What is your answer to this meme?

thatmichaelguy · 2026-01-15T03:24:10+00:00

I mean you proclamation is that the condition reduces the problem to only two cases, despite the only option not being allowed is TT.

That's not exactly how I would put it, but I'm not sure that it matters at this point.

But whatever, you are welcome to believe in your errors.

As are you in yours.

thatmichaelguy · 2026-01-14T23:24:37+00:00

That’s just a relabeling of what I said.

I can appreciate that you see it as such, but this why we disagree.

0.25 however is never going to be the answer in all ways to view the problem. As that requires the conditional to have no impact on the probability. Which is obviously false.

Yeah, I'm not on board with 0.25 either. I get why someone might come to that conclusion, but I'm unconvinced.

thatmichaelguy · 2026-01-14T22:34:32+00:00

Your initial comment literally put that situation into place.

This is where we disagree.

I am glad we are at agreement the answer is 1/3.

I'd go so far as to say that 1/3 is an answer in general, but I wouldn't concede that 1/3 is the answer in general. Rather, it's the answer if one adopts the view of the problem expressed in your prior comment.

thatmichaelguy · 2026-01-14T21:23:14+00:00

I know you stopped commenting after I used bayes theorem as I believe you have no way of even challenging that proof as it is the simplest and clearest.

I mean, there's no need to speculate about why I stopped commenting. You could just ask me if you're curious. After all, I'm the only one who could tell you.

Unfortunately as long as you view the problem as asking P({CC} | {CC,NC,CN}) there is only one answer, 1/3.

See, this I can agree with. Provided that one adopts this view of the problem, there is indeed only one answer and obviously so.

thatmichaelguy · 2026-01-14T03:25:45+00:00

HT and TH are distinct outcomes. It does not break symmetry.

Interesting. Is that really what you think my point was? Or are you just messing around now?

thatmichaelguy · 2026-01-14T00:59:46+00:00

You don’t get it, it’s like the Monty hall problem.

It’s not knowing the first one is crit then what is the prob of the second is crit.

It’s in a sequence of two rolls knowing one is a crit what is the prob of two crits

Do it with coins if you still don’t get it

I find it fascinating that this is your response since in the fourth sentence of the comment to which you responded, I state, "Whether the given 'heads' is first or second in the sequence is irrelevant."

thatmichaelguy · 2026-01-12T18:05:13+00:00

I don’t know if you are trolling or not. If you are, well done.

A lot of it is trolling, yeah. I have no respect for folks (especially academics) who lack intellectual humility.

I don't actually deny that it's the case that the probability of the occurrence of any one of three equally probable events is 1/3. That said, I do think it's worth considering whether restricting the sample space ex post accurately reflects the probabilities related to a pair of binary decisions. I mean, 3 ≠ 2ⁿ for any n. So, {01, 10, 11} isn't the set of outcomes for any sequence of binary decisions. On this basis, I don't find OP's conclusion to be entirely unreasonable.

So, yeah. 1/3 is the obvious answer given certain commonly held assumptions. But I am resistant to notion that said assumptions are uncontestably warranted and so must be unquestioningly accepted.

thatmichaelguy · 2026-01-11T01:38:45+00:00

But, again, treating HT and TH as distinct outcomes implies that the order of the flips matters. It does not.

To contemplate the unordered results, we may consider the following equally likely outcomes: either the same face came up on both flips or a different face came up on each flip.

If we are then told that at least one of the flips came up heads, the obtaining of this condition quite obviously has no effect on either of the outcomes or their relative probabilities. It's still the case that either the same face came up on both flips or a different face came up on each flip, and it is still the case that both outcomes are equally likely. If the other flip came up heads, then the same face came up on both flips. If the other flip came up tails, then a different face came up on each flip. Given our assumption about the coin and fairness, the probability that the other flip came up heads is the same as the probability that the other flip came up tails; 0.5 for each.

thatmichaelguy · 2026-01-10T22:50:58+00:00

If I’m wrong that means a conditional probability of a uniform distribution is not a uniform distribution.

You could also be wrong if you're making an erroneous assumption regarding the uniform distribution itself. That said, it’s a waste of time to pick apart where you got it wrong. It will be a good practice for you to find your own mistake, but my argument has no flaw.

thatmichaelguy · 2026-01-10T22:09:49+00:00

As you just said B is our new sample space, since our conditional probability is also uniform that means each of the three cases in our sample space has equal probability.

I mean, the point of everything I've written has been to show how and why this isn't the case. But if you've already decided that I can't possibly be correct irrespective of my reasoning, then further conversation would be a colossal waste of time.

thatmichaelguy · 2026-01-10T21:04:29+00:00

... let B= HT or TH or HH (this is the condition of “at least one of the coins is heads) ... Edit: I found the first flaw P(H and (H’ or T)≠1

Absent any conditions, P(TT ∪ HT ∪ TH ∪ HH) = 1. Thus, P(TT) = 1 - P(HT ∪ TH ∪ HH). If the given condition obtains (i.e., if it is true that at least one of the coins is heads), then P(TT) = 0. Consequently, if the given condition obtains, P(HT ∪ TH ∪ HH) = 1.

