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

Statistics and Probability is a fundamental knowledge that should be inside each Software Engineer, and here I will show you some of its wonders.

Some of you are already saying “hi” and posting pictures to boost your possibility to win the giveaway, but, how will it improve your chances? and how will I calculate? Keep reading to find out.

In probability the chance of an event to happen is measured between 0 and 1, with 0 being an impossible occurrence (it will never happen) and 1 being an occurrence that will always happen, anything in between is said to be the probability of an occurrence to happen, the closer to one the more chances it has to happen.

It’s all pretty basic stuff, but lets put it in the context of our giveaway.

Let’s imagine for a second that the giveaway ended and there’s 200 participants, this means that if you were to win you have to be randomly picked out of this crowd of 200, so you have a one out of 200 chances of winning, which is 0.005 probability. Not much is it? but there’s more so don’t worry.

Our contest randomly picks 5 people out of the 200 participants, this means that you have to be picked either one of this times, so you have one of 200 chances of winning, 5 times, so it’s (1/200)*5, or 5/200, which is 0.025, much better right? but this formula is plain wrong.

The thing is, your chances of winning are repeated 5 times and are exclusive: if you’ve already won, there’s no chance of you winning again as you are excluded from the population, so the population of participants decrements by one, but since one winner is already picked, then there’s one less chance of you to win, so both your chances and the population decrease every time until you have more chances of winning, so the formula is left like this:

1/200 + 1/199 + 1/198 + 1/197 + 1/196 = 0.025254

Looks nice but… the probability of you NOT winning is 1 – the probability of you winning, so you have a probability of loosing of 0.9747, which is really, really, really high.

This is usually where past giveaways end, but ours is a bit more complex than that.

You see, I added one more variable to the table: a contest. Users can post images and they get graded between 0 and 10, the higher the grade, the better, users with no picture will end up with a score of a 1 out of 10. Users need to have their picture win and to be picked from the crowd randomly to win the prize.

If your score is an 8 out of 10 then you have a probability of 0.8 that your picture will win, after that you have to cope with the fact that you have to be randomly selected from the crowd.

This means that your chances of winning are solely based on two variables: your picture’s score and the probability of you being randomly picked from the crowd.

So you have two probabilities that need to be met for you to win: your score, and being randomly picked.

This gives us 4 different occurrences that can happen, assuming your score is 8/10:

— Your picture wins (probability of 0.8), you are picked from the random crowd (probability: 0.02525) -> you win

— Your picture wins (probability of 0.8), you are not picked from the random crowd (probability: 0.9747) -> you loose

— Your picture looses (probability of 0.2), you are picked from the random crowd (probability: 0.02525) -> you loose

— Your picture looses (probability of 0.2), you are not picked from the random crowd (probability: 0.9747) -> you loose

So how do we calculate your final probability of winning? well, we use the bayes formula, which goes something like this: the probability of the event we want happening divided by the summary of the probabilities of all events.

The probability of you winning is the probability of your picture winning intersected with the probability of you being picked, which is 0.8 * 0.02525 = 0.0202

Knowing this, we go on to calculate your actual probability:

0.8*0.02525 / (0.8*0.9747 + 0.2*0.02525 + 0.2*0.9747) = 0.02061

Now it may seem like a low probability, but if we calculate the probability of someone with a picture of a score of 0.5 we get 0.01278, and the probability of someone winning without posting any picture is 0.00253.

You may think “Acid, I used to have a probability of winning of 0.025 with the other giveaways, now I have 0.02061 if I get a score of 8 out of 10”.

This is true, but someone with an 8 out of 10 has a much bigger chance than someone with a 5 or no picture at all.

I hope this has helped clarify how the contest works and how the picture matters.

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

Thanks for the infos and explanation…

Bored eh?

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.

So math aroused.

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

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

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 🙂

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

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

I don’t like cat pictures as I find them all to be the same thing, you see one, you’ve seen them all.

So Gifs are allowed?

images, gifs, video (as long as it’s short)

So do we have to have a picture?

if you read carefully, you don’t have to have a picture, but your chances are much bigger if you do have one than if you don’t

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

yes, you are still able to be picked, you just have less probability

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

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

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.

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 😉

101 megamix was removed from the store A LONG time ago

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).

“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

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.

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.

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.

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).

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.

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).

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.

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).

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.