Re: Re:
roundabout said:
Alex Simmons/RST said:
Hugo Koblet said:
Alex Simmons/RST said:
I'll grant that while operating with the dreaded Man Flu I may have made a bit of a probability calculation error but I don't think you guys have got it right either. I'm much closer than you are...
If you had been correct, then given the actual weather at the race over the last 20 years, how likely is it there will be a race day of 40C or more at any given race?
Here are the actual records of maximum temperatures in Adelaide for each and every day the TDU has been raced since 1999 (except for today's final stage since that max isn't available yet - but it's not going to be 40C+):
Amazingly </sarcasm mode>, the race has not suffered super hots days all that often, but don't let the facts spoil a good misinformation campaign.
Sorry but you are way off. The calculations I presented are correct, there's nothing to discuss. The question you're asking in this post is another calculation which can be solved by calculating the standard deviation.
Your calculations might be correctly performed but since they don't represent reality then something is wrong.
The actual prevalence of super hot days during the TDU is far less frequent compared with your suggested prediction. 20 years and 112 days of weather data is not an insignificant sample.
A 40C+ day has occurred on just 5/112 days (4.5%) and during 3 of the 20 tours (15%). That's way less frequently than you have suggested.
The average max temp on race days is right on the January average (just marginally higher in fact), so it's not like we have some kind of strange data bias going on.
So what are the chances this represents 20 years of below average temperature anomalies versus incorrect assumptions used in your calculations?
I could be wrong, since I don't do these things for a living, but as I remember it, this
If the premise (2 days in January are "super hot") then the chance of at least one "super hot" day in a 6-day race is (1-(29/31)^6)*100 = 33%.
Should be changed to
If the premise (2 days in January are "super hot") then the chance of one day being "super hot" in a 6-day race is (1-(29/31)^6)*100 = 33%.
If that is the formula used.
Edit: but maybe that was the meaning intended
No, what you're describing here is the likelyhood of
exactly one day being super hot - not the likelyhood of
at least one day being super hot.
If you want to calculate the probability of at least one day being super hot, then my calculation is correct. If you want to calculate the probability of exactly one day being super hot, you have to use binomial distribution.
With the premise presented (ie 2/31 chance of a day in January being super hot) you get the following:
Chance of 0 days being super hot: 67%
Chance of 1 day being super hot: 27,8%
Chance of 2 days being super hot: 4,8%
Chance of 3 days being super hot: 0,44%
Chance of 4 days being super hot: 0,023%
Chance of 5 days being super hot: 0,00063%
Chance of 6 days being super hot: 0,00000072%
Adding up these numbers you get 33% of at least one day being super hot (or you could just substract 67% (the chance of 0 days being super hot) from 100%.