Thread: Today's puzzle , thank you kindly strictly no nonsense please

I'll get the ball rolling .
The winner will receive a 10% voucher for teh Wongamat insect buffet cart

WHERE DO THE LEAD VAMPIRES RESIDE.

Transylvania. Romania

That post is too short

Nope good attempts but no cigar, I am almost certain their willbe clues not provided by me soon, I promise to provide the solution if you are all stumped.

Spotify?
California?

Originally Posted by Shutree

Spotify?
California?

good tries but no, pretty sure a clue available in next 24 hours

Wood to be or not to be?

Solution once POTUS winner confirmed

@B or knot to B a vampire with lead in his pencil

PENCILVANIA

Originally Posted by david44

What is the optimal strategy to maximize your chance of hiring the best applicant?

change your process

^ Good answer.


Pick the best of the five or six, and then hire the next person you interview who is better than them.

I would interview three, then keep interviewing until I found someone better than the first three.
Unless I died of boredom first.
Or unless a woman with the largest breasts I had ever seen walked in with a warm smile, that would clinch it. I believe in this new disTrumpian era it's quite okay for a bit of misogyny? Half the women in America voted for it, I believe we should respect their choice.

Glad nobody cheated
Eulers number is the key

The secretary problem demonstrates a scenario involving optimal stopping theory that is studied extensively in the fields of applied probability, statistics, and decision theory. It is also known as the marriage problem, the sultan's dowry problem, the fussy suitor problem, the googol game, and the best choice problem. Its solution is also known as the 37% rule

For more on the counter intuitive theory and why I was banned from casinos in Australia france and Switzerland in an era before counter surveillance. It was my grasp of not just maths but probability rather than e.Oh and yes I did piss it all up and had a very nice time thank you.

Deriving the optimal policy



The optimal policy for the problem is a stopping rule. Under it, the interviewer rejects the first r − 1 applicants (let applicant M be the best applicant among these r − 1 applicants), and then selects the first subsequent applicant that is better than applicant M. It can be shown that the optimal strategy lies in this class of strategies.[citation needed] (Note that we should never choose an applicant who is not the best we have seen so far, since they cannot be the best overall applicant.) For an arbitrary cutoff r, the probability that the best applicant is selected is
P(r)=∑i=1nP(applicant i is selected∩applicant i is the best)=∑i=1nP(applicant i is selected|applicant i is the best)⋅P(applicant i is the best)=[∑i=1r−10+∑i=rnP(the best of the first i−1 applicantsis in the first r−1 applicants|applicant i is the best)]⋅1n=[∑i=rnr−1i−1]⋅1n=r−1n∑i=rn1i−1.The sum is not defined for r = 1, but in this case the only feasible policy is to select the first applicant, and hence P(1) = 1/n. This sum is obtained by noting that if applicant i is the best applicant, then it is selected if and only if the best applicant among the first i − 1 applicants is among the first r − 1 applicants that were rejected. Letting n tend to infinity, writing x as the limit of (r−1)/n, using t for (i−1)/n and dt for 1/n, the sum can be approximated by the integral
P(x)=x∫x11tdt=−xln⁡(x).Taking the derivative of P(x) with respect to x, setting it to 0, and solving for x, we find that the optimal x is equal to 1/e. Thus, the optimal cutoff tends to n/e as n increases, and the best applicant is selected with probability 1/e.
For small values of n, the optimal r can also be obtained by standard dynamic programming methods. The optimal thresholds r and probability of selecting the best alternative P for several values of n are shown in the following table.[note 1]

n	1	2	3	4	5	6	7	8	9	10
r	1	1	2	2	3	3	3	4	4	4
P	1.000	0.500	0.500	0.458	0.433	0.428	0.414	0.410	0.406	0.399

The probability of selecting the best applicant in the classical secretary problem converges toward 1/e≈0.368.
Alternative solution

[edit]
This problem and several modifications can be solved (including the proof of optimality) in a straightforward manner by the odds algorithm, which also has other applications. Modifications for the secretary problem that can be solved by this algorithm include random availabilities of applicants, more general hypotheses for applicants to be of interest to the decision maker, group interviews for applicants, as well as certain models for a random number of applicants.[citation needed]
Limitations

