45 of the draws are made of 6 consecutive numbers (1-2-3-4-5-6, then 2-3-4-5-6-7, etc, until 45-46-47-48-49-50)

A single draw has a 45 out of 15,890,700 chances of being 6 consecutive numbers

A single draw has a probability of 1-(45/15,890,700) of NOT being 6 consecutive numbers

Assuming 1000 draws (lotteries) per day, in a year we expect probability (1-(45/15,890,700))^(1000*365) = 36% that none of the lotteries draw 6 consecutive numbers

So there is a 64% probability that at least one lottery will draw 6 consecutive numbers in a year. If there are 1000 draws per week (instead of per day) the probability is still 17% that this will happen in a year.

So this South African draw is kinda expected.

Of course, as others here already pointed out that there are far more "suspicious" patterns (e.g. other arithmetic progressions like 5-10-15-20-25-30, primes, ...). And of course they often play by different rules, etc.

If they are 200 draws per week, the probability is 26% over 10 years.

https://www.wolframalpha.com/input/?i=binomial+calculator&as...

This might not be in the news because it's so rare. It might just be a slow news day.

No. This is not particularly newsworthy. But for most HN users and this event the answer is: no, but I did see it on HN and it will probably get re posted cyclicaly ;)

You're being offered 5 to 1 odds on a probability that you just estimated to be significantly better than a coin flip. Are you in? Or does your gut tell you that 6 perfectly sequential numbers is crazy unlikely?

Just curious, if a random HN commenter said, there's about a 64% chance that at least one human being will grow to 12 feet tall in 2021, would you take that wager also? I mean, there are a ton of people on the planet, and it only needs to happen once.

https://news.ycombinator.com/item?id=25306489

At the same time, I'm not a thief -- to be entirely certain that you want to make this bet, have you read the entire thread, where alternate probabilities (far less than 64%) are proposed? You're aware that the original assumption of 1000 lottery draws per day was a guess from thin air with no research nor basis in fact? You're aware that mrb's calculations overstated the number of possible conforming sequences and undercounted the total number of potential outcomes? If you acknowledge that you're aware of all of this and still want to proceed, let's do it, no joke, I'm 100% in.

(That is - as is pointed out upthread, such 'weird' draws will occur naturally on occasion. It would, however, be quite suspicious if it happened twice in a row.)

I am sure one of the things the lottery looks into is how many coupons with 'weird' numbers chosen like this week's winning numbers have been picked in the past - of course, if you get a result which makes you suspect shenanigans AND the number of coupons with such a strange sequence of numbers is way up from normal, then perhaps that suspicion is well founded.

https://www.gamblingsites.org/blog/how-many-different-lotter...

Looks like a reasonable analysis. Their total is 180.

  You have: 50^6 / (50*2) / (1000/day)
  You want: yr
    * 427.79832

And wouldn't the odds really be: p = (1/49 * 1/48 * 1/47 * 1/46 * 1/45) for the balls to be drawn in order in one lottery?

The odds for them not to be drawn in order are then: 1 - p.

If there are 1000 lotteries per day, the odds it doesn't happen in a year are: ynp = (1-p)^(365 * 1000).

The yearly probability would be: 1-ynp.

If the balls can be drawn in any order, but they have to end as 6 ascending balls - then I think the probability is much higher, right?

No. 50^6 is about 938x bigger than (50 choose 6), which is the correct chance of picking 6 of 50 numbers.

> And wouldn't the odds really be: p = (1/49 * 1/48 * 1/47 * 1/46 * 1/45) for the balls to be drawn in order in one lottery?

This is not correct either. There are 45 sequential possibilities of winning numbers: 1-2-3-4-5-6, 2-3-4-5-6-7, ... 45-46-47-48-49-50. So "45 / (50 choose 6)", about 647 times more likely than that calculation for p.

Of course, if you start adding in other striking patterns like 2-4-6-8-10-12, the chance of a "fishy" draw becomes more and more likely.

