## Poll

21 votes (45.65%) | |||

14 votes (30.43%) | |||

6 votes (13.04%) | |||

3 votes (6.52%) | |||

12 votes (26.08%) | |||

3 votes (6.52%) | |||

6 votes (13.04%) | |||

5 votes (10.86%) | |||

12 votes (26.08%) | |||

10 votes (21.73%) |

**46 members have voted**

(i) Sir Albert wants to make the trip in minimum time.

(ii) "Staying overnight" means, for arguments sake, the day consists of travel, where you depart at 6am and arrive at 6pm; and wish to stay from 6pm until 6am the following day at the city. Assume no-one moves between the hours of 6pm and 6am.

(iii) Rations for a day mean either at 6am you must have enough food for that day which you will steadily consume by 6pm, or at 6am you have some food and during the day will pick up more food so that by 6pm you have eaten a day's food. For arguments sake you have eaten half a days' food by midday.

(iv) A maximum of six days' of food means at 6am, or any other time, you cannot have more than six days' of food on you.

(v) Helpers have to eat, they cannot deliver food and then die.

Days 1 and 2 Albert and Brian, each with 6 days' of food, set out from Base and travel 50 miles. Let's call this intermediate point "The Village". At this stage each have 4 days' of food left.

On Day 3 Albert takes 2 days' of food from Brian, and now has 6 days' of food. Brian has 2 days' of food.

Days 3 to 5, 6 to 8. Albert has enough food to travel to the city (now only 75 miles away, 3 days), stay overnight, and return (3 days) to the village.

Days 3 to 4. Brian has 2 days' of food so returns to Base.

Days 5 to 6. Brian stays at Base.

Days 7 to 8. Brian starts with 6 days' of food and sets out to the Village. Brian arrives at the village at the end of day 8 with 4 days' of food.

Days 9 to 10. Brian gives Albert 2 days' of food and they both return to Base.

Day 1: After having set out towards the city, all 3 explorers are at the 25 Mile post, Chuck gives 1 ration each to Albert & Bill.

Day 2: Albert & Bill are at Mile Post 50 while Chuck went back to the coast. Bill gives one of his rations to Albert, leaving him with 4.

Day 3: Albert and Bill have made it to Mile Post 75. Bill gives Albert one of his rations, leaving him with 2. Chuck spends the entire day at the beach.

Day 4: Albert has trekked alone to Mile Post 100 while Bill, having headed back to the coast is at MP 50. Chuck spends another day at the beach. This time in a cabana, nursing his sunburn from the previous day.

Day 5: Albert has reached the city! Chuck started toward the city with a full compliment of rations and meets Bill at MP 25 where he shares his rations.

Day 6: Albert spends the day archaeology-ing the Lost City while Bill & Chuck make their way back to the coast.

Day 7: Albert makes it back to MP 100, having only 2 rations left. Bill & Chuck both have a beach day.

Day 8: Albert makes it to MP 75 while Bill, with a full compliment of rations, makes it to MP 25

Day 9: Albert and Bill meet at MP 50 where Bill shares 2 of his 4 remaining rations with Albert.

Day 10: Albert & Bill make to MP 25.

Day 11: Everyone has made it safely to the coast.

Quote:charliepatrickI have made a few assumptions...

(i) Sir Albert wants to make the trip in minimum time.

(ii) "Staying overnight" means, for arguments sake, the day consists of travel, where you depart at 6am and arrive at 6pm; and wish to stay from 6pm until 6am the following day at the city. Assume no-one moves between the hours of 6pm and 6am.

(iii) Rations for a day mean either at 6am you must have enough food for that day which you will steadily consume by 6pm, or at 6am you have some food and during the day will pick up more food so that by 6pm you have eaten a day's food. For arguments sake you have eaten half a days' food by midday.

(iv) A maximum of six days' of food means at 6am, or any other time, you cannot have more than six days' of food on you.

(v) Helpers have to eat, they cannot deliver food and then die.The answer I get is two, Albert and (say) Brian.

Days 1 and 2 Albert and Brian, each with 6 days' of food, set out from Base and travel 50 miles. Let's call this intermediate point "The Village". At this stage each have 4 days' of food left.

