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.

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Letâ€™s assume that Stefan sent Postcard â„–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 Â±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%).

Letâ€™s also assume that Stefan sent Postcard â„–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 â„–2 arrives at the addressee is slightly larger 6138 Â±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 â„–1 and Đˇard â„–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 = 0.000012866666667â€¦ (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=[6126â€¦6150], then there it will be

P for [6126â€¦6150] = (1.0+1.0)/((6150-6126)*15,000) = 0,000005555555556â€¦ (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

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 Â±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.