DERIVATION  OF  CRYONICS  PROBABILITIES

 

Brook Norton;  27 Nov 93

 

Introduction:

Sections 1 - 6 present the background principles and nomenclature used in following Sections.

 

Each of Sections 7 - 13  correspond to an output data column in the spreadsheet.  These Sections calculate the odds of  events occurring, GIVEN that have you have already deanimated.   They do not including the possibility that you may be cured of old age prior to deanimation.  It has been arranged this way because I believe most people think in these terms when considering cryonics.  They say "What are my odds of  being reanimated if I'm frozen in liquid nitrogen?" Sections 7 - 13 address that question.

 

Section 14 looks at the bigger picture, by calculating the odds that you will defeat Death by either being cured of old age before deanimation or by being reanimated.  Section 14 addresses the question "What are my odds of  achieving indefinite lifespan by any means?"

 

All probability calculations used in this analysis are based on 2 fundamental rules given in Sections 1 & 2.

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Section 1:

Rule 1) This rule describes events that are independent of each other.  That is, if one event occurs, it does not affect the odds of another event occurring.  The probability that ALL  OF  THE  EVENTS  WILL  OCCUR  is the product of the probabilities that each event will occur.

 

Example:  Say that within a 20 yr period, the odds of a fire are 1/20, an earthquake 1/100, and a tornado 1/25.  The odds that ALL 3 will occur are:  (1/20)*(1/100)*(1/25) = .00002 or 1/50,000.

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Section 2:

Rule 2) This rule describes events that are mutually exclusive.  That is, if one event occurs, no other event can occur.  The sum of the probabilities of each event is 1.0.

 

Example:  Suppose you roll a die.  The probabilities regarding which number is rolled are:

#1(1/6) + #2(1/6) + #3(1/6) + #4(1/6) + #5(1/6) + #6(1/6) = 1.0.

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Section 3:

By combining rules 1) & 2) you can determine the odds that:  Given several independent events, what are the odds that NONE  OF  THE  EVENTS  WILL  OCCUR ?

 

Say that A, B, and C are the odds that each event will occur.  In example 1), A = 1/20, B = 1/100, C = 1/25.  For each event, there are 2 mutually exclusive outcomes; either the event occurs or it does not occur.  (odds that event occurs) + (odds that event does not occur) = 1.0.

 

Now say that A', B', and C' are the odds that each event will NOT occur.  A + A' = 1.0, and similarly for B and C.  It follows that A' = 1 - A.  In example 1), A(1/20) + A' = 1.0, and A' = 1 - A(1/20) = 1 - .05 = .95.

 

Now, using rule 1), we can say: the odds that NONE of the events will occur is the product of the probabilities that each event will not occur.  Odds that NONE of the events will occur = (A')*(B')*(C') = (1-A)*(1-B)*(1-C).  In example 1), odds that no occurrence of fire, earthquake or tornado will occur is (1-1/20)*(1-1/100)*(1-1/25) = (.95)*(.99)*(.96) = .903.

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Section 4:

It is of interest to know what the odds are that  ANY  EVENT  WILL OCCUR  AT  LEAST  ONCE.  From rule 2), (odds of any event occurring) + (odds of no event occurring) = 1.0.

 

Now, from above statement, and using result of Section 3, it follows that:

Odds of ANY (independent) event occurring = 1-(1-A)*(1-B)*(1-C).  In example 1), the odds that at least 1 of the 3 disasters will occur at least once in a 20 yr period = 1-(.95)*(.99)*(.96)=1-.903=.097.

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Section 5:  Single time frame and marching time frames:

 

If events occurring in 1 time period (or time frame) are independent of events occurring in other time frames, Section 4 can be used to determine the odds of an event occurring at any time during several time frames.

 

The odds that an event will occur at least once over the course of several time frames = 1 - (1 - odds of occurring in time frame #1)*(1 - odds of occurring in time frame #2)*(1 - odds of occurring in time frame #3)*(etc.).  In example 1), the odds that a fire will occur at least once in sixty years = 1-(1-1/20, for first 20 yr period)*(1-1/20, for second 20 yr period)*(1-1/20, for third 20 yr period) = 1-(.95)*(.95)*(.95)=1-.857=.103.

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Section 6:  Glossary of terms:

 

For convenience, the abbreviations in this section will be used in following derivations and discussions.

 

dead.bynow  - odds that patient will be destroyed by this time.

dis.1 to dis.5 - user input odds for each of 5 independent disasters.  Odds of occurrence in last 20 yrs.

dis.20 - odds that any disaster occurred in the last 20 yr time frame.

dis.bynow - odds that any disaster occurred at any time prior to now.

frozen - odds that patient will remain frozen till this time.

no.dis.20 - odds that no disasters occur in last time frame

no.tech.20 - odds that reanimation technology not developed in last time frame.

rean.20 - odds that patient will be reanimated in last 20 yr time frame.

rean.bynow - odds that patient will be reanimated by this time.

tech.20 - affordable reanimation technology developed in last 20 yr time frame.

tech.bynow - user input for: affordable reanimation technology available by this time.

tech&dis - reanimation technology and disaster occur in last time frame, either first

tech.after.dis - reanimation technology and disaster occur in last frame, disaster first

tech.first.20 - Reanimation technology is developed in last time frame prior to any disaster occurring in last frame.  Technology does occur.  Disaster may or may not have occurred.

tech.no.dis - reanimation technology developed and no disaster occurs in last frame

tech.prior.dis - reanimation technology and disaster occur in last frame, technology first

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Section 7:  dis.20 - odds that any disaster occurred in the last 20 yr time frame.

