# Estimating the age of the Universe, and getting it all wrong (1 Viewer)

#### The_Doc_Man

##### Immoderate Moderator
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Here is my proof that it is a straight 50/50 shot. In this PDF of a spreadsheet, I have done 1/3 of the exhaustive analysis. Here's how you read it.

The "Car" column shows which door hides the car. I have started from door #1. If I did door #2 and door #3, I would have two more groups of the same kind, 18 rows of which 10 are precluded by rules.

The "Pick" column shows which door the player picks. I have equal numbers of picks for cases 1, 2, and 3.

The "Reveal" column shows which door Monty reveals when he offers the option to switch. I have equal numbers of picks there, too.

The "Switch" column is 1 for switching, 0 for not switching.

The "Win" and "Lose" columns are simple - 1 in the Win column or Lose column but not both.

But... the game that Monty plays has rules. It can never occur that if you picked the car, he would reveal the car and ask you to switch. So in the game, that row can't happen. It ALSO can never occur that he would reveal your pick first and then invite you to switch. He would ALWAYS give you the chance to switch before revealing your door, so the rows where he immediately reveals your pick are ALSO precluded by the rules.

The bottom line is just that you have equal chances for all possible scenarios, of which there are 8 when the car is behind door #1. The same situations would apply for car behind #2 or #3.

#### Attachments

• MontyHall.pdf
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#### Jon

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I remember testing the switching hypothesis empirically using a set of playing cards. In a very short period of time, you will find that switching is the best strategy. Doing so helps you confirm or deny the hypothesis. If anyone else here cares to give it a try, we can have it confirmed by peer review!

Once you are confronted with a disparity between the theory and empirical reality, you then have to have a closer look at your hypothesis.

#### JonXL

##### Member
A human lifespan in the UK averages out to about 80. So, to get the best estimate of a human age, the most accurate prediction with the minimum error, given no other information, would be 80/2, or 40. If you were placing bets, and you said someone was 20, with no other information, and I said 40, I am more likely to be closer to the true age than you are, since more of the population would be closer to my estimate. Make sense?

Not quite. For an issue like that, the average age is not really relevant; what you want instead is the most common age. For the UK, that age is 55. You are more likely to be accurate guessing 55 as an age than you are guessing any other age.

In 2020, there were approximately 938.7 thousand people who were aged 55 in the United Kingdom, the most of any age group. The next most common age was 53, and 54 respectively, with 934.7 and 932.6 thousand people this age in 2020. - https://www.statista.com/statistics/281174/uk-population-by-age/

#### Jon

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Not quite. For an issue like that, the average age is not really relevant; what you want instead is the most common age. For the UK, that age is 55. You are more likely to be accurate guessing 55 as an age than you are guessing any other age.
Why is the average age not relevant and why would the most common age make the minimum sized error?

This is mine:

Which looks like it will give the minimum error, on average?

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#### The_Doc_Man

##### Immoderate Moderator
Staff member
I remember testing the switching hypothesis empirically using a set of playing cards. In a very short period of time, you will find that switching is the best strategy. Doing so helps you confirm or deny the hypothesis. If anyone else here cares to give it a try, we can have it confirmed by peer review!

Once you are confronted with a disparity between the theory and empirical reality, you then have to have a closer look at your hypothesis.

Ah, but there is a catch: How you count the "1st guess was right" case is crucial to the math. That case has TWO interpretations and one of them is NOT consistent with the actual game that Monty Hall played. I did an exhaustive enumeration for the "car is in door #1" case and the "WIN" result was 4 to 4 (win vs. lose) out of eight playable scenarios and ten that were barred by the rules of the game. Of those cases, 2 wins were due to staying and 2 were due to switching. Look carefully at my PDF of a spreadsheet in that post because all of the answers you need are right there. And of course you can do the same for door #2 or door #3 as holdingthe car, where you just have to shuffle the results. It will be 12 to 12 for stay vs switch. Further, the numbers for "lose" are ALSO even - another indication that it is a 50-50 shot.

Here is the crux: In the case where you picked right on the first try, he will STILL give you a chance to switch - but in the "1st guess right" case he has two possible picks to reveal as a non-winner. Do you count that IN THE TELEVISED GAME as 1 pick or 2 picks (both of which are winners if you DON'T switch)? If you don't count that correctly, one of the "wins" goes away and therefore skews the results. But strict interpretation of the possible cases says that is TWO wins because of the different choices.

