While Qualifying for the European GP in 1997, Jacques Villeneuve, Michael Schumacher, and Heinz-Harald Frentzen got identical laps to the nearest one thousandth of a second (1:21.072). What are the odds of that happening?Here is a video:https://youtu.be/jSiJW0sw9SM
>>16947860well the gap between 1st and 4th is only 60ms, and the guy in first doesnt count. so i guess its the odds of getting three identical times within a 60ms window, so like, 1/500?
>>16947860>HakkinenDo angloids really?
>>16947860100%. It happened.
>>16947860you will see the certain sign of end timesand his name is fakengay
Here are the other scores btwhttps://www.formula1.com/en/results/1997/races/670/europe/qualifying/0
>>16948243UTF-8
Same best lap, same gap, same speed, even more impressive
bump
>>16947860>>16947931Here's how I worked it.Throw out the minutes and seconds and focus on the milliseconds.There are a thousand milliseconds, or strips, inside of 1 second.Assuming any strip has a chance of being picked, the first driverhits the 72nd strip with probability [math] 1\over 10^3 [/math].The second and third drivers, independently, hits the same stripand so the probability becomes [math] 1\over 10^9 [/math].The complement to this (at most 2 drivers hit the same strip)is [math] 10^9-1\over 10^9 [/math], thus the odds of threedrivers having the same best lap is [math] 1:10^9-1 [/math].
>>16957207Thank you!
>>16957207>>16957724Yes, of course.Things to note:>if the time is down to ten-thousandths of a second, it's more likely the 3 drivers would have different times>2 drivers with the same best lap times are more common than with 3, but still rare (recently, Verstappen and Russell, 2024 and 2025; both qualifiers)>if it took 48 years including 1950 to get this record, we expect to see it again in another 48, naturally...?>nope, this expectation has grown to 76 years (if the record doesn't happen this year) plus one year for each year without this record...so, we'll see in 2102 after lunch
>>16958443>>16958444Trip checked >we'll see in 2102I'll be 98 years old lol. Well, Murray Walker nearly made it to that age, maybe I can too. Again, thank you for your response.
I did the maths, not sure if I'm right thoughFor the sake of simplicity, let's assume every car is equal and has equally skilled drivers.For qualifying (https://www.formula1.com/en/results/1997/races/670/europe/qualifying/0), the time margin is 1:21.072-1:24.301, which is 3,229 milliseconds, thus 3,229 possibilities. The first one is not the anomaly, so we calculate the odds of someone *else* (Michael Schumacher in this case) hitting this specific millisecond, which is 1/3,229, or a 0.0003096934035305048% chance. This is the probability of a tie happening once.To calculate the probability of something happening twice (Heinz-Harald Frentzen tying it again), you multiply the probability with itself, if we assume the probabilities of any tie between any racer is equal (in reality, it isn't; different cars, different skill level, etc.). Thus, 1/3,2292 = 1/10,426,441, or a 0.000000959100042% chance. A ~millionth of a chance (when rounded).But we can calculate more accurately if we use the fastest laps of the *race* (https://www.formula1.com/en/results/1997/races/670/europe/fastest-laps), which widens the margin from 1:21.072-1:24:301 to 1:21.072-1:26:434. This now leaves us with 5,362 possibilities.Running the same formula...1/5,362 for a single tie (Jacques Villeneuve and Michael Schumacher), or a 0.00018649757% chance.1/5,3622 for the tie to happen twice, which is 1/28,751,044, or a 0.0000000347813457% chance. This is 3 hundred millionths.
>>16960474Sorry those didn't show the proper 2. It was supposed to be the squared 2, not the normal-sized one. I don't know if the formula is correct but the maths should be.
>>16960474Uniformity is a naïve estimate, with unrealistically high variance. For given driver and car, lap times are better thought of as an additive positive error to some ideal lap. I don't know how best to model that error, but I would start with a log-normal distribution. In any case, taking the fastest lap per driver and considering only the top [math]k[/math] reduces variance yet more.
>>16960474I don't think that's correct.The main issue is taking the fastest and slowest lap times to make the sample space.This makes the probability dependent on how big the gap is, which becomes clearif the slowest driver was within, say, 20 milliseconds from the lead and everyone elsein the middle [for example].Adding on to the gap with the actual race lap times changes the gap and also theentire context of the probability, since qualifiers have one driver on track at a timevs. the race where all drivers are on the track at once, and therefore can't go asfast as they want.Do check >>16957207, since it does not have the issues I mentioned.
Choose 8 random times in a 600ms window. The probability that at least the top 3 times are equal according to Monte Carlo it about 1:39000.
>>16960491>>16960553>>16961486Yeah I'm kind of retarded. Thank you all for your responses.>>16957207This is probably the best answer here.
>>16957207Where does the -1 come from?
>>16947860>oh baby a triple>0 results- 2 points
>>16962605The -1 is from the calculation of the complement.It's exactly the numerator from the probability [math] 1\over 10^9 [/math].Thus: [math] 1- {1\over 10^9} = {10^9 \over 10^9}-{1\over 10^9} = {10^9-1} \over 10^9 [/math].
>>16962605The -1 is from the calculation of the complement.It's exactly the numerator from the probability [math] 1\over 10^9 [/math].Thus: [math] 1-{1\over 10^9}={10^9\over10^9}-{1\over10^9}={10^9-1\over10^9} [/math].
>>16963960>>16963961Thanks. So about 1 in 1 billion? Jeezus...
>>16947860tfw
>>16947860I like how everyone throws a guess but no one mentions the instruments themselves.