 

The user inputs the odds for each of 5 independent disasters.  The user specifies the odds that each disaster will occur in the last 20 yr time period.

 

dis.20 = 1-(1-dis.1)*(1-dis.2)*(1-dis.3)*(1-dis.4)*(1-dis.5)

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Section 8:  dis.bynow - odds that any disaster occurred at any time prior to now.

 

At time = 0:  You may be destroyed very shortly after deanimation due to such disasters as autopsy, deanimation in remote location, suspension procedure is inadequate to preserve necessary brain information, etc.  The odds of this occurring are a user input.

 

For time frames after time=0:  (from Section 4) dis.bynow = 1-(1- odds that any disaster occurred prior to last time frame)*(1-odds that any disaster occurred in last time frame)

= 1-(1-dis.bynow, at start of last time frame)*(1-dis.20).

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Section 9:  frozen - odds that patient will remain frozen till this time.

 

The patient will remain frozen if reanimation technology does not exist yet and he has not been destroyed.

(from Section 3) frozen = (1-tech.bynow)*(1-dis.bynow)

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Section 10:  tech.20 - affordable reanimation technology developed in last 20 yr time frame.

 

The development of reanimation technology can be described in terms of mutually exclusive events (see Section 2).  (odds that tech. was developed prior to last time frame) + (odds that tech. was developed in last time frame) + (odds that tech. will be developed after last time frame) + (odds that tech. will never be developed) = 1.0.  Changing nomenclature:

(tech.bynow, at start of last time frame) + (tech.20) + (tech. in future) + (tech. never) = 1.0

Restated another way, with first 2 terms combined:

(tech.bynow at end of last time frame) + (tech. in future) + (tech. never) = 1.0

Comparing above 2 equations we see:

(tech.bynow, at start of last time frame) + (tech.20) = (tech.bynow at end of last time frame)

or

tech.20 = (tech.bynow at end of last time frame) - (tech.bynow at start of last time frame)

 

For example, suppose the user input the odds of 20% (or .20) for reanimation technology occurring by start of last 20 yr period, and 30% chance of occurring by end of the 20 yr period.  The odds that reanimation technology was developed in last 20 yr period (tech.20) = .30 - .20 = .10.  You could say "The odds that the technology was developed by the end of the last time frame are 30%.  However, there is only a 10% chance that the technology was developed in the last 20 yrs, because there is a 20% chance that the technology was developed prior to the last time frame.

 

One must be careful to interpret this term (tech.20) correctly.  "Tech.20" is derived from the user input "tech.bynow" which is structured so that the user estimates each time frame without any knowledge as to whether the reanimation technology was or was not achieved in previous time frames.  "Tech.bynow" IS an estimate of whether reanimation technology has been developed at any point between time = 0 and the end of the last time frame.  "Tech.20" IS an estimate of whether reanimation technology was developed in the last 20 yr time frame.  "Tech.bynow" and "Tech.20" ARE NOT an estimate of whether reanimation technology occurred in the last time frame, given that the technology had not yet occurred at the start of the time frame.

 

An equivalent probability analysis could be constructed where the user did input odds of reanimation technology being developed in last 20 yrs, given that it had not been previously developed.  However, this analysis is not constructed that way.

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Section 11:  rean.20 (not to be confused with "tech.20") - odds that patient will be reanimated in last 20 yr time frame.

 

To be reanimated in last 20 years, the following 2 conditions must occur:

1) No disasters can have occurred up till the beginning of the last time frame.

2) Reanimation technology is developed in last time frame prior to any disaster occurring in last time frame.

 

From Section 1: rean.20 = (1-dis.bynow, at start of frame)*(tech.first.20);     call this equation A, or eqA

 

(Note that it was tempting to construct this argument where condition 1 above is: "patient is still frozen at beginning of the last time frame."  However, being frozen is not independent of reanimation technology, condition 2, and so a more complex approach would have to be used than that of EqA.)

 

From Section 2: Technology and disasters occurring in last time frame can be expressed in 3 ways (eqB, eqC, and eqD below) in terms of mutually exclusive events.

