Easy math puzzles
Poll
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| | 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
October 25th, 2020 at 7:05:14 PM
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Let’s say a basketball player has a “hot hand,” meaning he is more likely to make a basket if he made the last one. Assume he has a 2/5 chance of making his next shot if he missed the last one and a 4/7 chance if he made the last one.
Over the the long term, what percentage of all shots is he expected to make?
Over the the long term, what percentage of all shots is he expected to make?
It’s all about making that GTA
October 25th, 2020 at 7:19:22 PM
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The shooter will have a "run" of one or more misses followed by a "run" of one or more makes, continuing with alternating runs of makes and misses.
The expected length of the run of misses is 2/5 x (1 + 2 x 3/5 + 3 x (3/5)^2 + ...)
= 2/5 x (1 + 3/5 + (3/5)^2 + ...)^2
= 2/5 x (5/2)^2
= 5/2
The expected length of the run of makes is 3/7 x (1 + 2 x 4/7 + 3 x (4/7)^2 + ...)
= 3/7 x (1 + 4/7 + (4/7)^2 + ...)^2
= 3/7 x (7/3)^2 = 7/3
The fraction of all shots made = (7/3) / (7/3 + 5/2) = (14/6) / (29/6) = 14/29 = 48.275862%
October 26th, 2020 at 7:04:50 AM
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CorrectQuote: ThatDonGuy
The shooter will have a "run" of one or more misses followed by a "run" of one or more makes, continuing with alternating runs of makes and misses.
The expected length of the run of misses is 2/5 x (1 + 2 x 3/5 + 3 x (3/5)^2 + ...)
= 2/5 x (1 + 3/5 + (3/5)^2 + ...)^2
= 2/5 x (5/2)^2
= 5/2
The expected length of the run of makes is 3/7 x (1 + 2 x 4/7 + 3 x (4/7)^2 + ...)
= 3/7 x (1 + 4/7 + (4/7)^2 + ...)^2
= 3/7 x (7/3)^2 = 7/3
The fraction of all shots made = (7/3) / (7/3 + 5/2) = (14/6) / (29/6) = 14/29 = 48.275862%
It’s all about making that GTA
October 26th, 2020 at 8:33:12 AM
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In the first round of a World Cup, the teams in Group G finished as follows:
:strip_icc()/pic5739052.png)
GF stands for "Goals For" and GA stands for "Goals Against."
A victory earns three points and a tie one; a loss earns no points.
Each team played the other three once.
What were the scores of all the matches?
Have you tried 22 tonight? I said 22.
October 26th, 2020 at 10:46:51 AM
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Pairing the games that were likely played on the same day (or over 2 consecutive days):
| Romania | 1 | - | 1 | Tunisia |
| Colombia | 0 | - | 2 | England |
| Colombia | 1 | - | 0 | Tunisia |
| England | 1 | - | 2 | Romania |
| Romania | 1 | - | 0 | Colombia |
| England | 2 | - | 0 | Tunisia |
Last edited by: Joeman on Oct 26, 2020
"Dealer has 'rock'... Pay 'paper!'"
October 26th, 2020 at 10:58:14 AM
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6 matches were played
Since each win has a total of 3 points (3 to the winner, 0 to the loser) and each draw has 2 (1 to each team), the total number of points = 18 - the number of draws
There are 17 points, so there was one draw; since Romania's and Tunisia's points are not multiples of 3, they had the draw
Romania and England each won 2 matches, Colombia won 1, and Tunisia won none
Colombia beat Tunisia, so it lost to England and Romania
Romania won both of its non-drawn matches
Colombia has a win, and it scored only one goal, so Colombia-Tunisia is 1-0
Since Tunisia scored only one goal, its draw with Romania was either 0-0 or 1-1
If Romania-Tunisia was 0-0:
Tunisia gave up 1 goal to Colombia and 0 to Romania, so it gave up 3 to England;
Tunisia scored 0 against both Colombia and Romania, so it scored 1 against England, and England-Tunisia is 3-1
England gave up 1 against Tunisia and 0 against Colombia, so it gave up 1 against Romania, which means Romania-England was 1-0
England scored 0 against Romania and 3 against Tunisia, so it scored 2 against Colombia
Romania scored 1 against England and 0 against Tunisia, so it scored 3 against Colombia
...but if Colombia gave up 0 against Tunisia and 2 against England, it only gave up 1 to Romania
This is a contradiction, so Romania-Tunisia was not 0-0, which means it was 1-1
Tunisia's only goal was against Romania, so it scored 0 against both England and Colombia
England gave up 0 to both Colombia and Tunisia, so it gave up its 2 goals to Romania
Romania scored 2 against England and 1 against Tunisia, so it scored 1 against Colombia
Romania gave up 1 goal against Tunisia and 0 against Colombia, so it gave up 1 against England
Colombia gave up 1 against Colombia and 0 against Tunisia, so it gave up 2 against England
England scored 1 against Romania and 2 against Colombia, so it scored 2 against Tunisia
Romania 2, England 1
Romania 1, Colombia 0
Romania 1, Tunisia 1
England 2, Colombia 0
England 2, Tunisia 0
Colombia 1, Tunisia 0
October 26th, 2020 at 5:10:07 PM
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Quote: Joeman
Pairing the games that were likely played on the same day (or over 2 consecutive days):
Romania 1 - 1 Tunisia Colombia 0 - 2 England Colombia 1 - 0 Tunisia England 1 - 2 Romania Romania 1 - 0 Colombia England 2 - 0 Tunisia
Quote: ThatDonGuy
6 matches were played
Since each win has a total of 3 points (3 to the winner, 0 to the loser) and each draw has 2 (1 to each team), the total number of points = 18 - the number of draws
