## Poll

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**46 members have voted**

There are 121 possible choices for (B's number, C's number).

Note that any result where B chooses 0 is the same as where B chooses 1, and any result where C chooses 0 is the same as where C chooses 1, so B = 0 and C = 0 can be ignored.

B will not choose 5 or higher as choosing 4 does better for some of C's choices and the same in the rest.

C will not choose 5 or higher as choosing 4 does better for some of B's choices and the same in the rest.

B will not choose 2 as the results are the same as for choosing 1, except when C chooses 1, in which case B does better.

C will not choose 1 as the results are the same as for choosing 2, except when B chooses 2, in which case C does better.

Thus, B is reduced to choosing 1, 3, or 4, and C is reduced to choosing 2, 3, or 4.

The probabilities of B winning for C's choices of 1, 3, and 4 are the same when B chooses 1 with probability 4/9, 3 with probability 1/3, and 4 with probability 2/3

The probabilities of C winning for B's choices of 1, 2, and 4 are the same when C chooses 2 with probability 1/2 and 4 with probability 1/2.

B tosses a coin, and chooses 1 with heads and 4 with tails

C rolls two six-sided dice; on rolls of 2-5 or 7 (probability 4/9), choose 2; on rolls of 6, 8, or 11 (probability 1/3), choose 3; on rolls of 9, 10, or 12 (probability 2/9), choose 4

Quote:ThatDonGuy

There are 121 possible choices for (B's number, C's number).

Note that any result where B chooses 0 is the same as where B chooses 1, and any result where C chooses 0 is the same as where C chooses 1, so B = 0 and C = 0 can be ignored.

B will not choose 5 or higher as choosing 4 does better for some of C's choices and the same in the rest.

C will not choose 5 or higher as choosing 4 does better for some of B's choices and the same in the rest.

B will not choose 2 as the results are the same as for choosing 1, except when C chooses 1, in which case B does better.

C will not choose 1 as the results are the same as for choosing 2, except when B chooses 2, in which case C does better.

Thus, B is reduced to choosing 1, 3, or 4, and C is reduced to choosing 2, 3, or 4.

The probabilities of B winning for C's choices of 1, 3, and 4 are the same when B chooses 1 with probability 4/9, 3 with probability 1/3, and 4 with probability 2/3

The probabilities of C winning for B's choices of 1, 2, and 4 are the same when C chooses 2 with probability 1/2 and 4 with probability 1/2.

B tosses a coin, and chooses 1 with heads and 4 with tails

C rolls two six-sided dice; on rolls of 2-5 or 7 (probability 4/9), choose 2; on rolls of 6, 8, or 11 (probability 1/3), choose 3; on rolls of 9, 10, or 12 (probability 2/9), choose 4

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Well reasoned, but sadly incorrect.

Player A wants to win the round.

Player C wants to lose the round.

Player B has options ... Or does he?

The only chances that B has of winning are:

- if B has the median score in the last round and that median score is 0,1, or (less desirably) 2.

- if C has the median score

The only chance that C has of winning is if B has the median score in the last round and if that median score (of B's) is 2 or higher. Therefore, C must pick higher than 3, and should probably pick 10.

B must pick 10.

C will lose if he picks anything less than 10 so C must pick 10, which gives both B and C a 50/50 shot at winning.

Quote:Ace2Slight variation on a previously posted problem:

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s $1 per flip and they are really bored, so they decide to do one million flips. After each flip, the loser pays the winner $1. The players may bring any bankroll they want, but if a player goes bankrupt he automatically loses the game. Tom is very wealthy and tells Henry he's bringing a million dollars to the game. Henry is not rich and decides to bring an amount that will give him 50% confidence of not going bankrupt. How much should Henry bring?

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Assuming the risk of bankrupt is DOUBLE* the risk of loss MORE THAN X unit :-

σ = 1, μ = 0, N = 1,000,000

P[ (-X - μ)/(N)^0.5/σ] >= 50%/2 >= 0.25

When probability is 25%, Z score = -0.6745 = -X/(N)^0.5/σ = -X/1000

X = 674.5

* I think this only applicable when σ = 1 and μ = 0, for games with high variance(video poker, 3CP . . . .), you can't just DOUBLE the risk, what do you think ?

Quote:gordonm888

The only chances that B has of winning are:

- if B has the median score in the last round and that median score is 0,1, or (less desirably) 2.

- if C has the median score

The only chance that C has of winning is if B has the median score in the last round and if that median score (of B's) is 2 or higher. Therefore, C must pick higher than 3, and should probably pick 10.

B must pick 10.

C will lose if he picks anything less than 10 so C must pick 10, which gives both B and C a 50/50 shot at winning.

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Correct!!

