[quote=“ShotgunRagtime;87964”]
@Clown
You’re right, I am fortunate indeed to be able to blow $20 and not feel any major financial falloff.[/quote]
It’s not about you spending money, I feel happy whenever I get something above the lowest rarity. I bought 10 elites a few days ago and got all silver, I’m wondering if I can beat my record of 30 silvers in a row from elites.
That’s another thing. I am financially very secure and could spend money. I just really don’t like gambling. I’m a hungry whale with nothing to eat basically.
[quote=“ShotgunRagtime;87991”][quote=“Litego;87985”][quote=“ShotgunRagtime;87964”]@Litego
Yes, it is. It’s pretty basic probability. For example, 7% chance of opening a gold x 11 cases = 77% total chance of a gold.[/quote]
Nope. According to your calculations you should have a 100% chance to get a cobalt or gold, did not happen. Conclusion, your math is wrong. Percentages aren’t additive.[/quote]
[quote=“xAvenger;87987”][quote=“ShotgunRagtime;87964”]@Litego
Yes, it is. It’s pretty basic probability. For example, 7% chance of opening a gold x 11 cases = 77% total chance of a gold.[/quote]
I think Litego is correct. If each case yields a seven percent chance of a gold, then if you buy 15 cases you’d be guaranteed a gold. The chance is per case, not cumulative.
[/quote]
You guys do not understand probability math. An ASSUMED 100% chance does not mean that it will actually happen. I could get all bronze or all cobalt, hypothetically.
If I opened 1,000 regular cases, I would have a 100% chance of getting a cobalt. There’s a 1/1,000 chance at a cobalt in any case. Multiplied by 1,000, you get 1, or 100% in this context.
That means that I SHOULD open a cobalt. Is it guaranteed? No, because every case is random.
[/quote]
1/1000 chance for a cobalt. Times that buy 100.
100/100,000 = 1/1000
Please learn probability math.
[quote=“Kaneki;88052”][quote=“ShotgunRagtime;87991”][quote=“Litego;87985”][quote=“ShotgunRagtime;87964”]@Litego
Yes, it is. It’s pretty basic probability. For example, 7% chance of opening a gold x 11 cases = 77% total chance of a gold.[/quote]
Nope. According to your calculations you should have a 100% chance to get a cobalt or gold, did not happen. Conclusion, your math is wrong. Percentages aren’t additive.[/quote]
[quote=“xAvenger;87987”][quote=“ShotgunRagtime;87964”]@Litego
Yes, it is. It’s pretty basic probability. For example, 7% chance of opening a gold x 11 cases = 77% total chance of a gold.[/quote]
I think Litego is correct. If each case yields a seven percent chance of a gold, then if you buy 15 cases you’d be guaranteed a gold. The chance is per case, not cumulative.
[/quote]
You guys do not understand probability math. An ASSUMED 100% chance does not mean that it will actually happen. I could get all bronze or all cobalt, hypothetically.
If I opened 1,000 regular cases, I would have a 100% chance of getting a cobalt. There’s a 1/1,000 chance at a cobalt in any case. Multiplied by 1,000, you get 1, or 100% in this context.
That means that I SHOULD open a cobalt. Is it guaranteed? No, because every case is random.
[/quote]
1/1000 chance for a cobalt. Times that buy 100.
100/100,000 = 1/1000[/quote]
this is wrong as well. By this logic, when trying to get a ‘6’ on a dice, throwing 6x times would be just as likely to get you a ‘6’ as throwing the dice once.
Obviously your chances to eventually get a cobalt increase with each new attempt.
a) probability doesn’t have a memory. You could have gotten 10 cobalt cases in a row, and the chances to get the next cobalt case on the next throw are just as likely as if you didn’t get those 10 cobalt cases.
b) 100 attempts at getting a random thing is better than one attempt, but not by 100 times.
Please learn probability math.
well said.
No they don’t. He is right, you aren’t.
it is
[quote=“Kaneki;88052”]
1/1000 chance for a cobalt. Times that buy 100.
100/100,000 = 1/1000
Please learn probability math.[/quote]
Probability to get a cobalt remains the same unless somehow not unlocking a cobalt increases the chance that the next one will be a cobalt. That’s like saying I didn’t roll 6 on a die 5 times in a row, therefore the sixth roll is a guaranteed six.
You’re chance don’t increase but restart because you try it a second time 
More you try, more restart so you have a possibility to win again. It clearly doesn’t improve your chance of winning any thing.
Yes, for example, if you have a 1 in 100 chance of winning a local lottery, then you would be guaranteed a win by purchasing 100 tickets, and that would make no sense. Each ticket actually has a 1% independent chance of winning.
Expectation is the key word. (I hope I translated it right…)
Tries * Chance = Expectation value
Lets say…
a)You toss a coin 200 times, each side has a 50% chance, no “faillandings”.
200tosses * 50% chance of tails = 100 times of expected tails as result.
