Statistics Class: How I Will Calculate Your Probability of Winning the Exploits Giveaway


I like beer.

51 Responses

  1. Shivox

    Thanks for the crash course, we need these kind of reminders to not forget the basics :p

  2. Epoke

    Thanks for the infos and explanation…

  3. MGRP

    Bored eh?

  4. Leires

    I may enter yet..I don’t want or need to win, I’m already rocking the Megamix ‘sploit, and I’ve already (out of sheer generosity on his part) ‘won’ the Sims 3 from Tanooki, but i may enter for fun..Maybe type up some silly story or something. Iunno yet. Regardles, grats to whomever wins this game of numbers. Somewhere, Trogglus Normalus is enjoying the numbers you’ve laid out, good sir.

  5. bearded

    So math aroused.

  6. ivo

    what about statistics on the chance of psplink tools patches for vita ?

  7. whitest0rm

    thats all well and good… but now i feel stupid coz that went in 1 eye and out the other xD

    so whens the draw actually going to be? coz theres 500 entries atm and i havent seen a single rated one yet haha

  8. Claudio

    Wow… This Transported me into that fuc*ing days on Math Class…and the most cool thing…I reminded all the stuff…!!! Cool and Thanks 🙂

  9. Jamaicakid

    Really would like to use this exploit. we can use the psn store at all so. wish i could win.

  10. perfig

    Thx for explaining in detail :D. Being looking the pictures and wondering: why are so many cats? where is the love for dogs? 🙁

  11. Nonlin

    So Gifs are allowed?

  12. So do we have to have a picture?

  13. alpmaster

    But you are still able to be picked without a Gif or image right?

  14. wielku

    Forgot to enter contest – yep still going to continue playing MHFU…

  15. Gie

    nice explanation, but you still have to review all of those posts… It flooded with gifs, makes me dizzy somehow..

  16. wisest.guy

    You could have used 7-9 or 8-9 out of 10 to make it more applicable, than to only use only 8/10. Other than that good article.

  17. ayushon

    hi wololo where do i download 101 megamix game ..i dont hav a single game on vita …theres no exploitinggame on store too please reply …i let my brother to subscribe u too 😉

  18. Yifan Lu

    First of all, are we assuming you just choose a random number for each picture? Because otherwise, it really isn’t a probabilistic system right? Aren’t there characteristics you’re looking for in these pictures? For example, if I submit some picture I know you will hate, I can be sure that I get a low score. Even if we assume the score is random, the sum of probability of everyone’s chances does not add up to 1. For example, assume you scored everyone with 8/10. You say the chance of one person winning is 0.02061. 0.02061 * 200 = 4.122 > 1. Your probability model is flawed.

    An alternative approach. Let everyone have a max of 10 “entries” where the entries are determined by the score they have. 8/10 means 8 entries. No image means 1 entry. Etc. Now just sum up the entries and your chance of winning is entries/total. That way, the ratio is biased towards the mean of the scores. If everyone get’s 8/10, everyone has the same change of winning. If everyone has 1/10 but one person has 10/10, they get a much higher change. The math is simple and the experiment is easy to perform (order by name, sum up all entries, pick a random number between 1 and the number of entries, subtract the number from each name until you hit zero and that person is selected).

    • anyone

      “An alternative approach. Let everyone have a max of 10 “entries" where the entries are determined by the score they have. 8/10 means 8 entries.”

      I KNOW, right? That is the automatically obvious way to do it. I was initially going to suggest 10 entries per person without a pic, with 11 to 20 entries for a person with a pic, but apparently Acid found the dumb way to do it.

      Haha, j/k Acid, you know I love you 0.o

    • Acid_Snake

      The system is not flawed, your assumptions are, you are assuming that both the scores and probability of being picked from the crowd are one in the same occurrence, when in fact they are two separate ones, which is basically what bayes works with. You can’t just add up two different occurrences with two different probabilities and expect it to still be 1.

      The first occurrence is the probability of your picture winning, which is individual and independent from the amount of users there are. If your score is 8 out of 10, your picture has a probability of winning of 0.8 and a probability of loosing of 0.2. 0.8 + 0.2 = 1

      Then the second occurrence is the probability of being randomly picked from the crowd of, lets say, 200 people. The probability of success here is roughly around 0.025 (5/200, as there’s 5 chances of you winning). And the probability of failing is obviously 1-0.025.