[edit]
The solution of the secretary problem is only meaningful if it is justified to assume that the applicants have no knowledge of the decision strategy employed, because early applicants have no chance at all and may not show up otherwise.
One important drawback for applications of the solution of the classical secretary problem is that the number of applicants n must be known in advance, which is rarely the case. One way to overcome this problem is to suppose that the number of applicants is a random variable N with a known distribution of P(N=k)k=1,2,⋯ (Presman and Sonin, 1972). For this model, the optimal solution is in general much harder, however. Moreover, the optimal success probability is now no longer around 1/e but typically lower. This can be understood in the context of having a "price" to pay for not knowing the number of applicants. However, in this model the price is high. Depending on the choice of the distribution of N, the optimal win probability can approach zero. Looking for ways to cope with this new problem led to a new model yielding the so-called 1/e-law of best choice.
1/e-law of best choice

[edit]
The essence of the model is based on the idea that life is sequential and that real-world problems pose themselves in real time. Also, it is easier to estimate times in which specific events (arrivals of applicants) should occur more frequently (if they do) than to estimate the distribution of the number of specific events which will occur. This idea led to the following approach, the so-called unified approach (1984):
The model is defined as follows: An applicant must be selected on some time interval [0,T] from an unknown number N of rankable applicants. The goal is to maximize the probability of selecting only the best under the hypothesis that all arrival orders of different ranks are equally likely. Suppose that all applicants have the same, but independent to each other, arrival time density f on [0,T] and let F denote the corresponding arrival time distribution function, that is
F(t)=∫0tf(s)ds, 0≤t≤T.Let τ be such that F(τ)=1/e. Consider the strategy to wait and observe all applicants up to time τ and then to select, if possible, the first candidate after time τ which is better than all preceding ones. Then this strategy, called 1/e-strategy, has the following properties:
The 1/e-strategy
(i) yields for all N a success probability of at least 1/e,(ii) is a minimax-optimal strategy for the selector who does not know N,(iii) selects, if there is at least one applicant, none at all with probability exactly 1/e.The 1/e-law, proved in 1984 by F. Thomas Bruss, came as a surprise. The reason was that a value of about 1/e had been considered before as being out of reach in a model for unknown N, whereas this value 1/e was now achieved as a lower bound for the success probability, and this in a model with arguably much weaker hypotheses (see e.g. Math. Reviews 85:m).
However, there are many other strategies that achieve (i) and (ii) and, moreover, perform strictly better than the 1/e-strategy simultaneously for all N>2. A simple example is the strategy which selects (if possible) the first relatively best candidate after time τ provided that at least one applicant arrived before this time, and otherwise selects (if possible) the second relatively best candidate after time τ.[4]
The 1/e-law is sometimes confused with the solution for the classical secretary problem described above because of the similar role of the number 1/e. However, in the 1/e-law, this role is more general. The result is also stronger, since it holds for an unknown number of applicants and since the model based on an arrival time distribution F is more tractable for applications.
The game of googol



In the article "Who solved the Secretary problem?" (Ferguson, 1989)[1], it's claimed the secretary problem first appeared in print in Martin Gardner's February 1960 Mathematical Games column in Scientific American:

Ask someone to take as many slips of paper as he pleases, and on each slip write a different positive number. The numbers may range from small fractions of 1 to a number the size of a googol (1 followed by a hundred zeroes) or even larger. These slips are turned face down and shuffled over the top of a table. One at a time you turn the slips face up. The aim is to stop turning when you come to the number that you guess to be the largest of the series. You cannot go back and pick a previously turned slip. If you turn over all the slips, then of course you must pick the last one turned.[5]

Ferguson pointed out that the secretary game remained unsolved, as a zero-sum game with two antagonistic players.[1] In this game:

• Alice, the informed player, writes secretly distinct numbers on n cards.
• Bob, the stopping player, observes the actual values and can stop turning cards whenever he wants, winning if the last card turned has the overall maximal number.
• Bob wants to guess the maximal number with the highest possible probability, while Alice's goal is to keep this probability as low as possible.

The difference with the basic secretary problem are two:

• Alice does not have to write numbers uniformly at random. She may write them according to any joint probability distribution to trick Bob.
• Bob observes the actual values written on the cards, which he can use in his decision procedures.

