Gelfand’s trick (Gelfand’s lemma) in positive characteristic?

I came across this preprint that claims in Lemma 1.1 that Gelfand’s trick (also known as Gelfand’s lemma) only works in characteristic zero:

Let $ H < G$ be finite groups. Suppose we have an anti-involution $ \sigma : G \rightarrow G$ that preserves all H double-cosets. Then over algebraically closed fields of characteristic zero (G, H) is a Gelfand pair.

It is not obvious to me where the characteristic of the ground field being zero is used in the proofs from Lang’s $ SL_2(\mathbb{R})$ book (Theorem 1 and Theorem 3 in Chapter IV), the introduction in this preprint, the last slides in these slides, or anything else that I’ve seen.

What causes Gelfand’s trick to fail in positive characteristic? In this setting, the groups are finite but I would also like to know the answer for the more general versions of Gelfand’s trick (i.e. also for locally compact groups with compact subgroups or even reductive groups over local fields with closed subgroups).

Is there any practical trick to mentally count in Gray code?

When I was fairly young, I taught myself to count in binary. I thought it would be a fun party trick to impress people. I soon found out that it was not.

Over the years I’ve come to appreciate Gray code/reflected binary code for its property of only flipping one bit for each increment/decrement of the underlying count. But I’ve always been bothered by the fact that, if I wanted to mentally take any arbitrary Gray code and add or subtract 1 from it, I’d have to either convert it to and then back from its numeric value, or construct a table to work out what the next code should be.

It seems to me that there should be some trick that a person with “average” short term memory and addition skills should be able to do to take any arbitrary value in Gray code and figure out which bit to flip to get the next value… But I’ve never found it.

Does such a thing exist?

Can Wifi probe requests be abused to trick clients into connecting to a fake AP?

I just read about WiFi probe requests and that it is possible to track clients by the MAC-Address in the request.

I was wondering if it would be possible to set up a malicious AP which responds “Yes, that’s me” to every probe request from clients, resulting in clients automatically connecting to that “known” network. A malicious AP could for example sniff the traffic from smartphones of people walking by whose devices automatically connected. Is that possible in theory?

What is the trick behind the code?

def inc_subseqs(s): """Assuming that S is a list, return a nested list of all subsequences of S (a list of lists) for which the elements of the subsequence are strictly nondecreasing. The subsequences can appear in any order.  >>> seqs = inc_subseqs([1, 3, 2]) >>> sorted(seqs) [[], [1], [1, 2], [1, 3], [2], [3]] >>> inc_subseqs([]) [[]] >>> seqs2 = inc_subseqs([1, 1, 2]) >>> sorted(seqs2) [[], [1], [1], [1, 1], [1, 1, 2], [1, 2], [1, 2], [2]]""" def subseq_helper(s, prev):     if not s:         return [[]]     elif s[0] < prev:         return subseq_helper(s[1:], prev)     else:         a = subseq_helper(s[1:], s[0])         b = subseq_helper(s[1:], prev)         return insert_into_all(s[0], a) + b return subseq_helper(s, 0) 

I’m having trouble getting the idea of this function. Why prev is set to 0 at the begining, and what’s the functionality of a and b.I’m new to programming, please help!

Probability of X being a trick coin (heads every time) after heads is flipped k amount of times


A magician has 24 fair coins, and 1 trick coin that flips heads every time.

Someone robs the magician of one of his coins, and flips it $ k$ times to check if it’s the trick coin.

A) What is the probability that the coin the robber has is the trick coin, given that it flips heads all $ k$ times?

B) What is the smallest number of times they need to flip the coin to believe there is at least a 90% chance they have the trick coin, given that it flips heads on each of the flips?

Here is my approach:

Let $ T$ be the probability that the robber has the trick coin

Let $ H$ be the probability the robber flips a heads k times in a row

$ Pr(T) = 1/25$

$ Pr(H|T) = 1$

$ Pr(T’) = 24/25$

$ Pr(H|T’) = 1/2$ when $ k=1$ , $ 1/4$ when $ k=2$ , $ 1/8$ when $ k=3$ … etc

$ Pr(T|H) = (1 * 1/2) / (1 * 1/2 + Pr(H|T’) * 24/25) = 1/13, 1/7, 1/4,…$ etc

So the Pr(T|H) answer changes for every k, do I answer with the formula? How can I answer A? How do I make a probability distribution when k can be infinite?

Also is B 8 flips? Since when k = 8, Pr(T|H) = 1/256.

Thanks for any help.

The 5-Second Trick For weight loss formula

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The 5-Second Trick For weight loss formula

Extension to “The best card trick”

Kleber’s Best card trick proceeds as follows: The mark (audience member) freely selects five playing cards from a standard deck of $ 52$ and passes these five to the magician’s assistant. The assistant studies those cards, returns one mystery card to the mark, and places the remaining four exposed cards, face up, in a sequence on a table. The magician then enters, inspects the sequence of exposed cards, and correctly announces the identity of the mystery card.

The trick works because of clever mathematics. The five selected cards must contain a suit that is represented by two (or more) cards. The assistant will choose one of these as the mystery card, and another of the same suit to be the first in the series of exposed cards. Thus the magician learns the suit of the mystery card.

Playing cards can be placed in a canonical order ($ \clubsuit A, 2, \ldots K, \diamondsuit A, 2, \ldots K, $ etc.), and thus the three remaining exposed cards can be placed in $ 3!$ possible order sequences. Thus the assistant can signal six candidate card values, counting from the value of the first card (modulo 13). That alone will not cover all 12 potential card values. Thus instead, the assistant selected as the first card in the sequence to be the the one whose value is fewer than seven steps before the other of the final suit (modulo 13), which is the mystery card; that way the $ 3!$ possible steps ensure that the mystery card can be reached from the first card in the exposed sequence.

Question: How many four-card exposed sequences can arise in such tricks?