There are 17 points, so there was one draw; since Romania's and Tunisia's points are not multiples of 3, they had the draw
Romania and England each won 2 matches, Colombia won 1, and Tunisia won none
Colombia beat Tunisia, so it lost to England and Romania
Romania won both of its non-drawn matches
Colombia has a win, and it scored only one goal, so Colombia-Tunisia is 1-0
Since Tunisia scored only one goal, its draw with Romania was either 0-0 or 1-1
If Romania-Tunisia was 0-0:
Tunisia gave up 1 goal to Colombia and 0 to Romania, so it gave up 3 to England;
Tunisia scored 0 against both Colombia and Romania, so it scored 1 against England, and England-Tunisia is 3-1
England gave up 1 against Tunisia and 0 against Colombia, so it gave up 1 against Romania, which means Romania-England was 1-0
England scored 0 against Romania and 3 against Tunisia, so it scored 2 against Colombia
Romania scored 1 against England and 0 against Tunisia, so it scored 3 against Colombia
...but if Colombia gave up 0 against Tunisia and 2 against England, it only gave up 1 to Romania
This is a contradiction, so Romania-Tunisia was not 0-0, which means it was 1-1
Tunisia's only goal was against Romania, so it scored 0 against both England and Colombia
England gave up 0 to both Colombia and Tunisia, so it gave up its 2 goals to Romania
Romania scored 2 against England and 1 against Tunisia, so it scored 1 against Colombia
Romania gave up 1 goal against Tunisia and 0 against Colombia, so it gave up 1 against England
Colombia gave up 1 against Colombia and 0 against Tunisia, so it gave up 2 against England
England scored 1 against Romania and 2 against Colombia, so it scored 2 against Tunisia
Romania 2, England 1
Romania 1, Colombia 0
Romania 1, Tunisia 1
England 2, Colombia 0
England 2, Tunisia 0
Colombia 1, Tunisia 0
Correct!
Well done.
For those interested, this was the actual Group G in the 1998 World Cup.
Coming Soon: Hockey Stats
For those interested, this was the actual Group G in the 1998 World Cup.
Coming Soon: Hockey Stats
-------------------------------------
After his team was eliminated, the Nigerian captain personally offered to refund all expenses that fans of his country paid for travel to the World Cup.
According to sources close to the player, he just needs their bank details and pin numbers to complete the transactions.
Have you tried 22 tonight? I said 22.
October 27th, 2020 at 8:53:22 AM
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It's toughie Tuesday...
Yahtzee is a game played with five dice. On your turn you may roll the dice up to three times trying to make various scoring combinations.
After each of the first two rolls, you may set aside some, none or all of the dice. After the second roll you may also re-roll set aside (along with any other) dice if you feel a better scoring possibility becomes available. After the third roll your turn ends and you must score the dice as best you can.
Five of a kind is called a Yahtzee. Suppose that you were on your last turn and only a Yahtzee would score points for you.
Assuming you make good play decisions, what is the probability of getting a Yahtzee with the three rolls of your turn?
Yahtzee is a game played with five dice. On your turn you may roll the dice up to three times trying to make various scoring combinations.
After each of the first two rolls, you may set aside some, none or all of the dice. After the second roll you may also re-roll set aside (along with any other) dice if you feel a better scoring possibility becomes available. After the third roll your turn ends and you must score the dice as best you can.
Five of a kind is called a Yahtzee. Suppose that you were on your last turn and only a Yahtzee would score points for you.
Assuming you make good play decisions, what is the probability of getting a Yahtzee with the three rolls of your turn?
Have you tried 22 tonight? I said 22.
October 27th, 2020 at 4:35:51 PM
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I get an answer of 6.1657%. This was a lot of work. I can't count permutations like I used to! In an effort to learn the table functionality better, I will post a table of my calculations later.
ETA: found an error and I'm now at 4.6029%
ETA again: think I found another one... working on it
Last edited by: rsactuary on Oct 27, 2020
October 27th, 2020 at 4:39:20 PM
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I get 0.045294072.
Question for the forum -- If you get more than one Yahtzee, do you get 100 points for every one beyond the first, plus whatever points you do get for it for some other entry? For example, 5-5-5-5-5 could count as a four of a kind.
This is how I play, but I get rebuked all the time that there is no 100-point bonus rule.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 27th, 2020 at 5:13:50 PM
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Here are the official rules:
https://www.hasbro.com/common/instruct/Yahtzee.pdf
It indicates that you do get a Yahtzee bonus for each Yahtzee after the first, assuming that you scored an original Yahtzee as 50.
One thing I didn't know... bonus Yahtzees have to go into the upper section of the scorecard first. Only if it is completely full can you score it in the lower section.
https://www.hasbro.com/common/instruct/Yahtzee.pdf
It indicates that you do get a Yahtzee bonus for each Yahtzee after the first, assuming that you scored an original Yahtzee as 50.
