the lottery 416 times, what are the chances I win at least once?
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1Does this answer your question? If I buy 2 lottery tickets do I double my chance of winning? – Apr 25 '23 at 01:23
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@user1147844 I don't think the link is a duplicate since there the win for different choices are not independent, while here I think independence for lotteries on different days is a reasonable assumption. – ronno Apr 25 '23 at 06:13
2 Answers
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Hint:
It is 1 - the probability of never winning.
Bram28
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That makes sense, but I cannot figure how to find out the probability of never winning. Can you assist? – PeteZ Apr 28 '17 at 16:11
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@PeteZ OK, suppose I play the lottery just once. What is the chance of not winning? – Bram28 Apr 28 '17 at 16:12
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In this case I have a one in four hundred chance of winning and 99.75% chance of not winning. – PeteZ Apr 28 '17 at 16:13
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@PeteZ Correct. You have a chance of $\frac{399}{400}$ of not winning when you play once. OK, what if you play the lottery twice: what is the chance of not winning? – Bram28 Apr 28 '17 at 16:14
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That's the part I'm unsure how to calculate. Is it still 99.75? Is it 99.5? Thank you for this by the way. – PeteZ Apr 28 '17 at 16:16
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@PeteZ OK, so now not winning means that 2 independent events need to happen: you don't win the first time, and you don't win the second time. They are independent, since winning or not winning one time does not effect the chance of winning or not winning a second time. And, when events are independent, and both need to happen, you multiply their probabilities. So, you get $\frac{399}{400}*\frac{399}{400}$ as the chance of not winning when playing twice. OK, so what is the chance of not winning when playing 3 times? – Bram28 Apr 28 '17 at 16:18
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1@PeteZ It is like throwing dice by the way. What is the chance of getting a 6 with 1 die? It is $\frac{1}{6}$. And what is the chance of getting two 6's when throwing two dice? It is $\frac{1}{6} * \frac{1}{6} = \frac{1}{36}$. You multiply. – Bram28 Apr 28 '17 at 16:22
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1ok so i think i understand. the chances of not winning at .9975 which will be raised to the 416th power (number of attempts) equaling 0.35299472315 which turns into approx 35%. 100%-35% = 65%. – PeteZ Apr 28 '17 at 16:36
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The answer is this: $$1-\left(\frac{399}{400}\right)^{416}$$
In a single lottery you have probability of $\frac{399}{400}$ not to win. if you play $416$ times and lose all of them thats $\left(\frac{399}{400}\right)^{416}$. now we found what is the probability not to win $416$ times, if we subtract that from $1$ we get what we want.
projectilemotion
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dvd280
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