Correct me if I am wrong, because it's been a while, but I'm guessing you are not accounting for these numbers by means of Bayesian probabilities? Wouldn't you need to have access not only to the average number of entries per giveaway but the average number of wins per user in proportion to the number of giveaways entered per user over time? Or if you don't have access to those numbers, still find a different approximate measure to account for the prior probability of not winning a giveaway. You could then have a Bayesian formula for predicting the chances of winning, which may be more accurate (?).
Again, apologies if I am completely off, as I said, it's been quite a while since I concerned myself with these kinds of questions. I would also appreciate further explanations or if you could point out errors in my reasoning.
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Yes, I did not take into account the amount of wins per user on purpose. That being said, I would hate to contradict without being a hundred percent sure (again, statistics :P), but it seems to me like taking the wins into account in this scenario would be falling into the 'gambler's fallacy'.
Intuitively, it seems like the amount of wins is relevant to the probability of your next win, but at least in the scope of the above explanation, they is no correlation between them. Gambler's fallacy is basically taking events into account to make assumptions that are not actually possible to explain with statistics. Such as believing that the possibility of tossing a coin and getting heads continually increases as long as you get tails. It looks like it makes perfect sense by observation but it doesn't translate to probability.
But again, as you said I might be completely off with your explanation, too. So please correct me if I'm wrong.
and the most civilized discussion on probability started on the forum
Oh as another quote from another teacher: a professor in the university once told a joke on statistics, saying that if you are afraid of the possibility of being on a plane that will be hijacked, hijack the plane yourself because the probability of two people intending to do this is very, very slim.
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Of course there are so many parameters to take into account and this is no where near an accurate calculation. I made a lot of assumptions that I pointed out. But nonetheless it gives a fair idea about the situation.
Each event is of course independent, it's like tossing a coin.
the probability of getting a head in the first try is 50%.
the probability of getting two consecutive heads is 0.5 * 0.5 = 0.25.
even though getting a head in these two discrete events is 50% probable on their own, the event of getting two heads is 0.25.
so in this analogy heads is losing a giveaway, tossing a coin is a giveaway with two entries and getting a tail is winning.
1-(1/2)^2= 1 - 0.25 = 0.75. in this scenario, entring two giveaways would give you 75% chance.
If you were rolling a dice instead of tossing a coin, expecting to roll a specific number once would be 1/6 probable. Rolling the same number twice would be 1/36 probable, as the probability is 1/6 x 1/6. So instead, if you were to calculate NOT getting that specific number would be 5/6 x 5/6, which is 25/36 probable. if you subtract this number from 36/36, the sum of all possibilities, you would get 11/36, which is the possibility of getting the number you want at least once.
If you want to prove it, getting your number on the first roll and getting something else on the second one is 1/65/6 = 5/36.
getting something else on the first roll and getting your number on the second one is 5/6 1/6 = 5/36.
getting your number on both rolls = 1/36
5/36 + 5/36 + 1/36 = 11/36.
if you want to prove further, the possible outcomes are (for the picked number: 1): 11, 12, 13, 14, 15, 16, 21, 31, 41, 51, 61.
11 desired outcomes out of 36.
Ta-da! so if you consider this as two discrete giveaways with exactly 6 entries, your chance of getting at least one of the games would be: 11/36.
though as an anti-thesis to myself, if you win a giveaway you wouldn't enter for the same game again. So my assumption is that each giveaway is for a different game. but again none of us will probably see 25000 different giveaways so that being said all these calculations are hypothetical and just stated as a fun fact to keep in mind before ranting.
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Pretty cool. The only problem is that if the sun went nova it wouldn't matter if it was day or night as we would all find out at the same moment, which is to say 8 minutes after it had already occurred. No detector would be able to tell us that it was actively occurring until after we already knew. Sorry, I'm a astronomer not a statistician. :-)
P.S. Our star will never go nova so no worries.
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we would know it 8 minutes after the explosion, if the effects of the explosion travelled at the speed of light. So actually it would probably take more than 8 minutes for us to find out. Sorry I'm neither an astronomer or a statistician, just an enthusiast for both :p.
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Yes, this is based upon an assumption on the composition of the outermost edge of the blast wave. Since supernovae are among of the most energetic events in the known universe, it's assumed that the main blast front (lightest of the denser non-energy particles) would follow just moments after the initial indications that the star had exploded. Even so, while the leading edge of the blast front would be pure energy it would still be enough to disrupt (cook / boil away) much or all of the atmosphere. So even if it was night on your side of the planet, there would probably be a good indication that something unfortunate had just occurred. :-)
In reality though, these hypothetical alien scientists would have centuries or millennia of prior notice that their star was going to go nova. The star would start to misbehave long before it actually exploded, so those final 8 minutes (or whatever, depending upon how far their hypothetical planet is from their hypothetical star) would not really matter much in the overall scheme of things.
Yay astronomy!