One thing I didn't know... bonus Yahtzees have to go into the upper section of the scorecard first. Only if it is completely full can you score it in the lower section.
October 27th, 2020 at 5:22:10 PM
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Quote: rsactuaryMy understanding is that you must use the Yahtzee as your turn in some other space (so that you never get more than 13 turns in a game) and then you also get the bonus.
We seem in agreement about that.
Quote:Another question for the forum: Can a Yahtzee be used as a full house?
My interpretation of the rules is that should be allowed.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 27th, 2020 at 5:23:37 PM
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I changed my post while you were quoting it. Official rules above.
October 27th, 2020 at 6:22:19 PM
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Quote: rsactuaryOne thing I didn't know... bonus Yahtzees have to go into the upper section of the scorecard first. Only if it is completely full can you score it in the lower section.
I think you're reading that rule wrong. The way I am reading it, you only have to score it in the upper section if that particular number's box is empty. For example, if you roll five 3s, then if the Threes box is empty, you have to score it there (as 15), but if the Threes box is already filled in, you can score it in the lower section rather than put a zero in one of the other upper section boxes. Of course, if all of the lower section boxes are filled as well, then you do have to score it as a zero in one of the remaining upper boxes.
October 27th, 2020 at 6:34:03 PM
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Quote: ThatDonGuyI think you're reading that rule wrong. The way I am reading it, you only have to score it in the upper section if that particular number's box is empty. For example, if you roll five 3s, then if the Threes box is empty, you have to score it there (as 15), but if the Threes box is already filled in, you can score it in the lower section rather than put a zero in one of the other upper section boxes. Of course, if all of the lower section boxes are filled as well, then you do have to score it as a zero in one of the remaining upper boxes.
got it. you are correct.
October 27th, 2020 at 6:42:40 PM
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Quote: rsactuary
I get an answer of 6.1657%. This was a lot of work. I can't count permutations like I used to! In an effort to learn the table functionality better, I will post a table of my calculations later.
ETA: found an error and I'm now at 4.6029%
ETA again: think I found another one... working on it
Your first adjustment appears to be Correct!
:strip_icc()/pic5743326.png)
You're right. This is a tough one. The next time you run into the Wizard, he owes you a beer.
---------------------------------
Coming soon: A guest host.
Have you tried 22 tonight? I said 22.
October 27th, 2020 at 7:25:48 PM
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Quote: GialmereQuote: rsactuary
I get an answer of 6.1657%. This was a lot of work. I can't count permutations like I used to! In an effort to learn the table functionality better, I will post a table of my calculations later.
ETA: found an error and I'm now at 4.6029%
ETA again: think I found another one... working on it
Your first adjustment appears to be Correct!
You're right. This is a tough one. The next time you run into the Wizard, he owes you a beer.
---------------------------------
Coming soon: A guest host.
I guess I did it the hard way, lol.
October 27th, 2020 at 8:05:54 PM
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Quote: rsactuaryOne thing I didn't know... bonus Yahtzees have to go into the upper section of the scorecard first. Only if it is completely full can you score it in the lower section.
Interesting. I didn't know that either.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 27th, 2020 at 9:05:56 PM
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ThatDonGuy corrected me.Quote: WizardInteresting. I didn't know that either.
October 28th, 2020 at 8:40:46 AM
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:strip_icc()/pic5743325.png)
Place each digit from 1 through 9 in the empty squares of the grid so that the three rows across and the three rows down form correct arithmetic statements.
All calculations (which involve only positive integers) should be performed using the correct order of operations. Parentheses have been provided where needed.
Have you tried 22 tonight? I said 22.
October 28th, 2020 at 10:45:06 AM
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In the left column, one of the first two numbers must be 6 or 9 as otherwise the result cannot be 3 (it cannot be 3 as the other would have to equal the third number)
The possibilities for the left column are:
1,6,2
6,1,2
1,9,3
9,1,3
2,6,4
6,2,4
2,9,6
9,2,6
4,6,8
6,4,8
The right column of the bottom row must be 1 or 2
The possibilities for the bottom row are:
4,3,2 - but 2 is already used in the left column where the last number is 4
5,3,1 - but 5 cannot be the left number
5,4,2 - but 5 cannot be the left number
6,4,1
6,5,2 - but 2 is already used in the left column where the last number is 6
7,5,1 - but 7 cannot be the left number
7,6,2 - but 7 cannot be the left number
8,6,1 - but 6 is already used in the left column where the last number is 4
8,7,2
9,7,1 - but 9 cannot be the left number
9,8,2 - but 9 cannot be the left number
The four possible solutions are:
2 x x
9 x x
6 4 1
The possibilities for the middle row are 9,1,8; 9,2,7; 9,3,6; 9,4,5; 9,5,4; 9,6,3; 9,7,2; 9,8,1 - but all of those use a number that is already there
9 x x
2 x x
6 4 1
The first number in the middle row cannot be 2 (the only result would be (2 - 1) / 1)
4 x x
6 x x
8 7 2
The possibilities for the middle row are 6,1,5; 6,2,4; 6,4,2; 6,5,1
6,1,5 and 6,5,1 are good; however, the only numbers remaining are 3 and 9, and neither 4 + 3 - 9 nor 4 + 9 - 3 = 2, so the first row is impossible
6 x x
4 x x
8 7 2
The possibilities for the middle row are 4,1,3 and 4,3,1, both of which are good
The two remaining numbers are 5 and 9; 6 + 5 - 9 = 2
For 4,3,1 in the middle row, the middle column becomes 5 + 1 - 7, which is not 3
The only solution is:
6 5 9
4 3 1
8 7 2
October 28th, 2020 at 1:04:29 PM
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Quote: Wizard
I get 0.045294072.