The problem, of course, is caused by incorrectly mapping the word problem to the math. Those people who were saying there is a mathematical edge to switching were solving the wrong problem.

#### Jon

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I am a little unsure what you mean by your explanation. All I know is that when I followed the rules of the game set forth for the Monty paradox, my empirical test confirmed that you should switch. Only after I did that, I "got it" and understood why. Beforehand, I was as puzzled as anyone else.

I've got many hypothesises that turn out to be false after I've been bitten by a reality sandwich. When you have "a priori" reasoning, you need to confirm it with empirical testing. Drug trials do this for a reason: the hypothesis could be false.

I'm not sure if you are saying that an empirical test would say switching doesn't make a difference, or that there is something faulty or ambiguous with the wording in the Monty game.

#### JonXL

##### Member
Why is the average age not relevant and why would the most common age make the minimum sized error?

View attachment 97258

This is mine:

View attachment 97259

Which looks like it will give the minimum error, on average?

Your question is basically "which line looks like it's closest to the middle?", but if the aim is to figure out which guess is more likely to be correct, finding the middle doesn't help - what we want is the mode. A better analogy:

Suppose a barn wall covered in square tiles of different colors - red, green, yellow. There are 850 red, 500 green and 700 yellow tiles placed randomly on the wall covering it entirely. You're friend is going to close his eyes and take a shot at the barn wall. Meanwhile you've been given £1mil to wager as to which color tile he will hit.

For which color do you place your bet?

ABE: I smell the need for a VB-based simulation...

#### The_Doc_Man

##### Immoderate Moderator
Staff member
I'm not sure if you are saying that an empirical test would say switching doesn't make a difference, or that there is something faulty or ambiguous with the wording in the Monty game.

Because of an ambiguity in the wording of the game, there is a valid way to understand the game where switching is equal in expectation to not switching - i.e. a 50-50 shot, not a 2/3 preference to switching.

Here are the rules that make this a 50-50 shot.
1. Monty will ALWAYS give you a chance to switch even if you picked the winning door on the first try.
2. Monty will ALWAYS reveal a "zonk" door and then allow you to stay or to switch with the remaining door.
3. Corollary of #1 - Monty will never reveal the prize door before giving you a chance to switch.
4. Corollary of #2 - Money will never reveal the door you picked initially.
5. Special case: If you picked the prize on the first try, Monty will randomly reveal one of the two doors, both of which are zonks. But he has no preference as to which of the two zonks he will reveal.

#### Jon

##### Access World Site Owner
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Your question is basically "which line looks like it's closest to the middle?", but if the aim is to figure out which guess is more likely to be correct, finding the middle doesn't help - what we want is the mode. A better analogy:

Suppose a barn wall covered in square tiles of different colors - red, green, yellow. There are 850 red, 500 green and 700 yellow tiles placed randomly on the wall covering it entirely. You're friend is going to close his eyes and take a shot at the barn wall. Meanwhile you've been given £1mil to wager as to which color tile he will hit.

For which color do you place your bet?

ABE: I smell the need for a VB-based simulation...
I think there is a big misunderstanding behind this so let me close the gap for you.

I'm not trying to estimate the age with the highest frequency. Instead, I am trying to get the most accurate guess.

Consider this thought experiment:

Let us say that the chart I posted above is the same, except for one entry: age 79. Let us say age 79 has the highest frequency, or the highest number of people at that age. If you were trying to estimate the age of someone in the entire group of people, you are likely to be farther from the truth if you choose someone at the extreme end, or in this case age 79. Your average error will be smaller if you take the average age, around 40. Does that make sense?

If you were trying to get the highest likelihood of being 100% correct in guessing someones age, then you would say age 79, but that is completely different than trying to have a best guess that minimises your errors.

The example of 3 different colours of tiles are not analogous because you cannot get an average of colours like you can of ages. One is numerical, the other isn't.

Edit: An extreme case might illustrate the point. Say something can be 1 million years old. And there are a few more items that are 1 million years old than all the other ages. Your best guess would not be 1 million years old! It would be the half way point.

#### Jon

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Because of an ambiguity in the wording of the game, there is a valid way to understand the game where switching is equal in expectation to not switching - i.e. a 50-50 shot, not a 2/3 preference to switching.

