(~~~ :rl (-)ReLoad reset d:wipe 'Gott file:load.package ; (~~~

# Gott

This program takes

• \$g\$ in \$]0, infty[\$, a random time gone since something began.
• \$p\$ in \$]0, 1[\$, the portion of time that you consider to have elapsed.

and produces the interval in which that something should end.

The method is taken from

William Poundstone; The Doomsday Calculation; Little, Brown Spark, 2019

The relevant part of this book can be read as

Amazon preview https://www.amazon.com/Doomsday-Calculation-Equation-Transforming-Everything-ebook/dp/B07J4WCSMR

If you prefer an audio introduction, watch on YouTube

The Doomsday Calculation: Book Trailer https://youtu.be/jr693Q6M8OY

Also, an interview with the author is available as Michael Shermer with William Poundstone — The Doomsday Calculation (SCIENCE SALON # 76) https://youtu.be/M0tHz4BdrvA

In pp.14-18 of Poundstone's book there is a description of a prediction J. Richard Gott III made regarding the fall of the Berlin wall.

The input data are

• In 1961 the wall was built.
• In 1969 Gott visited the wall, which he considers that
to have been a random moment in time. • He proposes to make a prediction with a 50% level of confidence,
which is equivalent to saying that at least 25% of the total duration is assumed to have elapsed.

so that

gone is 1969 - 1961 = 7 years.
portion of the time gone is a half the 50% or 1/4.

which produces the interval \$[3, 24]\$, which translates to

• There is a 50% chance that the wall falls sometime between
\$1969 + 3 = 1972\$ and \$1969 + 24 = 1993\$ .

The wall actually fell in 1989.

Since the method is controversial, read the book before applying it to your personal problems. Pay attention to the randomness of gone and time scale invariance.

# Program

Consider something of interest that began at time start. Take the origin 0 of the time axis to be a random time now. Assuming that the end does come, it has to be within finite time from now. Let that maximum time be the unit \$1\$. Since the maximum time is unknown, the time unit 1 is unknown in any time scale.

```    now                                 maximum time      |                                       | ...--+---------+---------+---------+---------+--...--> time      0                                       1 ```

Consider the \$100 , q\$% confidence interval within which the period ends.

```                      100 q %     now   |<--- confidence interval --->|   max      |    |                             |    | ...--+----v****+*********+*********+****v----+--...--> time      0  (1-q)/2                   1-(1-q)/2  1 ```

Let \$p := (1 - q)/2\$. The *end falls somewhere within the confidence interval \$[p, 1-p]\$ with probability \$q\$.

```    now   |<------------ q ------------>|   max      |    |                             |    | ...--+----v****+*********+*********+****v----+--...--> time      0    p                            1-p   1 ```

The end will come with probability 1 sometime after 0, but with probability \$q\$ only after \$p\$ and before \$1-p\$. This means that with probability \$q\$ the time elapsed between start and now equals \$p\$.

Let the time gone be \$g :=\$ now \$-\$ start. This quantity comes with a time scale since now and start are measured in some unit such as days or years. Now let \$ell\$ be the time corresponding to \$p\$ in the scale of \$g\$. Then \$ell/g = p/(1-p)\$ so that \$\$ ell = frac{p}{1-p} , g \$\$

Similarly, the time corresponding to \$1-p\$ in the scale of \$g\$ is \$\$ u = frac{1-p}{p} , g \$\$

Now the end will come neither until the next moment of *now, which is 0, nor after the earliest end within the confidence interval which is at \$p\$.

To sum up, with \$100 , q\$% confidence \$l := g , r\$ and \$u := g / r\$, with \$r := p/(1-p)\$.

```start          l                             u   |            |<------------ q ------------>|   V            |                             | --+--...--+----v****+*********+*********+****v----+--...--> time           0    p                            1-p   1 ```

Implementing this method, I make a function called *gott which takes start, now, gone and returns l and u, where

l: shortest time from now that the event may take place
u: longest time from now that the event may take place

both with \$100 , q\$% confidence.

Helpers.

~~~:f:rot-       (-_abc-cab) f:rot f:rot ; :n:to-f       (n-_-n) n:to-float ; :.            (-)                    dump-stack   #0 f:depth  lt? [ nl 'f_  s:put f:dump-stack  ] if   #0 f:adepth lt? [ nl 'fa_ s:put f:dump-astack ] if ; :f:reset      (-__..-)        f:depth  [       f:drop ] times ;   :f:areset     (-__-__..-)     f:adepth [ f:pop f:drop ] times ;   :r            (..-__..-__..-) reset f:reset f:areset ;   :f:complement (-__n-n)_1-f .1. f:swap f:- ; ~~~

Here is the program.

~~~{{     :f:short      (-__p-r)_r=p/(1-p)_where_p=portion,_p=<1/2     f:dup f:complement f:/ ;   :f:lo         (-_gp-s)_g=gone f:short f:* ;   :f:hi         (-_gp-l)        f:short f:/ ; ---reveal---   :f:future.lo  (-__ngp-l)_n=now f:lo f:+ ;   :f:future.hi  (-__ngp-h)       f:hi f:+ ;   :f:future     (-__ngp-lh)       #3 [ f:dup f:push f:rot- ] times f:future.lo       #3 &f:pop                  times f:future.hi ; }} ~~~

## Berlin wall

This problem has been taken from Poundstone's book. In pp.14-18 there is a description of a prediction J. Richard Gott III made regarding the fall of the Berlin wall.

