Math rules!

Earlier this month I posted that I had never missed running on May 9th and 12th over all the years I've been doing this (1979-Present).  I wondered what the odds were for that.  Ken Tripp, the "stat man" did the math while riding the commuter rail home one evening.  I have no idea of any of thisis true or not but damn it is cool.  Great stuff!

From Ken:

You have been running for 13,310 days.  Of those days, you have missed 1,685 days.  If the days are independent and random (which they will not be), then the probability of missing any day is 0.127 (12.7%percent chance).  This also means that you have an 0.873 probability of running.

 So, the probability of running on May 9 is 0.873 (days are independent so date doesn’t matter).  The probability of running May 12 is also 0.873.  The probability of running May 9^th AND May 12^th is 0.837x0.837=0.763  So the probability of you running on May 9^th and May 12^th is 76.2%.  Now for over 37 years, it is still multiplicative like the 2 day calculation.  The prob(running on May 9^th and May 12^th for 37 years)=0.763^37=4.46584E-05( 0.004% or 4 out of 1000).  Clearly, it is a pretty small probability for this to occur.

Going forward, the probability of you running a complete year is 0.827^365=1.22146617E-43.

 

*Probability lesson:  If you flip a coin twice, what is the probability of getting 2 heads.  The possible outcomes are HH, HT, TH, or TT.  The prob of 2 heads = 1 (HH)/4(total number of possible outcomes).  This problem could be written as Prob(Head on Flip 1) and Prob(Head on Flip 2)=1/2*1/2=1/4.