Question for the forum -- If you get more than one Yahtzee, do you get 100 points for every one beyond the first, plus whatever points you do get for it for some other entry? For example, 5-5-5-5-5 could count as a four of a kind.
This is how I play, but I get rebuked all the time that there is no 100-point bonus rule.
Yahtzee game strategy has an element of "asset" allocation in it. If you have two rounds to go and don't have 5's or Yahtzee, and your first roll is 5-1-1-4-6 what do you do? go for Yahtzee and draw to 1-1 or go for 5's and draw to your single 5? Which choice gives you a higher EV? (assume that 3 fives are needed for the 35 point bonus).
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 28th, 2020 at 2:49:57 PM
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What do you get when you divide a pumpkins circumference by the pumpkins diameter?
At my age, a "Life In Prison" sentence is not much of a deterrent.
October 28th, 2020 at 2:53:41 PM
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Pumpkin pi
It’s all about making that GTA
October 28th, 2020 at 6:08:46 PM
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Quote: ThatDonGuy
In the left column, one of the first two numbers must be 6 or 9 as otherwise the result cannot be 3 (it cannot be 3 as the other would have to equal the third number)
The possibilities for the left column are:
1,6,2
6,1,2
1,9,3
9,1,3
2,6,4
6,2,4
2,9,6
9,2,6
4,6,8
6,4,8
The right column of the bottom row must be 1 or 2
The possibilities for the bottom row are:
4,3,2 - but 2 is already used in the left column where the last number is 4
5,3,1 - but 5 cannot be the left number
5,4,2 - but 5 cannot be the left number
6,4,1
6,5,2 - but 2 is already used in the left column where the last number is 6
7,5,1 - but 7 cannot be the left number
7,6,2 - but 7 cannot be the left number
8,6,1 - but 6 is already used in the left column where the last number is 4
8,7,2
9,7,1 - but 9 cannot be the left number
9,8,2 - but 9 cannot be the left number
The four possible solutions are:
2 x x
9 x x
6 4 1
The possibilities for the middle row are 9,1,8; 9,2,7; 9,3,6; 9,4,5; 9,5,4; 9,6,3; 9,7,2; 9,8,1 - but all of those use a number that is already there
9 x x
2 x x
6 4 1
The first number in the middle row cannot be 2 (the only result would be (2 - 1) / 1)
4 x x
6 x x
8 7 2
The possibilities for the middle row are 6,1,5; 6,2,4; 6,4,2; 6,5,1
6,1,5 and 6,5,1 are good; however, the only numbers remaining are 3 and 9, and neither 4 + 3 - 9 nor 4 + 9 - 3 = 2, so the first row is impossible
6 x x
4 x x
8 7 2
The possibilities for the middle row are 4,1,3 and 4,3,1, both of which are good
The two remaining numbers are 5 and 9; 6 + 5 - 9 = 2
For 4,3,1 in the middle row, the middle column becomes 5 + 1 - 7, which is not 3
The only solution is:
6 5 9
4 3 1
8 7 2
Correct! Impressive.
-------------------------------
My father loved crossword puzzles so much that it stipulated in his will that he should be buried six down and three across.
[Which reminds me, a guest host will be slithering out of the grave to fill (or would that be shovel me) in tomorrow.]
Have you tried 22 tonight? I said 22.
October 29th, 2020 at 8:11:06 AM
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Heh, Heh! Greetings kiddies! It's me, your host in horror... The Crypt Keeper! Welcome to The Crypt of Terror!
That idiot Gialmere asked me to sit in for him this Halloween while he enjoys the fall ... off a cliff that is! Heh, heh! I hope he didn't fall to pieces, or that he at least picked up his pieces after the fall! Hey, heh, heh!
I have a really chilling tale from my collection of spine-tinglers to relate to you! So, lie back in your caskets! Tuck yourselves in with your shrouds! Comfy? Good! Then I'll begin! I call this terror tale ... Cadaverinth Labyrinth!
:strip_icc()/pic5739037.png)
It's time to carve a jack-o'-lantern but first you need to get a pumpkin. As you enter the pumpkin patch, however, you find it to be a confusing, corn maze labyrinth, filled with disturbing Halloween decorations and pumpkins at every intersection. Can you work your way through the maze? Or will you stay lost ... forever?
Travel along the colored paths from Start to Goal. At each pumpkin, change to a path of a different color from the color you were just on -- any different color. U-turns are prohibited.
Answer with: Start, F ... M, Goal.
Last edited by: Gialmere on Oct 29, 2020
Have you tried 22 tonight? I said 22.
October 29th, 2020 at 8:17:25 AM
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Please confirm that B and J are correct? There's no choice at these path points. Weird.
October 29th, 2020 at 8:29:45 AM
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Heh, heh. Do we have our first cadaver candidate kiddies? Will our lost rsactuary make it through alive? Or will he become part of the disturbing labyrinth decor? The Memorial Rsactuary Mortuary Interchange! Heh, heh!