On Day 3 Albert takes 2 days' of food from Brian, and now has 6 days' of food. Brian has 2 days' of food.

Days 3 to 5, 6 to 8. Albert has enough food to travel to the city (now only 75 miles away, 3 days), stay overnight, and return (3 days) to the village.

Days 3 to 4. Brian has 2 days' of food so returns to Base.

Days 5 to 6. Brian stays at Base.

Days 7 to 8. Brian starts with 6 days' of food and sets out to the Village. Brian arrives at the village at the end of day 8 with 4 days' of food.

Days 9 to 10. Brian gives Albert 2 days' of food and they both return to Base.

Quote:JoemanI got a different answer than CP, but I made the assumption that Albert (Shouldn't it be Dr. Albert Falls? +10 points if you get the reference!) will spend an entire day exploring the Lost City. This makes my expedition a day longer.

Albert will need to recruit two more members for his expedition, Bill and Chuck.

Day 1: After having set out towards the city, all 3 explorers are at the 25 Mile post, Chuck gives 1 ration each to Albert & Bill.

Day 2: Albert & Bill are at Mile Post 50 while Chuck went back to the coast. Bill gives one of his rations to Albert, leaving him with 4.

Day 3: Albert and Bill have made it to Mile Post 75. Bill gives Albert one of his rations, leaving him with 2. Chuck spends the entire day at the beach.

Day 4: Albert has trekked alone to Mile Post 100 while Bill, having headed back to the coast is at MP 50. Chuck spends another day at the beach. This time in a cabana, nursing his sunburn from the previous day.

Day 5: Albert has reached the city! Chuck started toward the city with a full compliment of rations and meets Bill at MP 25 where he shares his rations.

Day 6: Albert spends the day archaeology-ing the Lost City while Bill & Chuck make their way back to the coast.

Day 7: Albert makes it back to MP 100, having only 2 rations left. Bill & Chuck both have a beach day.

Day 8: Albert makes it to MP 75 while Bill, with a full compliment of rations, makes it to MP 25

Day 9: Albert and Bill meet at MP 50 where Bill shares 2 of his 4 remaining rations with Albert.

Day 10: Albert & Bill make to MP 25.

Day 11: Everyone has made it safely to the coast.

Correct!

Very impressive since both of you seem to have out-thought the puzzle's creator. Here is the official solution...

Total 5 persons are required.

It is given that each person can only carry enough rations for six days. And there are 5 persons. Hence, total of 30 days rations is available.

First Day: 5 days of rations are used up. One person goes back using one day of rations for the return trip. The rations remaining for the further trek is for 24 days.

Second Day: The remaining 4 people use up 4 days of rations. One person goes back using 2 days of rations for the return trip. The rations remaining for the further trek is for 18 days.

Third Day: The remaining 3 people use up 3 days of rations. One person goes back using 3 days of rations for the return trip. The rations remaining for the further trek is for 12 days.

Fourth Day: The remaining 2 persons uses up 2 days of rations. One person goes back using 4 days of rations for the return trip. The rations remaining for the further trek is for 6 days.

Fifth Day: The remaining person uses up one day of rations. He stays overnight. The next day he returns to the coast using 5 days of rations.

Thus, total 5 persons, including you are required.

So CP solves with a mere two person team for an overnight, and JM solves with a three person team with and extra day for archaeology-ing.

Good show!

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

Look for his name to drop in a "soon to be released" major film.

An escaped criminal has stolen a spaceship, and has just instantaneously jumped 1 light year away.

On each successive jump she will only be able to jump half the distance of the immediately preceding jump due to heat build-up. Jump engines always take 1 hour to recharge.

My ship can only jump 1/2 light year, but it has a better cooling system so the jump distances drop-off more slowly. My maximum jump distances follow the sequence: 1/2 light year, 1/3 light year, 1/4 light year, and so on.

I can find her with my subspace-tracker. If she is closer than my maximum jump distance I can get close enough to remotely disable her jump drive and capture her. My jump engines are fully charged.