Very good.

To have any shot at winning, C (who currently has the most overall points) cannot win this final round. Therefore, C should pick a number that is extreme — that is, either 0 or 10. If C were to pick 0, then B could win the round by writing down 1, in which case B would be the overall winner with 9 points. Seeing that logic, let’s suppose C were to write down 10. What would B do?

If B writes a number less than 3, then A will win the round and the entire game. If B writes a number greater than 3 but less than 10, then B will win the round but C will win the overall game. If B writes down 3 (like A), then the winner will either be A (if A wins the round) or C (if B wins the round). So B’s only logical remaining move is to write down 10. In this case, B and C will flip a coin to see who wins the round. The overall winner is whoever between them did not win the round.

To summarize, both B and C should have chosen 10, giving them both a 50 percent chance of mediocre victory.

A number of readers alternatively suggested that all three players should pick 3, in which case the winner of the round would be picked randomly. If A won the round, then A would be the overall winner. If B won the round, then C would be the overall winner. And if C won the round, B would be the overall winner. That gave each player a one-in-three chance of winning. As we just saw, by both choosing 10, players B and C can improve their chances to one in two. So choosing 10 was the better strategy for both B and C.

Several solvers, including Jenny Mitchell and Erin Sy, also observed that B and C both choosing 10 represented the lone Nash equilibrium of the game, where no individual player can do better by switching strategies.

And so, in this particular Game of Mediocrity, it paid to be extreme.

----------------------------------------------------

Quote:ssho88

* I think this only applicable when σ = 1 and μ = 0, for games with high variance(video poker, 3CP . . . .), you can't just DOUBLE the risk, what do you think ?

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If by "this" you mean the step of multiplying by 50%, I believe that works with any standard deviation, but the mean must be zero.

Quote:WizardQuote:ssho88

* I think this only applicable when σ = 1 and μ = 0, for games with high variance(video poker, 3CP . . . .), you can't just DOUBLE the risk, what do you think ?

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If by "this" you mean the step of multiplying by 50%, I believe that works with any standard deviation, but the mean must be zero.

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After further thought(step of multiplying by 50%), I agree with you that mean must be 0 AND standard deviation can be any value.

B will not choose 5 or higher as choosing 4 does better for some of C's choices and the same in the rest.

C will not choose 5 or higher as choosing 4 does better for some of B's choices and the same in the rest.

Neither of those is correct. I think I had my result matrix switched between B and C for values > 3.

B will not choose 2 as choosing 1 has the same result, except that if C chooses 2, B's probability of winning is 1 instead of 3/4.

For all other possible choices for B, C cannot improve the probability of winning by choosing something other than 10, so C always chooses 10.

The only way B has a chance to win if C chooses 10 is if B also chooses 10.

Given: Four points (usually called vertexes) are placed randomly in space.

Any line that connects two vertexes is called an "edge." Thus, with four randomly-placed points there are six possible edges. Each edge is defined to have the same probability, p, of occurring/existing.

Write an algebraic expression (or ensemble of algebraic equations) for the probability that edges will exist such that they form one or more loops.

- Note that if two edges cross at a point that is not a vertex that this is not a corner in a closed loop. In other words, if the four vertexes were cities, and the edges are roads connecting cities, then at any point where two roads cross it should be assumed that one road has a bridge over the other road.

Extra credit for a plot (x-y graph) of the probability of at least one loop occurring as a function of p (the probability that an edge occurs between any two points.)

Quote:gordonm888I have been looking at random graph theory, which is a field in discrete mathematics. Notably, combination math is central to this math field. Here is the simplest (but non-trivial) conventional random graph problem I can think of.

Given: Four points (usually called vertexes) are placed randomly in space.

Any line that connects two vertexes is called an "edge." Thus, with four randomly-placed points there are six possible edges. Each edge is defined to have the same probability, p, of occurring/existing.

Write an algebraic expression (or ensemble of algebraic equations) for the probability that edges will exist such that they form one or more loops.

- Note that if two edges cross at a point that is not a vertex that this is not a corner in a closed loop. In other words, if the four vertexes were cities, and the edges are roads connecting cities, then at any point where two roads cross it should be assumed that one road has a bridge over the other road.

Extra credit for a plot (x-y graph) of the probability of at least one loop occurring as a function of p (the probability that an edge occurs between any two points.)

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If fewer than three edges exist, there cannot be any loops.

If three exist, they must connect each other; there are 4 ways to do this.

If four exist, either three of them connect each other, with the fourth path connecting one of those three vertices to the fourth one, or all four do.

There are four possible 3-loops, each with 3 possibilities for the fourth edge, and three possible 4-loops (1234, 1243, 1324 - note 1342, 1423, and 1432 are the same as 1243, 1324, and 1234).