Have you tried 22 tonight? I said 22.
October 29th, 2020 at 8:59:44 AM
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Quote: Gialmere
Heh, Heh! Greetings kiddies! It's me, your host in horror... The Crypt Keeper! Welcome to The Crypt of Terror!
That idiot Gialmere asked me to sit in for him this Halloween while he enjoys the fall ... off a cliff that is! Heh, heh! I hope he didn't fall to pieces, or that he at least picked up his pieces after the fall! Hey, heh, heh!
I have a really chilling tale from my collection of spine-tinglers to relate to you! So, lie back in your caskets! Tuck yourselves in with your shrouds! Comfy? Good! Then I'll begin! I call this terror tale ... Cadaverinth Labyrinth!
It's time to carve a jack-o'-lantern but first you need to get a pumpkin. As you enter the pumpkin patch, however, you find it to be a confusing, corn maze labyrinth, filled with disturbing Halloween decorations and pumpkins at every intersection. Can you work your through the maze? Or will you stay lost ... forever?
Travel along the colored paths from Start to Goal. At each pumpkin, change to a path of a different color from the color you were just on -- any different color. U-turns are prohibited.
Answer with: Start, F ... M, Goal.
Really love this thread!
F, E, D, C, B, A, J, I, H, Q, R, L, K, R, Q, H, I, J, A, B, C, D, E, F, O, N, M, Goal
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 29th, 2020 at 9:00:45 AM
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Start, F, E, D, C, B, A, J, I, H, Q, R, K, L, R, Q, H, I, J, A, B, C, D, E, F, O, N, M, Goal
Note K and L can be switched
This one took me a while, and at one point, I was convinced there was no solution.
Note that the last move is M-Goal, which is yellow, so the previous move is N-M (orange), then O-N (black), then F-O (orange), so you have to get to F from E or G.
Getting to F from G requires getting to G from P, and to P either directly from O or from O through N, both of which require using a black path from O, so you have to get to O from F using an orange path, but you are then at where you started (F from E or G), which means you have to move from E to F.
If the second move is F-G, this forces F-G-P-O-F or F-G-P-N-O-F, both of which get you to the position after the first move (moving to F along an orange path), so the second move is F-E. Having to move F-E and later having to move E-F means the route returns on itself at some point.
F-E-N leads to F-E-N-O-F, which is a loop, so F-E is followed by E-D
D-M leads to M-N-O-F-E-D, which is a loop, so E-D is followed by D-C, which results in D-C-B-A-J-I-H
H-G leads to G-P-O-F or G-P-N-O-F and we are back where we started, so the next move is H-Q
Q-P leads to P-O-F or P-N-O-F and we are back where we started, so the next move is Q-R
Both R-K-L-R and R-L-K-R allow the route back to F to be reversed
October 29th, 2020 at 10:39:08 AM
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I'm going to use the following notation, to make sure I am not cheating.
> yellow path
>> orange path
>>> black path
Start >> F > E >> D >>> C > B >> A > J >>> I > H >> Q >>> R > K >>> L >> R >>> Q > P > G >>> F >> O >>> N >> M > End
It seems the key is you can make a circular U-turn if you can get to K, L, or R by making a loop through that circle.
My method was to draw out a map with three colors of pens and work backwards drawing arrows on which ways would lend to the end. Eventually I found a way. I think there is at least one other.
> yellow path
>> orange path
>>> black path
Start >> F > E >> D >>> C > B >> A > J >>> I > H >> Q >>> R > K >>> L >> R >>> Q > P > G >>> F >> O >>> N >> M > End
It seems the key is you can make a circular U-turn if you can get to K, L, or R by making a loop through that circle.
My method was to draw out a map with three colors of pens and work backwards drawing arrows on which ways would lend to the end. Eventually I found a way. I think there is at least one other.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 29th, 2020 at 10:42:28 AM
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Quote: WizardI'm going to use the following notation, to make sure I am not cheating.
> yellow path
>> orange path
>>> black path
Start >> F > E >> D >>> C > B >> A > J >>> I > H >> Q >>> R > K >>> L >> R >>> Q > P > G >>> F >> O >>> N >> M > End
It seems the key is you can make a circular U-turn if you can get to K, L, or R by making a loop through that circle.
My method was to draw out a map with three colors of pens and work backwards drawing arrows on which ways would lend to the end. Eventually I found a way. I think there is at least one other.
You can’t go from:
Q>P>G
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 29th, 2020 at 12:13:20 PM
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Quote: Gialmere...Answer with: Start, F ... M, Goal.
If you work backwards there's only one way to arrive at M and it must be via FONM.
To get to the end you have to arrive at F along either the yellow or black.
Arriving along the black (GF) means you had to come from O or N, so that's can't work.
Arriving along the yellow (EF) then work backwards a long way, round the outside route, to H (and possibly G P N or Q P O F or Q R). You can also go the other way, i.e. out from F via E to H or G; the trick is how to turn round. This is achieved by the loop RLK (which can be taken either way round).
So one route is Start, F, round the outside to H, then Q R (loop RLK), then return from R to F, now you can do ONM Goal.
Since P and G have been missed out, if you wanted to visit every town you might want to go round twice! Start, F-H, G P O F then as above.
To get to the end you have to arrive at F along either the yellow or black.
Arriving along the black (GF) means you had to come from O or N, so that's can't work.
