Wikipedia:Reference desk/Mathematics

Are there different methods for an elliptic curve point to a suitable hyperelliptic curve cover than Weil descent?

I ve a curve defined on an extension field but with a point whoes coordinate lies in the base prime field (same coordinate as the prime field version of the curve).

As you know, in the case of applying index calculus, this is largely regarded as impossible as the Weil descent decrease the prime degree (which simplify discrete logarithms computations). But are there really no other methods to lift suchs points to an hyperelliptic curve?

My purpose would be for pairing inversion. I m meaning I can invert type 3 pairings on hyperelliptic curves, so it would be usefull in terms of computational Diffie Hellman if I can move the computations of pairings from bn or bls curves to hyperelliptic curves. ~2026-11394-20 (talk) 23:13, 8 March 2026 (UTC)

Ignorant Monty Hall

Imagine you're faced with a Monty Hall problem, but unlike in the canonical example, the host is ignorant. (In other words, as far as the host knows, #3 could have been the car.) When he asks you whether you want to switch from 1 to 2, is your chance of winning at all affected by the choice? I know the correct answer for the canonical problem is "yes, switch", but I've never understood why, so I don't know whether the host's ignorance would affect the answer. Nyttend (talk) 06:53, 12 March 2026 (UTC)

An easy way to analyze the problem is as follows. We assume that the initial placement is random and that the contestant had rather win a car than a goat. Before any doors are opened, there are three placement patterns with equal probabilities:

1. C G G
2. G C G
3. G G C

Because of the perfect symmetry we may restrict the analysis, without loss of generality, to the case that the contestant initially picks door 1. The contestant knows at this point that the probability there is a car behind the chosen door is 1 in 3; if they further completely ignore what the game host is doing and stick to their guns whatever, they'll end up going home with the big prize with probability 1/3. In the original problem, with an informed game host, the latter now has the choice of opening either door 2 or door 3. Again, without loss of generality, assume they open door 3. The contestant should realize that if they don't switch, their chance of winning is (as we saw before) unaffected: it is still 1/3. Since the car is not behind door 3 and the probabilities need to still add up to 1, the initial 2/3 probability of doors 2 and 3 combined remains now with door 2 alone.

With an ignorant game host, opening door 3 and revealing a goat eliminates pattern 3. We have, by symmetry, with equal probabilities:

1. C G G
2. G C G

Clearly, there is no point in switching. This is in fact the naive analysis most people apply to the original Monty Hall problem, not taking into account that in the setting the game host is careful not to open the door with a car behind, breaking the naive symmetry.  ​‑‑Lambiam 08:48, 12 March 2026 (UTC)

I think the main tripping point on thinking about the "ignorant host" is that the host can pick the car. Imagine you pick door 1. The host opens door 2. It has the car. The host didn't know it was there. But, you do. Do you want to switch? Who cares? Both doors 1 and 3 are losing doors and you aren't allowed to switch to the door the host opened. In the end, the ignorant host is picking a door that has a 1 in 3 chance of being the car. You are doing the same. There is no further information. You are just finding out if you are a loser a little quicker when the host shows the car. ~2026-16820-81 (talk) 19:17, 18 March 2026 (UTC)

The analysis may depend on the actual formulation of the problem. Take the wording of the original question as published in 1990 in vos Savant's "Ask Marilyn" column in Parade (see Monty Hall problem) but change the words "who knows what's behind the doors" into "who doesn't know what's behind the doors". Then the Ignorant Monty Hall problem becomes this:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who doesn't know what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

This is the problem version analyzed above, with some additional assumptions, such as that you, the contestant, prefer a car over a goat.  ​‑‑Lambiam 07:35, 19 March 2026 (UTC)

While this is not directly responsive to the question, it may be of interest here to point out that the standard analysis of the problem assumes that Monty was always going to show you a door, other than the one you picked, with a goat behind it, or at least that the probability of him doing so did not depend on whether you had picked the car.

That is the "simplest assumption" (the one that these problems are usually written with) but there is no actual warrant for that assumption in the original Parade article. It is possible, say, that Monty had been asked to reduce prize awards because they were costing too much, and that he had decided to show the door with the goat and offer to switch only if you had already picked the car. In that case you should definitely not switch.

When this was pointed out to vos Savant, she claimed in the second Parade piece that that was a different problem, and we may certainly take her at her word that it wasn't the problem she was thinking of. But I saw nothing in the original wording that excludes it.

