11 straight phasma extra attacks.

That's a 1 in 10^9 ish chance. Add in mine never proc ing and we have a 1 in 10 19ish chance of happening. I should buy a lotto ticket.

Replies

  • Eaywen wrote: »
    That's a 1 in 10^9 ish chance. Add in mine never proc ing and we have a 1 in 10 19ish chance of happening. I should buy a lotto ticket.

    Well, that is the "in" thing to do nowadays.
    IGN: Malmsteen's Comet
  • Nothing like 1 in 10^9 chance, it's actually 1.76%, still long odds but almost a 2% chance. If only the Powerball lottery had such good odds.

    Based on a 16% proc rate of an assist times 11 in a row, you get the following formula:
    (16/100)11 =1.76%

    Ally Code: 945-699-762
  • Rheen
    269 posts Member
    Nothing like 1 in 10^9 chance, it's actually 1.76%, still long odds but almost a 2% chance. If only the Powerball lottery had such good odds.

    Based on a 16% proc rate of an assist times 11 in a row, you get the following formula:
    (16/100)11 =1.76%

    Actually it's much lower. It would be (0.16)^11

    Which is about 1 in 568,000,000.

    So yes, less likely than winning the powerball lottery.
    Guide to Beating Galactic War from Team Instinct
    TeamInstinct.net -- Community Site and Home to Team Instinct
  • Yeah. I did a message one of the CG guys about this yesterday happening to me as well. Exactly 11 straight procs before it failed to. Insane odds and figure Phasma has to be bugged
  • Perhaps the roll doesn't go down exactly as written.
  • Baal
    602 posts Member


    Stay out of casinos, bud. :)
    Rng has no memory.
    It's 16% each spin
    16% is 16% and 11 is well within common expectations.
    It's actually a completely different animal than powerball (a unique event) odds.

    The notion of any connection between the outcome of a spin of the wheel and any past outcomes on that same wheel, and that therefore something that has not happened recently becomes 'overdue' is known as "The Gambler's Fallacy". Similarly, there is the "doctrine of the maturity of chances” (aka “Monte Carlo fallacy”) that falsely assumes that each play in a game of chance is connected with other events and that therefore a series of outcomes of one sort should be balanced in the short run by other possibilities. Both these fallacies have been exposed by studies showing that over a long series of plays, a gambler playing any of the supposed betting systems will lose exactly the same percentage of initial purse as a player betting at random.
  • Baal wrote: »

    Stay out of casinos, bud. :)
    Rng has no memory.
    It's 16% each spin
    16% is 16% and 11 is well within common expectations.
    It's actually a completely different animal than powerball (a unique event) odds.

    The notion of any connection between the outcome of a spin of the wheel and any past outcomes on that same wheel, and that therefore something that has not happened recently becomes 'overdue' is known as "The Gambler's Fallacy". Similarly, there is the "doctrine of the maturity of chances” (aka “Monte Carlo fallacy”) that falsely assumes that each play in a game of chance is connected with other events and that therefore a series of outcomes of one sort should be balanced in the short run by other possibilities. Both these fallacies have been exposed by studies showing that over a long series of plays, a gambler playing any of the supposed betting systems will lose exactly the same percentage of initial purse as a player betting at random.

    It is not quite this simple. The probability of any single assist is 16%, but calculating the probably of multiple events in a row is another matter. The question isn't "what is the probability that Phasma will enable an assist" but rather "what is the probability that Phasma will enable another assist after causing 10 previous assists in a row". The probability must then be calculated all together leading to the incredibility low probability described above. Rheen has it right.

    Imagine the likelihood of flipping 11 heads in a a row. That seems incredibly unlikely (but minutely possible)--11 assists is much less likely than that--16% doesn't even come close to describing that event.

    You are quite correct with the gambler's fallacy and the monte carlo fallacy, but they do not apply when we have known probabilities and can readily calculate the likelihood of particular events. This fallacy is a psychological error, not a math issue.

