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

The_Doc_Man

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Thank you, @Cronk.
@Jon - here is the updated game results grid, this time exhaustive and with further analysis.

I repeat the game rules just so we agree (and if you believe I have misstated a rule, PLEASE be specific.)

1. Car placement: The car IS behind one of three doors. A non-prize (zonk) is behind each of the other two doors. In the updated grid, the car placement is in the "Car" column.

2. Pick: You select one of the three doors. Your selection is in the "Pick" column.

3. Reveal: Monty reveals to you the contents of a door you did not pick, and this door WILL be a zonk. The door chosen for the reveal is listed in the "Reveal" column.

However, the "reveal" choice is constrained by these rules:
3.a - Monty will never reveal the door you picked
3.b - Monty will never reveal the door hiding the car (see "Resolution" later.)
3.c - IF and ONLY IF you picked the correct door, Monty can choose which of two zonks he will show you and those choices have equal probability. Since you have no way of knowing WHY Monty revealed a door, you cannot draw an inference for why one was picked vs. another. I.e. it is questionable to apply the principle of restricted choice since you don't know whether Monty HAD a choice. (In fact, more often than not, he DOES NOT have a choice.)
3.d - Because of rules 3.a and 3.b, certain combinations of picks cannot occur - and there is a 0 in the "Possible" column for those cases, along with a comment in the "comments" column as to which rule was violated. NO statistics can be derived from rows that violate game rules. I have also highlighted the "forbidden" cases in light yellow.

4. Switch: You now get the offer to switch from the "Pick" door to the remaining door that was neither picked nor revealed. In the PDF/Spreadsheet, the "Switch" column is 1 if you switched or 0 if you stayed with your original choice.

5. Resolution: You either WIN or LOSE the car. If you won, it was either because you stayed or switched. The four remaining columns of the PDF/Spreadsheet show you wins and losses and also show you wins because of switching and wins because of staying.

Analysis:
You have 3 doors for a car, 3 doors for a pick, 3 doors for a reveal, and 2 choices (stay/switch). That is 3 x 3 x 3 x 2 = 54 combinations.
There are 54 rows in the spreadsheet, one row for each possible combination.
You have 27 cases where you switched (= 54/2) and 27 where you stayed (0 in the Switch column)
The car appears behind each door 18 times (=54/3).
Each possible pick appears 18 times.
Each possible reveal (regardless of the rules) appears 18 times.
All combinations of Car, Pick, Reveal, and Switch have been represented, so this is an exhaustive layout.

The rules block cases where the reveal overlaps either the pick or the car location. A reveal can be any one of 3 numbers, so you would expect this overlap to occur 54/3 or 18 times. In the grid there are 18 cases where the reveal overlaps the car location. There are also 18 cases where the reveal would overlap the pick location. Of these, six cases break both rules at once. 18 + 18 = 36. then 36 - 6 overlaps = 30 impossible cases, leaving 24 cases out of the 54 possible combinations. The sums of "Possible" rows show that you have 24 possible situations. The rules have eliminated all other cases. This result therefore is consistent.

The sums for wins and losses are 12 and 12 so all 24 situations are accounted for, showing you have equal odds of winning or losing. The sums for wins by staying vs. wins by switching are 6 and 6 so all 12 winning situations are accounted for, and you have equal odds for staying or switching. Quod erat demonstradum.
 

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The_Doc_Man

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So I think the point of contention is that while you initially agree that door 1 only has a 1/3 rd probability of containing the car, there is no additionally information available that alters that probability, because the door revealing in group 2 is independent of the probability of door 1.

Ah, but there IS new information. You are looking at the remaining doors where you know one of them is still unrevealed. You know nothing about it. But you don't know about the door you picked, either. It is still in the group of "unrevealed doors." It is my contention that this asymmetric redistribution of probabilities is incorrect. You have no reason to know that your pick is less probable than the remaining door.

Let's ask the question differently: Please JUSTIFY the asymmetric redistribution of probabilities to an unknown door.
 

Jon

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In a similar manner to the tenth toss of a coin where the previous 9 have been heads, the odds for the tenth toss is 50/50. After one of three doors has been opened, there are now 2 doors and it is a 50/50 probability that the car is behind either.
Would you concede that your hypothesis (and therefore understanding) is wrong if empirical results show different odds?
 