How many jumps will it take me to capture the escapee assuming I can even catch her?

Since the follower has a full charge, he will instantly jump 1/2, so the escapee has an initial head start of 1 - 1/2 = 1/2. To close the gap use the equation 1 - .5^x + .5 = Ln(x) + .58 - 1. Solve for x and verify.

It will take 6 hours/jumps

Criminal: 1.998046875

Gialmere: 2.019877345

Edit. Followers first jump after recharge will be 1/3 not 1/2

Where x is hours, the escapee’s position is 1 - .5^x. The follower’s position is estimated as Ln(x) + .58 - 1.5

Since the follower has a full charge, he will instantly jump 1/2, so the escapee has an initial head start of 1 - 1/2 = 1/2. To close the gap use the equation 1 - .5^x + .5 = Ln(x) + .58 - 1.5. Solve for x and verify.

It will take 9 hours

Another way to estimate it is: the escape is never going to travel more than 1 additional light year since 1/2 + 1/4 ....expanded infinitely equals 1.

Take 1/3 + 1/4 ...1/11 = 1.52 light years. 9 terms . 9 hours to close the gap

I get the same answer...Quote:WizardI have a feeling I am misunderstanding the question, but here goes...

I show the two ships will pass in the 10th jump. Here is how far each will have traveled after the 10th jump:

Criminal: 1.998046875

Gialmere: 2.019877345

Quote:WizardI have a feeling I am misunderstanding the question, but here goes...

I show the two ships will pass in the 10th jump. Here is how far each will have traveled after the 10th jump:

Criminal: 1.998046875

Gialmere: 2.019877345

Quote:charliepatrickI get the same answer...Quote:WizardI have a feeling I am misunderstanding the question, but here goes...

I show the two ships will pass in the 10th jump. Here is how far each will have traveled after the 10th jump:

Criminal: 1.998046875

Gialmere: 2.019877345...it doesn't really matter what the escapee does as they get closer and closer to 2; in fact you can give them a few days headstart and get the same answer! After N jumps the chaser will eventually reach 2 light years (as 1+1/2+1/3.... is an infinite series). As it happens 1/2+1/3+1/4+...+1/10 = 1.928968 and 1/2+..+1/11 = 2.019877; so after 10 jumps the chaser reaches 2 light years and hence the escapee.

Correct!

Yeah. Ten jumps.

This is another Leslie Green problem so apologies for any confusion. You continue a sequence by jumping without letting the jump-drive cool off. Waiting an hour begins a new sequence. It would be interesting to look at various jump-drives and calculate when you're better off letting the drive cool.

-Flatland is a plane extending infinitely in all directions. It has an infinite number of airfields no two of which are exactly the same distance apart.

-At a point in time a plane will take off from each airport and land at its nearest neighbouring airport.

-What is the maximum number of airplanes that may land at any single airfield?

Quote:Gialmere-Flatland is a plane extending infinitely in all directions. It has an infinite number of airfields no two of which are exactly the same distance apart.

-At a point in time a plane will take off from each airport and land at its nearest neighbouring airport.

-What is the maximum number of airplanes that may land at any single airfield?

5

Let O be an airport, and A the nearest airport to O among the airports that have O as their closest airport.

Let d be the distance from A to O.

Let B be another airport that has O as its closest airport.

If angle AOB is less than 60 degrees, then A is closer to B than O is, which contradicts the fact that O is the closest airport to B.

If AOB is exactly 60 degrees, then either A is closer to B than O is, or AOB is an equilateral triangle, but this contradicts the fact that no two airports are the same distance apart.

Therefore, AOB is greater than 60 degrees.

Let C be a third airport that has O as its closest airport; BOC is also greater than 60 degrees.

There cannot be six such airports as the sum of the angles would be greater than 60 degrees.

Construct an equilateral triangle (OAX) to the East of OA. If B were at X, then the distance to A and O would be equal. So B has to be slightly moved from X (and further away from A than 1 mile). (Aside if distances could be equal then you would just construct a hexagon, and six planes could land at O, they might not, but mathematically they could.)