If five or more exist, then at least one 3-loop must exist

P = 4 p^3 (1-p)^3 + 15 p^4 (1-p)^2 + 6 p^5 (1-p) + p^6

= 4 p^3 + 3 p^4 - 12 p^5 + 6 p^6

You are an expert counterfeiter, and you specialize in forging one of the most ubiquitous notes in global circulation, the U.S. $100 bill. You’ve been able to fool the authorities with your carefully crafted C-notes for some time, but you’ve learned that new security features will make it impossible for you to continue to avoid detection. As a result, you decide to deposit as many fake notes as you dare before the security features are implemented and then retire from your life of crime.

You know from experience that the bank can only spot your fakes 25 percent of the time, and trying to deposit only counterfeit bills would be a ticket to jail. However, if you combine fake and real notes, there’s a chance the bank will accept your money. You have $2,500 in bona fide hundreds, plus a virtually unlimited supply of counterfeits. The bank scrutinizes cash deposits carefully: They randomly select 5 percent of the notes they receive, rounded up to the nearest whole number, for close examination. If they identify any note in a deposit as fake, they will confiscate the entire sum, leaving you only enough time to flee.

How many fake notes should you add to the $2,500 in order to maximize the expected value of your bank account? How much free money are you likely to make from your strategy?

Quote:GialmereA Riddler racket puzzle by Jason Ash...

You are an expert counterfeiter, and you specialize in forging one of the most ubiquitous notes in global circulation, the U.S. $100 bill. You’ve been able to fool the authorities with your carefully crafted C-notes for some time, but you’ve learned that new security features will make it impossible for you to continue to avoid detection. As a result, you decide to deposit as many fake notes as you dare before the security features are implemented and then retire from your life of crime.

You know from experience that the bank can only spot your fakes 25 percent of the time, and trying to deposit only counterfeit bills would be a ticket to jail. However, if you combine fake and real notes, there’s a chance the bank will accept your money. You have $2,500 in bona fide hundreds, plus a virtually unlimited supply of counterfeits. The bank scrutinizes cash deposits carefully: They randomly select 5 percent of the notes they receive, rounded up to the nearest whole number, for close examination. If they identify any note in a deposit as fake, they will confiscate the entire sum, leaving you only enough time to flee.

How many fake notes should you add to the $2,500 in order to maximize the expected value of your bank account? How much free money are you likely to make from your strategy?

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I can't think of any way to integrate combinations particularly easily, so this appears to be a Brute Force problem

Let N be the number of fake bills; the object is to maximize the following value:

N x C(25, celing((25 + N)/20)) / C(25 + N, ceiling((25 + N)/20))

since you are drawing ceiling((25 + N)/20) bills from a set of 25 + N, and want to get one of the combinations of ceiling((25 + N)/20) from a set of 25.

The number of counterfeit bills needed is 15; the expected value is $576.923

Quote:GMI disagree. You don't seem to take into account that the probability of the bank not detecting any fake notes depends on the number of fake and genuine notes in the sample selected by the bank and I don't think there is a nice formula for it.

Actually, I did forget to take something into account - the fact that there is a possibility that fake notes won't be detected.

Quote:GMI disagree. You don't seem to take into account that the probability of the bank not detecting any fake notes depends on the number of fake and genuine notes in the sample selected by the bank and I don't think there is a nice formula for it.

Let n be the number of fake notes you are trying deposit. Let a be the number of notes selected by the bank, which is (n+25)/20 rounded up. The probability that exactly b of these notes are fake is binomial(n,b)*binomial(25,a-b)/binomial(n+25,a), then the conditional probability of the bank not noticing any of the fake notes is (3/4)^b. Hence the probability of the bank not noticing the fake notes is the sum of (3/4)^b*binomial(k,b)*binomial(25,a-b)/binomial(n+25,a) from b=0 to b=a, and the expectation of the amount you are able to deposit is 100(n+25) times this probability. I calculated the expectation for all n up to 200 and the maximum is at n=55, $3756.89, and the expected profit is $1256.89.

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For 55 fakes plus 25 genuine bills = 80 total bills, we know the bank will select 4 bills and that of the 55 fakes, they can only tell that 14 of them (25% rounded up) are actually fakes. They can't tell that the other 41 are actually fake.

