Too bad you didn't spend 19 cents. Cards are always useful for those really cheap deals. :P
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It was not much effort though. I anyways enter giveaways, I just searched polarity and entered a lot of them and then did the math which was simple as well (look at the original post edit)
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I haven't been around SG long, but I'm pretty sure many of us don't make super efficient use of our time :)
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It's not the point of the game or neither the price of the game. It's about the math :)
I had to have a game that has low entry points and a lot of giveaways or else it wouldn't work out.
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I had only one giveaway that was over 1000 (1001 to be exact). Mostly the giveaways ranged between 100-1000. You could do an approximation of the probability of winning a copy of any game if you enter a certain number of times.
For example let's take a solid 500 entries for each giveaway:
(500-1)/500 is the probability of loss and just put the number of entered giveaways as the power (let's say 256) ((500-1)/500)^256=0,6 of probability of not winning. So it seems you have to enter a lot of giveaways with larger entries to reach a chance of 50-50
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I don't know the math but did you factor this, per giveaway %, in when you did the math for the 61% outcome you posted on the topic?
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I took every single probability of not winning of each giveaway and multiplied them resulting in the probability of losing all of them.
The winning % actually includes the combinations of many wins and the combinations of winning all the giveaways, but since I limited the problem to just winning and losing then the multiple win scenarios are also treated as desirable outcomes.
If I wanted a realistic approach to the problem I would need to consider the fact that if I win a giveaway I will stop entering them thus this will decrease the win % and make some of the outcomes impossible (multiple wins of giveaways). So in this way the winning % is not to be taken seriously, but the losing % still holds.
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Probability is a b****. I used to work at the local broadcaster and we had storage server which had a very low of probability of failure of an entire node. When I was a rookie it actually happened, once in a lifetime event :D
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Who told that there are just 1-6 numbers on die sides? :-P
Or if there are 1 to 6 points on the sides how do you be sure that 7-th point on 6 point side will not appear somehow (from a dirty table for example)?
Or how abount different numbering systems?
Its just human limitations and simplifications but everything is possible in the universe! Look more widely on everything. :)
P.S.: Had you studied probability theory in the university or wherever?
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What is the probability of winning the lottery if you do not have a lottery ticket ? 0
What is the probability of picking a blue ball in a sack only filled with red balls ? 0
What is the probailiy of picking the ace of spades from a deck of cards without aces ? 0
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You can be granted with a winning ticket. You even may not know it. <>0
If your hand is in blue paint - any ball you pick will be blue. <>0
What if because of absence of aces 2 are considered as aces in this deck? <>0
Ofcourse in math probability theory if there is no such event in chosen multiplicity - probability of it is 0.
But math sucks! Universe rules :)
Is Schrödinger's cat dead or alive in the box? ;)
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You are just twisting these events into different events.
I said that you do not have a lottery ticket. If you are granted a ticket without knowing it, then you do have lottery ticket and it is a different event.
I did not say the 2 are considered aces, that is a different event.
As for the ball covered in blue painting...well one could argue that it is still a red ball, only covered in blue painting.
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Your setting is simply invalid. That is not even included in probability.
Probability is a way of assigning every "event" a value between zero and one, always not equal to.
You may want to see the description yourself.
http://en.wikipedia.org/wiki/Almost_surely
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Probability tell us what-ever can happen will happen
I think you're the one misunderstanding probabilty
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You are right, it will happen in infinite time. For short span of time it might occur, but is unlikely to happen, thus you can consider it a once in a lifetime event. But theoretically you are correct.
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I was trying the same, didn't get it and didn't buy it, opportunity lost.
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Yay, mathematical proof that winning is possible, but unlikely ;)
Oh, and if you got the game on the cheap, I'd recommend keeping it for yourself and playing it.
It's short and quite rough around the edges, but what is in there works quite well for a first-person puzzler.
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talk about dedicaton, though I admit to holding off on getting some games just to see if how long it'd take to win them.
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Why are you combining multiple giveaways? I mean, if you enter 1 million giveaways with 500 entries the probability (0.002) will always be the same because they are independent events. Please explain your experiment/logic to me as if I were a child.
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the probability of something happening over and over again. For each giveaway the probability will still be the same, but at some point losing all giveaways would become less probable than winning any of the giveaways.
A coin toss is a good example (let's assume the coin is not rigged) - it may occur that you throw the coin and get 10 times in a row heads, but with every next toss is comes more and more probable that the next throw will be tails. But it is still probable to throw a coin 100 times in a row and get heads.
To sum: if you enter a lot of giveaways it becomes more and more probable that you will win one, but it still possible to never win any of them.
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I think you're contradicting yourself, you said:
The probability is 50% and will always be.
"if you enter a lot of giveaways it becomes more and more probable that you will win one": No, you are only buying a lot of lottery tickets for different (independent) raffles. The probability will always be 1/(number of tickets sold). You're only "buying" your ticket, your chance to win, not increasing the probability of winning.