I think this is one of the most fascinating aspects of the problem. It might be a little counterintuitive that, under the "standard assumptions", you have a 2/3 chance of winning the car if you switch. But how much more counterintuitive is it that this result depends on Monty's mental state, or plans, or rules? --Trovatore (talk) 20:28, 19 March 2026 (UTC)

A long time ago, I had wondered if there was a nice way to encode the various possibilities into Bayesian priors. I think there is, and it makes your point quite neatly.

Fix the contestant's initial choice as door 1, and let ${\displaystyle A}$ be the event that Monty opens one of the other doors, reveals a goat, and offers the switch. Let ${\displaystyle C}$ be the event that the contestant initially chose the car. Then before Monty acts we have ${\displaystyle P(C)=1/3}$ and ${\displaystyle P(\neg C)=2/3}$.

Now write

${\displaystyle \alpha =P(A\mid C),\qquad \beta =P(A\mid \neg C).}$

By Bayes,

${\displaystyle P(C\mid A)={\frac {\alpha /3}{\alpha /3+2\beta /3}}={\frac {\alpha }{\alpha +2\beta }},}$

and therefore

${\displaystyle P(\neg C\mid A)={\frac {2\beta }{\alpha +2\beta }}.}$

So switching is better exactly when ${\displaystyle 2\beta >\alpha }$.

That packages the whole issue into Monty's policy. In the "standard" problem, Monty always opens a goat door and always offers the switch, regardless of whether the contestant initially picked the car, so ${\displaystyle \alpha =\beta =1}$. Then

${\displaystyle P(\neg C\mid A)={\frac {2}{3}},}$

which is the usual result.

In the "ignorant Monty" variant, if he just picks one of the other two doors at random, then ${\displaystyle \alpha =1}$ (if you picked the car, both unchosen doors have goats) but ${\displaystyle \beta =1/2}$ (if you picked a goat, one of the two unchosen doors has the car, so he reveals a goat only half the time). Then

${\displaystyle P(C\mid A)=P(\neg C\mid A)={\frac {1}{2}},}$

so there is no advantage either way.

And in your "reduce prize awards" scenario, where Monty offers the switch only when you already picked the car, we have ${\displaystyle \alpha =1}$, ${\displaystyle \beta =0}$, so conditional on getting the offer, you should definitely stay.

So I think your observation is exactly right: the answer is not determined by the bare fact that Monty opened a goat door and offered a switch. It depends on the selection rule generating that event. "Monty's mental state" is perhaps a colorful way of putting it, but in probabilistic terms it is the likelihood model. Sławomir Biały (talk) 09:55, 20 March 2026 (UTC)

In your 'ignorant Monty Hall' if you get the car if the host opens its door and can choose another door if there was a goat behind then your chance of getting a car is 2/3 - exactly like the normal Monty Hall problem. NadVolum (talk) 19:59, 20 March 2026 (UTC)

I see now that our article has a table with the heading "Possible host behaviors in unspecified problem", which has had the label "Ignorant Monty" for one of the entries since 22 June 2009.  ​‑‑Lambiam 21:38, 20 March 2026 (UTC)

'American' and metric values

I've read a few articles recently containing mathematical references and was disappointed (frustrated?) that only the metric value was shown. I'm 81, never learned metric, so having both shown is valuable to me, and I'm not competent to leave the page to try to go calculate it. Is this a new decision by Wikipedia?, or just the way this particular contributor does it? (I believe I've also seen a few entries where there was no metric equivalent. 🙂)

I'm sure this is a really silly question/request, but I use Wikipedia a LOT (often 5+ hours in a day), and it'd really help to have both values shown. (I believe I've even seen a few with the opposite issue - no metric values.)

Thanks for all the great articles - in particular one (read it several years ago) about a European man in the 1700s?, 1800s?, who walked from Europe to northern Africa - pretending (for safety) to be a local - and wrote about his travels. Awesome! (Gotta try to find that one again.) Again, THANKS. Julia L. '~2026-17920-45 (talk) 16:43, 21 March 2026 (UTC)'

@~2026-17920-45: We have a guideline at Wikipedia:Manual of Style/Dates and numbers#Unit conversions. Conversions are still very common but guidelines aren't always followed and we have millions of articles. If you give links and say which values it's about then we can examine the situation. PrimeHunter (talk) 19:38, 21 March 2026 (UTC)

Perhaps the man you are thinking of is Friedrich Hornemann? GalacticShoe (talk) 19:47, 21 March 2026 (UTC)