    As for the lottery, you have the same odds of winning if you play or if you DON'T play. Save your money kids!
  • DarthHernia
    200 posts Member
    edited January 2016
    Nothing like 1 in 10^9 chance, it's actually 1.76%, still long odds but almost a 2% chance. If only the Powerball lottery had such good odds.

    Based on a 16% proc rate of an assist times 11 in a row, you get the following formula:
    (16/100)11 =1.76%
    Baal wrote: »
    16% is 16% and 11 is well within common expectations.

    ...Somebodies weren't paying attention in 9th grade math class. Or maybe you are just American?

    It should be (16/100)^11. That's 16% to the power of 11, not times 11.
    Rheen wrote: »
    Actually it's much lower. It would be (0.16)^11

    Which is about 1 in 568,000,000.
    To be precise 568,434,188.608 in one chance. You probably would have won the lottery.
  • Nothing like 1 in 10^9 chance, it's actually 1.76%, still long odds but almost a 2% chance. If only the Powerball lottery had such good odds.

    Based on a 16% proc rate of an assist times 11 in a row, you get the following formula:
    (16/100)11 =1.76%
    Baal wrote: »
    16% is 16% and 11 is well within common expectations.

    ...Somebodies weren't paying attention in 9th grade math class. Or maybe you are just American?

    It should be (16/100)^11. That's 16% to the power of 11, not times 11.
    Rheen wrote: »
    Actually it's much lower. It would be (0.16)^11

    Which is about 1 in 568,000,000.
    To be precise 568,434,188.608 in one chance. You probably would have won the lottery.

    He's right. It's 0.16 X 0.16 X 0.16 etc
    My name is cosmicturtle333, aka CT-333, aka Threes.
  • Baal
    602 posts Member
    You are attributing memory to a system which has none. The odds don't go down depending on past results. It's roughly a 1/6 chance each roll, entirely independent of past history.

    You're basing the odds on this:

    1) 1/6
    2) 1/36
    3) 1/216
    4) 1/1,296
    5) 1/7,776
    6) 1/46,656
    7) 1/279,936
    8) 1/1,679,616
    9) 1/10,077,696
    10) 1/60,466,176
    11) 1/362,797,056

    When the actual odds are:

    1/6 each independent roll.

    Or

    1) 1/6
    2) 6/36
    3) 36/216
    4) 216/1,296
    5) 1,296/7,776
    6) 7,776/46,656
    7) 46,656/279,936
    8) 279,936/1,679,616
    9) 1,669,616/10,077,696
    10) 10,077,696/60,466,176
    11) 60,466,176/362,797,056

    Or simply 1/6.

    The history has no influence. Odds are never worse than 1/6. It's a common fallacy to believe odds change based on history. They don't.

    Again, this is an entirely different animal than powerball. There's never less than a 1/6 chance of getting a particular roll on a die even on the 11th try.
  • Baal wrote: »
    You are attributing memory to a system which has none. The odds don't go down depending on past results. It's roughly a 1/6 chance each roll, entirely independent of past history.

    You're basing the odds on this:

    1) 1/6
    2) 1/36
    3) 1/216
    4) 1/1,296
    5) 1/7,776
    6) 1/46,656
    7) 1/279,936
    8) 1/1,679,616
    9) 1/10,077,696
    10) 1/60,466,176
    11) 1/362,797,056

    When the actual odds are:

    1/6 each independent roll.

    Or

    1) 1/6
    2) 6/36
    3) 36/216
    4) 216/1,296
    5) 1,296/7,776
    6) 7,776/46,656
    7) 46,656/279,936
    8) 279,936/1,679,616
    9) 1,669,616/10,077,696
    10) 10,077,696/60,466,176
    11) 60,466,176/362,797,056

    Or simply 1/6.

    The history has no influence. Odds are never worse than 1/6. It's a common fallacy to believe odds change based on history. They don't.

    Again, this is an entirely different animal than powerball. There's never less than a 1/6 chance of getting a particular roll on a die even on the 11th try.

    This is still not true. The probability of one assist is 16% (which I suspect is not actually true in the game). The probability of two in a row is 1/6 * 1/6 = 1/36.