### 50% confidence

The input items are:

• .1961. *start : floating time when it began
• .1969. now : floating present time
• .0.25 portion : floating portion gone, between 0 and 1

It is crucial that now may be considered a random moment in time line after start. This is a special case of the Copernican principle.

1961 is when the wall was built. 1969 is the time in which Gott visited the wall. He considers this point to be a random time in the period of existence of the wall. 0.5 is the confidence level; the prediction is made so that there is a 50% chance that the end of the wall will fall within the time interval to be produced below.

~~~r 'Berlin_wall_1/2 s:put nl #1969 n:to-f (now)       f:dup #1961 n:to-f  (now_now_start)   f:-         (now_gone)   .1. .4. f:/         (now_gone_portion)   f:future . nl nl ~~~

gives

f 1971.666667 1993.000000

meaning that there is a 50% chance that the wall will fall sometime between 1972 and 1993.

"His 1967 prediction was that there was a 50 percent chance that the wall would stand at least 2.67 years after his visit but no more than 24 years." (p.16)

### 95% confidence

To recalculate with a confidence level of 95% as in p.19 of the book,

~~~r 'Berlin_wall_95% s:put nl #1969 n:to-f (now)       f:dup #1961 n:to-f (now_now_start)   f:-          (now_gone)  .0.025             (now_gone_portion)   f:future . nl nl ~~~

which gives

f 1969.205128 2281.000000

or between 1969 and 2281; reasonable but uninteresting.

## Diana and Charles

Now try the relationship duration between Diana and Charles that comes as the first example in the book, p.3.

~~~r 'Di_&_Charles s:put nl .1993 f:dup          (now_now) .1981 .7 .12 f:/ f:+ (now_now_start)     f:- (now_gone) .0.1                 (now_gone_portion) f:future.lo . nl nl ~~~

giving

f 1994.268519

or after 1994. future.hi is irrelevant, considering the couple's life span. Note that the value of Gone is set to 0.1 rather than 0.05, even though the level of confidence is stated to be 90%.

Regarding this result the book says

"Gott's formula predicted a 90 percent chance that the royal marriage would end in as little as 1.3 more years."

which is misleading: if a 90% confidence level is assumed the split will not take place at least until 1.3 years from "now."

"The split was formalized on August 28, 1996."

## The Third Reich

Poundstone book p.82.

• 1934-09 The Third Reich proclaimed.
• 20 months before, Hitler rised into power.
• 95% confidence.

~~~r '3rd_Reich s:put nl .1934 .9 .12 f:/ f:+ f:dup f:dup (now_now_now) .20 .12 f:/ f:-                  (now_now_start)     f:- (now_gone) .0.05 .2. f:/                    (now_gone_portion) f:future . nl nl ~~~

f 1934.792735 1999.750000

"A Copernican would have predicted the Nazi state to survive somewhere between another two weeks and another sixty-five years (at 95 percent conficence). The Third Reich lasted another eleven years."

# Personal applications

Dates are in YYMMDD.

## 190816--191126 UL project

190726 Brasilia Time (BRT)     Project proposal from a software company FS to      a chemical company UL, both in Brazil, upon a request from UL. 190815 Received a message from FS that UL has     not yet made a decision.     19 days have passed since the proposal.

~~~r 'UL_BR s:put nl .0. .19. .0.2 (now_gone_portion} f:future . nl nl ~~~

f 4.750000 76.000000

So there is an 80% chance that UL will not give its decision within 5 days.

190822 No notice yet, as predicted.

Since 76 days are about 2.5 months, the chance that a notification will arrive by the end of November is 80%.

191112 Project cancelled due to a change in market prices.

This is also as predicted.

## 190816 HE worker evaluation

190722 HE, a Japanese company, asked me to comment on     its employee's work, so they can make a decision on her. 190724 HE sent me related data. 190730 I turned in my report to HE. 190801 HE asked me to wait for their decision. 190816 Today.

~~~r 'HE_on_an_employee s:put nl .16. f:dup (now_now)   .-1. (now_now_start) f:-        (now_gone)  .0.2 (now_gone_portion) f:future . nl nl ~~~

f 20.250000

So there is an 80% chance that HE will not send me their decision before 190820.

190822 Received HE's decision on its employe.

This is as predicted.

## 190922 War Japan

~~~r 'War_JP s:put nl .2019. f:dup .1945. (now_now_start) f:- (now_gone) #2 [ f:dup-pair f:push f:push ] times (fa:gone_now_gone_now) .0.05            (now_gone_portion) f:future.lo f:pop f:pop .0.1 (now_gone_portion) f:future.lo f:pop f:pop .0.2 (now_gone_portion) f:future . nl ~~~

f 2022.894737 2027.222222 2037.500000 2315.000000

There are 90, 80, and 70% chances that Japan will not go into a war until 2023, 2027, and 2038, respectively.