      For you to win, both occurrences need to be met: P(B|A). From there on you just apply bayes and you’re good to go.

      • Yifan Lu

        What do you mean by “winning” a picture? Let’s say everyone got 10/10. Then everyone “wins” so what does it mean? And I’m not assuming that’s the same, I proposed a clearer system in which they are the same.

        • Acid_Snake

          I probably used a wrong naming there, maybe “approved” is a better word. Your image has to be randomly approved for you to win, aside from being randomly picked from the crowd, and the probability of it being approved is simply your score, not much, not less.

  19. Yifan Lu

    Oh, and also, what is your formula for getting
    0.8*0.02525 / (0.8*0.9747 + 0.2*0.02525 + 0.2*0.9747) = 0.02061

    Assuming A = event that your picture is good and B = event that you are chosen randomily, we get from your formula:
    P(A)P(B) / P(A)P(Bc) + P(Ac)P(B) + P(Ac)P(Bc)
    assuming A and B are disjoint
    P(A|B)P(B) / P(A|Bc)P(Bc) + P(Ac|B)P(B) + P(Ac|Bc)P(Bc)
    I don’t see what equation that represents

    If you want to use Baynes’ theorem you can get
    P(B|A) = P(A|B)P(B) / P(A|B)P(B) + P(A|Bc)P(Bc)
    P(B|Ac) = P(Ac|B)P(B) / P(Ac|B)P(B) + P(Ac|Bc)P(Bc)
    P(B) = P(B|A) + P(B|Ac)
    P(B) = P(A|B)P(B) / P(A|B)P(B) + P(A|Bc)P(Bc) + P(Ac|B)P(B) / P(Ac|B)P(B) + P(Ac|Bc)P(Bc)
    because A and B are disjoint
    P(B) = P(A)P(B) / P(A)P(B) + P(A)P(Bc) + P(Ac)P(B) / P(Ac)P(B) + P(Ac)P(Bc)
    using your example above
    P(B) = 0.8*0.02525 / 0.8*0.02525 + 0.8*0.9747 + 0.2*0.02525 / 0.2*0.02525 + 0.2*0.9747
    P(B) = 0.05050

    However, it’s still a bit flawed because when designing an experiment, you need to make sure that the pdf used for event A is valid. That means if you have to score everyone manually with the knowledge of everyone else’s score. Otherwise, the universe would be broken and you could have something like two people with 100% chance of winning (in theory).

    • Acid_Snake

      where the heck did you get this from:
      P(B) = 0.8*0.02525 / 0.8*0.02525 + 0.8*0.9747 + 0.2*0.02525 / 0.2*0.02525 + 0.2*0.9747

      P(B) is the probability of being randomly picked from the crowd, and it’s totally independent on the probability of your picture winning, it’s as easy as 5/200 if there’s 200 people in the crowd and 5 chances of winning, although it’s a bit more like 1/200 + 1/199 + 1/198 + 1/197 + 1/196, since the same person can’t win twice.

      • Acid_Snake

        that is assuming that what you call P(B) is the probability of being randomly picked from the crowd.

        Still I don’t see where the flaw is. P(B) and P(A) are two different probabilities. To win, both probabilities must be met, which is a join: P(A)^P(B).

        • Yifan Lu

          If what you say is true, then the sum of everyone’s P(A) must add up to one. Everyone’s P(A) does add up to one. But two people can get 0.8 for P(B).

          Where I got that equation from is the definition of Bayes’ theorem (look at Wikipedia), which I showed step by step of me applying it. However, I don’t know where you got your equations from.

          • Acid_Snake

            no, the probability P(A) (assuming we are talking about the scores) are individual, not collective, as I already said, it makes no sense in a statistical way to sum them up. The only collective probability is P(B) (if we assume it’s the probability of being randomly picked from the crowd).

          • Acid_Snake

            To clear it up a bit, let’s take as an example flipping a coin. My probability of getting head is 1/2, your probability of getting head is 1/2, and some other guys probability of getting head is 1/2. Does it make sense to add up all those probabilities? no! as they are individual and not related at all, you would end up having 3/2, which we both know is impossible. Same applies to the scores here: it either wins, or not, whatever another person’s probability is is not related at all.