Changing Lanes
February 2nd, 2010 at 6:11:55 PM
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If you changed lanes in your car without looking at your blind spot,
what are the chances of hitting the car next to you if you:
Change Lanes 10000 times
There is a car next to you 20% of lane changes
what are the chances of hitting the car next to you if you:
Change Lanes 10000 times
There is a car next to you 20% of lane changes
February 2nd, 2010 at 6:47:33 PM
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Are you saying you NEVER check your blind spot? Please get the hell off the road!
At any rate, the simple math suggests that you've an accident 2,000 times - and still haven't learned your lesson!
But seriously, is the math really that simple? Instinct tells me that it's not.
Even if it is, there's a variable you didn't mention.
How often will the other driver react fast enough to avoid the collision?
At any rate, the simple math suggests that you've an accident 2,000 times - and still haven't learned your lesson!
But seriously, is the math really that simple? Instinct tells me that it's not.
Even if it is, there's a variable you didn't mention.
How often will the other driver react fast enough to avoid the collision?
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February 2nd, 2010 at 8:08:07 PM
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I always check my blind spot!!
Assume that the other driver avoids the accident 70% of the time.
Assume that the other driver avoids the accident 70% of the time.
February 3rd, 2010 at 4:36:24 AM
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Ok, so you change lanes 10000 times, and there is a 20% chance there is a car next to you, and a 70% chance that they will avoid you.
Are you asking:
Number 2 is easy. 10000 x 20% x 30% = 600 accidents is the most likely number.
Notice that the answer is 6% of the number of iterations.
Number 1 has tiny bit more math involved.
The chances of having NO accident is...
(0.8 + (0.2 x 0.7))10000
=
0.9410000
(Notice that the base number is 94%, or 100% - 6%, the answer to number 2.)
=
~1.899 x 10-269
That is, 80% chance of no car next to you, plus 20% x 70% chance that there is a car, but they swerve (ie. a 94% chance of no accident), iterated 10000 times.
Take that number from 1, and you have a 99.99999...% (with about 269 nines after the decimal) chance you will have at least one accident.
Does this sound reasonable?
Are you asking:
- How many times will you have at least 1 accident in 10000 lane changes?
- Or how many accidents will I have total?
Number 2 is easy. 10000 x 20% x 30% = 600 accidents is the most likely number.
Notice that the answer is 6% of the number of iterations.
Number 1 has tiny bit more math involved.
The chances of having NO accident is...
(0.8 + (0.2 x 0.7))10000
=
0.9410000
(Notice that the base number is 94%, or 100% - 6%, the answer to number 2.)
=
~1.899 x 10-269
That is, 80% chance of no car next to you, plus 20% x 70% chance that there is a car, but they swerve (ie. a 94% chance of no accident), iterated 10000 times.
Take that number from 1, and you have a 99.99999...% (with about 269 nines after the decimal) chance you will have at least one accident.
Does this sound reasonable?
-Dween!
