Work out the number of throws in a row giving a specified number, say 6, to give the odds of 1 in 10^3000, which was the probability worked out in the previous thread for evolution of the wing to occur independently twice.
1 throw has a probability of 1 in 6.
2 throws have odds of 1 in 6^2
.
.
.
n throws have odds of 1 in 6^n
Hence 6^n=10^3000.
n log 6 = 3000 log 10
n = 3855
So the chance of this happening is the same as throwing 3855 sixes in a row
How long does it take to roll a dice? Lets say I takes 5 seconds. We have a person that can do this non stop for as long as you like. How long would it take this person the roll the dice to be sure they were able to get 3855 sixes in a row.
If it takes 5 seconds to throw a dice, you could throw it 3855 times in
3,855 x 5 = 19,275 secs = 5.4 hours.
How many sets of the 3855 rolls will I need. Since the probability I'm trying to model is 1 in 6^3855, I need to roll 6^3855 sets.
How long is that going to take? It will take 5.4 hrs x 6^3855 = a number too big for my calculator.
Change to base 10.
10^x = 5.4, so x = log 5.4 / log 10 = 0.732
And 10^0.732 x 10^3000 = 10^3000.732 hrs
I now need to throw the dice continuously for that period of time. How many years is this.
1 year is 24 x 365 = 8760 hrs = 10^3.9 hrs
10 years = 87,600 hrs = 10^4.9 hrs
100 years = 876,000 = 10^5.9 hrs
1000 years = 8,760,000 hrs = 10^6.9 hrs
This is getting boring. Lets go for a million years
1,000,000 = 8,760,000,000 hrs = 10^9.9 hrs
A billion years
1,000,000,000 = 8,760,000,000,000 = 10^12.9 hrs
Notice how slowly the index is changing.
Remember I need to get to 10^3000.732 hrs
How about a billion billion years
1,000,000,000,000,000,000 = 8,760,000,000,000,000,000,000 hrs = 10^21.9 hrs
I just remembered. The universe is only about 14 billion years old ie. 14,000,000,000 years. I ran out of time and am no where near the number of throw I need to be sure that I get the desired outcome.
The number of years required is going to be:
10^3000.732 / 10^3.94 = 10^(3000.732-3.94) = 10^2997 approximately.
I will show this number in the window below.