B has to be over 1 mile from A, hence outside a circle with radius 1.

B also has to be closer to O than A, hence is South of a line that runs East-West through a point 1/2 mile North of O.

B can't be too far round (clockwise) otherwise we can't get the other airfields in.

It therefore transpires that B needs to be just further round than X, and for arguments sake can be 1 mile from O. Radially this means BOA is greater than 60 degrees.

Repeating this you can get five airfields A thru E, lying on or near the points of a pentagon.

As has been explained you can't get six airfields (i) as each angle like BOA is greater than 60 degrees and (ii) they would have to lie on a hexagon.

Hence the answer is 5.

Quote:ThatDonGuy

5

Let O be an airport, and A the nearest airport to O among the airports that have O as their closest airport.

Let d be the distance from A to O.

Let B be another airport that has O as its closest airport.

If angle AOB is less than 60 degrees, then A is closer to B than O is, which contradicts the fact that O is the closest airport to B.

If AOB is exactly 60 degrees, then either A is closer to B than O is, or AOB is an equilateral triangle, but this contradicts the fact that no two airports are the same distance apart.

Therefore, AOB is greater than 60 degrees.

Let C be a third airport that has O as its closest airport; BOC is also greater than 60 degrees.

There cannot be six such airports as the sum of the angles would be greater than 60 degrees.

Quote:charliepatrickLet's assume the airfield (O) where the planes land is at the origin and one of the airfields (A) is 1 mile due North of O. Now working clockwise from A, try and determine where the 2nd airfield B might be such that the plane at A still finds O nearest, and the plane at B also finds O the nearest.

Construct an equilateral triangle (OAX) to the East of OA. If B were at X, then the distance to A and O would be equal. So B has to be slightly moved from X (and further away from A than 1 mile). (Aside if distances could be equal then you would just construct a hexagon, and six planes could land at O, they might not, but mathematically they could.)

B has to be over 1 mile from A, hence outside a circle with radius 1.

B also has to be closer to O than A, hence is South of a line that runs East-West through a point 1/2 mile North of O.

B can't be too far round (clockwise) otherwise we can't get the other airfields in.

It therefore transpires that B needs to be just further round than X, and for arguments sake can be 1 mile from O. Radially this means BOA is greater than 60 degrees.

Repeating this you can get five airfields A thru E, lying on or near the points of a pentagon.

As has been explained you can't get six airfields (i) as each angle like BOA is greater than 60 degrees and (ii) they would have to lie on a hexagon.

Hence the answer is 5.

Correct!

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

What is infinity plus infinity?

2 infinity and beyond!

On a camping trip, a young boy sees the reflection of a mountain peak 10 meters from a lake shoreline. At the visitor center he had learned that the difference between the height of the lake and that of the peak is 1,800 meters.

If the boy stands 1.5 meters tall, how far away is he from the mountain?

Quote:Gialmere

On a camping trip, a young boy sees the reflection of a mountain peak 10 meters from a lake shoreline. At the visitor center he had learned that the difference between the height of the lake and that of the peak is 1,800 meters.

If the boy stands 1.5 meters tall, how far away is he from the mountain?

Quote:GialmereOn a camping trip, a young boy sees the reflection of a mountain peak 10 meters from a lake shoreline. At the visitor center he had learned that the difference between the height of the lake and that of the peak is 1,800 meters.

If the boy stands 1.5 meters tall, how far away is he from the mountain?

Quote:Gialmere

On a camping trip, a young boy sees the reflection of a mountain peak 10 meters from a lake shoreline. At the visitor center he had learned that the difference between the height of the lake and that of the peak is 1,800 meters.

If the boy stands 1.5 meters tall, how far away is he from the mountain?

https://imge.to/i/PZmvH1

1800/x = 1.5/10

x = 12000

The distance from the boy to the mountain = 10 + x = 12010m

Puzzle: The two leading mathematicians in the kingdom, Alice and Bob, have run afoul of their tyrannical king. Rather than behead them outright, the king decides to prolong their misery by locking them in separate dungeons, so that any communication between them is impossible.

