If I have odds of 1 out of 1000 for each attempt, how many attempts must I make to give me a...?

Question

...50% rate of success?
I don't wish to run into something like the birthday paradox while trying to design odds for a rare event to happen. A player may craft an item in the game I am writing. The odds of being successful in making an artifact(rare item), is 1 in 1000. I wish to make sure I understand that when i add:

1:1000 + 1:1000 + 1:1000 + ...

how often i should expect an item to be made. Let me phrase it another way. If someone made 50 attempts each with a 1:1000 chance, what are their combined odds that at least ONE of them is successful?

Thanks,

Brian

Answer 1

1:1000 = 0.1% of that event happening.

So if you had 500 attempts, the probability of that event happening is 500 * 0.1% = 50%.



But if you had made 499 attempts and in none of them the event happened, the probability of that event happening on the 500th attempt is still 0.1%.

Answer 2

crrob's answer was spot on.

1 - 0.999^50 = 4.87%

1 - 0.999^693 = 50%



You also asked, "how often i should expect an item to be made", which is a different question than asking when the chance is 50%. You shouldn't *expect* the item to be made in 693 attempts.



The answer to that question is 1000 attempts (hence the chances of it are 1 in 1000). 1000 attempts gives you a 63.23% chance, which might seem like a random number but it's really not, there is a fundamental reason for it which I won't go into here (if you care that much, message me).



Another thing: you said "I wish to make sure I understand that when I add: 1:1000 + 1:1000 + ...".

Adding isn't the right way to think about it in this situation. Adding only works when you're talking about different possibilities of a SINGLE event, not multiple events. For instance, when rolling a die, the probability of rolling a 1 or a 2 is 1/6 + 1/6. But when flipping a coin twice, the probability of it landing on heads at least once is NOT 1/2 + 1/2. Rolling the die once is a single event; flipping the coin twice is two separate events. Each attempt at making an item is a separate event.



*Edit*

Lol, why are so many people still answering the question, and saying the same exact thing me and pdq and crrob said?

Answer 3

To have a 50% rate of success, you would need 693 attempts. This is calculated by determining the probability of not winning that many times. That is, if n is the number of attempts then the equation is



(1 - 1/1000)^n = 0.5

giving you

n = ln (0.5) / ln (1 - 1/1000)

n = 692.8



If someone made 50 attempts then the odds of winning at least once would be 4.88% Again this is calculated by calculating the probability of not winning any at all. That is if p is the probability of winning at least 1 prize at 1:1000 odds in 50 attempts then the equation is:



p = ( 1 - ( 1 - 1/1000 )^50)

p = 0.0488



Whatever makes you think any of this involves adding the odds together?

Answer 4

I don't know the proper theory behind it, but I wrote a simple program that tested 1000 times if a random number between 1 and 1000 was equal to 2 (since there is a 1 in 1000 chance of the result being 2) and in the end the number 2 appeared at most 3 times. When I did 10.000.000 (ten million) tests instead of 1000, the number 2 came up around 10000 times ( actually 9885, 9993 etc times ), roughly giving a 1/1000 success/attempt ratio.



So, the combined odds are just that, 1 in 1000.

Answer 5

crrob has it perfect. "Best Answer" should go to him.



To add 1/1,000 + 1/1,000 + 1/1,000 +..... is not the correct way to calculate your over all odds. If it were, then you would say that you have a 100% chance of winning this if you attempted this 1,000 times. I think you realize this is not true. There IS a chance that you WON'T win if you attempted this 1,000 times.



As crrob mentioned, the correct formula is 999/1,000 raised to the power of "x", where "x" is the number of times you want to play. This will give you the odds AGAINST you winning. If you take out a calculator and do the math, you'll see his "693" number is in fact the point at which the odds against you winning finally fall below 50%. (49.999% to be almost exact.)

Answer 6

About 5%. If you want to ensure a 50% chance of at least one attempt being successful, then you need to try 693 times.



A way to look at it is to say if I try 50 times, what are the odds that I fail every time? That would be 999/1000 * 999/1000 * 999/1000... 50 times so (999/1000)^50 = 95.1%. So that odds of succeeding at least once is 100%-95.1% or about 5%.



Hope that helps.

Answer 7

If there are N possible outcomes and each outcome has a probability p=1/N, why aren't the frequencies equal? Shouldn't every outcome show up once in N tries? I was about to say it will never happen. It would be wrong mathematically. Let's take the simplest game of chance: coin tossing. There are 2 possible outcomes, with an equal probability of ½. P(heads)=P(tails)=1/2. The probability to get 'heads' AND 'tails' in 2 (N=2) trials is: ½ x ½ = ¼ (25%). Therefore the event is not impossible. Reality confirms it. Toss a coin 1000 times. In 25% of the cases, there will be 250 (1/4) tosses in the sequence 'heads-tails'. (Of course, as in any probability phenomenon, the figures mean closely around.)

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Answer 8

you have success with probabiliy p=1/1000



you want to compute the probability of obtaining at least one success repeating the

test k = 50 number of times.



The general formula is obtained in this way (assuming that the various tests are independent)



the probability of having at least one success, let me call it S(p,k) is equal to 1 minus the probability

of having all (k) failures. The probability of a failure is 1 minus the probability of a success, that is

1-p and thus the probability of having all failures is (1-p)^k that is (1-p) to the k-th power.

Summarizing the probability to have at least one success is

S(p,k) = 1 - (1-p)^k



in your case p=1/1000 and k = 50, for example, so this probability is equal to

1 - (1-0.001)^50 = 1 - 0.999^50 = 0.0488 which means a little less than 5 %



hope this helps.

Answer 9

It's not hard to figure out. 500 attempts at 1:1000 is 50% because 500 is 50% of 1000. 50 attempts is 1 in 20. 1 attempt would be 0.1%. Get it?

Answer 10

heuristics of statistics... No matter how many attempts the chance is 1:1000 Each time.

Answer 11

Uh, 1 in 1000. each attempt is a seperate event, according to the way I read your question. The 1 in 1000 does not change. No matter how many attempts, the odds do not change.

Answer 12

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Answer 14

1

Answer 15

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