Mark:
ALASKAN GRAVES
14, 15, and 18
At 90% the answer is 14, at 95% the number is 15, at 99% the number is 18.
"Spreadsheets are wonderful!"
Your use of a spreadsheet is the same as Rembrandt with a brush, a tool to produce a beautiful result.
This is a basic binomial distribution problem. Given n people and a probability of death of p the probability of x deaths is n!/(x!*(n-x)!)*px*(1-p)(n-x). In this particular case the formula is 1000!/(x!*(1000-x)!)*.01x*.99(1000-x).
The table below shows the probability of specific numbers of death using this formula:
Code:
Numberof Deaths (x) Probability total=x Probability total<=x
0 0.00004317 0.00004317
1 0.00043607 0.00047924
2 0.00220019 0.00267943
3 0.00739322 0.01007265
4 0.01861375 0.02868640
5 0.03745311 0.06613951
6 0.06273711 0.12887663
7 0.08998657 0.21886319
8 0.11282407 0.33168726
9 0.12561333 0.45730059
10 0.12574021 0.58304080
11 0.11430928 0.69735009
12 0.09516152 0.79251160
13 0.07305328 0.86556489
14 0.05202279 0.91758768
15 0.03454173 0.95212941
16 0.02147955 0.97360896
17 0.01255845 0.98616742
18 0.00692759 0.99309501
19 0.00361663 0.99671164
20 0.00179188 0.99850352
The table shows that if 14 graves were dug then the probability of not running out would be 0.91758768 (the smallest number greater or equal to 90%). So they must dig 14 graves to have a 90% of not running out. To have a 95% chance they must dig 15 and to have a 99% chance they must dig 18.
Normal Distribution Solution
We can expect that the number of deaths to approximate a normal distribution curve. The mean is 1000*.01=10. The standard deviation is (1000*(.01)*(.99))1/2 =~ 3.1464 . Let g be the number of graves such that the probability of running out is less than 90%. Let d be the number of deaths.
Pr(d <= g+.5) = 0.90 (The 0.5 is added because the number of deaths must be an integer)
Pr(d-9.5 <= g-10) = 0.90
Pr((d-9.5)/3.1464 <= (g-10)/3.1464) = 0.90
Pr(Z <= (g-9.5)/3.1464) = 0.90
(g-9.5)/3.1464 = 1.28
g =~ 13.527
The number of graves must be an integer, thus we round up to 14.
RANDOM NUMBERS
If I had to guess, I would have guessed e, since it pops up everywhere and seems to be in the ballpark. I came up with a method (it involves computing the volumes of hypercubes with corners cut off) that yields e.
I didn't get past a 3-D cube, but extrapolated what seemed to be the pattern for the formula.
Eeeek! Why is life so unfair? Your guesses are better than my answers. Are you sure Shari is not giving you a helping hand?
The answer is e = ~ 2.7182818
Lets rephrase the question to a more general case, what is the expected number needed for the sum to exceed x, where x<=1?
Obviously at least one number will be needed. If the first number is less than x then another will need to be drawn with a lesser total needed to finish.
So E(x) = 1 + integral from 0 to x of E(y) dy.
Take the derivative of both sides...
E'(x) = E(x)
What function is a derivative of itself? E^x of course! So the answer is e^1 = e.
caravan > raven
"Try's wishin' he was Orville Reddenbacher"
I learned of his death when I had CNN on TV in the background. Bernard Shaw came on and reported Orville Reddenbacher had died. In his lifetime Orville had found a way to grow better Corn to make better Popcorn. He died Mid-September, 1995 of natural causes in his California condominium. He choked to death on a competitor's Popcorn while entertaining several bikini clad women in his hot tub. In keeping with his family's wishes, his remains were micro waved on high for three minutes.
What! Three minutes, I would hardly be warmed up. Plus the girls in the hot tub seem a trifle overdressed. However, there are worse ways to go.
Shari wrote of Jane Austin novel "Emma". I could read her books all day, such polite gentle times.
ABCDE * 4 = EDCBA. Solve for A,B,C,D, and E where each is a unique integer from 0 to 9
