Nice analysis, now we have a thread to link everytime someone comes around and complains that they have not yet won one or more giveaways.

2 months ago

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I've seen enough people complaining about XCOM cheating on rolls to know that some people will never understand probability.

2 months ago

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That's wrong since he made the calculation for 0 lvl but your lvl 4 so the person's who join is lesser and there's group giveaways and region locked giveaways that have less entries too also we don't know does he entered randomly or entered to the games he wants

If am not wrong

2 months ago*

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The 1834 giveaways used in the analysis above are mostly a representative selection of what is available at level 0, I did not make any strong limits on game type or such.

Any SteamGifts participant can look at their own giveaway history to compile their own personal sample of the number of entries, and compute WinChance from those as described in Part 2 above. (If you do, feel free to comment here! I want to stress that the equation to compute it is an average of 1/K, that is not the same as 1/"Average of K".)

2 months ago

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If u didn't calculated all of them manual i can suggest using/improving a bot and collecting the giveaway data's at https://www.steamgifts.com/archive and creating analysis for each lvl

2 months ago

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Estimating level-wise probabilities based on the archive might be possible in the future, but that would indeed need a lot of page crawling since the archive page itself does not list the number of entries and winners per ended giveaway, they are only listed on the page of each giveaway.

If one wants to evaluate winning chances over a period of time (e.g. over one month of SteamGifts participation), the needed number is how many giveaways the user is able to enter over that time.

Estimating that number is somewhat difficult. Things that will affect the number include:

1) A user might not be actively participating all the time over the month

2) The point price of the giveaways will limit how many giveaways one can enter over the month

3) Self-imposed limits: for example, if you only enter giveaways with a limit on the number of entries, you will have a higher WinChance, but on the other hand there will be less potential giveaways available over the month.

4) Higher SteamGifts levels, invite-only giveaways etc. will increase availability.

In this sense, the 1000 giveaways used above represent "1000 giveaways, over however much time it takes you to enter that many". One can e.g. check from their Entered history how many ended giveaways they have entered over a particular time like one month (or how far back in history they need to go to reach a particular number like 1000 giveaways).

2 months ago*

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Hi everyone,

I performed a simple analysis of the winning chance. My numbers are based on level-0 giveaways, but the principle should work more generally as well. Please feel free to comment if I made an error or left out some important consideration; and of course feel free to discuss otherwise.

In any individual giveaway, if K is the number of entries at the time the giveaway closes, your chance of winning will be 1/K, for example 1/500 if there were 500 entries. However, the number of entries varies a lot between giveaways. The question I analyze is: how many keys are you likely to win over 1000 giveaways that you enter?

## Part 1: data analysis

To analyze this question, I have collected the number of entries in 1834 ended level-0 giveaways that I participated in. The number of entries has a wide variation. Its overall statistics are:

Mean number of entries: 1574.20

Median number of entries: 1285.0

Sample standard deviation of the number of entries: 1222.98

More precise giveaway statistics:

Below is a histogram of the number of entries over the giveaways, divided into brackets of 100. For example "[900, 1000): 0.06925" means that about 6.9 percent of giveaways had at least 900 entries and less than 1000 entries.

[1, 100): 0.00218

[100, 200): 0.00273

[200, 300): 0.01309

[300, 400): 0.02454

[400, 500): 0.02890

[500, 600): 0.03544

[600, 700): 0.03490

[700, 800): 0.03653

[800, 900): 0.05943

[900, 1000): 0.06925

[1000, 1100): 0.07579

[1100, 1200): 0.05943

[1200, 1300): 0.07034

[1300, 1400): 0.05943

[1400, 1500): 0.04962

[1500, 1600): 0.04253

[1600, 1700): 0.04417

[1700, 1800): 0.03272

[1800, 1900): 0.02290

[1900, 2000): 0.02890

[2000, 2100): 0.02126

[2100, 2200): 0.01963

[2200, 2300): 0.02181

[2300, 2400): 0.01145

[2400, 2500): 0.01200

[2500, 2600): 0.01254

[2600, 2700): 0.00654

[2700, 2800): 0.00872

[2800, 2900): 0.01036

[2900, 3000): 0.00763

[3000, 3100): 0.00382

[3100, 3200): 0.00709

[3200, 3300): 0.00327

[3300, 3400): 0.00491

[3400, 3500): 0.00545

[3500, 3600): 0.00436

[3600, 3700): 0.00436

[3700, 3800): 0.00164

[3800, 3900): 0.00164

[3900, 4000): 0.00218

[4000, infinity): 0.03653

## Part 2: Winning probability in an individual future giveaway

Now let's consider what will happen in the future, over 1000 upcoming giveaways that we have not seen or entered yet. Let's assume that the distribution above will hold over those giveaways. Let's also assume for simplicity that the number of people entering each giveaway is a statistically independent variable from the number of people entering any other giveaway.

Then the success probability in one of the future giveaways is a weighted average, over the possibilities of how many people will enter:

p("I win") = sum

{K=1,2,3,...} p("I win" | "K people enter") * p("K people enter") = sum{K=1,2,3,...} (1/K) * p("K people enter")An estimate of the above is simply the average of (1/K) over the sample of giveaways that I gathered.

The resulting estimate is: WinChance = 0.000970906

## Part 3: Number of wins in 1000 giveaways

Now that we have the winning chance, we can model what will happen in 1000 independent future giveaways: this is a binomial distribution (https://en.wikipedia.org/wiki/Binomial_distribution) with 1000 trials and success probability equal to WinChance above.

The probability to win zero (0) keys is the probability mass of that binomial distribution at zero. The probability to win 1 or more keys is 1 minus the probability to win 0 keys. The probability to win 2 or more keys is 1 minus the cumulative probability to win 1 or less keys (cumulative distribution function of the binomial distribution at 1); probability to win 3 or more is 1 minus the cumulative probability at 2; and so on. This gives the following probabilities over 1000 giveaways:

Probability of winning 0 keys: 0.379

Probability of winning 1 or more keys: 0.621

Probability of winning 2 or more keys: 0.2535

Probability of winning 3 or more keys: 0.0749

Probability of winning 4 or more keys: 0.0172

Probability of winning 5 or more keys: 0.003

Probability of winning 6 or more keys: 0.0005

Probability of winning 7 or more keys: 0.000068

Note that these are estimates of probabilities only in future giveaways - whatever happened in past giveaways that already ended does not affect the probability in this analysis.

It is possible I missed or miscomputed something - feel free to comment.

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