What makes WHM special? Nothing anymore.

[J-Dax]

Lol this is not the gambler fallacy.
Flip a coin in sets of 10.
Count how many sets where heads appears at least once.
Count how many times you get an entire set of tails.
by your logic they should be 50/50. but that is not the case

This is the case each flip is its own separate event, previous or subsequent flips do not effect outcomes. When you flip the coin once, you have a 50/50 chance of getting heads. The next flip is also a 50/50 chance of getting heads, weather you had heads before or not.

[winsock]

This is the case each flip is its own separate event, previous or subsequent flips do not effect outcomes. When you flip the coin once, you have a 50/50 chance of getting heads. The next flip is also a 50/50 chance of getting heads, weather you had heads before or not.

The previous flips to not impact subsequent outcomes. This is true. But you're looking at the chance of something occurring within a certain number of attempts
To simplify, let's say AST flips coins. For an encounter, you only have time to flip your coin 3 times. Your possible flips for the encounter are:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

There is an equal chance for any of the above combinations to be the combination you get during the run. "TTT" has the same chance of appearing as "HTH" or "THT" or "HHH". However, there are 7 outcomes that would allow you to get at least 1 H during the encounter. 7/8 outcomes means you have a 87.5% chance of getting at least 1 H over the course of 3 flips.

[J-Dax]

The previous flips to not impact subsequent outcomes. This is true. But you're looking at the chance of something occurring within a certain number of attempts
To simplify, let's say AST flips coins. For an encounter, you only have time to flip your coin 3 times. Your possible flips for the encounter are:
HHH
HHT
HTH
HTT
THH
THT
TTH
TTT

There is an equal chance for any of the above combinations to be the combination you get during the run. "TTT" the same chance of appearing as "HTH" or "THT" or "HHH". However, there are 7 outcomes that would allow you to get at least 1 H during the encounter. 7/8 outcomes means you have a 87.5% chance of getting at least 1 H over the course of 3 draws.

Here are the probabilities.
Three cards are favorable, three cards aren't. For simplification we will equate favorable to heads and unfavorable to tails.

The probability of getting heads in three attempts is 85%. The inverse must also be true meaning the probability of getting tails is also 85% It is just as likely that after three attempts I will pull a favorable outcome as it is I will pull an unfavorable one or 50/50.


This all still is off the original topic.

[winsock]

This is vastly oversimplifying, in theory it sounds great, in reality this is not what happens. AST has 6 cards, if three of them are desirable and three undesirable. Then having an 87.5% chance to get a favorable result means the inverse is true giving me an 87.5% chance of not getting the card I want. So the reality being that its just as likely to gain a favorable outcome as an unfavorable one, it therefore follows that believing anything else to be true is gamblers fallacy.

This all still is off the original topic.

This is what happens in reality.
The desired result is for the coin to be heads at least 1 time when making 3 attempts. The inverse would be the coin never landing on heads. The likelihood of that happening is 1/8 or 12.5% because there is only one outcome "TTT" that did not contain any heads.

If you dont want it simplified, I'll unsimplify it. Let's use a 6 sided die as the AST deck has 6 cards. There are 3 even numbers and 3 odd numbers just like in the scenario of 3 desired and 3 undesired cards. You roll it twice. What is the probability of rolling an even number at least once?
There are 36 possible outcomes (6x6).
27 of said possible outcomes will contain at least one even number:
2,1 2,2 2,3 2,4 2,5 2,6
4,1 4,2 4,3 4,4 4,5 4,6
6,1 6,2 6,3 6,4 6,5 6,6
1,2 1,4 1,6
3,2 3,4 3,6
5,2 5,4 5,6

27/36 = 75% chance

[J-Dax]

Snip

Explaining why the probability is 1/2 for a fair coin
We can see from the above that, if one flips a fair coin 21 times, then the probability of 21 heads is 1 in 2,097,152. However, the probability of flipping a head after having already flipped 20 heads in a row is simply 1⁄2. This is an application of Bayes' theorem.

This can also be seen without knowing that 20 heads have occurred for certain (without applying of Bayes' theorem). Consider the following two probabilities, assuming a fair coin:

probability of 20 heads, then 1 tail = 0.520 × 0.5 = 0.521
probability of 20 heads, then 1 head = 0.520 × 0.5 = 0.521
The probability of getting 20 heads then 1 tail, and the probability of getting 20 heads then another head are both 1 in 2,097,152. Therefore, it is equally likely to flip 21 heads as it is to flip 20 heads and then 1 tail when flipping a fair coin 21 times. Furthermore, these two probabilities are equally as likely as any other 21-flip combinations that can be obtained (there are 2,097,152 total); all 21-flip combinations will have probabilities equal to 0.521, or 1 in 2,097,152. From these observations, there is no reason to assume at any point that a change of luck is warranted based on prior trials (flips), because every outcome observed will always have been as likely as the other outcomes that were not observed for that particular trial, given a fair coin. Therefore, just as Bayes' theorem shows, the result of each trial comes down to the base probability of the fair coin: 1⁄2.

https://en.wikipedia.org/wiki/Gambler%27s_fallacy