Each morning, a guard is to enter the corresponding dungeon and toss a coin so that the prisoner in that dungeon can see the outcome. Then the prisoner will be asked to guess the outcome of the coin toss in the other dungeon (i.e., Alice has to guess the outcome of the toss witnessed by Bob, and Bob has to guess the outcome of the toss witnessed by Alice). If at least one of the two prisoners guesses correctly, they will live to see another day. Otherwise they will be put to death forthwith.

It would seem that the mathematicians are doomed. But as they are being led away in chains Alice and Bob manage to confer for a brief moment and they agree on a strategy that will delay their execution indefinitely. What is the strategy?

Quote:gordonm888Here's one that I thought was interesting.

Puzzle: The two leading mathematicians in the kingdom, Alice and Bob, have run afoul of their tyrannical king. Rather than behead them outright, the king decides to prolong their misery by locking them in separate dungeons, so that any communication between them is impossible.

Each morning, a guard is to enter the corresponding dungeon and toss a coin so that the prisoner in that dungeon can see the outcome. Then the prisoner will be asked to guess the outcome of the coin toss in the other dungeon (i.e., Alice has to guess the outcome of the toss witnessed by Bob, and Bob has to guess the outcome of the toss witnessed by Alice). If at least one of the two prisoners guesses correctly, they will live to see another day. Otherwise they will be put to death forthwith.

It would seem that the mathematicians are doomed. But as they are being led away in chains Alice and Bob manage to confer for a brief moment and they agree on a strategy that will delay their execution indefinitely. What is the strategy?

One prisoner will guess the outcome SAME as outcome witnessed by other prisoner, and other prisoner will guess the outcome NOT THE SAME as outcome witnessed by other prisoner ?

Quote:ssho88One prisoner will guess the outcome SAME as outcome witnessed by other prisoner, and other prisoner will guess the outcome NOT THE SAME as outcome witnessed by other prisoner ?

Please remember to put answers in spoiler tags.

Quote:WizardPlease remember to put answers in spoiler tags.

My bad

Quote:WizardQuote:GialmereOn a camping trip, a young boy sees the reflection of a mountain peak 10 meters from a lake shoreline. At the visitor center he had learned that the difference between the height of the lake and that of the peak is 1,800 meters.

If the boy stands 1.5 meters tall, how far away is he from the mountain?12,000 meters.

Correct!

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

Smokers are great people to go camping with.

You can easily outrun them if a Bear attacks.

Quote:ssho88Quote:gordonm888Here's one that I thought was interesting.

Puzzle: The two leading mathematicians in the kingdom, Alice and Bob, have run afoul of their tyrannical king. Rather than behead them outright, the king decides to prolong their misery by locking them in separate dungeons, so that any communication between them is impossible.

Each morning, a guard is to enter the corresponding dungeon and toss a coin so that the prisoner in that dungeon can see the outcome. Then the prisoner will be asked to guess the outcome of the coin toss in the other dungeon (i.e., Alice has to guess the outcome of the toss witnessed by Bob, and Bob has to guess the outcome of the toss witnessed by Alice). If at least one of the two prisoners guesses correctly, they will live to see another day. Otherwise they will be put to death forthwith.

It would seem that the mathematicians are doomed. But as they are being led away in chains Alice and Bob manage to confer for a brief moment and they agree on a strategy that will delay their execution indefinitely. What is the strategy?

One prisoner will guess the outcome SAME as outcome witnessed by other prisoner, and other prisoner will guess the outcome NOT THE SAME as outcome witnessed by other prisoner ?

CORRECT Good job ssho88!

Each prisoner seemingly has the objective of guessing the other coin flip correctly in order to stay alive - thus, my first reaction was that learning the outcome of their own coin flip is irrelevant to the prospect of a successful guess of the other coin flip.

But of course, my first reaction was wrong. This is a case of two people guessing about two coin flips with the objective of avoiding the scenario where both guesses are wrong.

