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

Acid_Snake

I like beer.

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51 Responses

  1. Shivox says:

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

  2. Epoke says:

    Thanks for the infos and explanation…

  3. Leires says:

    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.

  4. bearded says:

    So math aroused.

  5. ivo says:

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

  6. whitest0rm says:

    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

  7. Claudio says:

    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 :)

  8. Jamaicakid says:

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

  9. perfig says:

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

  10. Nonlin says:

    So Gifs are allowed?

  11. So do we have to have a picture?

  12. alpmaster says:

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

  13. wielku says:

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

  14. Gie says:

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

  15. wisest.guy says:

    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.

  16. ayushon says:

    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 ;)

  17. Yifan Lu says:

    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 says:

      “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 says:

      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 says:

        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 says:

          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.

  18. Yifan Lu says:

    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 says:

      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 says:

        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 says:

          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 says:

            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 says:

            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.

          • Yifan Lu says:

            I think the main point of confusion is what exactly is P(A). I said above “A = event that your picture is good” which is what I believe you mean by P(A). However, if we go back to my original argument, A is NOT a random event, so you can’t model it with a probabilistic model. None of this applies at all since the score you assign to pictures are not random. That being the case, the most fair way to pick the winner is as I’ve said above of assigning different number of tickets to people and pick a winner like a weighted lottery.

            Also your coin flipping example doesn’t work either because we are not talking about two people flipping coins since there’s only ONE winner. If you are flipping coins and you say “heads means you win” there can be two winners. Which is exactly what I’ve been trying to say above.

          • Acid_Snake says:

            if P(A) is the probability that your picture is good, and your picture has a score of 8 out of 10, then this probability is 0.8, as simple as that. After this just go ahead and generate a random number between 0 and 1 (not counting these), if P(A) is bigger or equal to this number, your picture is good.

  19. ayushon says:

    |\_|\
    (*-*)
    (TT)
    ! !

  20. Hejk says:

    I dont understan lol but its cool…

  21. ivo says:

    u forget to add the 99% of people not intrested lol

  22. anyone says:

    But what if Acid_Snake is a fool with bad taste in memes, who will bias his contest in a childish fashion? Then he’ll just look like a jerk, showcase his bad taste in humor (oh, sorry, humor is subjective unless you’re begging a child for his favor), and write a silly post about the probability of winning his silly contest?

    I entered your contest for fun, but you rating everybody somewhat harshly is just embarrassing to you. I say this as someone who has not been rated; I just think you’re a bit silly.

    You should probably add a “makes me angry” factor for people you want to exclude, now, huh? =)

    So basically your contest is worse than every other contest on the site, ever, since we have to guess at what someone with bad taste will like. I say this after reviewing most of your rating thus far, which – shockingly! – were more harsh as you rated more pics.

    lol

    • Acid_Snake says:

      “I don’t agree with your ratings so I’m gonna call you names to make myself sound smarter, when in fact it makes me sound like a dumbass”
      I just summed up your wall of text.

  23. snikt says:

    wow Yifan Lu !
    :)
    love it

  24. Adams Myth says:

    Pff, feeble attempt at a “stats” class.

    I’m sorry, your other articles are good.

    You didn’t get this one right at all.

    • Acid_Snake says:

      I would kindly have a nice discussion with you, but if you’re only gonna be like “lol you suck, I win” instead of doing an actual debate with actual data, then there’s not much to discuss other than that you’re a dumbass, unlike yifanlu above who provided good points to his thoughts and allowed for a more mature debate.

      • Adams Myth says:

        “Ooohhh I can’t take criticism so I’m going to start namecalling! BUT I want a mature debate.”

        Lifanyu already made my point so there’s no reason for me to repeat it.

        I expressed my opinion about your article. Take it or leave it.

  25. ivo says:

    what about psplink on other devices … ps3 ps4 one etc ?

  26. HITMAN says:

    Will posting multiple comments gives me a higher chance of winning :P

    • Acid_Snake says:

      the contest ended, but to answer your question: no. The script I use to choose the winners automatically filter people who make more than one post as well as filtering my own posts.

  27. Lionhearttal says:

    i hope i get it i really wanna experience the full potential of my vita lol

  28. Guardian says:

    It sounds good! Yeah… to me statistics and probability is the math that bores mathematicians, but that is another story, my question is how is the picture going to be evaluated? How is the winner going to be picked? Am I the only one who feels like in nested if statement?
    If(picture == winner)
    {
    WinningChances = WinningChances + PictureScore
    }//can’t think of something else right now

  29. AR says:

    i guess result of this competition will be emailed to those 5 winners

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