Here are the rules that make this a 50-50 shot.
1. Monty will ALWAYS give you a chance to switch even if you picked the winning door on the first try.
2. Monty will ALWAYS reveal a "zonk" door and then allow you to stay or to switch with the remaining door.
3. Corollary of #1 - Monty will never reveal the prize door before giving you a chance to switch.
4. Corollary of #2 - Money will never reveal the door you picked initially.
5. Special case: If you picked the prize on the first try, Monty will randomly reveal one of the two doors, both of which are zonks. But he has no preference as to which of the two zonks he will reveal.
Doc, I am finding that far too confusing for me to understand. Can you refer to the specific bit of wording in the Monty paradox that is ambiguous? If you can narrow it down to that one item or few words (assuming there is only one), then I can see why you think it is ambiguous.

#### Cronk

##### Registered User.
Statistically, you can calculate a probability that the actual average age of a mass of people (say UK population) by determining the age of each in that random sample. (Census is a 100% or near enough sample).

Jon, your method of taking a sample of guesses about the age of the universe is akin to sampling a number of guesses about the age of a single person in a windowless room from the outside. If you average the guesses or get the mid point of a bell curve, it's still based on a guess.

#### The_Doc_Man

##### Immoderate Moderator
Staff member
Jon, I don't know to make it any simpler. Look at the PDF I attached earlier. You will see that it is exhaustive for the "car is behind curtain #1" case. It has every combination of pick, reveal, and switch that you can make - and it shows that, based on the rules I stated, there is no preference for stay/switch. You can do the same thing for the other cases (car behind door 2 or car behind door 3) and the concepts will remain the same.

The key to understanding this issue (using my viewpoint) is that there is a statement folks usually make saying that the chance that the car is behind the door you didn't pick suddenly shoots up to 2/3 because the door that got revealed becomes 0 probability. The fallacy is that you cannot asymmetrically redistribute probabilities without cause.

When you have this problem, at ALL TIMES the probability of the car being behind a door is 1 (by definition of the problem). So the sum of the probabilities must equal 1. After you pick a door but it isn't revealed yet, probabilities remain as they were, 1/3 per each of 3 doors. The fact of you having made a pick DOES NOT affect the probability of the car's location.

Now Monty opens a door and shows a zonk. The probabilities MUST still add up to 1 because there IS a car back there somewhere, so the higher math types say that all of the probability that vanished from the revealed door must have gone to the other doors. It is here that mathematicians do it wrong.

The door you picked has NOT been revealed yet. One other door hasn't been revealed yet. But if you say that the unpicked door probability just became 2/3, you did something wrong because you added all of the probability from the revealed door to one door - without justifying this action. You SHOULD have added 1/2 of the original probability of the "reveal" door to each of the still-closed doors, or 1/6 to the picked door and 1/6 to the remaining door. If you do, their probabilities should be 1/2 each.

#### Cronk

##### Registered User.
Similarly, after a run of 9 heads, the odds of the next toss of coming down heads is 50%, not 1 in 2^10, because the first 9 results are now given.

#### Jon

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Statistically, you can calculate a probability that the actual average age of a mass of people (say UK population) by determining the age of each in that random sample. (Census is a 100% or near enough sample).
Yes, you can do sampling for approximation. It doesn't make much difference in this context though because the principles are the same. Take the halfway point or the average for a good guess. Doesn't matter whether it is a sample of the population or the whole population.

Jon, your method of taking a sample of guesses about the age of the universe is akin to sampling a number of guesses about the age of a single person in a windowless room from the outside.
I am not taking a sample of guesses. I am taking ONE guess. But what decision criteria do you use to minimise your likely error if you only have one guess? That is where you take the average age of all the people in the population.

If you average the guesses or get the mid point of a bell curve, it's still based on a guess.
Casinos work on guesses and the house nearly always wins. They guess they will come out ahead. But if they are just guessing, how come they nearly always do come out ahead? => Probabilities and statistics.

There seems to be considerable misunderstanding about this. My hypothesis (although tongue in cheek), was that if you had ONE GUESS, what would be your best stab at the answer that would give the MINIMUM ERROR?

