Easy math puzzles
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46 members have voted
September 8th, 2020 at 7:47:35 PM
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Quote: ChesterDog
This I've seen before--it's 312211, but I won't explain it.
Correct!
This is one of those trick "think outside the box" types of puzzles. When spoken aloud, each element in the sequence (except the first) describes the digits that the previous element contains. So the second element tells you that the first element contains one 1 etc.
---------------------------
I just thought of a great new Fibonacci sequence joke.
It's as good as my last two Fibonacci sequence jokes combined.
Have you tried 22 tonight? I said 22.
September 8th, 2020 at 9:24:50 PM
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Quote: GialmereQuote: ChesterDog
This I've seen before--it's 312211, but I won't explain it.
Correct!This is one of those trick "think outside the box" types of puzzles. When spoken aloud, each element in the sequence (except the first) describes the digits that the previous element contains. So the second element tells you that the first element contains one 1 etc.
...
By the way, here's a Family Guy bit that illustrates the puzzle:
September 9th, 2020 at 11:53:32 AM
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There is a single 21-foot high vertical palm tree growing in the middle of a large flat volcanic Hawaiian island.
On a certain day, the sun will rise at 6:00 AM and set at 6:00 PM.
At noon on that day, the sun will be directly overhead and the trunk of the palm tree will cast no shadow.
Bored waiting for the next solar eclipse, the Wizard, who was visiting the island, decided to witness the event. He camped in the open nearby on the night before this special day and was awakened in the morning when the sun rays reached his eyes, which were 13 feet due west from the tree.
What time did he awake?
Round your answer to the nearest minute.
Have you tried 22 tonight? I said 22.
September 9th, 2020 at 12:40:06 PM
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9:53
It’s all about making that GTA
September 9th, 2020 at 12:59:14 PM
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I think the Wizard is nearer Tuvalu and is still asleep!
In Hawaii the day length is 12 hours on 27th September, whereas today it is 12 hours in Funafuti ( https://www.timeanddate.com/sun/tuvalu/funafuti ) and the solar noon is exactly 12:00.
In Hawaii the day length is 12 hours on 27th September, whereas today it is 12 hours in Funafuti ( https://www.timeanddate.com/sun/tuvalu/funafuti ) and the solar noon is exactly 12:00.
Assuming the sun rays are blocked by the tree until the sun reaches the desired angle. The sun will be at an angle of 0 degrees at 6am and 90 degrees at 12 noon, this is six hours or 360 minutes. Thus it takes four minutes tfor the sun to side by one degree.
The height of the tree is 21ft, the distance to the wizard is 13st, so the tangent of the angle is 21/13. Using an arctan calculator ( https://www.omnicalculator.com/math/arctan ) this means the angle is 58.24 degrees.
This means the sun took 58.24 * 4 minutes to rise to that angle = 232.96. This is about 3 hours 53 mins.
So the sun woke the Wizard up at about 09h53.
(Actually the sun WILL wake up the wizard at 09h53 - since it is coming up to 8am on Tuvalu!)
The height of the tree is 21ft, the distance to the wizard is 13st, so the tangent of the angle is 21/13. Using an arctan calculator ( https://www.omnicalculator.com/math/arctan ) this means the angle is 58.24 degrees.
This means the sun took 58.24 * 4 minutes to rise to that angle = 232.96. This is about 3 hours 53 mins.
So the sun woke the Wizard up at about 09h53.
(Actually the sun WILL wake up the wizard at 09h53 - since it is coming up to 8am on Tuvalu!)
September 9th, 2020 at 7:04:09 PM
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Quote: Ace29:53
Quote: charliepatrickI think the Wizard is nearer Tuvalu and is still asleep!
In Hawaii the day length is 12 hours on 27th September, whereas today it is 12 hours in Funafuti ( https://www.timeanddate.com/sun/tuvalu/funafuti ) and the solar noon is exactly 12:00.Assuming the sun rays are blocked by the tree until the sun reaches the desired angle. The sun will be at an angle of 0 degrees at 6am and 90 degrees at 12 noon, this is six hours or 360 minutes. Thus it takes four minutes tfor the sun to side by one degree.
The height of the tree is 21ft, the distance to the wizard is 13st, so the tangent of the angle is 21/13. Using an arctan calculator ( https://www.omnicalculator.com/math/arctan ) this means the angle is 58.24 degrees.