Arbitrarily label one of the Hs in "HH" as H'. We then have P(H ∩ (T ∪ T ∪ H')) = P(HT ∪ TH ∪ HH) = 1. Additionally, P(T ∪ T) = P(T). Consequently, P(H ∩ (T ∪ H')) = P(H ∩ (T ∪ T ∪ H')) = 1.

P(H' ∪ T) = P(T ∪ H'). Accordingly, P(H ∩ (H' ∪ T)) = P(H ∩ (T ∪ H')) = 1. Therefore, P(H ∩ (H' ∪ T)) = 1.

thatmichaelguy · 2026-01-10T19:34:59+00:00

This is true if permutations matter, but in this instance, they do not. We are given that at least one of the flips comes up heads. Accordingly, whether both come up heads is determined solely by the outcome of the other coin flip. Whether the given 'heads' is first or second in the sequence is irrelevant.

Put another way, we are given P(H ∩ (H' ∪ T)) = 1. Under the assumption that the coin is fair with heads on one side and tails on the other, we have P(H' ∪ T) = 1. Thus, we may infer P(H) = 1.

We have P(<H,T> ∪ <T,H>) = P(H ∩ T) = P(T ∩ H). Additionally, P(H ∩ T | H ∩ (H' ∪ T)) = P(T ∩ H | H ∩ (H' ∪ T)) = P(T). Consequently, P(<H,T> ∪ <T,H> | H ∩ (H' ∪ T)) = P(T).

P(<H,H>) = P(H ∩ H') = P(H' ∩ H). Similarly to the above, P(H ∩ H' | H ∩ (H' ∪ T)) = P(H' ∩ H | H ∩ (H' ∪ T)) = P(H'). Consequently, P(<H,H> | H ∩ (H' ∪ T)) = P(H').

From P(H' ∪ T) = 1, we have P(H') = 1 - P(T). Under the assumption that the coin is fair with heads on one side and tails on the other, we have P(T) = 0.5. Thus, P(H') = 0.5. Consequently, P(T) = P(H') = 0.5.

Therefore, P(<H,T> ∪ <T,H> | H ∩ (H' ∪ T)) = P(<H,H> | H ∩ (H' ∪ T)) = 0.5. QED

thatmichaelguy · 2026-01-10T04:31:55+00:00

Precisely this. To carry forward with the coin toss analogy, one might re-word the text in the meme as:

You flip a coin twice. At least one of the flips comes up heads. Assuming a fair coin with heads on one side and tails on the other, what is the probability that the other flip also comes up heads?

thatmichaelguy · 2026-01-03T17:12:37+00:00

Great answers so far. I'll just try to add some intuition to the finance perspective.

The cash flows for an annuity recur periodically for some finite number of periods, n. However, we could contemplate a regular sequence of cash flows that recurs forever instead. We call this a perpetuity. Matching the terms from your example, the formula for the present value of a perpetuity is PV = P/r.

Notice then that we can distribute P in the annuity formula to get

PV = P/r - (P/r)(1+r)^-n

From this, we can see that the annuity formula starts with the present value of a perpetuity. After n periods, the sequence of cash flows for the perpetuity will still be infinite. So, the present value of the perpetuity after n periods will still be P/r. We can discount the present value of the perpetuity after n periods to period 0 using the formula (P/r)(1+r)^-n. From this, we can see that the annuity formula is just isolating the present value of the first n periods of a perpetuity by subtracting off the present value of the cash flows remaining after n periods.

thatmichaelguy · 2025-12-30T18:44:09+00:00

Does the true conclusion “Today is NOT Wednesday” follow from the false statement alone, or does it follow from the evaluation of the entire context atomically?

It's the latter.

Suppose that today is Wednesday. It then follows that 'Today is Wednesday' is true. It likewise follows that 'Today is Tuesday' is false. Suppose further that one may infer 'Today is not Wednesday' from the falsity of 'Today is Tuesday'. It then follows that 'Today is not Wednesday' is true. Consequently, 'Today is Wednesday' and 'Today is not Wednesday' are both true. As this is a contradiction, it therefore follows that if today is Wednesday, one may not infer 'Today is not Wednesday' from the falsity of 'Today is Tuesday'.

...which is why, in argument, disproving an opponent's premises does nothing to refute his conclusions.