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so to wrap things up, how many steamgifts points would you need to ensure that you have 90% chance to win a game within 8 minutes after the explosion of the Sun?
I LIKE TRAINS (OF THOUGHT)
By the way, since the nova explosion will disintegrate a notable portion of the star, it would change the mass and the gravitational pull of the star. So I'm guessing they would experience the explosion at that instant from the change in the orbit - if not from the arriving debris - wouldn't they?
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As for the first part, the OP (or another statistician) would need to answer that as that is not my forte. ;-)
As for gravity, and as crazy it sounds, based upon the general theory of relativity gravity waves still move at the speed of light. So we would not feel any gravitational disruption until after the blast wave already reached us. I have a hard time wrapping my mind around this one as it would seem that gravity should be instantaneous, but based upon observations that gravity is also a wave then it must also obey the laws of the universe, which is to say that it cannot exceed the speed of light. Under the theory of "special" relativity, it cannot exceed the speed of the fastest thing in the universe through normal space (which is still probably the speed of light.)
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Interesting, I didn't know that. From what I remember the gravitation was responsible for bending the curvature of space-time itself and it would explain a lot of phenomena from the bending of light onto black holes to the infinitesimal differences in the speed of the flow of time in different altitudes on the earth. I'll read more.
By the way I AM the OP :p.
P.S. I never thought I would use the word 'infinitesimal' in a sentence.
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As for the bending of space, assuming that it is possible for space to twist in upon itself (a wormhole) then all bets are off as for the speed of light or gravity as it should theoretically be possible for someone to travel through time and experience a gravity wave or some such wave phenomenon before it actually reached them, and also for such a wave to travel through the wormhole itself and reach you before it reached you. So all previous assumptions are based upon interactions of waves only through normal space-time.
Gravity does measurably bend space-time but an orbiting astronaut arrives back on Earth a tiny amount younger than he or she should actually be due to their higher relative speed compared to the ground back home. The warping of space-time due to this greater speed is so small (infinitesimal!) :-) as to be only detectable via an atomic clock and unfortunately only in a single direction (forward, not backward.) But yes, to really bend space-time and create a gravitational lens then you need something quite a bit heavier such as a star or black hole.
An interesting place our universe is. To me it really gives a different perspective on how utterly irrelevant things are on this planet in the overall scheme of things.
Anyway I just found you a link that should sufficiently blow your mind for the rest of the day.
Heh and indeed you are the OP... I hadn't even noticed. :-)
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False. Even after you entered 10 000 giveaways with 100 entires each your winning chance would still be 1%.
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i explained the logic twice, i won't do it for the third time...
for each giveaway with 100 entries, your chance of winning is 1% no matter how many times you enter. I'm talking about the cumulative probability of winning at least once, read it carefully.
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Cool, but I have nearly 1600 entries and haven't won anything. :c
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Hello,
Yesterday I've seen a post ranting about how the OP didn't get the game after 9 entries. So I wrote a reply about the probability of winning a game but before I could send my reply the thread was closed. I think some people could find it interesting in case they are not familiar with it, so I'm writing it once again as a thread of its own.
the probability of an event happening at least once in x times is calculated by subtracting the probability of not happening at all, from 1.
Because, for instance, the possibility of something happening at least once in three times is equal to the sum of happening once, twice and thrice. If you add "not happening" at all to this sum, you end up having the universal set, which is 1.0.
The even in this context is winning a giveaway. if in general the average number of entries to a giveaway is roughly 300, then the possibility of not winning is 299/300 = 0.997, or 99.7%. if you had entered two giveaways, however, the chance of winning at least one of them would be:
1-(0.997*0.997) = 0.993. So trying twice would increase your chance by 0.004, or 0.4%.
If you tried 100 times, the chance of winning would be:
1 - (299/300)^100 ~= 28.4%
for 300 times:
1 - (299/300)^300 ~= 63.3%
for 1000 times:
1 - (299/300)^300 ~= 96.5%
So, even if you tried a thousand times, there is a chance of not winning at all by 3.5%.
For more demanded games:
100 trials on 1000 entries:
1 - (999/1000)^100 ~= 9.5%
5000 trials on 1000 entries:
1 - (999/1000)^5000 ~= 93.3%
10000 trials on 2500 entries:
1 - (2499/2500)^10000 ~= 98.17%
25000 trials on 5000 entries:
1 - (4999/5000)^25000 ~= 99.33%
So, as you see, even if you try to get games of borderlands' caliber for 25000 times, there is a chance that you won't get a single one of them. You have to be very, very unlucky, but after all there is a chance that it would happen, just like there is a slim chance of getting it on the first try.
The bottomline is, there is no point in getting upset over not winning anything. It's just a roll of a dice. with 5000 faces :p.
When I was in high school, our theory of knowledge teacher once said: "The nature does not have probabilities. An event either occurs or does not occur. Probability is the measure of our uncertainty of the universe.", in simpler words, there is no way of knowing :), keep entring.
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