🎯 Countdown to 68 million registered cards

Here we keep records of the way traveled to the level of 68 million postcards registered.
And we make predictions about the moment when we cross this level.
Anyone can make as many predictions as they want at any time.

On the Postcrossing home page, we can find out how many postcards are registered in the game at the moment:

And here we can see monthly statistics on registered postcards.
These data will help to calculate or guess the moment of passing the level of 68 million.

Please let’s make our predictions of date and time in UTC format.

Well, based on the results of this countdown, we will see whose predictions were the most accurate, taking into account the distance from which they were made. And I will send a prize postcard to the most talented diviner.

The counting method is outlined here.

Let’s go!

6 Likes

Ok, if no one wants to take the first step, then I’ll start :slight_smile:
August 13, 09:21 (UTC)

2 Likes

12 August at 12.08 UTC

2 Likes

August 15, 13:05 (UTC)

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August 7, 7:07 UTC :notes: :notes: :notes:

2 Likes

I guess without some changes in the bigger picture the trend of our small one here won’t change either.

983,730 to go

15.08.2022
15:08 UTC

a card from to FI to SE

2 Likes

16.8. at 9.02 UTC

2 Likes

06.September.2022, 16:58 UTC

2 Likes

Since this issue was discussed earlier, I prepared a small explanation on how to calculate the chances of @Stevyy 's sent cards to get bingo when passing the millionth level.

Read more...

Let’s assume that Stefan sent Postcard №1 from Austria to USA and it should arrive at the day x=6132 (time 00:00). But since the moment of its arrival is probabilistic in nature, we plot this moment as a Probability Density Function with a maximum value at the moment x=6132 (blue).

The mail exchange between Austria and USA is very well established, so the spread of dates is small 6132 ±6 days.
That means that the Card â„–1 will arrive at its USA addressee in the date range from 6126 to 6138 with a probability close to 1.0 (100%).

Let’s also assume that Stefan sent Postcard №2 from Austria to Madagascar and it should arrive at the day x=6138 (time 00:00). We plot this moment as a Probability Density Function with a maximum value at the moment x=6138 (magenta).
Madagascar is much further away and the spread of time during which Card №2 arrives at the addressee is slightly larger 6138 ±12 days.
That means that the Card â„–2 will arrive at its Madagascar addressee in the date range from 6126 to 6150 with a probability close to 1.0 (100%).

Let us assume that we know for certain that the passage of the next million level will occur in the interval x=[6133, 6134] (this is the interval between the purple dotted lines).
The gray areas under the Probability Density Function curves on this interval are the Probabilities that Card №1 and Сard №2 will reach their addresses in USA and Madagascar accordingly.
Areas Squares are calculated as integrals of Probability Density Functions on this interval x=[6133, 6134].
And in our case, these probabilities are:
P1=0.150 (15%) and P2=0.043 (4.3%).

We know that N=15,000 postcards will be registered during a day at this season.

Then the probability that one of these Stefan cards hits the bingo will be
P for [6133, 6134] = (P1+P2)/N = (0.15+0.043) /15,000 = 0.000012866666667… (0.0013%)

If there are more postcards claiming to get bingo, then new terms will appear in the numerator of the expression.

The wider the estimated interval, the higher the probabilities for each postcard, but also the greater the total number of postcards that will be registered in this wide time interval.
If we estimate such a probability on the interval x=[6126…6150], then there it will be
P for [6126…6150] = (1.0+1.0)/((6150-6126)*15,000) = 0,000005555555556… (0,0006%)

Since the nature of the randomness of the moment of passing the millionth level is not uniform, but also normal (similar to a very narrow and high mountain), in order to obtain the most reliable estimate of the chances, it is necessary to take as narrow an interval as possible, on which the probability of passing the millionth level is as close as possible to 1.0 (100 %).

Congratulations to those who have read this far :slight_smile:
I hope this was informative and not too abstruse.

Stefan (@Stevyy),
If towards the end of this countdown you tell me when your cards are supposed to arrive and you give me roughly a spread in ±days for each one, then with an understanding of when we pass the next million level, I can estimate your chances to get bingo with the method described above.

1 Like

I guess one could calculate an estimate arrival time of every single card, given the averages from country (region) to country (region), and then create a range of probability within one hour, two hours, three hours …

What I still do not understand:
Let’s say I have 15 cards that can arrive with a high chance on a day with 15,000 registrations. So my chances for that day are 1:1000.
Why is this chance not changed to 1:100 if only 1500 cards are left.
I know it does not change (otherwise I would end up with a guaranteed hit eventually), but I can’t find the correct wording.

1 Like

Yes, you are right. We need to study the statistics of postcards sent by you from Austria to different countries and estimate the average travel time for each destination country and the spread.

This is an interesting challenge.

Read more...

Let’s consider a similar case, but more practically understandable.

Suppose we have 15,000 boxes. Someone randomly placed 15 coins in them. One coin per box.
Then we start opening the boxes. It so happened that we opened 13,500 boxes and we found nothing inside. This is in principle possible.

We have 1,500 closed boxes left, in which there are exactly 15 coins.
Let’s call an outsider who knows nothing about what we’ve been doing here. We ask the outsider to open one of the 1,500 boxes. What are the chances of this person opening the box with coin?

Do these chances depend on who opens the box, we or an outsider? (I guess no)
Do these chances depend on how the outsider or we choose the box? (I guess no)

The trick is which event we are investigating.

  1. We estimates the probability that we will open 13,500 boxes and find nothing in them.
  2. We estimate the probability that we will open a box with a coin with a probability of 15/1500 without any conditions.
  3. We estimate the probability that ((we will open 13,500 boxes and find nothing in them) and (we will open a box with a coin)).
    These are related events, and here we need to apply the conditional probability formula.

To be honest, I’m a little confused myself :slight_smile:
But I tend to think that the more empty boxes there are at the beginning, the higher the chances of finding a coin in the next box.

In the case of postcrossing, we will deal not with discrete events and combinatorics, but with continuous values given by curves and their integrals.

If you want to completely lose sleep, then I suggest you read about the Monty Hall paradox.

Yes, I want to lose sleep

Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?

3 Likes

I guess we will be half way 67,500,000 on July 05 at about noon.

I also thought about this game show with the three doors here (when I was not finished reading your post) :smiley:
And here it’s always better to switch, but that math magic is beyond my skills.

My guess for the next miilion, lets say at 20. August, 16: 45 UTC.

1 Like

I admit that we will go halfway a little earlier, namely on July 04, closer to midnight on July 05.

This is not a prediction, but a kind of calibration mark by which we can judge how everything is going according to plan.

Please feel free to make your own predictions or refine previous ones.

I leave in force my earlier (1st) prediction regarding the passage of the level of 68 million.

I haven’t given a guess yet but have just done some quick maths and come up with 13 Aug 2022 at 04:19 UTC
However, as everything slows in the Northern hemisphere over summer holidays, I am going to add an arbitrary 12 hour delay, so my guess is:

13 Aug 2022 at 21:19 UTC DE to US

2 Likes

July 04 in the evening, I believe that at about 21:15 we will go halfway.

And I leave my previous forecast for 68 million unchanged.

1 Like

We got halfway there on July 04 at about 15:20 (UTC). I was only wrong by +6 hours.
I leave my initiating prediction regarding the 68 million passing unchanged.

3 Likes

Halfway there.

491,581 to go

15.08.2022
15:08 UTC

a card from to FI to SE

1 Like

Since June 22, my Magic Crystal Ball has been recording an increase in the intensity of registrations.

August 3 2022
20:20 UTC
from Germany to Taiwan

2 Likes