Arriving along the yellow (EF) then work backwards a long way, round the outside route, to H (and possibly G P N or Q P O F or Q R). You can also go the other way, i.e. out from F via E to H or G; the trick is how to turn round. This is achieved by the loop RLK (which can be taken either way round).
So one route is Start, F, round the outside to H, then Q R (loop RLK), then return from R to F, now you can do ONM Goal.
Since P and G have been missed out, if you wanted to visit every town you might want to go round twice! Start, F-H, G P O F then as above.
October 29th, 2020 at 3:18:29 PM
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Please forgive me if you've seen this puzzle before which I just stumbled across.
It comes from https://mindyourdecisions.com/blog/2018/07/12/can-you-solve-amazons-hanging-cable-interview-question/ (which also has the solution).Quote:A cable of 80 meters (m) is hanging from the top of two poles that are both 50 m from the ground. What is the distance between the two poles, to one decimal place, if the center of the cable is 10 m above the ground?
October 29th, 2020 at 3:19:39 PM
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Love a catenary.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 29th, 2020 at 5:24:20 PM
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Quote: unJonReally love this thread!
F, E, D, C, B, A, J, I, H, Q, R, L, K, R, Q, H, I, J, A, B, C, D, E, F, O, N, M, Goal
Quote: ThatDonGuy
Start, F, E, D, C, B, A, J, I, H, Q, R, K, L, R, Q, H, I, J, A, B, C, D, E, F, O, N, M, Goal
Note K and L can be switched
This one took me a while, and at one point, I was convinced there was no solution.
Note that the last move is M-Goal, which is yellow, so the previous move is N-M (orange), then O-N (black), then F-O (orange), so you have to get to F from E or G.
Getting to F from G requires getting to G from P, and to P either directly from O or from O through N, both of which require using a black path from O, so you have to get to O from F using an orange path, but you are then at where you started (F from E or G), which means you have to move from E to F.
If the second move is F-G, this forces F-G-P-O-F or F-G-P-N-O-F, both of which get you to the position after the first move (moving to F along an orange path), so the second move is F-E. Having to move F-E and later having to move E-F means the route returns on itself at some point.
F-E-N leads to F-E-N-O-F, which is a loop, so F-E is followed by E-D
D-M leads to M-N-O-F-E-D, which is a loop, so E-D is followed by D-C, which results in D-C-B-A-J-I-H
H-G leads to G-P-O-F or G-P-N-O-F and we are back where we started, so the next move is H-Q
Q-P leads to P-O-F or P-N-O-F and we are back where we started, so the next move is Q-R
Both R-K-L-R and R-L-K-R allow the route back to F to be reversed
Quote: charliepatrickIf you work backwards there's only one way to arrive at M and it must be via FONM.
To get to the end you have to arrive at F along either the yellow or black.
Arriving along the black (GF) means you had to come from O or N, so that's can't work.
Arriving along the yellow (EF) then work backwards a long way, round the outside route, to H (and possibly G P N or Q P O F or Q R). You can also go the other way, i.e. out from F via E to H or G; the trick is how to turn round. This is achieved by the loop RLK (which can be taken either way round).
So one route is Start, F, round the outside to H, then Q R (loop RLK), then return from R to F, now you can do ONM Goal.
Since P and G have been missed out, if you wanted to visit every town you might want to go round twice! Start, F-H, G P O F then as above.
Heh, heh! Well, that's my story for today boils and ghouls!
It seems that three of you took the Correct path and are now happily carving your pumpkins!
As for the rest of you still lost in the labyrinth, I guess you'll just join in with the rest of the decomposing decorum!
Well, as I always say ... The morgue the merrier! Heh, heh, heh!
---------------------------------
Heh, heh! Speaking of carving pumpkins, the WOV mods are having a carving party tonight and I've asked if I can make the final cut...
Heh, heh! I'll beheading your way again tomorrow. See you then kiddies!
Have you tried 22 tonight? I said 22.
October 29th, 2020 at 5:56:28 PM
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Quote: charliepatrickPlease forgive me if you've seen this puzzle before which I just stumbled across.
It comes from https://mindyourdecisions.com/blog/2018/07/12/can-you-solve-amazons-hanging-cable-interview-question/ (which also has the solution).Quote:A cable of 80 meters (m) is hanging from the top of two poles that are both 50 m from the ground. What is the distance between the two poles, to one decimal place, if the center of the cable is 10 m above the ground?
That is a very skinny catenary!
Whenever I see I problem like this, I always check first to see if has a degenerate solution.
The poles are 0.0 m apart.
October 29th, 2020 at 7:29:00 PM
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Quote: unJonYou can’t go from:
Q>P>G
Dang. Well, now I get the fun of doing it again.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 29th, 2020 at 10:00:51 PM
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F to E to D to C to B to A to J to I to H to Q to R to L* to K to R to Q to H to I to J to A to B to C to D to E to F to O to N to M to Goal
* One can also go the other direction around the roundabout to K to L to R
* One can also go the other direction around the roundabout to K to L to R
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 30th, 2020 at 3:07:59 AM
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Quote: charliepatrickPlease forgive me if you've seen this puzzle before which I just stumbled across.