So I get the expected value for a deposit of 80 total bills of:

combin(25+41,4) / combin(80,4) * 80 = 36.4557 bills = $3645.57

Quote:GMLet n be the number of fake notes you are trying deposit. Let a be the number of notes selected by the bank, which is (n+25)/20 rounded up. The probability that exactly b of these notes are fake is binomial(n,b)*binomial(25,a-b)/binomial(n+25,a), then the conditional probability of the bank not noticing any of the fake notes is (3/4)^b. Hence the probability of the bank not noticing the fake notes is the sum of (3/4)^b*binomial(k,b)*binomial(25,a-b)/binomial(n+25,a) from b=0 to b=a, and the expectation of the amount you are able to deposit is 100(n+25) times this probability. I calculated the expectation for all n up to 200 and the maximum is at n=55, $3756.89, and the expected profit is $1256.89.

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Correct!!

Well done.

The full counterfeit solution (counterfeit bills, that is … the solution itself is very real) comes to us from this puzzle’s submitter, Jason Ash:

Your expected gain is the weighted average of expected profit from successful deposits and expected losses from bank seizures. With 55 fake notes, there is a 47 percent chance — a combination of the chances of the bank selecting our fake notes for examination and of the examination revealing any fake notes — we avoid detection and collect a profit of $5,500, the value of the fake notes we were able to sneak into circulation. (Remember we started with $2,500, so only the fake notes count as profit.) There is a 53 percent chance we are caught by the bank and lose the $2,500 in real dollars we used as decoys. Therefore, our expected profit is (0.47)(5,500) – (0.53)(2,500) = 1,256.

How can we be sure no other strategy produces a higher profit? For example, suppose we combine 30 fake notes with 25 real notes instead. The bank will select three notes for its audit — that is, 5 percent of 55, rounded up to the nearest whole number. Depending on our luck, the bank could choose all three fake notes, all three real notes, or some combination in between.

Let’s assume the bank randomly chooses two fake notes and one real note for its audit. This occurs with roughly 41.5 percent probability, given by the formula (3 choose 2)(30/55)(29/54)(25/53). Each fake note is detected 25 percent of the time, which means at least one fake note from the pool of two is detected 1−0.752

= 43.75 percent of the time. We use similar logic to solve for the likelihood and detection rate of the audits with three fake, one fake and zero fake notes. The overall detection rate is equal to the weighted average across each potential audit.

For the 30 fake, 25 real strategy, the probability of success is 64.3 percent — again, we need to either avoid their selecting our fake notes for examination or, if they do select them, avoiding identification of them as such — and the probability of detection is 35.7 percent. Therefore, the expected profit is (0.643)(3,000) – (0.357)(2,500) = 1,038. That’s a decent payday, but we can do better. The chart below shows the profit for strategies with up to 200 fake notes, and it illustrates that the maximum is achieved when we use 55 of them.

We can see two patterns above. First, as we move to the right, we enter the “greedy danger zone,” in which the bank becomes more likely to discover our fraud and seize the starting capital, resulting in larger and larger expected losses. Second, we see “sawtooth” behavior caused by the bank’s practice of auditing 5 percent of deposited notes. The effect is significant: 55 is the ideal answer because if we deposit 80 total notes, the bank will audit four. If we use 56 fake notes for a total of 81, then the bank audits five notes instead, which cuts the expected gain in half!

A life of crime only pays if you’re good with numbers. But with great power comes great responsibility, Riddler Nation.

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Quote:ThatDonGuyQuote:gordonm888I have been looking at random graph theory, which is a field in discrete mathematics. Notably, combination math is central to this math field. Here is the simplest (but non-trivial) conventional random graph problem I can think of.

Given: Four points (usually called vertexes) are placed randomly in space.

Any line that connects two vertexes is called an "edge." Thus, with four randomly-placed points there are six possible edges. Each edge is defined to have the same probability, p, of occurring/existing.

Write an algebraic expression (or ensemble of algebraic equations) for the probability that edges will exist such that they form one or more loops.

- Note that if two edges cross at a point that is not a vertex that this is not a corner in a closed loop. In other words, if the four vertexes were cities, and the edges are roads connecting cities, then at any point where two roads cross it should be assumed that one road has a bridge over the other road.

Extra credit for a plot (x-y graph) of the probability of at least one loop occurring as a function of p (the probability that an edge occurs between any two points.)

link to original post

If fewer than three edges exist, there cannot be any loops.

If three exist, they must connect each other; there are 4 ways to do this.

If four exist, either three of them connect each other, with the fourth path connecting one of those three vertices to the fourth one, or all four do.

There are four possible 3-loops, each with 3 possibilities for the fourth edge, and three possible 4-loops (1234, 1243, 1324 - note 1342, 1423, and 1432 are the same as 1243, 1324, and 1234).

If five or more exist, then at least one 3-loop must exist

P = 4 p^3 (1-p)^3 + 15 p^4 (1-p)^2 + 6 p^5 (1-p) + p^6

= 4 p^3 + 3 p^4 - 12 p^5 + 6 p^6

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Correct. Very good. I wondered whether this problem statement had broken the thread because of the long silence that followed it's posting.