Does this mean it is like that in every case this experiment is done? No. The expectation value however is 100.
b)You roll the dice. 30 times and youd like to know how often youre expecting to get a 3.
Each side has a chance of 1/6, so does the 3.
30rolls * 1/6 chance = 5
The expected amount of threes rolled is 5. Will it happen? Likely but not absolutely certain.
Ok these were examples with results over 1. Could also say over 100%…
Lets go to our cases:
7% of gold,11 cases tried => Expectation of 77% or 0.77 to get gold.
In about 3 of 4 tries, someone will get a gold. In less than 1 of 4 tries, someone will not, like the OP.
3% of cobalt, 11 cases => Expectation of 33% to get a cobalt.
And now the magic to calculate the expectation of getting 1 single gold or cobalt card. We add the expectation values:
0.77 + 0.33 = 1.1 or 110%
We could also look at this a bit differently:
Lets just add the chances of cobalt and gold of a single case. 3% +7% =10%
and multiply it with the amount of tries (cases) later. 10% *11 =110% or 1.1 of expected gold or cobalt cards.
Of course noone can have a 1/10 of card… But the full one is nice to have!
We see, we can expect to get either a cobalt or gold card from these 11 cases with 10%, but again it is not certain.
Would be more fun with calculating the chances for getting 2 or 3 from 11 cases 
On Topic:
I feel for you OP. So far I refused to give them money for this model. Id send a cheque over to SD for their job-well-done, but this gambling is just a no-go.
i brought 30 normal cases two days ago.i opened them all in a row.
i got 3 iron and a bronze rest lead. then cried.
[quote=“laudatoryLunch;88151”][quote=“Szakalot;88132”]
Obviously your chances to eventually get a cobalt increase with each new attempt.
[/quote]
No they don’t. He is right, you aren’t.
it is
I phrased it badly, sorry. The chances increase the more attempts there are. 100 attempts is more likely to succeed than a single attempt. He was proposing that 100 attempts has the same chance of success as a single one, which is obviously wrong.
[quote=“Szakalot;88193”][quote=“laudatoryLunch;88151”][quote=“Szakalot;88132”]
Obviously your chances to eventually get a cobalt increase with each new attempt.
[/quote]
No they don’t. He is right, you aren’t.
it is
I phrased it badly, sorry. The chances increase the more attempts there are. 100 attempts is more likely to succeed than a single attempt. He was proposing that 100 attempts has the same chance of success as a single one, which is obviously wrong.[/quote]
No it isn’t. Again, just for you. Every case has its own dice roll. If you open 1 you have the percentage X, and if you open 100, you have 100X/100…
Just look at it like this each case consists of 100%
11 cases=1100%
Lets say a cobalt has a 0.7% chance (idk the actual number).
0.7x11=7.7%
7.7/1100 is the chance on a cobalt, that’s how I see it 
Imo instead of expectation like you described, it is more useful to talk about probability. It is not likely at all that you will get exactly 5x rolls of 3s in 30rolls. Just because each side is just as likely to happen does not mean that all sides will be happening equally. While this would not be an issue with huge amounts of throws, at measly 30 you will get a lot variance.
How likely are you to get a cobalt card from 1,10,100,1000 cases?
The easiest way to go about it is to consider how UNLIKELY it is. In one throw you have 99.9% chance of NOT getting a cobalt case. In two throws its 99.9% of not getting it on the first try and 99.9% of not getting it on the second try so: 99.9%*99.9% = 99.8%
In 10 throws you have (99.9%)^10 = 99.0% chance of not getting the case yet. Indeed, by having 10, instead of a single attempt you have increased your chances roughly tenfold.
In 100 throws its about 90% This still follows the same pattern.
In 1000 throws its about 36% of not getting a cobalt case. So only 64% of actually getting one. Yes, even though the drop rate is 0.1% per case.
[quote=“sensitiveJellyfish;88202”]Just look at it like this each case consists of 100%
11 cases=1100%
Lets say a cobalt has a 0.7% chance (idk the actual number).
0.7x11=7.7%
7.7/1100 is the chance on a cobalt, that’s how I see it :p[/quote]
sorry, you can’t just add percentages like that : )
What you are saying is that no matter whether i throw a coin once, or 50 times, or a million times, the chances of getting ONE HEAD or ONE TAIL throw are 50% for THE ENTIRE throwing exercise.
Yes, the probability doesn’t have memory, each individual throw has a certain fixed chance, irrespective of all other throws. However, if you consider all the throws at once, you can consider the probabilities from each throw together (that is not the same as adding them all up : ) )
Throwing a coin always is a 50% chance lol, no matter how many times u throw it you’ll always have only 2 options.
yes, but we are talking about only scoring heads/tails once in multiple throws. You open multiple cases, each case can drop a cobalt. These probabilities have to be considered together.
Do you disagree that you are more likely to get heads AT LEAST ONCE if you keep throwing the coin?