Strategic analysis

[edit]
Alice first writes down n numbers, which are then shuffled. So, their ordering does not matter, meaning that Alice's numbers must be an exchangeable random variable sequenceX1,X2,...,Xn. Alice's strategy is then just picking the trickiest exchangeable random variable sequence.
Bob's strategy is formalizable as a stopping ruleτ for the sequence X1,X2,...,Xn.
We say that a stopping rule τ for Bob is a relative rank stopping strategy if it depends on only the relative ranks of X1,X2,...,Xn, and not on their numerical values. In other words, it is as if someone secretly intervened after Alice picked her numbers, and changed each number in X1,X2,...,Xn into its relative rank (breaking ties randomly). For example, 0.2,0.3,0.3,0.1 is changed to 2,3,4,1 or 2,4,3,1 with equal probability. This makes it as if Alice played an exchangeable random permutation on {1,2,...,n}. Now, since the only exchangeable random permutation on {1,2,...,n} is just the uniform distribution over all permutations on {1,2,...,n}, the optimal relative rank stopping strategy is the optimal stopping rule for the secretary problem, given above, with a winning probabilityPr(Xτ=maxi∈1:nXi)=maxr∈1:nr−1n∑i=rn1i−1Alice's goal then is to make sure Bob cannot do better than the relative-rank stopping strategy.
By the rules of the game, Alice's sequence must be exchangeable, but to do well in the game, Alice should not pick it to be independent. If Alice samples the numbers independently from some fixed distribution, it would allow Bob to do better. To see this intuitively, imagine if n=2, and Alice is to pick both numbers from the normal distribution N(0,1), independently. Then if Bob turns over one number and sees −3, then he can quite confidently turn over the second number, and if Bob turns over one number and sees +3, then he can quite confidently pick the first number. Alice can do better by picking X1,X2 that are positively correlated.
So the fully formal statement is as below:

Does there exist an exchangeable sequence of random variables X1,...,Xn, such that for any stopping rule τ,Pr(Xτ=maxi∈1:nXi)≤maxr∈1:nr−1n∑i=rn1i−1?

Solution

[edit]
For n=2, if Bob plays the optimal relative-rank stoppings strategy, then Bob has winning probability 1/2. Surprisingly, Alice has no minimax strategy, which is closely related to a paradox of T. Cover[6] and the two envelopes paradox. Concretely, Bob can play this strategy: sample a random number Y. If X1>YY}">, then pick X1, else pick X2. Now, Bob can win with probability strictly greater than 1/2. Suppose Alice's numbers are different, then conditional on Y∉[min(X1,X2),max(X1,X2)], Bob wins with probability 1/2, but conditional on Y∈[min(X1,X2),max(X1,X2)], Bob wins with probability 1.
Note that the random number Y can be sampled from any random distribution, as long as Y∈[min(X1,X2),max(X1,X2)] has nonzero probability.
However, for any ϵ>00}">, Alice can construct an exchangeable sequence X1,X2 such that Bob's winning probability is at most 1/2+ϵ.[1]
But for n>22}">, the answer is yes: Alice can choose random numbers (which are dependent random variables) in such a way that Bob cannot play better than using the classical stopping strategy based on the relative ranks

^ So... big tits then!

This made me scratch my old brain for a minute, have go a deceptively tricky one

3 men walk into a hotel and ask for a triple bed room for the night. The receptionist says, "That'll be £30 please." Each man hands over a £10 note, and they head up to their room. The receptionist then realises their mistake: that the room only costs £25. So they hand the porter five £1 coins and says, "Could you go up and give them their change?"
The porter realises that five pound coins can't be split three ways, and pockets two of the £1 coins, then gives the men £1 each.
Now, from the £30 originally paid, each of the men has paid £9 ( 3 x £9 = £27 ) and the porter has pocketed £2.
BUT £27 + £2 = £29. So what has happened to the missing £1?

Originally Posted by Mendip

^ So... big tits then!

That is the only answer that makes sense to me.

^^ I just checked me back pocket and Manny The Mathematician says that it looks like you're double counting the 2.

27=25+2

^ It thirdenly does

Hint deconstruct the grammar and the numbers

It takes a while but I am sure some genius a published author or one of the fare err sex KAtie MsKit or Nev can wrangle it

Originally Posted by david44

This made me scratch my old brain for a minute, have go a deceptively tricky one

3 men walk into a hotel and ask for a triple bed room for the night. The receptionist says, "That'll be £30 please." Each man hands over a £10 note, and they head up to their room. The receptionist then realises their mistake: that the room only costs £25. So they hand the porter five £1 coins and says, "Could you go up and give them their change?"
The porter realises that five pound coins can't be split three ways, and pockets two of the £1 coins, then gives the men £1 each.
Now, from the £30 originally paid, each of the men has paid £9 ( 3 x £9 = £27 ) and the porter has pocketed £2.
BUT £27 + £2 = £29. So what has happened to the missing £1?

count the pounds. 1+1+1+2+25.

Originally Posted by nidhogg

count the pounds. 1+1+1+2+25.

We have a winner! nice on Oggy do you or others have any puzzles can be Visual, riddle or brainteaser ?

I have discovered mental exercise is a superb way to put offf Alzheimers, gardening , car washing etc

A realy annoying quickie for which I have the solution

What came first, the chicken or the egg?