You are on your way to visit your Grandma, who lives at the end of the valley. It's her birthday, and you want to give her the cakes you've made.

Between your house and her house, you have to cross 7 bridges, and as it goes in the land of make believe, there is a troll under every bridge! Each troll, quite rightly, insists that you pay a troll toll. Before you can cross their bridge, you have to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.

How many cakes do you have to leave home with to make sure that you arrive at Grandma's with exactly 2 cakes?

Al and Ben are standing back to back next to a railroad track. When the front of a train passes them, Al starts to walk in the opposite direction of the train, while Ben starts to walk in the same direction as the train. Each stops walking when the back of the train passes him.

If the two of them walk at the same speed, and Al walks exactly 30 feet, and Ben walks exactly 45 feet, how long is the train?

Assuming I am reading this right, the number of bridges does not matter.

If you start with 2 cakes, each troll will take 1 cake, then give it back.

The answer is 2 cakes

The train moves its length - 30 feet in the time it takes Al to walk 30 feet

The train moves its length + 45 feet in the time it takes Ben to walk 45 feet

The train moves 75 feet in the time it takes either to walk 15 feet, so it moves 150 feet in the time it takes Al to walk 30 feet

150 feet = its length - 30 feet, so the train's length is 180 feet

What a great puzzle...Quote:GialmereHere's a few easy Monday puzzles...

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You are on your way to visit your Grandma, who lives at the end of the valley. It's her birthday, and you want to give her the cakes you've made.

Between your house and her house, you have to cross 7 bridges, and as it goes in the land of make believe, there is a troll under every bridge! Each troll, quite rightly, insists that you pay a troll toll. Before you can cross their bridge, you have to give them half of the cakes you are carrying, but as they are kind trolls, they each give you back a single cake.

How many cakes do you have to leave home with to make sure that you arrive at Grandma's with exactly 2 cakes?

So at each bridge you arrive with two cakes, pay one cake in toll, and receive one back. Thus you can start with two cakes!

No need to work through seven iterations!

We also know in the first two seconds the back of the train reached 30ft from the starting position, hence it was 30+2*75 foot long = 180ft.

Boss told me that as a security guard, it's my job to watch the office. I'm up to season 4.

Quote:DieterBoss told me that as a security guard, it's my job to watch the office. I'm up to season 4.

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Don't be ridiculous - The Office only had 2 seasons (and a Christmas special).

Oh, you're thinking about that American nonsense, aren't you?

I came up with the same answer everyone else did.

May I suggest an alternate version where grandma gets 3 cakes?

Is there a way to express the answer as a function of c cakes and t trolls?

Quote:WizardRE: Puzzle #1

I came up with the same answer everyone else did.

May I suggest an alternate version where grandma gets 3 cakes?

Is there a way to express the answer as a function of c cakes and t trolls?

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Assuming C is the number of cakes you have at the end, you start with 2^T C - 2^(T + 1) + 2 cakes

If you work backwards, and assume you have 2^T C - 2^(T + 1) + 2 cakes with T bridges remaining, then, at the previous bridge, you had 2^T C - 2^(T + 1) + 1 before you got one back, which meant you had twice as many, or 2^(T + 1) C - 2^((T + 1) + 1) + 2, when you reached that bridge.

Quote:ThatDonGuy

Assuming I am reading this right, the number of bridges does not matter.

If you start with 2 cakes, each troll will take 1 cake, then give it back.

The answer is 2 cakes

The train moves its length - 30 feet in the time it takes Al to walk 30 feet

The train moves its length + 45 feet in the time it takes Ben to walk 45 feet

The train moves 75 feet in the time it takes either to walk 15 feet, so it moves 150 feet in the time it takes Al to walk 30 feet

150 feet = its length - 30 feet, so the train's length is 180 feet

Quote:charliepatrickAfter the last bridge you have to have two cakes, The troll gives you one, so after the troll had taken their toll you needed one cake. Hence prior to the bridge you had two cakes.

So at each bridge you arrive with two cakes, pay one cake in toll, and receive one back. Thus you can start with two cakes!

No need to work through seven iterations!Let's assume the walkers travel at 15 ft/sec and the train is travelling right to left. After 2 seconds the back of the train is 30 feet from the starting point (on the right) as that is how far Al has walked, call this "X". In the next second Ben manages to complete his 45ft walk so is 45 feet to the left of the starting point, call this "Y"; at this time the back of the train passes him. So adding 45ft and 30ft gets the distance (YX) and is how far the back of the train travelled in the third second. Thus the train travels 75ft/sec.

We also know in the first two seconds the back of the train reached 30ft from the starting position, hence it was 30+2*75 foot long = 180ft.