Normally, there would be these scenarios:

Alice and Bob both guess correctly (25%)

Alice and Bob both guess incorrectly -and die! (25%)

One of Alice and Bob guess correctly and the other guesses incorrectly (50%)

Using ssho88's strategy this is converted to: 0% probability of the guesses being both correct or both incorrect, and 100% probability of one guess being correct and the other guess being wrong. And notice that both Bob and Alice's guesses are still each correct only 50% of the time -so we are not violating any of Mike Shackleford's principles of coin flips.

Mathematically, I would define this as a group of two coin flips. The group is symmetric if both coin flips are the same and is antisymmetric if both coin flips are different. If you know the outcome of Alice's coin flip then the question of Bob's outcome becomes one of whether the group is symmetric or antisymmetric. And vice versa from Bob's point of view. Thus, this redefinition of the problem allows both Alice and Bob to take advantage of the information that they are each independently given -the outcome of their own coinflips - and to each re-define the uncertainty they face into a single common question -is the group of two coin flips symmetric or antisymmetric? Thus their prior agreement to make opposite guesses about the group symmetry enables them to assure that one of them will always be correct.

I briefly considered whether these principles could be used in sports betting to pick parlay bets that hedge each other. Alas, I think the answer is no. Apparently. sports books are not as crazy as the tyrannical king of this puzzle.

This is a good puzzle and curious that any solution works. This is pretty much paraphrasing what Gordon wrote, but I think it will help me retain it if I write it out.

Let person A guess the same as what he sees and B guesses the opposite.

Person A will be right if both flips are the same. Meanwhile, person B will be right if the two flips are opposite.

ONE of them has to be right -- the two flips will be the same or opposites.

No matter how many times you do this, the combined success rate will be always 50%.

The Good, the Bad and the Ugly--three intelligent gunfighters--decide to settle their differences in a three-way duel.

The Ugly is a poor shot, hitting his target only 1/3 of the time. The Bad is better, hitting his target 2/3 of the time. The Good is a sure shot hitting 100% of the time. They take turns shooting, first the Ugly, then the Bad, then the Good, then back to the Ugly, and so on until only one is left.

What is the Ugly's best course of action? And his chances of survival?

Yes I remember this one, nice puzzle/answer!Quote:Gialmere...The Good, the Bad and the Ugly--

Quote:charliepatrickYes I remember this one, nice puzzle/answer!

Counter intuitively Ugly wins 25/63, Bad 24/63 and Good 14/63.

Correct!

Any solves on the Ugly's strategy?

Quote:GialmereQuote:charliepatrickYes I remember this one, nice puzzle/answer!

Counter intuitively Ugly wins 25/63, Bad 24/63 and Good 14/63.

Correct!

Any solves on the Ugly's strategy?

The solution is for Ugly to shoot into the air.

If Bad aims at Ugly and kills him, Good then kills Bad, so Bad will aim at Good instead.

If Bad kills Good, Ugly has a 3/7 chance of winning (1/3 chance of killing Bad; 2/3 chance of missing x 2/3 chance of Bad killing Ugly)

If Bad misses, Good then kills Bad, as he then has a 2/3 chance of winning rather than just 1/3; Ugly has a 1/3 chance of winning in this case.

However, if Ugly shoots into the air, what stops the other two from doing the same thing?

Quote:ThatDonGuyQuote:GialmereQuote:charliepatrickYes I remember this one, nice puzzle/answer!

Counter intuitively Ugly wins 25/63, Bad 24/63 and Good 14/63.

Correct!

Any solves on the Ugly's strategy?

The solution is for Ugly to shoot into the air.

If Bad aims at Ugly and kills him, Good then kills Bad, so Bad will aim at Good instead.

If Bad kills Good, Ugly has a 3/7 chance of winning (1/3 chance of killing Bad; 2/3 chance of missing x 2/3 chance of Bad killing Ugly)

If Bad misses, Good then kills Bad, as he then has a 2/3 chance of winning rather than just 1/3; Ugly has a 1/3 chance of winning in this case.

However, if Ugly shoots into the air, what stops the other two from doing the same thing?

Quote:ThatDonGuy...if Ugly shoots into the air, what stops the other two from doing the same thing?

I kind of like the film solve...

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

Did you hear about the man who was shot 200 times with an upholstery gun?