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#### Jon

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Similarly, after a run of 9 heads, the odds of the next toss of coming down heads is 50%, not 1 in 2^10, because the first 9 results are now given.
Correct. Not in disagreement with that.

To be clear, my example of one guess is not talking about a sequence of guesses and so the 9 heads example you give is not related to this example.

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#### Jon

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Let me do a simple example with only 3 people in a population.

Example:

You have 3 people in a population and you have one guess to minimise your error. The actual ages are: 1, 50, 100.

You guess age 1 and I guess age 50. Who is more likely to get the most accurate guess (or the one with minimum error)?

Scenario 1: Person was actually age 1. You made the more accurate prediction. You win.

Scenario 2: Person was actually age 50. I was more accurate. I win.

Scenario: 3: Person was actually 100. Because 50 is closer to 100 than 1, I was more accurate and therefore I win.

So, you can see that I have a chance of being more accurate than you 2/3 rds of the time. You are more accurate 1/3 rd of the time. So, I am twice as likely to be accurate than you are. Therefore, choosing the average will minimise your error.

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#### Cronk

##### Registered User.
I wouldn't pick the age as being 1. For the UK, I'd pick 41 which is closer to the medium age

Using people, the maximum is say a bit over 100 with a few exceptions. Going back to the topic of this thread, how do you determine a maximum as the age of the universe?

Another example. I have a box with a piece of string in it. What's your guess as to how long it is? The average of what?

#### JonXL

##### Member
I'm not trying to estimate the age with the highest frequency. Instead, I am trying to get the most accurate guess.

Perhaps it is that I don't understand the purpose of trying to guess a number that "minimises ... errors" instead of trying to guess the number most likely to be correct.

However, if you want to average closer to the right answer (even if it means less-often getting the right answer), then guessing 40 does that trick. Here are the results of the simulation I mentioned earlier run 1,000,000 times using the spread of ages in the UK from 0 to 80:

When you guess 40 and run the scenario, you are, on average, the closest to the correct age for all the random individual ages (averaging only 18.97 years off). When you guess 55, you are more likely to be correct (1.47% of the time vs 1.39% for 40 - the next most likely).

I think the issue is there are two things at play with this analogy: the average age of a person and the distribution of ages throughout the population. I am still not sure I quite understand why you've chosen the method you've chosen and what it is meant to represent. But within the bounds of the analogy and parameters you have set, your answer of 40 will "minimise[] your errors".

Jon

#### Jon

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I wouldn't pick the age as being 1. For the UK, I'd pick 41 which is closer to the medium age
And that is why the midpoint or an average improves your odds of being closer to the truth.

Using people, the maximum is say a bit over 100 with a few exceptions. Going back to the topic of this thread, how do you determine a maximum as the age of the universe?
If there is no end to time, then it can be infinitely old. I've argued what happens in that case in earlier posts. But let us say for the sake of argument that time stops at some point, and we assume that is when the Universe ends (if it does). Still, the total potential age of the universe is estimated to be exceedingly large.

Look at these potential ages to see that they dwarf the current estimated age of the universe:

Total potential universe age = current age + future age

In one example, Total potential universe age = 13.8 billion years + 100 quintillion years. Consequently, the midway point to get the best statistical estimate of the age of the Universe would not be 13.8 billion years, but way higher. Most models assume the current age is no where near the midpoint.

Since the estimated future ages are much larger than the current estimated age, if you accept the midpoint or average argument for getting the best guess on the universes age, then you can see that the current age estimates are statistically improbable.

Another example. I have a box with a piece of string in it. What's your guess as to how long it is? The average of what?
With your box example, am I inside or outside the box? It makes a difference.

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#### Jon

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Perhaps it is that I don't understand the purpose of trying to guess a number that "minimises ... errors" instead of trying to guess the number most likely to be correct.
The purpose in trying to guess a number that minimises your errors is so that you are closer to the true age more often than not, given one guess. If you chose a number which has the highest frequency of age, you are less likely to be closer to the true age. I believe based on what you have said, we are in agreement.

If someone asked you, "What is the life expectancy of a man in the UK?", would you choose the average age of death (or even the median (midpoint)), or the age with the highest frequency? I think the answer is obvious.

The principle, as it relates to guesstimating the age of the Universe, is that choosing the midpoint will give you a statistically more accurate chance of being closer to the truth than choosing a non-midpoint. Did that make sense?

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