It's very, very far, between non-replacement (~1.36x), order (5!=120x), the fact that final ball is out of 20 instead of out of 50 (2.5x), and the number of possible sequences which would be notable (at least 16x, but actually probably a fair bit more because recognizable patterns are pretty broad).

The upshot of which is that estimating at 50^6 is off by multiple orders of magnitude...

For 5 drawn in a specific order from a pool of 50, we're talking about 254251200 ways 5 numbers can be drawn, with a 0.0000004% chance.

In this case it was actually 6 numbers, which is about 10x less likely to match a combination (orderless) and 50x less likely to match a permutation (with order).

First definition. Let's count a sequence as ascending if the sorted set of drawn balls is ascending. E.g. 18,15,14,19,16 with a powerball of 17 would be counted as ascending.

I found that it was easiest to start with the odd one out and work from there. The powerball can be chosen in 20 ways. If the powerball is 1, then the sequence must start at 1 (1 combination). If it is 2, then the sequence can start with either 1 or 2 (2 combinations), 3 means 3 combinations, 4 means 4 combinations. For 5 and above, there are 5 combinations (the upper part of the range has no similar issue, since the normal balls go all the way to 50). We get a total of 85 combinations.

Now every combination of the 5 normal balls can be permuted in 5! ways, so we have 85 * 5! = 10200 ascending draws.

The total number of possible draws are 50 * 49 * 48 * 47 * 46 * 20 = 5085024000.

So the probability of an ascending draw is 10200 / 5085024000 ~= 2 in a million.

In probability, when you roam outside of the 0-1 (or 0-100%) range, you can guarantee with 775% certainty that something went wrong somewhere. ;)

You cannot have a probability P[X ≥ 1] >1.

There is no 'probability per year' (if it were meaningful year would have to be <1 for a >1 result anyway) - that time frame is built into 'X', it's 'inside' the probability.

It's been clear throughout this thread what was meant.

Probably something like reciprocal half-life; 775% / yr is about 1 / 1.5 month ie, happens about once every month and a half.

There’s still some chance it doesn’t happen in one year.

To the other folks ending up with some wild results, there is a basic checksum on probability: If you compute the probability of an event happening at greater than 100% you've borked something up.

Bug: They drew 5 balls from one pool of 50 with an independent draw from another smaller pool of 20, so you need ~~50 ncr 6~~ (50 ncr 5) * 20, not 50 ncr 5.

Nit: I would rephrase your answer that there is a 0.85% chance that it happens one or more times in any given year. There remains a (vanishingly small) chance that it happened on every single random draw during the year.

There's 6 possible triple dice combinations and there are 6^3 possible dice combinations, so it's 6/(6^3).

If you wanted to know the odds of rolling 3 dice in order, you could roll: 1, 2, 3 OR 2, 3, 4 OR 3, 4, 5 OR 4, 5, 6 - which is 4/(6^3) - which is not 6c3.

Why is it different with the lottery? Or did I get the dice wrong?

Or are you calculating that the balls can be drawn in any order?

The probability of guessing all 6 balls in a single lottery is 1 in (50 ncr 6). So, the probability of losing is 1 - 1/(50 ncr 6). The probability of losing every time is (1 - 1/(50 ncr 6)^(n_games), where n_games = 365 * 1000. Therefore, the probability of winning at least one game is (1 - (1 - 1/(50 ncr 6)^(n_games)).

But the main point was the methodology. (1 - (1 - chance_of_sequence)^(n_draws)).

Couldn't there be 1 2 3 4 5 6 AND 2 3 4 5 6 7 AND 3 4 5... Doesn't this give you 45?

You get 14 possible draws if the order the balls are drawn in matters (14-20 being the highest, with 20 drawn last), and 20 permutations if it does not (since any straight above 20-25 is not possible). The math is much different for drawing if the order matters though. There also would probably be some funky stuff going on for the higher straights where order doesn’t matter, since for a 20-25 straight the 20 ball must be the special ball. For a 19-24 straight either 19 or 20 must be, etc. Really you’re looking at calculating “Chance that the first five balls can create a straight with a number between 1-20, and then a 1/20 chance that straight actually happens.”