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Jon

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Ah, but there IS new information. You are looking at the remaining doors where you know one of them is still unrevealed. You know nothing about it. But you don't know about the door you picked, either. It is still in the group of "unrevealed doors." It is my contention that this asymmetric redistribution of probabilities is incorrect. You have no reason to know that your pick is less probable than the remaining door.

Let's ask the question differently: Please JUSTIFY the asymmetric redistribution of probabilities to an unknown door.
Doc, I have thought of a way to explain it....

But we have to do it in Socratic style.

Imagine this: you have one man in a boat floating around in the Atlantic, standing behind his door! You also have 1.5 billion people in China standing behind doors (iron curtain?). One of the people in this group is infected with Covid! No wait, that is probably not a good one to go with. One of them is holds a rare coin.

1. Do you agree that the probabilities are that it is nearly certain that the coin resides in China?

2. If you agree to #1, do you also agree that if you opened all the doors in China, you are almost certain to find the coin there?

3. Do you agree that the physical act of opening the door does not move the coin, and therefore it won't alter the probability that it is almost certainly in China?

4. If you agree to #2 and #3, and are almost certain the coin is in China, do you also agree that if you opened all the doors in China, but stopped opening the very last door that the remaining closed door almost certainly holds the coin?

=> Why would the physical act of opening doors alter the probability of it being in China?
 
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Jon

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Aside from my explanation above, I propose a test.

Who wants to play the Monty Hall game here in public on the forum? It would be best from someone who thinks I am wrong and they are right. Its a battle! You see, I am prepared to put my conviction on the line. Are you?

It's like taking candy from a baby...

I know @Cronk and @The_Doc_Man seem convinced of their beliefs. Maybe one or both of them would like to play?
 
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The_Doc_Man

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1. Do you agree that the probabilities are that it is nearly certain that the coin resides in China?
Yes, before you open any doors, the odds are that the coin is in China.

2. If you agree to #1, do you also agree that if you opened all the doors in China, you are almost certain to find the coin there?
Yes, ALMOST certain to find the coin there. ALMOST certain - before I opened those doors. But not 100% certain. (What was your other recent Watercooler "age estimation" topic about? Certainty and guessing?)

3. Do you agree that the physical act of opening the door does not move the coin, and therefore it won't alter the probability that it is almost certainly in China?

Part a - Opening a door does not move the coin.
Part b - I disagree regarding the odds not being altered.

4. If you agree to #2 and #3, and are almost certain the coin is in China, do you also agree that if you opened all the doors in China, but stopped opening the very last door that the remaining closed door almost certainly holds the coin?

If this is specifically asking about the very last door in China to be opened, so that only the door in my boat and that one door in China remain untested, the odds are now 50/50. The coin didn't move. It has always been where it was. However, the probabilities adjust with each opened door that reveals no coin. There is no justification for asymmetric reassignment of probabilities.

Let's vary the Monty Hall game to allow a "Pick" reveal. You make a pick. In an astounding gesture of generosity, Monty reveals the door you picked to be a zonk and says, "Now which of the two remaining doors will you pick?" What are the odds that you can now pick the door hiding the car from behind one of the two unpicked (but also unrevealed) doors? (If your answer is ANYTHING other than 50% then we have no basis for discussion because we aren't even talking the same reality.) Now explain to me how that is materially different from Monty showing me a zonk that I didn't pick and then allowing me the "Switch" choice, in effect a second chance to pick one of two unrevealed doors. The fact of me having picked a door HAS NOT EFFECT on probability. The fact of a "reveal" is what alters probability. The "Pick" might as well have been a "dart board" choice.

By the way, I'm not going to play your proposed game. I went through ALL of the possibilities and enumerated the results. By playing the game through all 54 possible combinations and excluding the cases that would violate the rules, I have shown that the result should be 50% with the given rules. If there is a problem with the rules as I've stated them, explain it to me. There is no approach you can take that will convince me short of finding a flaw in the way I stated the rules or finding an error in my enumeration. If you can identify the procedural or semantics error in the rules then I will reconsider.
 

Jon

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I only propose 10 rounds. I've already shown the flaw, many times, but you don't believe me. So if you don't believe me, and are convinced that you are indeed correct, the best way to protect that belief and not be proven wrong is to not do an empirical test. If you say my explanations are wrong and you refuse to play the game out to see the results, then you cannot be convinced.

But empirical results should convince you. I am happy to play the game on here with anybody, whether they think I am wrong or not! If you are right, what are you afraid of? Come on challengers!