Don't worry, a lot of people thinks this way, that's why lottery and gambling are great businesses.
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For the single event it is 0.5 probability, but if you view a sample of n it is not actually 0.5. Sorry for the wording before.
Each fair coin toss is 0.5, but the probability of combinations of multiple coin tosses/events is not 0.5.
As I pointed out in a later comment this math is based on Bernoulli trial. I would recommend looking at the example at the end of the article.
Over a very large sample winning a lottery can be done, but if you compare the % of winning in a lottery and a SG giveaway they are way way way different. Some link on numbers, can't vouch for the correctness of those numbers but the jackpot probability is very small compared to winning an average SG giveaway. And my point is that if you enter a large sample of giveaways the probability of the event of losing all of them becomes less and less probable.
I think you would be a person who would find the Monty Hall problem interesting. The problem usually brings up quite a stir and by you logic the problem is incorrectly solved and the probability for it should be 0.5.
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Monty Hall problem is conditioned by the host's choice, so the probability changes when he opens one of the doors because he knows where the prize is. It has nothing to do with an innocent coin toss.
I'm not sure if Bernoulli trial is aplicable on Steamgifts giveaways, maybe some mathematician could help us, but I'm almost sure your numbers are wrong, I mean THESE numbers:
"I got that the probability was 0,383538... circa 38,35% that I don't win any of the giveaways. Meaning that there was chance of 61,65% to win a copy of polarity."
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In the case of Polarity the probability of winning is not entirely correct for a SG setting - once you win a copy you can't/shouldn't enter any giveaways and win them. The winning probability includes the probability of at least winning one giveaways, meaning it includes also the probability of winning all of the giveaways or some of them. This math holds if you enter giveaways for different games.
However, the probability of losing all the giveaways should be correct. It is the probability of the event where all the single giveaway events end with loss.
Easy example:
Let's say you enter giveaways (for different games) which have all 500 entries.
Winning chance for 1 giveaway would be 1/500 and losing it 499/500.
If you enter 2 giveaways the probabilities are 499/500 x 499/500 for losing both of them and the probability of at least winning 1 is the sum of the probabilities: 499/500 x 1/500 (probability of losing the first giveaway and winning the second) 1/500 x 499/500 (winning first losing second) 1/500 x 1/500 (winning them both). At least winning 1 is also the 1-(probability of losing both of them).
But if you enter 250 giveaways with 500 entries the probability to lose is:
(499/500)^250=0.61 which means 61% chance to lose every single one of them.
However for 250 giveaways with 300 entries the probability to lose is:
(299/300)^250=0.43 which means there is a 57% chance to win at least one giveaway.
But to put in perspective: Do you really enter 200-300 giveaways for some game? For polarity it is easy, but for some other game it would take a lot of time (there has to be a lot of giveaways as well). Also, 61% and 43% aren't that great probabilities, it's pretty close to 50/50 :)
I might not be a mathematician, but I'm getting a PhD in Information and Communication Technology and probabilities and probability density functions should be right down my alley :)
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Ya I just took a quick glance and your math seems off. I'll double check when I get home and have some sleep in me but it should probably be something to the effect of: x1/y1 () x2/y2 () .... xn/yn where x is the number of entries you have per giveaway (in this case x is equal to one for all cases) and y is the number of entries total on a particular giveaway.
At the end of the day, you are looking at gamblers fallacy: while the unlikelyhood of you not winning increases per giveaway lost, each event act independently of each other and thus the outcome of one has no baring on any subsequent giveaway.
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Excuse format and possible errors; at work on mobile and it's 4:30am here -.-. In lieu of mutiple sign I used empty closed braces.
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I calculated the probability of NOT winning a giveaway which is the 1 - the probability of winning.
My approach is based on Bernoulli trial
As far as I understand gamblers fallacy, it is the notion that if something occurs a lot of times it must be balanced with the other option a lot of times. This is easily avoidable by calculating the probabilities.
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So, over the course of 3-4 weeks I entered 256 Polarity giveaways and won none of them and here is the math for it:
I calculated for each giveaway the probability of not winning. (Numberofentries-Numberofcopiesgivenaway)/Numberofentries
I multiplied all the resulting values and got the probability of not winning any of the giveaways.
I got that the probability was 0,383538... circa 38,35% that I don't win any of the giveaways. Meaning that there was chance of 61,65% to win a copy of polarity.
I was actually a bit surprised that the number was that big, but it doesn't seem very much off. Anyway, if there is any mistakes in my logic I'm happy to mend the solution.
Oh and here is an obligatory giveaway.
EDIT:
I actually bought it as a gift to my inventory (upon winning I would have given away the copy in my inventory). I entered so many giveaways purely out of curiosity and math.
Also getting the data is pretty simple: Giveaways - View entered - Search Polarity - Copy paste the tables into spreadsheet - some formula magic and we have the numbers :)
EDIT2:
more giveaways:
first
second
EDIT3
third
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