I wish keys to open chest were cheaper, how about $1.50 or even $.099 instead of $2.50.
I wish keys to open chest were cheaper, how about $1.50 or even $.099 instead of $2.50.
it is not 0,98% but 98% = 0,98 (% already stands for the 1/100)
He obviously presented the chance to get no courier at all when opening 70 chests. There is nothing misleading about that.
Still OP was unlucky, I got 3 modifiers with about 7 keys from chests that could drop modifiers.
Last edited by mindfaQ; 09-18-2012 at 06:42 AM.
specs for bugreports:
i5-4200U
AMD Radeon HD 8750M
win 7
Are you saying a person who opens 70 chests has the same chance to obtain a strange after being finished than a person who opens 1?
The first guy's probabilities are correct.
Before opening his 70 chests, the person has a 70% chance.
With each failed chest attempt though, his chance decreases.
If the chance of getting unusual is 1% = 0.01, then with 70 tries the chance is: 1 - (0.99^70) = 50.5% just to keep mathematics intact.
No, he's saying that previous chances don't affect your next one.Are you saying a person who opens 70 chests has the same chance to obtain a strange after being finished than a person who opens 1?
Wrong.Before opening his 70 chests, the person has a 70% chance.
Wrong.With each failed chest attempt though, his chance decreases.
We're talking here about the probability to obtain a strange in 70 chests. This probability is forcibly greater than the probability to obtain a strange in 1 chest.
If we were to take the chance of one individual chest, say obtaining a strange on chest 20 out of the 70, then the chance would be the chance of one chest. If you take every possible outcome of events listing the 70 obtained, there is a smaller proportion of combinations with no stranges (0 stranges out of 70) proportionally to without stranges than with a single chest (0 stranges out of 1), because this time you need one strange on 70 events, not one strange on a single event. The strange can be in any of 70 places.
The probability to calculate is of having one strange on 70 results, not one on one result, the problem and probability changes completely. The fact that the chance on every individual chest is equal isn't relevant insofar as calculating the total probability.
Last edited by jim109109; 09-21-2012 at 05:47 PM.
Read (and understand :P) this http://en.wikipedia.org/wiki/Gambler%27s_fallacy and thank me later!
Here's an example. With two dice of 1-3, the chance to obtain 1 is 1/3.
The chance to obtain 1 in two throws is however greater.
If you consider all possibilities in 2 throws, {1 1, 1 2, 1 3, 2 1, 2 2, 2 3, 3 1, 3 2, 3 3}
There are 5/9 outcomes containing 1, thus a 5/9 chance, which is greater than 1/3.
You can also calculate this by (2/3)^2 [chance not to obtain one]. Which is 4/9, therefore 5/9 chances to obtain a 1.
Gambler's fallacy states that after throwing the first dice, a gambler would maybe think the chance on the second changes, which it doesn't. Once one of the two dices are thrown, the probability to obtain 1 becomes 1/3 again.
That's why I was clear to say that the chance is greater before he has opened his 70 chests.
Last edited by jim109109; 09-21-2012 at 08:13 PM.
I think the simplest way to explain all this for people is:
Every unlock has the same chance, no matter how many you open. But, because there is always a chance, then the more you open, the less probable it becomes that you don't finally land on the favorable outcome. But no matter how many you open, it can never be guaranteed, only increasingly more likely into infinity. You could flip a coin a million times and never have it land on heads, even though the heads chance is 50%; but the percent chance of this happening is infinitesimally small, and becomes smaller the longer you go on.
100 tries at something with only a 1% chance is still very likely to never land on that 1%.
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