  2*(1/49)*(1/48)*(1/47)*(1/46)*365*1000*(45/50) = 12.9%

Edit: but if the numbers don't have to be drawn in order (e.g. 8-5-9-6-7 is OK), then the odds are much higher still:

  2*(4/49)*(3/48)*(2/47)*(1/46)*365*1000 = 344.5%

Most lotteries are order-insensitive, and typically present the results in ascending numeric order. The actual draw often happens in some order (e.g. numbered balls being drawn from a hopper), and it'd be even more unusual if the numbers were actually drawn in consecutive order, but drawing 5-2-3-1-4 would typically be presented as 1-2-3-4-5 and would still be remarked upon as unusual.

Doing this, your final figures should be 12.1% and 96.8%.

For the expected number of times it would happen though, you are right, it would be three or four times per year on average for the second case.

Edit: What I did was take all the possible combinations of 5 balls (5!) by the number of different sets that could be drawn (50-5, based on lowest number), over the total possible draws (50!). I think perhaps what that does it not account for overlap between sets (1-5 and 2-6), inflating the number somewhat, which is why it's a bit more than twice the probability than you got for any possible sequence of 5.

  function rn() {
      return parseInt(Math.random() * 100);
  }


  function drawing(){
      let result = [rn(), rn(), rn(), rn(), rn()]; 
      return result.sort();
  };


  function sequential(arr){
      for(let i = 0; i < arr.length-1; i++){
          if(arr[i] + 1 != arr[i+1]){
              return false
          } 
      } 
      return true
  }

  let counter = 0; 

  for (let i = 0; i < 10000000; i ++){
      if (sequential(drawing())){
          counter ++;
      } 
  }; 

  console.log(counter);

  16

According to the article, this event happened for South Africa Powerball, where 5 numbers are chosen out of 45, and 1 number chosen out of 20: https://en.wikipedia.org/wiki/South_African_National_Lottery...

That same Wikipedia page does some of the math for us: The chance of one combination being chosen is 1/42,375,200. So if we count all possible sequential combinations, N, we'll know that the chance of a single winning combination being sequential is N/42,375,200.

Say the powerball comes out as any number 6 <= M <= 20. There are 6 ways the numbers 1-45 could be picked such that M is part of the sequence. That's 90 ways total. If the powerball is 5, there are only 5 ways, same continuing down to a powerball of 1 where there's only 1 combination of numbers 1-45 where it could be part of the sequence. 90 + 5 + 4 + 3 + 2 + 1 brings us to 105 as our N.

So this single event had a probability of 105/42,375,200, or 1/403,573. This means that for similar lotteries one would expect to see a sequence after about 200,000 picks.

EDIT: If you only count events where the powerball is the high number, as happened this time, N goes down to 15, making the odds 1/2,825,013, so one would expect such a sequence after about 1.4 million picks.

1. How often does this happen? This is a question about expected value, and the answer could be anything zero or above.

2. What are the chances that this will happen within a year? This is a question about probability, and the answer must lie between zero and one. There is no such thing as "a 775% chance it would happen yearly".

Note that for N close to 0, 1-N is also a good approximation to 1-e^(-N).

For large N, it's generally more convenient to talk about the expectation rather than the probability of 0 hits—I'm sure many readers implicitly converted 775% to the expectation in their heads.

Most people cannot do this correctly; the most obvious interpretation of a "775% chance" is that it represents a 25% chance of seven occurrences and a 75% chance of eight occurrences, with no other possibilities.

The problem gets even worse when you have expectations less than one. If the expected number of occurrences is 80%, what are the odds of getting any occurrences at all? They're less than 80% as long as it's possible to have more than one occurrence.