Edit: Maybe there is only Doc, Cronk and me reading this thread, so nobody cares anyway!
 

Jon

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I just did a test myself. Results are in the pic below. Sample size was 50. In hindsight, I think 10 rounds was too small to get a realistic estimate. The longer you play the game, the more the average will tend to 2/3rds odds of switching being the winning strategy. In other words, the results will converge towards the expected odds.

The test for 50 goes gave 68% odds of switching being the best strategy. I thought @Cronk and @The_Doc_Man might find that interesting. Of course, it could just have been a "lucky" result, because 1/3rd of the time on a specific event, switching is not the best strategy.



1641764617216.png
 

The_Doc_Man

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Jon, I'm not protecting anything. I'm convinced that the devil is in the details of how you simulate the game.
 

Jon

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@The_Doc_Man, it is good that you are participating in this discussion because whoever is right or wrong, it is interesting that there is debate on it, and highlights the difficulty in understanding probability. It is rather akin to Quantum physics that can also twist the brain in knots. I also like it when you structure things in steps, as it breaks this down into digestible chunks for my ageing brain.

Two questions:

1. So, in essence I believe you are saying that if you simulated the game 100,000 times following the rules you state, the strategy of always switching would be approximately 50% wins, 50% losses. And therefore, there is no benefit to switching from your original choice. Correct?

2. Accordingly, my hypothesis that always switching would give you a win rate of 67% is indeed false. Yes?
 
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The_Doc_Man

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Actually, I'm still working on understanding this because last night I built a Monty Hall simulator and am not getting the results consistent with my previous analysis. However, according to my simulator (and my analysis), your win rate overall is 50% based on random choices to swap or stay. I have to work a little on adding a "forced swap/stay choice" simulation to what I have. built

Part of the problem is how you base the percentages. Using a forced choice of swap (or stay) leads to the question of how you declare the wins.

The part that is blowing me away is that 50% win rate in the simulator, That is exactly what I predicted. However, the ratio of swap wins to stay wins is NOT 50% and that is paradoxical.
 

Jon

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Actually, I was going to say I was about to get a simulator built via the VBA section of this forum. It is only a few lines of code to create one. That is why I asked you the 100,000 simulations question in post #90. And perhaps you sussed I might do that!

If we are to solve this riddle, the best way is via the Scientific Method. Or, as follows:


Source: Wikipedia

An untested hypothesis is pseudo science, hence my insistence on testing.

your win rate overall is 50% based on random choices to swap or stay
I am a little confused on what you are saying here. The Monty Hall paradox is all about whether or not switching gives you an advantage. It was never about randomly switching or staying. So, the only test you really need in your simulator is this: what happens if you decide to switch every time (or never switch)? That is the acid test.

Your prediction was 50%, but the answer is 67%. Try it!

When you get your answer, feel free to come back to me here and say I was right all along. Give credit where credit is due! :D

Ultimately, this proves that probability is complicated, can often be counter-intuitive and it is easy to get your knickers in a twist.

However, the ratio of swap wins to stay wins is NOT 50% and that is paradoxical.
Has your swap wins rate converged on 67%? If it hasn't, there is something wrong with your code! In any case, it would be interesting to see your code so I can check for logical flaws!

My version was going to be something like this:

Code:
//Loop through the rounds
For round = 1 to 100,000

// Which door on this round hides the car?
currentDoor = rnd(2)+1

// Count the number of times you win based on your type of strategy and never switching
If currentDoor = 1 then strategyAlwaysChooseDoor1 =  strategyAlwaysChooseDoor1 + 1
If currentDoor = 2 then strategyAlwaysChooseDoor2 =  strategyAlwaysChooseDoor2 + 1
If currentDoor = 3 then strategyAlwaysChooseDoor3 =  strategyAlwaysChooseDoor3 + 1
If currentDoor = rnd(2)+1 then strategyChooseRandomDoor = strategyChooseRandomDoor + 1

Next round

// The answers for sticking are...
Print strategyAlwaysChooseDoor1/100,000
Print strategyAlwaysChooseDoor2/100,000
Print strategyAlwaysChooseDoor3/100,000
Print strategyChooseRandomDoor/100,000

// And the answers for switching are...
Print 1 - strategyAlwaysChooseDoor1/100,000
Print 1 - strategyAlwaysChooseDoor2/100,000
Print 1 - strategyAlwaysChooseDoor3/100,000
Print 1 - strategyChooseRandomDoor/100,000

The above should give you the results of never switching for all possible strategies. i.e. always choosing the same door, or always choosing randomly, and for always switching. You should be getting 0.33 and 0.67 accordingly. Bear with me though, since I haven't written any VBA code in 10 years and my syntax is polluted with React + JavaScript! With a little modification, my pseudo code above should be close.