Surgeons say he is now fully recovered.

A chess queen attacks all squares along its path horizontally, vertically and diagonally.

I would like to place at least 2 white queens and at least 2 black queens on a 5x5 chessboard, such that the queens on either side are not attacking any opposing queens.

What is the maximum number of queens I can place on the board?

Quote:Gialmere

A chess queen attacks all squares along its path horizontally, vertically and diagonally.

I would like to place at least 2 white queens and at least 2 black queens on a 5x5 chessboard, such that the queens on either side are not attacking any opposing queens.

What is the maximum number of queens I can place on the board?

I can see my way to:

First of all consider if you put four White Queens, one at each corner. This would allow four Black Queens, each being a Knight's move away from the corners.

Now remove two of the White Queens, this will leave the previous four Black Queens asis, but also allows three additional Black Queens along the middle section of the row (or column) that lies between the two removed Queens. This adds three more Black Queens.

Hence two Whitte Queens together with 4+3 Black Queens = 9.

Q - - - Q

- - q - -

- q - q -

- - q - -

- q q q -

- - Q - Q

q - - - -

- q - q -

q - - q -

- q - q -

Quote:charliepatrickI can get to nine.

First of all consider if you put four White Queens, one at each corner. This would allow four Black Queens, each being a Knight's move away from the corners.

Now remove two of the White Queens, this will leave the previous four Black Queens asis, but also allows three additional Black Queens along the middle section of the row (or column) that lies between the two removed Queens. This adds three more Black Queens.

Hence two Whitte Queens together with 4+3 Black Queens = 9.Q - - - Q

- - q - -

- q - q -

- - q - -

- q q q -

Agree. I missed the back middle.

Quote:charliepatrickI can get to nine.

First of all consider if you put four White Queens, one at each corner. This would allow four Black Queens, each being a Knight's move away from the corners.

Now remove two of the White Queens, this will leave the previous four Black Queens asis, but also allows three additional Black Queens along the middle section of the row (or column) that lies between the two removed Queens. This adds three more Black Queens.

Hence two Whitte Queens together with 4+3 Black Queens = 9.I think also worksQ - - - Q

- - q - -

- q - q -

- - q - -

- q q q -- - Q - Q

q - - - -

- q - q -

q - - q -

- q - q -

Correct!

And a nice second solve.

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

I just had lunch with a champion chess player.

Took her 8 minutes to pass the salt.

You're a contestant on Let's Make a Deal and playing a new game called Monty Hall's Revenge. You are shown 100 bags...

"These bags," Wayne Brady explains, "each contain a different amount of money. I'll let you open a bag, and then you may decide to pick that bag, or choose another one. Once you have passed up a bag, you may not go back to it. You may keep the money in the bag you finally choose. If, however, you manage to choose the bag with the most money, you will also win this!"

You watch, along with the rest of the crowd, as a door opens revealing the car of your dreams (a 1981 Chrysler Station Wagon).

What strategy should you take to give yourself the best chance of winning the car?

Quote:WizardA classic puzzle. Usually referred to as the "100 secretaries" problem, or something like that. I'm quite familiar with it so will stay out for now.

I am familiar with this one as well - but have never seen an explanation as to just why the solution is correct.

I have also seen an alternative version, where Monty says, "Oh, and I will give you one hint: if you use the {insert the "best strategy" here} method, you will not select the bag with the most money" - what is the best strategy in this case?

Quote:ThatDonGuyI am familiar with this one as well - but have never seen an explanation as to just why the solution is correct.

I can't explain why it works either, but the principle works for any number of boxes.

Quote:I have also seen an alternative version, where Monty says, "Oh, and I will give you one hint: if you use the {insert the "best strategy" here} method, you will not select the bag with the most money" - what is the best strategy in this case?

The way I've heard it phrased is the first person to play used the optimal strategy and didn't find the highest prize. Now it's your turn -- what is your strategy?

Quote:WizardThe way I've heard it phrased is the first person to play used the optimal strategy and didn't find the highest prize. Now it's your turn -- what is your strategy?

Yes, but in the Monty Hall case, the audience (and, presumably, the second player) would have seen what was in the first few bags in advance.