It comes from https://mindyourdecisions.com/blog/2018/07/12/can-you-solve-amazons-hanging-cable-interview-question/ (which also has the solution).Quote:A cable of 80 meters (m) is hanging from the top of two poles that are both 50 m from the ground. What is the distance between the two poles, to one decimal place, if the center of the cable is 10 m above the ground?
For catenary, y = a * cosh (x/a)
when a = 10, y = 10cosh(x/10)
y=50, cosh(x/10) = 5, x = 22.924
Distance between the two poles = 2 *22.924 = 45.848, is it correct ? Then what is the cable length 80m for ?
October 30th, 2020 at 3:53:39 AM
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Quote: ssho88Quote: charliepatrickPlease forgive me if you've seen this puzzle before which I just stumbled across.
It comes from https://mindyourdecisions.com/blog/2018/07/12/can-you-solve-amazons-hanging-cable-interview-question/ (which also has the solution).Quote:A cable of 80 meters (m) is hanging from the top of two poles that are both 50 m from the ground. What is the distance between the two poles, to one decimal place, if the center of the cable is 10 m above the ground?
For catenary, y = a * cosh (x/a)
when a = 10, y = 10cosh(x/10)
y=50, cosh(x/10) = 5, x = 22.924
Distance between the two poles = 2 *22.924 = 45.848, is it correct ? Then what is the cable length 80m for ?
No. a <> 10
For the special case of a catenary hung between two points the same height:
a = (0.25 * L^2 - h^2) / 2h
With L = 80 and h = 50 - 10 = 40
ETA: I must be misremembering as that looks like a=0
ETA2: ha. I get it now and was not misremembering the formula. The chain is hanging straight down and the poles are 0 apart.
a = (0.25 * L^2 - h^2) / 2h
With L = 80 and h = 50 - 10 = 40
ETA: I must be misremembering as that looks like a=0
ETA2: ha. I get it now and was not misremembering the formula. The chain is hanging straight down and the poles are 0 apart.
Last edited by: unJon on Oct 30, 2020
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 30th, 2020 at 8:08:08 AM
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Heh, Heh, Welcome back gambling ghouls! Welcome once again to the Crypt of Terror! This is your host of howls, the Crypt Keeper, ready to narrate another nauseating tale from my collection! Just drag over that burlap sack and sit down! It's nice and soft! The corpse in it isn't quite stiff yet! Heh, heh! Comfy? Good! Now listen to this yelp yarn I call... HEADS OR SCALES?
:strip_icc()/pic5739051.png)
You are the lab assistant to a mad scientist. He's working on reanimating the dead (of course) and so you've been busy digging up corpses from the local graveyard and stealing abby normal brains from the morgue. Finally, a storm fills the sky with lighting. The experiment can begin!
While the scientist preps his electrodes, you begin making some baseline scale measurements. The first three scales balance well enough, but the fourth one...
Scales A, B and C are balanced. How many Frankenstein heads should you put on the question mark side of scale D to balance it as well?
Have you tried 22 tonight? I said 22.
October 30th, 2020 at 8:59:42 AM
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5 Frankie Heads should balance one Skull.
Scale A: S + F = B -or- S = B - F
Scale B: S = P + F
Combining: B - P = 2F
Scale C: 2B = 3P -or- B = 3/2P
Substituting into the equation above:
3/2P - P = 2F -or- P = 4F
Substituting for P in the Scale B equation:
S = 4F + F -or- S = 5F
"Dealer has 'rock'... Pay 'paper!'"
October 30th, 2020 at 5:00:46 PM
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Quote: Joeman5 Frankie Heads should balance one Skull.
Scale A: S + F = B -or- S = B - F
Scale B: S = P + F
Combining: B - P = 2F
Scale C: 2B = 3P -or- B = 3/2P
Substituting into the equation above:
3/2P - P = 2F -or- P = 4F
Substituting for P in the Scale B equation:
S = 4F + F -or- S = 5F
Well boils and ghouls, as the saying goes: Five heads are better than four! Heh, heh!
I see that one of you dead heads was Correct with his headhunting skills! You know, I've always thought that my girlfriend has a good head on her shoulders...
I just wish I could keep it there! Heh, heh, heh!
It's back to the crypt for me kiddies! I have a busy night tomorrow! If I don't see you out trick-or-treating with your burlap sack then...
Have you tried 22 tonight? I said 22.
November 2nd, 2020 at 9:54:05 AM
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It's easy Monday...
You're sightseeing in Vegas when you come across one of those robotic pizza places. Unfortunately, there's a glitch in the programming during the slicing sequence.
When you tell the machine how many cuts you want it to make across the pie, for each cut, the computer randomly (and independently) selects two points on the circumference of the crust and slices a chord connecting the points.
If you told the machine you wanted it to make 3 cuts, how many slices of pizza would you end up with on average?
Assume that the machine will never cut the same chord more than once, nor will all 3 chords intersect at the same point (like they're actually supposed to).
You're sightseeing in Vegas when you come across one of those robotic pizza places. Unfortunately, there's a glitch in the programming during the slicing sequence.
When you tell the machine how many cuts you want it to make across the pie, for each cut, the computer randomly (and independently) selects two points on the circumference of the crust and slices a chord connecting the points.
If you told the machine you wanted it to make 3 cuts, how many slices of pizza would you end up with on average?
Assume that the machine will never cut the same chord more than once, nor will all 3 chords intersect at the same point (like they're actually supposed to).