Come over to the Dark Side Doc, the waters warm!
 
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The_Doc_Man

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OK, here are the simulation results: 3 separate sets... first set - player allowed to randomly swap. second set - locked to always swap. third set - locked to never swap.

1. (Random swap) - 10,000 games, 4984 wins. Of those wins, 3351 were by swap, 1633 were by stay. So... 50% win, 33% by swap, 16% by stay.
2. (Locked to swap) - 10,000 games, 6596 wins (all by swap) - so roughly 2/3
3. (Locked to stay) - 10,000 games, 3292 wins (all by stay) - so roughly 1/3

I still ABSOLUTELY and CATEGORICALLY disagree philosophically with these results. (But then again, I supported Trump, so the liberals out there will dismiss me as inconsequential anyway.)
 

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Doc, are you sure your simulation software is excluding the revealed door? Can you show your simulation software?
 

The_Doc_Man

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I believe it is correctly doing the things it should do. I can post it and I might just do that, but there is something else in the test bed file that I want separate, so I'll have to clean that up first.

OK, let's put cards on the table here. This was actually an intellectual exercise anyway, but I am still convinced that the wrong question is being asked OR that the question is being asked in the wrong way. I know that according to Bayes's Theorem the 2/3 probability for switching is the expected result and my tests roughly confirm it.

Often you get counter-intuitive results from problems, but my goal is actually to decide why the result is counter-intuitive. Because as far as I can tell from that spreadsheet I posted earlier, the odds are quite even a priori. Worse, one of the results I predicted is working correctly - the 50% win when you randomly decide between swap and stay.

The issue has to be in the mapping between the word problem and the math problem. I'm not alone in that, by the way. If you Google the "Monty Hall" problem you will get hundreds of results that suggest the issue is at least partly in the way you ask the question and the way you count the parts of the problem. I'm in good company.
 

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A tardy reply from me...

2. (Locked to swap) - 10,000 games, 6596 wins (all by swap) - so roughly 2/3
3. (Locked to stay) - 10,000 games, 3292 wins (all by stay) - so roughly 1/3
As I have been saying all along.

Because as far as I can tell from that spreadsheet I posted earlier, the odds are quite even a priori.
I did explain the precise reason why your conclusion from your spreadsheet was erroneous because of the allocation of "counts" rather than the allocation of "probabilities". If you re-read what I said through the prism of my reasoning, as opposed to yours, it may give insight and understanding. You may disagree with my reasoning, but my reasoning does lead to the correct empirical results.

I still ABSOLUTELY and CATEGORICALLY disagree philosophically with these results.
Disagreeing with reality is up to you. The alternative is to view this as a learning experience that indeed probability can be counter-intuitive and can catch anybody out.

What I will say is that it was good of you to conduct your own mini-scientific experiment since it shows your quest to find the truth of the situation, by following the well-proven utility of the scientific method.
 

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I did have a further thought about the midway point argument. It is not really necessary. If the Universe can potentially be infinitely old, then any finite number that estimates the age of the Universe is likely to be infinitely inaccurate.
 

oleronesoftwares

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I have been monitoring this thread, I doubt if the exact age of the universe can be gotten.
 

The_Doc_Man

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This is just another impractical question. Knowing the age of the universe gets me nowhere most of the time anyway. I have no idea how we would put the right number of candles on the cake, for example.

I DO know that for my 40th birthday, my friends put 40 candles on the cake, lit them, and melted the icing. They also set of the smoke detector alarm. So maybe you can forgive me for not wanting to celebrate.
 

The_Doc_Man

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Disagreeing with reality is up to you. The alternative is to view this as a learning experience that indeed probability can be counter-intuitive and can catch anybody out.

I'm well aware of that problem. The issue I have is that there is a MAJOR disconnect between my detailed analysis that feeds the odds and the actual outcome of the test.

Jon, I'm working on something I learned long ago form a professor Max Lande', noted for his work in electron spin resonance. He said to us a simple statement: Sometimes when something makes no sense, it is because you asked the wrong question. I'm still working on the right question. I know the odds.
 

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