That’s an interesting puzzle. (a) assume the audience will help (b) assume the audience won’t. (not sure if this will affects the answer)Quote:Wizard...The way I've heard it phrased is the first person to play used the optimal strategy and didn't find the highest prize. Now it's your turn -- what is your strategy?

Quote:charliepatrickThat’s an interesting puzzle. (a) assume the audience will help (b) assume the audience won’t. (not sure if this will affects the answer)

Well, if the first player sees the bag that turns out to have the most money and decides not to keep it, then everybody in the audience will know exactly which bag is the winning bag.

Quote:charliepatrickYes if the biggest was in the first part then you’re correct, but what if the best one was yet to be found as the first person stopped too early. So it’s something like 1-x plus something.

In that case, the audience is of no help, assuming that Monty did not reveal any of the amounts other than the last one chosen, and that the second contestant knows what strategy the first one used.

King Charlemagne, his cousin Hogier, Alexander the Great, Julius Caesar, Queen Elizabeth I, King David, Judith from the bible, Rachel also from the bible, Sir Lancelot, his half-brother Sir Hector, Pallas Athena, and the French soldier La Hire

...grouped together? (Most people own this resource.)

Hmmm...12 people, four of whom are women, and four of the eight men appear to be rulers.

They wouldn't be, by any chance, the face cards on a deck of cards, would they?

Quote:ThatDonGuy

Hmmm...12 people, four of whom are women, and four of the eight men appear to be rulers.

They wouldn't be, by any chance, the face cards on a deck of cards, would they?

Ding! Ding!

Winner!!

The names were occasionally printed on the old cards, but today these personalities aren’t really considered, standardized or used in any way.

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So the next time someone says you spend too much time playing cards, you tell them that you're simply a student of history.

Quote:charliepatrickThat’s an interesting puzzle. (a) assume the audience will help (b) assume the audience won’t. (not sure if this will affects the answer)

So the second player has two options...

Scenario A - Player #2 looks at the first N/e = 37 then picks the second box higher than that seen so far (he knows the first one encountered wasn't highest).

Scenario B - Player #2 assumes the best box is in the first 37 and uses the same strategy for just those - Pr success ~= 1/e * 37/100 = .137286

(i) Player #1 - finds the best in the first 37 boxes, so cannot win (and therefore goes right through the boxes). Pr = 37/100

(This is the only case where the audience can help, let's assume they don't - so #2 will also lose.)

(ii) Player #1 - finds the 2nd best in the first N, only 1st is left, so will always win.

(iii) Player #1 - finds the 3rd best in the first N, so 1st and 2nd left, so has a 50/50 chance to win; if lose the 2nd competitor will always win.

(iv) Player #1 - finds the 4th best in the first N, 1st 2nd 3rd left, so has 1/3 chance to win, if lose the 2nd competitor has a 1/2 chance to win.

(lxx) 1-70 left, so has 1/70 chance to win, if lose the 2nd competitor has a 1/69 chance to win.

Assuming Player No2 adopts this strategy, the probabiity without knowing than Player No1 didn't win, is ....

Pr #1 or #2 wins if the 1st is in the first half = 0.

Pr #1 wins = Pr(finds the 2nd in the first 37)*1 + Pr(finds the 3rd in the first 37)*1/2 + Pr(finds the 4th in the first 37)*1/3 etc.

Pr #2 wins = Pr(finds the 2nd in the first 37)*0 + Pr(finds the 3rd in the first 37)*1 + Pr(finds the 4th in the first 37)*1/2 etc.

This works out to be .233740 (which is close to 1/e/(1-1/e)).

However the above includes situations where Player #1 would win. e.g. if the 2nd highest was in the first 37, Player No2 wouldn't be playing. So we have to exclude cases where Player #1 won. This is 1/e, so we have to multiply the probability by (1-1/e).

Gosh this means the chances that Player #2 wins as (.371631) about 1/e!!!

If the audience helps then you can add best box was in the first 37, and that's 1/e as well, so gives 2/e.

Isn't Life Strange - Moody Blues