[My first sequence]

11, 12, 13, 22, 15, ?, 17, 32, 23, 111, ?

Fill in the ?'s

[Complete the Series 2]

How about

11, 12, 13, 22, 15, ?, 17, 32, 23, 111, ?

[How can you weigh a living fish?]

if there are no restrictions other than those imposed,

measure the height of the water, then remove the water and put it into another aquarium, allowing you to calculate it's volume.

you can now derive the fish's weight based on the height (depth), length, and width of the initial aquarium.

the fish'll probably die, though, so you'll have to be quick with your calculations to detemine its weight while it's still alive.

[Numbered Foreheads]

By looking at the OP, you can infer that all the logician's are "perfectly logical".

That is, they along will follow proper logic, and they can expect the other two to do so as well.

In this way, this is similar to the Blue Eyed/Brown Eyed puzzle.

OK,

with that out of the way, here's the problem: By allowing the option

of "giving up", we introduce the observer effect into play (Giving up

might change the conditions that made one conclude giving up initially).

The answer provided still may be correct, but it would need improved justification.

Consider the statement,

If

everyone has a 3, (or if only C has a 2,) and no one gives up, then A

and B will each be able to name their number on the third time they are

asked.

The logic that determines C to give up

relies on then scenario where no can give up, which isn't the case, and

thus the argument breaks down.

Additionally, A and B wouldn't be

able to name their number on the third time, only A would per the OP

rules. So wouldn't B give up at some point?

I haven't

attempted the deeper logic; to answer the original problem: as each

logician doesn't knows if he's a 2 or a 3, the scenarios of 332, 323,

and 332 would need to be solved and would reveal the answer to the OP as

all logicians viewpoints would be considered.

A helpful tidbit: a

logician will say "I don't know" if and only if there exists a

situation (from his viewpoint) in which he will be able voice his

number. So every "I don't know" you conclude would require such an

example. What? Yeah. It seems the proof would be quite involved.

(1) The original problem states that the question is repeated until each

of the three logicians has either named his number or given up. So B is

not forbidden from winning after A; in fact, by the rules of the OP,

all three could theoretically win. In this respect the problem is

different from many other similar puzzles which permit only one winner,

so I should probably have emphasized this more in the original wording.

(2)

The "observer effect" is an interesting aspect, but it is not

problematic in this case due to note (1) above. No one who gives up

(assuming they are perfect logicians, of course) can ever regret doing

so; at most, he may change which of the other logicians may name their

number afterward (and we can assume the logicians aren't petty enough to

care about that ). Having said that, you have me wondering how a similar problem would play out if only the first

logician to name his number could win, since then hypothetically there

could be a situation where the winner is indeterminate (whoever gets

frustrated and gives up first helps the other person to win). It could

get interesting.

(3) Aside from that, what I posted was

only the "short version" of the solution, with many details left out.

The logic of the argument is correct (despite relying on a

hypothetical), it just needs backup. I left out the details

deliberately, so readers could work the problem out for themselves and

yet have a way to check their answer. I may post a more complete

solution later, if no one else does.

From your simple proof:

If

everyone has a 3, (or if only C has a 2,) and no one gives up, then A

and B will each be able to name their number on the third time they are

asked. C will never be able to.

So would the following be an accurate summarization?

If

C keeps saying "I don't know" then both A and B will know their number

on the third round based solely on the conclusions from each other

saying "I don't know" two times?

As a result of this, C's number

is irrelevant and this same outcome (A and B know on the third round)

would occur if he is a 2 or a 3?

Note that (if this

isn't the case) the observer effect does come into play as C decided to

give up based on the conclusions from a hypothetical where no one gives

up, rendering his decision moot.

Yes, that's exactly right.

With a few notes...

C's number is not altogether irrelevant: if it were anything other than

3 or 2, then the game would play out differently. Also, were no one to

give up, A and B would not learn their numbers strictly from each

other's responses, but from each other's responses in combination with

C's responses. In the end, only C would be left hanging if no one gave

up. So in anticipation of this, it makes sense for C to give up, and by

doing so he provides A and B with an additional clue that helps them

deduce their numbers sooner than they otherwise would have.

My point is C's number is irrelevant with the given circumstances (that A and B are both 3, which limits it to 2,3).

More importantly, I do believe A and B determine their numbers based strictly on their responses if C always say "I don't know".

This is all the thought process of A. He can deduce his number solely based on what he thinks B thinks he thinks..etc (without a "C thinks...").