Quote:WizardRE: Puzzle #1

I came up with the same answer everyone else did.

Correct!!

If you don't know math, you'll starve to death.

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The picture above shows two semicircles. The lighter region (inside the larger semicircle but outside the smaller one) has an area of 7.

What’s the area of the darker region?

Quote:Gialmere...What’s the area of the darker region?

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The logic involves creating right angled triangles (one for each semi circle) that can be formed used the diameter as the hypotenuse. Consider the large one first, its hypotenuse is 2r. Now chop the triangle in half (so two of the sides of it are radii) - clearly its area is half the original one. Then rotate it by 45 degrees. This will now line up so its hypotenuse matches the diameter of the smaller semi-circle.

Thus by similar triangles within similar semi circles, the area of the smaller semi-circle is half the larger one.

Let A and B be the endpoints of the "base" of the larger semicircle, O the midpoint of AB, C and D the endpoints of the "base" of the smaller semicircle, and M the midpoint of CD.

Since the orientation of the smaller semicircle is not specified, assume its base is parallel to the base of the larger one. (This may be a requirement in order for the smaller semicircle's endpoints to touch the larger one as well as the smaller one to touch the midpoint of the larger base; I'll look into that...)

Edit:Actually, its orientation is irrelevant - since CD is a chord of AB, the line from the larger semicircle's center to the midpoint of a chord is always perpendicular to the chord, since the two triangles formed are SSS-congruent)

This means CMO is a right angle.

CM, DM, and OM are all radii of the smaller semicircle, so CM = OM, which makes OMC an isosceles right triangle with hypotenuse sqrt(2) x CM.

This is also the radius of the larger semicircle.

Since the ratio of the radii of the two semicircles is sqrt(2), the ratio of their areas is (sqrt(2))^2 = 2, which means the area of the larger semicircle outside of the smaller semicircle must be the same as the area of the smaller semicircle.

Therefore the area of the darker region = 7.

When you've googled the answer, then if you can understand the answer, then you are still high in my estimation.

If you can prove the answer is 1/6, then you are Alan M

The question....

There are 10 pieces of paper face down on a table in front of you. Each piece of paper has a number on it. Literally ANY REAL NUMBER. Can be a decimal fraction, can be tiny, can be massive, can be rational or irrational, though irrational would make it messy, can be ANY real number.

You didn't write the numbers and you have no idea what they are!

You will turn the pieces of paper over, 1 by 1. You must stop when you think you have the highest number. Simple enough?

The final piece of paper you choose to flip over is your final number; you can’t flip 6 pieces of paper over and then go back and choose the 3rd piece of paper as your number. What you opick, you stick with as your final answer.

What strategy would you use in this game to maximise the chances of having the highest number?

If you use the best strategy, what is your probability of being right.

Clue: Use a random number generator to play this game.!!!!!

There is a fun corollary to this question...

There are ten hot women at a party and you decide, because you're an old mathematician, to marry the most sexually aggressive one. Till death do you part.

You can have a day of unbridled sex with as many of them as you want, individually, of course. But only once per woman (or not at all), You will rate them yourself out of 100.

The last one you have sex with, you will marry and live out the rest of your days with.

What is the best strategy to give you the best probability of marrying the 'best' one.

What is the probability you do actually marry the 'best' one.

Who says maths has no practical value?

Quote:OnceDearHere's a cripplingly hard maths puzzle. If you can work it out, you are a better man than I.

When you've googled the answer, then if you can understand the answer, then you are still high in my estimation.

If you can prove the answer is 1/6, then you are Alan M

The question....

There are 10 pieces of paper face down on a table in front of you. Each piece of paper has a number on it. Literally ANY REAL NUMBER. Can be a decimal fraction, can be tiny, can be massive, can be rational or irrational, though irrational would make it messy, can be ANY real number.

You didn't write the numbers and you have no idea what they are!

You will turn the pieces of paper over, 1 by 1. You must stop when you think you have the highest number. Simple enough?

The final piece of paper you choose to flip over is your final number; you can’t flip 6 pieces of paper over and then go back and choose the 3rd piece of paper as your number. What you opick, you stick with as your final answer.

What strategy would you use in this game to maximise the chances of having the highest number?

If you use the best strategy, what is your probability of being right.

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This has been asked already.

Multiple times.

I never could decipher how the solution was determined

Reject the first N / e values, where N is the total number (in this case, 100), then accept the first value that is higher than all of the values revealed up to that point. For N = 10, reject the first 3, then, starting with the fourth, accept the first number that is higher than all of the numbers before it.

Harder problem:

You are the second person in line to do this. The first one says that he will implement the above strategy. Before you make your choice, you are told that the first person's answer was incorrect. What is your strategy now?