    If you believe your math is correct, try to flip heads on a coin 11 times in a row. I guarantee you the likelihood of that is not 50%.
  • Nerds
  • Sejalx wrote: »
    Nerds

    Nerd = financially well off :) Modern economy is all about STEM!
  • Baal
    602 posts Member
    You can populate the distribution infinitely and any number of crazy things will assuredly happen but the odds don't change.

    You could have rolled 99 hits in a row, and the 100th still has the same chance as the 99th, the 50th, and the 1st.

    I think you're confusing probabilities with predictability. They are not interchangeable concepts even with a largest finite representative samples.

    Flipping a coin is 1/2, that never changes. That doesn't mean that the odds of 11 repeats in a row is 50% but each flip is 50%, and is never greater or lesser than 1/2, it is not influenced by previous results.

    The odds are not 1/2048. The odds are 1/2, 11 times...believing anything else is falling victim to the gambler's fallacy.

    I used a program to simulate coin flips and landed an 11streak in the 3rd batch of 100 numbers for what it's worth.


    Maybe this will help:

    "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"



  • Baal wrote: »

    Flipping a coin is 1/2, that never changes. That doesn't mean that the odds of 11 repeats in a row is 50% but each flip is 50%, and is never greater or lesser than 1/2, it is not influenced by previous results.

    This is correct, but you are not addressing the right question. The question is not about the probability (not odds) of any one assist. That is 1/6 (which again, I am sure this isn't really the case). The question is: what is the probability of getting 11 assists in a row. That is not 1/6, not even close. The probability is so low, it is almost a guarantee that one of these is at work: 1) the RNG is not actually random, 2) more variables are at work then have been accounted for, 3) that the probability is not 1/6 in the code, or 4) that initial poster is lying. My money is on 2 and 3 being the culprit here.

    Here is text that may help:

    "Example: Why is it unlikely to get, say, 7 heads in a row, when each toss of a coin has a ½ chance of being Heads?

    Because we are asking two different questions:

    Question 1: What is the probability of 7 heads in a row?

    Answer: ½×½×½×½×½×½×½ = 0.0078125 (less than 1%).

    Question 2: Given that we have just got 6 heads in a row, what is the probability that the next toss is also a head?

    Answer: ½, as the previous tosses don't affect the next toss."

    You are talking about question 2, this thread is about question 1.
  • Qeltar
    4326 posts Member
    edited January 2016
    @Cyclonus is right.

    Probably of getting 11 assists in a row after having already received 10 assists: 1/6. (People don't believe this, though, feeling the other result is "overdue" after a long string. It's called the Gambler's Fallacy: https://en.wikipedia.org/wiki/Gambler's_fallacy).
    Probably of getting 11 assists from the start of a match: a number so low that while possible, it suggests a possible bug.
    Quit 7/14/16. Best of luck to all of you.
  • Whoops, I did misspeak earlier in the thread. I posted: "what is the probability that Phasma will enable another assist after causing 10 previous assists in a row". This is the wrong wording for sure! My apologies if this cause confusion (it was late at night!). All of the math is still correct though. It should read: "what is the probability that Phasma enabled 11 assists in a row".
  • Just wanted to chime in quick to say that I do think she may be bugged. I read on here somewhere that until they made the AI stop sucking they buffed the chance of the AI getting procs (like bonus attacks on duku, debuff apply chance, dodge etc.). Is that true? Because maybe something is going on there with extra attacks.

    I can't tell if I'm just getting unlucky, but the last week or so I can't move in the PVP ladder because everyone has Phasm as their team lead and they get so many bonus attacks I'm dead by turn 2 O.o
  • Mathematical Proof: Divide pi by the golden ratio and then add the Fibonacci sequence to every other prime digit right of the decimal point. Assign each digit to a different person on the server and have them flip a coin. Take those values and use kth nearest neighbor analysis each time, where k is RNGed. Run a regression analysis on the results from each server, and take the p-value divided by the R-squared. That's the answer.
  • Cyclonus wrote: »
    Baal wrote: »

    Flipping a coin is 1/2, that never changes. That doesn't mean that the odds of 11 repeats in a row is 50% but each flip is 50%, and is never greater or lesser than 1/2, it is not influenced by previous results.