Have you tried 22 tonight? I said 22.
November 2nd, 2020 at 3:18:10 PM
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Choose one endpoint of one cut and number it 1; number the others clockwise 2 through 6
There are 15 ways to make the cuts, each of which is equally likely:
1-2, 3-4, 5-6 - 4 slices
1-2, 3-5, 4-6 - 5
1-2, 3-6, 5-4 - 4
1-3, 2-4, 5-6 - 5
1-3, 2-5, 4-6 - 6
1-3. 2-6, 4-5 - 5
1-4. 2-3, 5-6 - 4
1-4, 2-5, 3-6 - 7
1-4, 2-6, 3-5 - 6
The 1-5 ways are mirror images of the 1-3 ways (5, 6, 5), and the 1-6 ways are mirror images of the 1-2 ways (4, 5, 4)
The mean = 75 / 15 = 5 slices
November 2nd, 2020 at 4:40:59 PM
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Quote: ThatDonGuy
Choose one endpoint of one cut and number it 1; number the others clockwise 2 through 6
There are 15 ways to make the cuts, each of which is equally likely:
1-2, 3-4, 5-6 - 4 slices
1-2, 3-5, 4-6 - 5
1-2, 3-6, 5-4 - 4
1-3, 2-4, 5-6 - 5
1-3, 2-5, 4-6 - 6
1-3. 2-6, 4-5 - 5
1-4. 2-3, 5-6 - 4
1-4, 2-5, 3-6 - 7
1-4, 2-6, 3-5 - 6
The 1-5 ways are mirror images of the 1-3 ways (5, 6, 5), and the 1-6 ways are mirror images of the 1-2 ways (4, 5, 4)
The mean = 75 / 15 = 5 slices
Correct!
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I like how my local pizza place cuts my pizza into 6 slices instead of 8.
I can't finish 8 slices.
Have you tried 22 tonight? I said 22.
November 3rd, 2020 at 8:12:38 AM
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It's toughie Tuesday. Time for goofy golf...
The ball bin at the front counter of a miniature golf course contains a number of colored golf balls, with equal numbers of each color. Adding 20 balls of a new color to the bin would not change the probability of the counterperson blindly drawing (without replacement) two balls of the same color.
Before the extra balls are added, how many golf balls are in the bin?
The ball bin at the front counter of a miniature golf course contains a number of colored golf balls, with equal numbers of each color. Adding 20 balls of a new color to the bin would not change the probability of the counterperson blindly drawing (without replacement) two balls of the same color.
Before the extra balls are added, how many golf balls are in the bin?
Have you tried 22 tonight? I said 22.
November 3rd, 2020 at 11:00:00 AM
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Quote: GialmereIt's toughie Tuesday. Time for goofy golf...
The ball bin at the front counter of a miniature golf course contains a number of colored golf balls, with equal numbers of each color. Adding 20 balls of a new color to the bin would not change the probability of the counterperson blindly drawing (without replacement) two balls of the same color.
Before the extra balls are added, how many golf balls are in the bin?
There is only one color in the bin and there are at least 2 of them.
I heart Crystal Math.
November 3rd, 2020 at 11:25:22 AM
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Quote: CrystalMathThere is only one color in the bin and there are at least 2 of them.
If there is only one color in the bin, the probability that the first two will be the same color is 1.
If 20 balls of a different color are then added - note the problem says that 20 balls "of a new color" are added - then the probability that the first two will be the same color < 1.
Instead of adding 20 balls of the new color, add N balls, where N - 1 is prime
Each color is equally likely to be the first ball drawn
The probability that the second ball is the same as the first is (B - 1) / (BC - 1)
Instead of adding 20 balls of the new color, add N balls; there are now BC + N balls
The probability that the first ball is one of the originals is BC / (BC + N);
the probability that the second ball is the same as the first is (B - 1) / (BC + N - 1)
The probability that the first ball is one of the new ones is N / (BC + N);
the probability that the second ball is the same as the first is (N - 1) / (BC + N - 1)
The overall probability that the first two balls are the same color is
(BC (B - 1) + N (N - 1)) / ((BC + N)(BC + N - 1))
These are equal when
(B - 1) / (BC - 1) = (BC (B - 1) + N (N-1)) / ((BC + N)(BC + N - 1))
(B - 1)(BC + N)(BC + N - 1) = (BC (B - 1) + N (N-1)) (BC - 1)
(B - 1) ((BC)^2 + 2 BCN + N^2 - BC - N) = (B^2 C - BC + N^2 - N) (BC - 1)
B^3 C^2 + 2 B^2CN + BN^2 - B^2C - BN - (BC)^2 - 2 BCN - N^2 + BC + N = B^3 C^2 - (BC)^2 + BCN^2 - BCN - B^2 C + BC - N^2 + N
2 B^2CN + BN^2 - BN - BCN = BCN^2
2 BCN + N^2 - N - CN = CN^2
2 BC + N - 1 - C = CN
C (N + 1 - 2B) = N - 1
N - 1 is prime, so either C = 1 or N + 1 - 2B = 1
If C = 1, then N + 1 - 2B = N - 1, so B = 1, but that means there is only one ball at the start
Therefore, N + 1 - 2B = 1 and C = N - 1
BC = (N / 2) (N - 1)
For N = 20, there are 10 x 19 = 190 balls at the start
