>What are the odds of that happening?
At least 1

>>1561946
too broad a question, you will need to be more precise

>>1561994
>>1561946
>that happening
there is a dozen ways this question could be answered
the fpbp anon is not wrong: the odds of it happening is 100%, given that it's already happened P(A|A) = 1
if you're asking the odds of a three-way tie eventually happening in a gp qualifier, the answer is it will certainly happen, given enough time. once again, 100%
now you could be asking the odds that three pilots tie for first place in this specific race
the odds that three pilots tie for any place in this race
the odds that these three specific pilots tie for first place in any race that season
the odds of observing a three-way tie for first place in any gp qualifier
etc.

>>1561946
Going only from your image, there's 8 drivers with times ranging from 1:21.72 to 1:21.656. I'll add a little wiggle room both ways and say that these 8 drivers always finish that circuit in 1:21.000 to 1:21.700.

That means there's 700 possible times the drivers can score.

With 8 drivers, there's an 8 in 700 chance that one of them will score 1:21.072.
There's a 7 in 700 chance that a second one will score 1:21.072.
And a 6 in 700 chance that a third one will score 1:21.072.

The odds of this happening are 8/700 * 7/700 * 6/700.
So 3 in 3062500 or a 0.000098% chance.

>>1562059
>ranging from 1:21.72 to 1:21.656
*1:21.072, just a typo, didn't make it to the math

>>1562059
very close to 1 in a million then

>>1562059
>1:21.000 to 1:21.700
Pretend that I said 1:21.000 to 1:21.699

>>1562049
>>1562059
>>1562060
>>1562073
Thank you so much for your reply, I appreciate it.

>>1562059
bold of you to assume…
… that there aren't more than 8 pilots (there are)
… that all pilots' time follow the same distribution (why would it?)
… that that distribution would be the goddamn uniform distribution (c'mon, man)
and even if all these assumptions were justified, your maths is wrong
first pilot would have probability 1/700 of scoring 1:21.072
second pilot would have 1/700 of scoring the same
again for the third
and then you have to consider the probability of the other pilots each scoring otherwise
that would be 699/700, but OP may have asked for the odds of a three-way tie in first place, in which case you would need to have all other pilots completing the lap in more than 1:21.072 (about 628/700 probability for each)
finally, you would need to multiply by 8C3 for obvious reasons
again, that would be how you would calculate it, given the bass-awkward assumptions you made

>>1562059
https://www.formula1.com/en/results/1997/races/670/europe/qualifying/0

>>1561946
https://medium.com/@fizzi36/jerez-1997-qualifying-analyzing-the-three-way-tie-d8b35e9880e

>>1562131
Thank you for replying. This is very interesting, I never thought that maybe it was an error. This is a bit disappointing :(