Then look at the rest and pick the first one which is higher than the one you've already seen.

Firstly if the best is in the first N, then you can't win.

If the second best is in the first N then you must win (as only the best is left and no other pick will apply)

If the third best is in the first N then it's 50/50 whether you encounter the 1st or 2nd during the second part.

Similar logic for 4th best (i.e. 1 chance in 3)

Similarly for 5th best (i.e. 1 chance in 4)

Thus what is the best value for N.

Best seen | Perms | Pr(win) |

1st | 1 | none |

2nd | 1 | 1/1 |

3rd | 1 | 1/2 |

4th | 1 | 1/3 |

5th | 1 | 1/4 |

6th | 1 | 1/5 |

7th | 1 | 1/6 |

8th | 1 | 1/7 |

9th | 1 | 1/8 |

10th | 1 | 1/9 |

28.290% |

Best seen | Perms | Pr(win) |

1st | 9 | none |

2nd | 8 | 1/1 |

3rd | 7 | 1/2 |

4th | 6 | 1/3 |

5th | 5 | 1/4 |

6th | 4 | 1/5 |

7th | 3 | 1/6 |

8th | 2 | 1/7 |

9th | 1 | 1/8 |

10th | 0 | 1/9 |

36.579% |

Best seen | Perms | Pr(win) |

1st | 36 | none |

2nd | 28 | 1/1 |

3rd | 21 | 1/2 |

4th | 15 | 1/3 |

5th | 10 | 1/4 |

6th | 6 | 1/5 |

7th | 3 | 1/6 |

8th | 1 | 1/7 |

9th | 0 | 1/8 |

10th | 0 | 1/9 |

Pr of winning | 39.869% |

Best seen | Perms | Pr(win) |

1st | 84 | none |

2nd | 56 | 1/1 |

3rd | 35 | 1/2 |

4th | 20 | 1/3 |

5th | 10 | 1/4 |

6th | 4 | 1/5 |

7th | 1 | 1/6 |

8th | 0 | 1/7 |

9th | 0 | 1/8 |

10th | 0 | 1/9 |

39.825% |

I believe for N the best strategy is SQRT(N) but couldn't prove it!

Quote:charliepatrickThis is known as the wives problem (or similar name) but I get much better chances than 1 in 6.

The plan is to look at the first N items and remember the highest (so far)

Then look at the rest and pick the first one which is higher than the one you've already seen.

Firstly if the best is in the first N, then you can't win.

If the second best is in the first N then you must win (as only the best is left and no other pick will apply)

If the third best is in the first N then it's 50/50 whether you encounter the 1st or 2nd during the second part.

Similar logic for 4th best (i.e. 1 chance in 3)

Similarly for 5th best (i.e. 1 chance in 4)

Thus what is the best value for N.

Best seen Perms Pr(win) 1st 1 none 2nd 1 1/1 3rd 1 1/2 4th 1 1/3 5th 1 1/4 6th 1 1/5 7th 1 1/6 8th 1 1/7 9th 1 1/8 10th 1 1/9 28.290%

Best seen Perms Pr(win) 1st 9 none 2nd 8 1/1 3rd 7 1/2 4th 6 1/3 5th 5 1/4 6th 4 1/5 7th 3 1/6 8th 2 1/7 9th 1 1/8 10th 0 1/9 36.579%

Best seen Perms Pr(win) 1st 36 none 2nd 28 1/1 3rd 21 1/2 4th 15 1/3 5th 10 1/4 6th 6 1/5 7th 3 1/6 8th 1 1/7 9th 0 1/8 10th 0 1/9 Pr of winning 39.869%

Best seen Perms Pr(win) 1st 84 none 2nd 56 1/1 3rd 35 1/2 4th 20 1/3 5th 10 1/4 6th 4 1/5 7th 1 1/6 8th 0 1/7 9th 0 1/8 10th 0 1/9 39.825%

I believe for N the best strategy is SQRT(N) but couldn't prove it!

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Probability of success with best strategy about 1 in 3

Strategy is to calculate N= Count of cards/e (That's e: Eulers number)

Reject the first N cards

Stop at the first, next card with a higher value than all those previously rejected.

https://www.youtube.com/watch?v=OeJobV4jJG0

Of course, according to a notorious youtube video, you don't choose a wife or even a date by how sexy they are or how great in bed they may be. The other major factor is how crazy they are. If they are too crazy, it doesn't matter how sexy they are, you will regret it forever.

I know this answer is unlikely to be completely correct but:

You choose a number by where it falls in the range defined by the previously generated numbers.

Given the stated maximum of 10 numbers, from the 5th number on, choose the number if it is higher than at least 4 of the previous numbers that were generated.