 

(tech.first.20)+(no technology developed in last frame)+(tech. and disaster occur in last frame, disaster first)=1.0

abbreviating:  (tech.first.20)+(no.tech.20)+(tech.after.dis)=1.0          eqB

 

(technology and disaster occur in last frame, technology first)+(technology developed and no disaster occurs in last frame)+(no tech. developed in last frame)+(tech. and disaster occur in last frame, disaster first)=1.0

abbreviating: (tech.prior.dis)+(tech.no.dis)+(no.tech.20)+(tech.after.dis)=1.0        eqC

 

(no.tech.20)+(technology and disaster occur in last frame, either first)+(tech.no.dis)=1.0

abbreviating: (no.tech.20)+(tech&dis)+(tech.no.dis)=1.0          eqD

 

Comparing eqB & eqC:  tech.first.20 = (tech.prior.dis)+(tech.no.dis)         eqE

 

Substituting eqE into eqA:  rean.20 = (1-dis.bynow, at start of frame)*[(tech.prior.dis)+(tech.no.dis)]  eqF

 

Comparing eqC & eqD:  tech&dis = (tech.prior.dis)+(tech.after.dis)      eqG

 

In eqG, it is given that reanimation technology and a disaster will both occur in the last time frame.  When they both occur, the odds are 50/50 as to whether technology or disaster occurs first.  Therefore:

tech.after.dis = tech.prior.dis           eqH

 

Substituting eqH into eqG:  tech&dis = (2)*(tech.prior.dis)

Rearranging:      tech.prior.dis = (.5)*(tech&dis)                  eqI

 

Substituting eqI into eqF:  rean.20 = (1-dis.bynow, at start of frame)*[(tech.no.dis)+(.5)*(tech&dis)]   eqJ

 

From Section 1:  tech.no.dis = (tech.20)*(no.dis.20)        eqK

 

From Section 2:  (no.dis.20)+(dis.20)= 1.0   or  no.dis.20 = 1-(dis.20)         eqL

 

Substituting eqL into eqK:  tech.no.dis = (tech.20)*[1-(dis.20)]                         eqM

 

From Section 1:  tech&dis = (tech.20)*(dis.20)       eqN

 

Substituting eqM and eqN into eqJ:

rean.20 = (1-dis.bynow, at start of frame)*{(tech.20)*[1-(dis.20)]+(.5)*(tech.20)*(dis.20)}

             = (1-dis.bynow, at start of frame)*[(tech.20)-(tech.20)*(dis.20)+(.5)*(tech.20)*(dis.20)]

             = (1-dis.bynow, at start of frame)*(tech.20)*[1-(.5)*(dis.20)]

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Section 12:  rean.bynow (not to be confused with "tech.bynow") - odds that patient will be reanimated by this time.

 

The odds of reanimation can be described in terms of mutually exclusive events (see Section 2).  (odds that reanimation occurred prior to last time frame) + (odds that reanimation occurred in last time frame) + (odds that reanimation will occur after last time frame) + (odds that reanimation will never occur) = 1.0.

 

Changing nomenclature:

(rean.bynow, at start of last time frame) + (rean.20) + (rean. in future) + (rean. never) = 1.0

Restated another way, with first 2 terms combined:

(rean.bynow at end of last time frame) + (rean. in future) + (rean. never) = 1.0

Comparing above 2 equations we see:

(rean.bynow at end of last time frame) = (rean.bynow, at start of last time frame) + (rean.20)

 

After many time frames, "rean.bynow" approaches a constant value.  This constant value is the odds that reanimation will occur.

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Section 13:  dead.bynow  - odds that patient will be destroyed by this time.

 

The destruction of the patient can be expressed in mutually exclusive terms (see Section 2):

At any given time, the patient must exist in 1 of 3 states: frozen, reanimated, or destroyed.

(remains frozen)+(has been reanimated)+(has been destroyed)=1.0

Abbreviating:  (frozen)+(rean.bynow)+(dead.bynow)=1.0

dead.bynow = 1-(frozen)-(rean.bynow)

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Section 14:  live - Overall odds that you will overcome old age and live indefinitely.

 

After many centuries, you can exist in only 1 of 3 states. (As time goes to infinity, odds of remaining frozen approach 0) (It is assumed for simplicity in this analysis that you aren't killed in an accident at any time after you are cured of old age or reanimated.)  The 3 states are mutually exclusive (see Section 2):

(cured of old age prior to deanimation)+(reanimated)+(destroyed)=1.0

Abbreviating:  (cured)+(reanimated)+(destroyed)=1.0  As time approaches infinity.

 

Rearranging:  (destroyed)=1-(cured)-(reanimated)                                Equation A  or  eqA

 

(cured) is a user input.

 

(reanimated) is the odds that you are not cured AND that you are reanimated. From Section 1:

(reanimated)=(1-cured)*(rean.bynow in last time frame)                       eqB

The subtle difference between (reanimated) and (rean.bynow) should be noted.  (rean.bynow) is the odds of being reanimated, GIVEN that you deanimated.  (reanimated) is looking at the bigger picture, at the odds you will be reanimated, taking into account you may never deanimate (since you might be cured of old age prior to deanimation).

 

Sub eqB into eqA:  (destroyed)=1-(cured)-(1-cured)*(rean.bynow in last time frame)       eqC

 

It is also true that as time approaches infinity:

(live)+(destroyed)=1.0   or

(live)=1-(destroyed)                                               eqD

 

Sub eqC into eqD:  (live)=1-1+(cured)+(1-cured)*(rean.bynow in last time frame)  or

 

(live)=(cured)+(1-cured)*(rean.bynow in last time frame)