C's 3.

A know's he's 2,3.

(A)CASE 2 B:3 A-2,3

4 A-1,2

(B)CASE 4 A:1 B-4,5

2 B-3,4

(A)CASE 1 B: 4 A-1,2

5 A-1! ~ {A would have voiced this}

(A would know A's 1) _Round 1

(B would know B's 4 [!5])

(A would know A's 2 [!1]) _Round 2

(B would know B's 3 [!4])

(A knows A's 3 [!2]) _Round 3

(B) CASE x A: <--> Suppose B considered he was x. Then B knows A's thinking would be:

[!x] <--> because he cannot be x (via proof by contradiction for the hypothetical posed directly above [equally spaced])

[Numbered Foreheads]

By looking at the OP, you can infer that all the logician's are "perfectly logical".

That is, they along will follow proper logic, and they can expect the other two to do so as well.

In this way, this is similar to the Blue Eyed/Brown Eyed puzzle.

OK, with that out of the way, here's the problem: By allowing the option of "giving up", we introduce the observer effect into play (Giving up might change the conditions that made one conclude giving up initially).

The answer provided still may be correct, but it would need improved justification.

Consider the statement,

If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked.

The logic that determines C to give up relies on then scenario where no can give up, which isn't the case, and thus the argument breaks down.

Additionally, A and B wouldn't be able to name their number on the third time, only A would per the OP rules. So wouldn't B give up at some point?

I haven't attempted the deeper logic; to answer the original problem: as each logician doesn't knows if he's a 2 or a 3, the scenarios of 332, 323, and 332 would need to be solved and would reveal the answer to the OP as all logicians viewpoints would be considered.

A helpful tidbit: a logician will say "I don't know" if and only if there exists a situation (from his viewpoint) in which he will be able voice his number. So every "I don't know" you conclude would require such an example. What? Yeah. It seems the proof would be quite involved.

(1) The original problem states that the question is repeated until each of the three logicians has either named his number or given up. So B is not forbidden from winning after A; in fact, by the rules of the OP, all three could theoretically win. In this respect the problem is different from many other similar puzzles which permit only one winner, so I should probably have emphasized this more in the original wording.

(2) The "observer effect" is an interesting aspect, but it is not problematic in this case due to note (1) above. No one who gives up (assuming they are perfect logicians, of course) can ever regret doing so; at most, he may change which of the other logicians may name their number afterward (and we can assume the logicians aren't petty enough to care about that ). Having said that, you have me wondering how a similar problem would play out if only the first logician to name his number could win, since then hypothetically there could be a situation where the winner is indeterminate (whoever gets frustrated and gives up first helps the other person to win). It could get interesting.

(3) Aside from that, what I posted was only the "short version" of the solution, with many details left out. The logic of the argument is correct (despite relying on a hypothetical), it just needs backup. I left out the details deliberately, so readers could work the problem out for themselves and yet have a way to check their answer. I may post a more complete solution later, if no one else does.

From your simple proof:

If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked. C will never be able to.

So would the following be an accurate summarization?

If C keeps saying "I don't know" then both A and B will know their number on the third round based solely on the conclusions from each other saying "I don't know" two times?

As a result of this, C's number is irrelevant and this same outcome (A and B know on the third round) would occur if he is a 2 or a 3?

Note that (if this isn't the case) the observer effect does come into play as C decided to give up based on the conclusions from a hypothetical where no one gives up, rendering his decision moot.

[Numbered Foreheads]

Ah, more twist and turns. Thanks for such a great puzzle.

I

erroneously interpreted a 'win' as being the first to name the hat

number. My mistake. Defining a win as being able to name a hat number

changes the games, since now C is motivated differently. The game now

goes as follows

Turn 1- A: I don't know

Turn 2- B: I don't know

Turn 3- C: I give up

Turn 4- A: My number is 3 (C would not have given up if A's number is 2)

Turn 5- B: My number is 3.

I think you've got it now.

If

everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked. C will never be able to. On the other hand, if one of (A, B) has a 2, then C's situation is much improved, and he can eventually deduce his number. So, the fact that C gave up informs both A and B that their numbers are identical (neither A nor B would know his number at this point in any other way).

Glad you enjoyed it.

By looking at the OP, you can infer that all the logician's are "perfectly logical".

That is, they along will follow proper logic, and they can expect the other two to do so as well.

In this way, this is similar to the Blue Eyed/Brown Eyed puzzle.

