So your tellin me u have a 25% chance to get 1 Epic in 8 ceramic packs? The drop rates for an epic are about 2.5% so thats not possible at all

We were talking about the probability to get an A or better. Of course, you’re right, the probability to get exactly one A is smaller, but not that small. Here you go:

Thats what Hutch states about chances in a ceramic:

Coming from that, chances to get an A as 1st card (and anything else for cards 2-5, including the case of two A‘s) are: 2.5% Chances to get an A as 2nd card (and anything else for cards 1 and 3-5, including the case of two A‘s) are: 0.25% Chances to get an A as 1st and as 2nd card (two in one pack) are: 2.5% * 0.25% = 0.00625% Thus, chances to get an A as 1st or 2nd card (excluding the case of two A‘s in one pack) are: 2.5% - 0.00625% + 0.25% - 0.00625% = 2.7375% = 0.027375

To expand that to 8 packs, we will use the complementary probability, i.e., the probability to get anything but a combination with exactly one A, as it’s easier to use: 1 - 0.027375 = 0.972625

For 8 packs, we need to adjust that to the power of 8: 0.972625^8 = 0.800873 ~ 80.09%

Like said, that’s the probability to get any combination of cards except these having exactly one A, in 8 packs. We are interested in the contrary, that’s: 1 - 0.800873 = 0.199127 ~ 19.91%

Usually everybody’s fine with a result even better than an A. That’s why we stated 25%. Calculated as follows:

Again, to calculate this we use the complementary probability, which is to get anything up to B in one pack:

mmmm i think no... if you buy one pack or 100 you have same chance to find an A class... you open the first pack and no A, the second have the same % to find an A class, not plus or minus... always the same... every pack... obviously if you buy 1 pack or 1 milion, you find more A class in 1 milion packs but not sure, you can buy 1 milion packs and find only C class because the chance to find an A class not depending on how many packs you buy!!!

mmmm i think no... if you buy one pack or 100 you have same chance to find an A class... you open the first pack and no A, the second have the same % to find an A class, not plus or minus... always the same... every pack... obviously if you buy 1 pack or 1 milion, you find more A class in 1 milion packs but not sure, you can buy 1 milion packs and find only C class because the chance to find an A class not depending on how many packs you buy!!!

You’re talking about the probability per pack, so one A in one pack. Right, that’s constant. But, we’re talking about the overall probability in a sequence of pack openings, so one A in multiplepacks. The latter will rise with the number of packs you’re opening. Of course, it will never reach 100%, it will never be „sure“ you get even one out of multiple packs.

but don't exist overall probability because it does not increase after that you buy the first pack, or the second or the 100th... it's not like a box with 100 [email protected] and you take one and then you have 99 [email protected] so the probability increase... so the % is always the same...

but don't exist overall probability because it does not increase after that you buy the first pack, or the second or the 100th... it's not like a box with 100 [email protected] and you take one and then you have 99 [email protected] so the probability increase... so the % is always the same...

Sorry.. I was in the middle of editing your post because the filter is somewhat weird like that when you fixed it first. lol.

Let's say you're flipping a coin. If you flip it once, you have a 50% chance of getting heads, and a 50% chance of getting tails. But what if you flip it 4 times, and all you care about is that at least one of those coin flips is heads? The probability is certainly much higher than just flipping the coin once.

The easiest way to approach this is to find the probability that you get zero heads (so all tails), and subtract that from 1. So for this example:

nope.. it's the same probability.. 50%... this logic cannot work in this way... every launch, every pack buy have same % ... don't increase if you launch the coin one time or 1000 times as for 1 Ceramic or 100 ceramic, always the same %

It has to though, otherwise there would be no point in buying more than one pack ever. You're still talking about individual pack probability, not the compounded probability of opening many packs.

You should try the coin flipping example, and see how often you get at least 1 heads out of 4 flips. I expect if you do that for 10 minutes and record your results, I you'll get pretty close to that 93.75% number. Just make sure you record after each set of 4, not each time you get heads or anything weird like that.

@mauro07 That is absolutely correct. Probability of independent events increase as the number of attempts increase.

The probability of getting the desired results EACH event remain the same, however if you are attempting to get one result in multiple attempts, the probability goes up with each added event.

It has to though, otherwise there would be no point in buying more than one pack ever. You're still talking about individual pack probability, not the compounded probability of opening many packs.

You should try the coin flipping example, and see how often you get at least 1 heads out of 4 flips. I expect if you do that for 10 minutes and record your results, I you'll get pretty close to that 93.75% number. Just make sure you record after each set of 4, not each time you get heads or anything weird like that.

****. Stop with the maths. You're giving me a man crush.

It has to though, otherwise there would be no point in buying more than one pack ever. You're still talking about individual pack probability, not the compounded probability of opening many packs.

You should try the coin flipping example, and see how often you get at least 1 heads out of 4 flips. I expect if you do that for 10 minutes and record your results, I you'll get pretty close to that 93.75% number. Just make sure you record after each set of 4, not each time you get heads or anything weird like that.

Wilf. Stop with the maths. You're giving me a man crush.