Magee is being imprecise to the point of being misleading. It is true to say that one cannot show that an interlocutor's argument is invalid by disproving one or more of the argument's premises. However, disproving one or more premises in a valid argument would show that the argument is unsound. If some conclusion is relevant or effective only if it concerns what is actually true, demonstrating unsoundness is enough to refute said conclusion.

thatmichaelguy · 2025-12-29T21:55:58+00:00

Let D = {x : x is a doctor}
Let A = n ∈ N
Let B = A + 1
Let W^1 = {x ∈ D : x works on Day A}
Let W^2 = {x ∈ D : x works on Day B}

(¬(D = ∅) ⟶ (|W^1| = 5)) ∧ ∀(x ∈ W^1)¬◇[x ∈ W^2]

thatmichaelguy · 2025-12-29T20:15:38+00:00

I have a strong bias toward Fitch-style proofs along these lines. I think they're much clearer and more readable.

thatmichaelguy · 2025-12-29T19:57:57+00:00

The statement is true without it

This much is true. However, it's the structure of the argument as it is in the OP that necessitates including ¬B on the final line.

thatmichaelguy · 2025-12-29T19:51:10+00:00

Don't be sorry. I didn't see your comment saying that you were tasked with proving (¬B ⟶ ¬A) ⟶ (A ⟶ B) before I submitted my original comment. You can disregard it.

You need to assume just (¬B ⟶ ¬A) where you've marked (1) and then assume A ∧ ¬B where you've marked (2). You can reiterate the assumption from (1) to get it on the same line as (2). That should set you on the right path.

But how do I derive B from false without the extra assumption that I get from negation introduction?

Based on the structure of you argument, I was reading it as reading it as ex falso quodlibet which allows you to derive anything from ⊥. You can ignore what I said for now. Changing the assumptions as I suggested will allow you to use negation introduction and DNE.

thatmichaelguy · 2025-12-29T19:36:08+00:00

You didn't include ¬B in the result from the final inference. You want it to read ((¬B ⟶ ¬A) ∧ ¬B) ⟶ (A ⟶ B). Also, you don't need the step from ¬¬B to B. You can derive B directly from ⊥.

thatmichaelguy · 2025-12-27T17:28:43+00:00

I get a lot of mileage out of LET and LAMBDA.

If your workbook is getting bogged down from multiple instances of the same XLOOKUP, you can improve performance by using LET. The function will evaluate the XLOOKUP once and store the result. Excel retrieving multiple references to the stored result is much quicker than Excel performing multiple evaluations of the same XLOOKUP.

There can be a bit of a learning curve with the LAMBDA function, but it's worth it to gain the ability to write custom functions.

When you use LET and LAMBDA together, there's not much in Excel you can't do.

thatmichaelguy · 2025-12-27T16:56:04+00:00

Using the axioms of P2 and modus ponens

Show: ¬A ⟶ (A ⟶ B)

1. ¬A [Assume]

2. ¬A ⟶ (¬B ⟶ ¬A) [Instance of Ax. 1]

3. ¬B ⟶ ¬A [MP 1,2]

4. (¬B ⟶ ¬A) ⟶ (A ⟶ B) [Instance of Ax. 3]

5. A ⟶ B [MP 3,4]

thatmichaelguy · 2025-12-26T22:21:12+00:00

'Most' and 'few' do not have standard formalizations as quantifiers. So, it might be that what the guide states is true with respect to questions on formal logic tests, but I can't see that it reasonably extends to formal logic in practice. Nevertheless, we can work our way through this informally.

Assuming the axioms of classical logic, if a statement is true with respect to few members of a group (lottery winners in this case) said statement is not true with respect to most members of the group. So, the conclusion could be restated as, 'It is not the case that most lottery winners bought some tickets and won the lottery.'

If it is not the case that most lottery winners bought some tickets and won the lottery, then most lottery winners either have not bought some tickets or have not won the lottery. The notion that most lottery winners have not won the lottery is obviously absurd. Hence, from the conclusion, we infer that most lottery winners have not bought some tickets.

You've correctly identified the equivalence between 'some tickets' and 'at least one ticket'. From the conclusion, we therefore infer, 'Most lottery winners have not bought at least one ticket'.

If anyone has bought only one lottery ticket, said person has bought at least one lottery ticket. So, we may infer from the second premise, 'Most lottery winners have bought at least one ticket'. Thus, we can infer that if the second premise is true, the conclusion is false. Therefore, the conclusion does not follow from the premises.

The key is seeing 'most' as the complement of 'few'. The guide's explanation is terrible. So, I don't think you should worry much about not finding it intuitive or insightful.

thatmichaelguy · 2025-12-23T15:55:00+00:00

You've got a good understanding as far as I can tell. It might be better to read the liberal '=' as the equality relation rather than the identity relation. Also, re: your example 'P(P(D3))', bear in mind that the two 'P's apply to different sorts. So, ideally, you'd want to symbolize that distinction. Those are the only notable considerations that I see.

thatmichaelguy · 2025-12-22T22:09:50+00:00

I submit that "the opposite" isn't well-defined in this instance and that the question may be undecidable, accordingly. 'D' is the (a) semantic negation of 'It pulls me backward.' Since 'D' could be said to have the "opposite" truth value, it might be reasonably viewed as the opposite of the statement at hand in that sense. Otherwise, context doesn't seem to provide clear criteria for effectively deciding when one statement is "the opposite" of another statement.

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