Today I "/?op=randomnode"'d to my own node, twice. Inspired by this, I hit this node and here's what came out because of that:

u=number of your nodes
e=number of nodes in e2
n=number of your nodes in random nodes nodelet
l=number of nodes in random nodes nodelet (13)
p=propability to get that number of your nodes in the nodelet

    (u/e)n*(1-u/e)(l-n)*l!
p = ------------------------
            (l-n)!n!

Now, that's very helpful, isn't it? Let's test it. As of right now, there are 735207 nodes in E2, of which 451 have a writeup by me. Placing them in the above formula, we see that likelihood to see exactly q of my nodes in the random nodelet are as following:

[chart 1]
  q | p 
-------
  1 | 0.007916
  2 | 0.000029
  3 | 0.0000000656148
  4 | 0.000000000100687
  5 | 0.000000000000111245
  6 | 0.0000000000000000910444
  7 | 0.0000000000000000000558839
  8 | 0.00000000000000000000000257265
  9 | 0.00000000000000000000000000877288
 10 | 0.00000000000000000000000000000215395
 11 | 0.00000000000000000000000000000000360577 
 12 | 0.00000000000000000000000000000000000368875
 13 | 0.000000000000000000000000000000000000000174169
-----------------------------------------------------
sum | 0.007945    

Now, those are just numbers that really don't tell you much. But since the propabilities seem to be fairly linear (well, in logarithmic way...), let's make a new propability measuring unit: rn, RandNodelet. 1 rn shall be propability to get one of your nodes in the nodelet, 6 rn for 6 of your nodes and so on. Of course, this isn't accurate because the chart above isn't really logarithmic, only close, but it'll do. Also, its value depends on person, so I'll use my own statistics here. Your Mileage May Vary; with 4000 writeups these figures are quite different - 13 rn with 4000 nodes would be about 10.3 rn on 450 nodes. Anyways, here are likelihoods for some events to happen you during one year, on my rn-scale):

[chart 2]

Getting into an elevator accident
2.7 rn
Dying on a dog bite
3 rn
Getting struck by a lightning
2 rn
Getting hit by a stellar object
5.7 rn
Getting killed along with the rest of mankind in a meteor strike
2.2 rn
Getting killed
1.5 rn
Winning approximately one million dollars in Finnish lottery by playing with one line every week
1.8 rn

Now, I can hear you thinking "geez, I'm dead already, I saw once two and one fifth of my nodes in the random nodes nodelet, and that's equals the chances of meteor strike. ACK, wait, that means we're all dead already!". Well, sorry to tell you this, but you're not going to die... yet. Those are likelihoods for something to happen once a year. You check random nodelets out more than once a year, don't you? Now, I checked my Navigator history for visits on pages at everything2.com. There were 677 in the range of 21 days, which makes approximately 32 visits/day. Now, each of these visits (well, excluding XML ticker visits, but they were few) generated a new random nodelet. Here's a chart for likelihoods that during these three weeks, at least once I had a number of my nodes (or more) in the nodelet:

[chart 3]
  n | p
-------
 1+ | 0.995486
 2+ | 0.019588
 3+ | 0.000044
 4+ | 0.0000000682416
 5+ | 0.0000000007447
[at which time my TI-89 ran out of horsepower...]
5->s:1-(1-sum(randlet(13,seq(x,x,s,13),735207,451.)))677
7.447E-11.
6->s:1-(1-sum(randlet(13,seq(x,x,s,13),735207,451.)))677
0.

As you can see, there isn't too much to worry yet; even three weeks of random-node-letting, which corresponds to 677 years of life on the events on chart 2, gave me only one-to-22727 chances of dying on a dog bite!

Conclusion: I had a point I forgot it already. But isn't mathematics wonderful? ;)

Sources: My brains, TI-89, Tiede 2000 (and through it, Discover 5/96 and Tilastotälli 1997), www.veikkaus.fi

A quick Probability math intro using the random node list on E2.

Or "should I be surprised to see my node?"

Please Note:Key Probability words (words that give us instructions) are in caps and that a year at A Level just got condensed into this write up.

But First the basic rule of probability:

    Number of ways it can happen
P = ------------------------------
    Number of possible outcomes

1 is a dead cert while 0 is Impossible, most probabilities are in that range and are naturally expressed as decimals. All other forms are for reading purposes only.

So if you flip a coin

            1
P(Heads) = ---
            2

1/2 = 50% = 0.5 (50/50 in layman's terms).

We shall now examine the probability of any given node showing AT LEAST ONCE in the list.

Some Variable names for formula purposes

e = number of nodes in e2 (e = 735207 for tie in purposes)
l = number of nodes in random nodes nodelet (13)
p = probability of getting any given node show in the nodelet

At the most basic level:

     l      13
P = --- = ------
     e    735207

This is not strictly true the exact result is a little more fiddly. The above example assumes that the number of spaces to fill (13) does not change and that the population to draw from is not diminished. This assumption can only be made if the sample is taken from an enormous (read assume infinite) population.

If we want one or the other we add thus with the coin P(HEADS) + P(TAILS) = 1/2 + 1/2 = 1 

So the true value for our maths is

13   12    11    10     9     8     7     6     5     4      3      2      1
-- + --- + --- + --- + --- + --- + --- + --- + --- + --- + ---- + ---- + ----
 e   e-1   e-2   e-3   e-4   e-5   e-6   e-7   e-8   e-9   e-10   e-11   e-12
or
  l+(l-1)+(l-2)...+(l-(l-1))
-------------------------------
le - ( (l-1)+(l-2)..(l-(l-1)) )
if l = 13 then l+(l-1)+(l-2)...+(l-(l-1)) = 91
 
So we can say:
     91
-------------
13e - (91-13)

This gives us a simple enough formula for future calculations when the value of e changes as it will have done as I am using a value from a write up already present.

Given e = 735207:

        91                     91          91
-------------------  =  ------------- = -------    
13*735207 - (91-13)     9557691 - 78    9557613

Which my calculator gives as:  9.52120576549814268478960175516627e-6.  
Which is: 0.00000952120576549814268478960175516627

Approximately: 0.00000952

This number represents the probability that any given node will show in the first slot OR the second OR the third OR... etc

If we had 451 nodes to our name we could add the number to itself 451 times and get the probability for any of the 451 nodes showing at least once in any of the 13 locations on the random node nodelet. (or for speed we can multiply)

0.00429406380023966235084011039157987 (451 nodes)
Approximately: 0.0043

That is better than 4/1000 (1/250) chance of seeing at least one of your nodes every time you refresh your page, so it's not nearly as surprising as you would at first think. For the interested: the odds are still 1/250 no matter how many times you load the page this is because each random occurrence (theoretically) in no way effects the next. The oft stated misunderstanding that "if you refresh 249 times the next one must be yours" is meaningless as the number 1/250 means that given infinite refreshes 1/250 would contain a required node (one of the 451).

The probability of seeing your node AND then seeing your node again is just slightly greater than: 0.00001843898, but each occurence has an indipendant 1/250 chance. Probability's always fun!




Thanks to Taliesin's Muse for the typo spotting.

Log in or registerto write something here or to contact authors.