Awwww thanks. But save some love for @prex, I was way too lazy to show all my work haha

So your tellin me u have a 25% chance to get 1 Epic in 8 ceramic packs? The drop rates for an epic are about 2.5% so thats not possible at all

We were talking about the probability to get an A or better. Of course, you’re right, the probability to get exactly one A is smaller, but not that small. Here you go:

Thats what Hutch states about chances in a ceramic:

Coming from that, chances to get an A as 1st card (and anything else for cards 2-5, including the case of two A‘s) are: 2.5% Chances to get an A as 2nd card (and anything else for cards 1 and 3-5, including the case of two A‘s) are: 0.25% Chances to get an A as 1st and as 2nd card (two in one pack) are: 2.5% * 0.25% = 0.00625% Thus, chances to get an A as 1st or 2nd card (excluding the case of two A‘s in one pack) are: 2.5% - 0.00625% + 0.25% - 0.00625% = 2.7375% = 0.027375

To expand that to 8 packs, we will use the complementary probability, i.e., the probability to get anything but a combination with exactly one A, as it’s easier to use: 1 - 0.027375 = 0.972625

For 8 packs, we need to adjust that to the power of 8: 0.972625^8 = 0.800873 ~ 80.09%

Like said, that’s the probability to get any combination of cards except these having exactly one A, in 8 packs. We are interested in the contrary, that’s: 1 - 0.800873 = 0.199127 ~ 19.91%

Usually everybody’s fine with a result even better than an A. That’s why we stated 25%. Calculated as follows:

Again, to calculate this we use the complementary probability, which is to get anything up to B in one pack:

Thus, chances to get the contrary, one A or better, are: 1 - 0.9651 = 0.0349 = 3.49%

To get the probability for a number of 8 packs we simply calculate: 0.9651^8 = 0.7526 = 75.26%

Again, this is the complementary probability thus the probability we are looking for is: 1 - 0.7526 = 0.2474 = 24.74%

This cant be right... if u have a chance to get an a of 2.7% there is no way u can have a chance of 25% of getting an a in 8 ceramics... even if u just multiply 2.7 with 8 and that would be asolutely bs And further to that.... im getting an a or s in about every 40 - 50 packs tho...

Edit: ok now i got it you were talking about the chance to get an a or better and not an a so your calculation does make sense😄

It has to though, otherwise there would be no point in buying more than one pack ever. You're still talking about individual pack probability, not the compounded probability of opening many packs.

You should try the coin flipping example, and see how often you get at least 1 heads out of 4 flips. I expect if you do that for 10 minutes and record your results, I you'll get pretty close to that 93.75% number. Just make sure you record after each set of 4, not each time you get heads or anything weird like that.

Wilf. Stop with the maths. You're giving me a man crush.

Awwww thanks. But save some love for @prex, I was way too lazy to show all my work haha

Sorry Prex, your working got me all worked up too.

## Comments

456✭✭✭✭588✭✭✭✭Thats what Hutch states about chances in a ceramic:

Coming from that, chances to get an A as 1st card (and anything else for cards 2-5, including the case of two A‘s) are: 2.5%

Chances to get an A as 2nd card (and anything else for cards 1 and 3-5, including the case of two A‘s) are: 0.25%

Chances to get an A as 1st and as 2nd card (two in one pack) are: 2.5% * 0.25% = 0.00625%

Thus, chances to get an A as 1st or 2nd card (excluding the case of two A‘s in one pack) are:

2.5% - 0.00625% + 0.25% - 0.00625% = 2.7375% = 0.027375

To expand that to 8 packs, we will use the complementary probability, i.e., the probability to get anything but a combination with exactly one A, as it’s easier to use: 1 - 0.027375 = 0.972625

For 8 packs, we need to adjust that to the power of 8: 0.972625^8 = 0.800873 ~ 80.09%

Like said, that’s the probability to get any combination of cards except these having exactly one A, in 8 packs. We are interested in the contrary, that’s: 1 - 0.800873 = 0.199127 ~

19.91%Usually everybody’s fine with a result even better than an A. That’s why we stated 25%. Calculated as follows:

probability 1st card to get anything up to B * prob. 2nd card * prob. 3rd card * prob. 4th card * prob. 5th card

= (100% - 2.5% - 0.65%) * (100% - 0.25% - 0.1%) * 100% * 100% * 100%

= (1 - 0.025 - 0.0065) * (1 - 0.0025 - 0.001) * 1 * 1 * 1

= 0.968 * 0.9965 * 1 * 1 * 1

= 0.9651 = 96.51%

Thus, chances to get the contrary, one A or better, are:

1 - 0.9651 = 0.0349 = 3.49%

To get the probability for a number of 8 packs we simply calculate:

0.9651^8 = 0.7526 = 75.26%

Again, this is the complementary probability thus the probability we are looking for is:

1 - 0.7526 = 0.2474 = 24.74%

1,220✭✭✭✭588✭✭✭✭one A in one pack. Right, that’s constant. But, we’re talking about the overall probability in a sequence of pack openings, soone A in multiplepacks. The latter will rise with the number of packs you’re opening. Of course, it will never reach 100%, it will never be „sure“ you get even one out of multiple packs.1,220✭✭✭✭2,019Wrenchmen › ✭✭✭✭✭1,220✭✭✭✭265✭✭✭at leastone of those coin flips is heads? The probability is certainly much higher than just flipping the coin once.1,220✭✭✭✭265✭✭✭2,019Wrenchmen › ✭✭✭✭✭573✭✭✭✭1,129✭✭✭✭✭265✭✭✭456✭✭✭✭And further to that.... im getting an a or s in about every 40 - 50 packs tho...

Edit: ok now i got it you were talking about the chance to get an a or better and not an a so your calculation does make sense😄

1,129✭✭✭✭✭#statslove4ever

221Hutch Staff › Wrenchman378✭✭✭2,019Wrenchmen › ✭✭✭✭✭273✭✭✭286✭✭✭