Quote:gordonm888I give OnceDear credit for writing an unusually entertaining problem statement.

Of course, according to a notorious youtube video, you don't choose a wife or even a date by how sexy they are or how great in bed they may be. The other major factor is how crazy they are. If they are too crazy, it doesn't matter how sexy they are, you will regret it forever.

I know this answer is unlikely to be completely correct but:

You choose a number by where it falls in the range defined by the previously generated numbers.

Given the stated maximum of 10 numbers, from the 5th number on, choose the number if it is higher than at least 4 of the previous numbers that were generated.

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I recognise that this has, indeed been seen before. Sorry.

Quote:charliepatrickIt's also 7.

The logic involves creating right angled triangles (one for each semi circle) that can be formed used the diameter as the hypotenuse. Consider the large one first, its hypotenuse is 2r. Now chop the triangle in half (so two of the sides of it are radii) - clearly its area is half the original one. Then rotate it by 45 degrees. This will now line up so its hypotenuse matches the diameter of the smaller semi-circle.

Thus by similar triangles within similar semi circles, the area of the smaller semi-circle is half the larger one.

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Quote:ThatDonGuy

Let A and B be the endpoints of the "base" of the larger semicircle, O the midpoint of AB, C and D the endpoints of the "base" of the smaller semicircle, and M the midpoint of CD.

Since the orientation of the smaller semicircle is not specified, assume its base is parallel to the base of the larger one. (This may be a requirement in order for the smaller semicircle's endpoints to touch the larger one as well as the smaller one to touch the midpoint of the larger base; I'll look into that...)

Edit:Actually, its orientation is irrelevant - since CD is a chord of AB, the line from the larger semicircle's center to the midpoint of a chord is always perpendicular to the chord, since the two triangles formed are SSS-congruent)

This means CMO is a right angle.

CM, DM, and OM are all radii of the smaller semicircle, so CM = OM, which makes OMC an isosceles right triangle with hypotenuse sqrt(2) x CM.

This is also the radius of the larger semicircle.

Since the ratio of the radii of the two semicircles is sqrt(2), the ratio of their areas is (sqrt(2))^2 = 2, which means the area of the larger semicircle outside of the smaller semicircle must be the same as the area of the smaller semicircle.

Therefore the area of the darker region = 7.

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Correct!!

Well done.

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At long last, Dakota Jones is close to finding the Lost Arc, a geometric antiquity buried deep in the sands of Egypt. Along the way, she discovered what she described as a “highly symmetric crystal” that’s needed to precisely locate the Arc. Dakota measured the crystal using her laser scanner and relayed the results to you. But nefarious agents have gotten wind of her plans, and Dakota and the crystal are nowhere to be found.

Locating the Arc is now up to you. To do that, you must recreate the crystal using the data from Dakota’s laser scanner. The scanner takes a 3D object, and records 2D cross-sectional slices along the third dimension. Here’s the looping animation file the scanner produced for the crystal:

What sort of three-dimensional shape is the crystal?

No pressure — Dakota Jones, nay, the entire world, is counting on you to locate the Lost Arc and ensure its place in a museum!

The inside, and the outside. :)

Quote:Gialmere

...

What sort of three-dimensional shape is the crystal?

No pressure — Dakota Jones, nay, the entire world, is counting on you to locate the Lost Arc and ensure its place in a museum!

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The crystal is a cube.

Quote:JoemanTwo.

The inside, and the outside. :)

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If I remember my "new math" right, a circle is a line and lines have two sides.

Quote:GialmereWhat sort of three-dimensional shape is the crystal?

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If you take a side of the triangle at this largest point and divide by sqrt(2), you get the length of the side of the cube.

It's a cube, being measured perpendicular to a line through two opposite vertices.

Quote:ChesterDogThe crystal is a cube.

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Quote:WizardI'm pretty sure it's a cube.

If you take a side of the triangle at this largest point and divide by sqrt(2), you get the length of the side of the cube.

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Quote:ThatDonGuy

It's a cube, being measured perpendicular to a line through two opposite vertices.

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Correct!!

Well done all.

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Professor: I am about to make a statement. If that statement is false, you will give me back $1. If it is true, you keep both $1 bills. Do you agree?

Student: I agree.

What should the professor's statement be so that the student must pay the professor $100?

I think Ed will especially like this one.

Note: I am looking to see if things like "the liar paradox" can be apply.

Quote:ksdjdj.Must the student pay the professor a net value of $100?

Note: I am looking to see if things like "the liar paradox" can be apply.

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Well, the student would be in a position to have to pay the $100 or reneg on the deal he accepted.

Yes, this is a Liar's Paradox type of problem.