Where can I get a random integer that is subject to third-party verification?

Question

random numbers

To determine the winner in the draw, you need to choose it from the list at random. But you need to do it in such a way that it is reliably known that there are no frauds.
For the time being, I came up with the idea of ​​taking a random integer somewhere on the Internet, which will appear there no earlier than the end of the competition, and will remain for a while so that those who wish can check. And just divide this number by an integer by the number of names in the list. The resulting remainder will be the number of the winner. Because The list is pre-compiled, everything is reliable.
I just can't think of a simple way to find this random integer. All that comes to mind is the number of characters in an article in some news publication that will appear on a certain day at a certain time on the main page. But it seems to me that it is not very convenient and beautiful.

Answer 1

The option with the number of characters in the article is good, visual, as for me - better than random.org. It's even better to take for example "Photo of the day" on yandex, fotki.yandex.ru/calendar/ and say that the winner number will be determined as the remainder of dividing the ID of the photo of the day by the day the contest ends (or you can take the size in bytes, but the ID is simpler and also by chance) on the number of participants. The photo of the day for any date can then be viewed at any time.

Answer 2

random.org

Answer 3

At the moment, the ONLY random number generator that is not subject to manipulation and verifiable (more precisely, with an accurate assessment of the guarantee in money or other understandable values) is bitcoin (or rather, any cryptocurrency, it’s just quite expensive).
Among the disadvantages - you can get any number of pseudo-random numbers based on several random variables (hashes from blocks or transaction id) that are formed (or rather received) no more than a sufficiently large interval, and the guarantee directly depends on their number. We are talking about tens of minutes and hours (6 confirmations ~ an hour) ... and most importantly, this interval is also subject to a random number.
The algorithm is very simple - you choose and publish - the selected cryptocurrency, block number(greater than the last one found), an algorithm (deterministic) for getting the number from the hash of the block and the number of blocks found after it (the more, the more expensive manipulations that increase the probability of 'cancelling' the value).
Then all participants wait for the appearance of this block and the required number of blocks after it (roughly speaking, confirmations) and calculate the number according to the specified algorithm.
Possible manipulations - the owner of sufficient computing power (for PoW) and for some cryptocurrencies a sufficient number of coins (for PoS), and now even more interesting technologies are being developed, for example Proof-of-Wasted-Space (presence of memory with random access) - can 'cancel ' found blocks, thus changing the detected random number (but to some other one, the value of which is also not predictable).

For example, let it be the block following the current block and 6 confirmations after it. Algorithm - the last byte in the hash of the block (i.e. the interval of the number 0..255)

Estimating the cost of an attack depends on the network parameters, for example, for bitcoin this is a reward in a block (article study of the required power estimation btcsec.com/double-spending/ ), at the moment it is quite expensive (estimated cost is at least 2% - now it is 72 terahesh - at the price of at least cex.io cloud mining at 0.12btc per GH/s ~ 8640btc or at $223/btc - ~$2 million , and this does not include the cost of maintaining this percentage).

Answer 4

After almost six months, I came up with what seems to me a fairly reliable algorithm that does not require any external data.
The only problem is relative complexity. It is necessary that ALL participants in the draw enter data twice. And in the worst case - several times.