    This is correct, but you are not addressing the right question. The question is not about the probability (not odds) of any one assist. That is 1/6 (which again, I am sure this isn't really the case). The question is: what is the probability of getting 11 assists in a row. That is not 1/6, not even close. The probability is so low, it is almost a guarantee that one of these is at work: 1) the RNG is not actually random, 2) more variables are at work then have been accounted for, 3) that the probability is not 1/6 in the code, or 4) that initial poster is lying. My money is on 2 and 3 being the culprit here.

    Here is text that may help:

    "Example: Why is it unlikely to get, say, 7 heads in a row, when each toss of a coin has a ½ chance of being Heads?

    Because we are asking two different questions:

    Question 1: What is the probability of 7 heads in a row?

    Answer: ½×½×½×½×½×½×½ = 0.0078125 (less than 1%).

    Question 2: Given that we have just got 6 heads in a row, what is the probability that the next toss is also a head?

    Answer: ½, as the previous tosses don't affect the next toss."

    You are talking about question 2, this thread is about question 1.

    The probability of 7 heads in a row is the same as this sequence happening: heads-tails-heads-tails-heads-tails-heads
    But nobody will be surprised when that happens.
  • I think you guys will have to redo your math >:)

    First the 16% doubles for any first order toons, but I don't know if the team your referencing had any besides phasma, but you at least have to factor in Phasma's higher percent.

    Second, and I could be wrong here, but it was my understanding is the 16/32% only applies to when you play the team and the percent is boosted when AI is controlling the same team. It was my understanding that this was done to account for how poorly the ai does (ie-not focus firing, poor use of specials, not accounting for heal immunities, etc).

    I don't believe the percentage for the assist for an ai team, has been made known, which is REALLY going to make this a tough calculation.

    Good luck! :)
  • Nothing like 1 in 10^9 chance, it's actually 1.76%, still long odds but almost a 2% chance. If only the Powerball lottery had such good odds.

    Based on a 16% proc rate of an assist times 11 in a row, you get the following formula:
    (16/100)11 =1.76%
    Baal wrote: »
    16% is 16% and 11 is well within common expectations.

    ...Somebodies weren't paying attention in 9th grade math class. Or maybe you are just American?

    It should be (16/100)^11. That's 16% to the power of 11, not times 11.
    Rheen wrote: »
    Actually it's much lower. It would be (0.16)^11

    Which is about 1 in 568,000,000.
    To be precise 568,434,188.608 in one chance. You probably would have won the lottery.

    How has nobody commented on the American slight?
  • Cyclonus wrote: »
    Baal wrote: »

    Flipping a coin is 1/2, that never changes. That doesn't mean that the odds of 11 repeats in a row is 50% but each flip is 50%, and is never greater or lesser than 1/2, it is not influenced by previous results.

    This is correct, but you are not addressing the right question. The question is not about the probability (not odds) of any one assist. That is 1/6 (which again, I am sure this isn't really the case). The question is: what is the probability of getting 11 assists in a row. That is not 1/6, not even close. The probability is so low, it is almost a guarantee that one of these is at work: 1) the RNG is not actually random, 2) more variables are at work then have been accounted for, 3) that the probability is not 1/6 in the code, or 4) that initial poster is lying. My money is on 2 and 3 being the culprit here.

    Here is text that may help:

    "Example: Why is it unlikely to get, say, 7 heads in a row, when each toss of a coin has a ½ chance of being Heads?

    Because we are asking two different questions:

    Question 1: What is the probability of 7 heads in a row?

    Answer: ½×½×½×½×½×½×½ = 0.0078125 (less than 1%).

    Question 2: Given that we have just got 6 heads in a row, what is the probability that the next toss is also a head?

    Answer: ½, as the previous tosses don't affect the next toss."

    You are talking about question 2, this thread is about question 1.

    Spot on and beautifully explained ☺ Probability tree anyone?
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