OK, with that out of the way, here's the problem: By allowing the option of "giving up", we introduce the observer effect into play (Giving up might change the conditions that made one conclude giving up initially).

The answer provided still may be correct, but it would need improved justification.

Consider the statement,

If everyone has a 3, (or if only C has a 2,) and no one gives up, then A and B will each be able to name their number on the third time they are asked.

The logic that determines C to give up relies on then scenario where no can give up, which isn't the case, and thus the argument breaks down.

Additionally, A and B wouldn't be able to name their number on the third time, only A would per the OP rules. So wouldn't B give up at some point?

I haven't attempted the deeper logic; to answer the original problem: as each logician doesn't knows if he's a 2 or a 3, the scenarios of 332, 323, and 332 would need to be solved and would reveal the answer to the OP as all logicians viewpoints would be considered.

A helpful tidbit: a logician will say "I don't know" if and only if there exists a situation (from his viewpoint) in which he will be able voice his number. So every "I don't know" you conclude would require such an example. What? Yeah. It seems the proof would be quite involved.

[Can you find a closed form equation for the sequence?]

I'm assuming you can't use floor or mod?

[Deal or No Deal]

This is a case of p(A|B). That is the probability that A is true GIVEN that B is true.

In this situation, it's A: having the winner GIVEN B:after 38 rounds of choosing a winner still remains.

The odds that the remaining case is a winner remains the same (this is crux of the Monty Python puzzle) as it would in the beginning: 1/40.

Thus, the probability of A (he has the winning case) is 39/40. He would be a fool to switch.

This isn't the case where the game show host purposely opened 38 of the empty cases. (in which case the probability would be switched). This is the case that B occurred (he himself picked 38 blanks).

Think about it: You have a 1/40 change that you pick the right case. If this happens the probability of B occurring is 1.

If the remaining case had the winner, the probability of B occurring would be (38/39 * 37*38 * ,,, * 1/2) which simplifies to 1/39.

This adds up to 1 + 1/39 = 40/39, with

p(B|A) = 1/( 40/39 ) = 39/40

p(B|!A) = ( 1/39 )/( 40/39 ) = 1/40

Thus, it is more likely that the events occurred as they did if you initially picked the winning case.

[Limit derivation]

Sorry messed up the OP.

lim(1-(1-1/x)^x) [x→ ∞] = (1-1/e) or lim((1-1/x)^x) [x→ ∞] = 1/e

The main question of how it is derived stlil stands.

[Limit derivation]

lim((1-1/x)^x) [x→ ∞] = (1-1/e)

How would one derive this limit?

[Mnemonic]

My bad. I got them mixed up. I guess both are B then.

[Mnemonic]

B-Attitude = beatitude = exalted happiness People with Type-A personalities are the ones that have beatitude.

[Mnemonic]

is B. Get it?

[Mnemonic]

(Type) A-Personality => ?-Attitude

[easy one]

Translate "A * B = C" into

2 * ( B + floor( (A + 1) / 2 ) ) - 1 = C

For A=7, B=8, this works out to 23

floor(x + 1) is more easily written as ceil(x)

[I can't or I won't say]

To solve this, one needs cardinal numbers.

The edge is an uncountable set of points with specific cardinality of Aleph₁.

This can be shown by creating a simple bijection y=tan(pi*(x-0.5)) of (0,1) to R.

Similarly, each plan contains an uncountable number of edges and the cube and an uncountable number of planes.

Thus, the cardinality of the set of points in the cube == Aleph₁*Aleph₁*Aleph₁ == Aleph₁

So the cube contains the same number of points as any of its edges.

[+-0!]

15=0--+

21=0+-+

[Classic eggs problem]

max # of drops = ceil( n/(2^(m-1)) ) + (m - 1) This is because m eqgs allow you a span of 2^(m-1) eqgs to find the correct floor (1 egg to get the span and the rest to get the precise floor). If the number larger than 2^(m-1), you'll have to drop at half a span higher than each multiple of 2^(m - 1) (+ 2^(m - 1) until you reach the span. In this scenario 3 eggs, you're span is 2^2=4; ceil (120/4 ) + 2 = 30 + 2 = 32 (You drop at 2,6,10 and so forth until you hit your span (of 4), then use the remaining two eggs to figure out the floor.

[Mindhunters - A set of ultimate short puzzles]

there is enough moonlight to see enough to drive and it has recently just snowed a tiny layer.