4 ms·

there are 6 balls drawn in the example, which would drop that number down to .26%, which would be once every 400 years. I didn't check the rest of your math tho

by ouid 6y ago

there are 6 balls drawn in the example, which would drop that number down to .26%, which would be once every 400 years. I didn't check the rest of your math though.

onlyrealcuzzo 6y ago
Cool, as a rough approximation, it is close to (1/50)^6, right? But couldn't the first ball be any ball? That doesn't really matter. So wouldn't it just be 1/50^5 for 6 balls? 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?
- naniwaduni 6y ago
  > Cool, as a rough approximation, it is close to (1/50)^6, right? But couldn't the first ball be any ball? That doesn't really matter. So wouldn't it just be 1/50^5 for 6 balls? 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...
- jonahx 6y ago
  > Cool, as a rough approximation, it is close to (1/50)^6, 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.

by ouid 6y ago

there are 6 balls drawn in the example, which would drop that number down to .26%, which would be once every 400 years. I didn't check the rest of your math though.

onlyrealcuzzo 6y ago

Cool, as a rough approximation, it is close to (1/50)^6, right? But couldn't the first ball be any ball? That doesn't really matter. So wouldn't it just be 1/50^5 for 6 balls? 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?

naniwaduni 6y ago
> Cool, as a rough approximation, it is close to (1/50)^6, right? But couldn't the first ball be any ball? That doesn't really matter. So wouldn't it just be 1/50^5 for 6 balls? 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...
jonahx 6y ago
> Cool, as a rough approximation, it is close to (1/50)^6, 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.