This means the sun took 58.24 * 4 minutes to rise to that angle = 232.96. This is about 3 hours 53 mins.
So the sun woke the Wizard up at about 09h53.
(Actually the sun WILL wake up the wizard at 09h53 - since it is coming up to 8am on Tuvalu!)
Correct!
-----------------------------
Have you tried 22 tonight? I said 22.
September 10th, 2020 at 8:16:26 AM
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:strip_icc()/pic5648674.png)
Determine the meaning of the 20 puzzles above.
1a:
1b:
1c:
1d:
2a:
2b:
2c:
2d:
3a:
3b:
3c:
3d:
4a:
4b:
4c:
4d:
5a:
5b:
5c:
5d:
Have you tried 22 tonight? I said 22.
September 10th, 2020 at 10:47:04 AM
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Quote: Gialmere
Determine the meaning of the 20 puzzles above.
1a:
1b: Deep rooted
1c: Paint by the numbers
1d: Against all odds
2a: Waiting on a table/ table waiter
2b: Times Square (though its not quite square,lol)
2c: Blinded by the light?
2d: Have fingers in many pies ?
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a: Stack of money?
4b: Y intercept
4c:
4d: Drawn and quartered ?
5a: Positive attitude
5b: Money is the root of all evil/sin
5c: Apple pie
5d: Forever ours
Last edited by: gordonm888 on Sep 10, 2020
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 10th, 2020 at 11:07:46 AM
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1a: Back to square one
2b: I thought this one could be X-Box, but I like Gordon's answer better.
4a: What I hope to be after a craps session... $1,000 up?
4d: Close quarters
2b: I thought this one could be X-Box, but I like Gordon's answer better.
4a: What I hope to be after a craps session... $1,000 up?
4d: Close quarters
"Dealer has 'rock'... Pay 'paper!'"
September 10th, 2020 at 11:12:18 AM
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Quote: Gialmere
Determine the meaning of the 20 puzzles above.
1a:
1b: Deep rooted
1c: Paint by the numbers
1d: Against all odds
2a: Waiting on a table/ table waiter
2b: Times Square (though its not quite square,lol)
2c: Blinded by the light?
2d: Have fingers in many pies ?
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a: Stack of money?
4b: Y intercept
4c: A higher force?
4d: Drawn and quartered ?
5a: Positive attitude
5b: Money is the root of all evil/sin
5c: Apple pie
5d: Forever ours
Where we conflict, I like Joeman's answers!
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 10th, 2020 at 11:41:36 AM
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1a: Back to square one
1b: Deep rooted
1c: Paint by numbers
1d: Against all odds
2a: Wait on tables
2b: Times Square
2c:
2d: A finger in every pie
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a:
4b: Y intercept
4c:
4d: Close quarters
5a: Positive attitude
5b: Money is the root of all evil
5c: Apple pie
5d:
Hints posted later if needed.
Have you tried 22 tonight? I said 22.
September 10th, 2020 at 12:16:55 PM
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5D: Forever and a day (24 "ours")
September 10th, 2020 at 12:29:25 PM
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Quote: ThatDonGuy
5D: Forever and a day (24 "ours")
Correct
1a: Back to square one
1b: Deep rooted
1c: Paint by numbers
1d: Against all odds
2a: Wait on tables
2b: Times Square
2c:
2d: A finger in every pie
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a:
4b: Y intercept
4c:
4d: Close quarters
5a: Positive attitude
5b: Money is the root of all evil
5c: Apple pie
5d: Forever and a day
Still Needed
2c:
3b:
4a:
4c:
Hints posted later if needed.
Have you tried 22 tonight? I said 22.
September 10th, 2020 at 1:10:37 PM
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Quote: GialmereQuote: ThatDonGuy
5D: Forever and a day (24 "ours")
Correct
1a: Back to square one
1b: Deep rooted
1c: Paint by numbers
1d: Against all odds
2a: Wait on tables
2b: Times Square
2c:
2d: A finger in every pie
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a:
4b: Y intercept
4c:
4d: Close quarters
5a: Positive attitude
5b: Money is the root of all evil
5c: Apple pie
5d: Forever and a day
Still Needed
2c:
3b:
4a:
4c:
Hints posted later if needed.
I think for 4a
4a: grandstand
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 10th, 2020 at 1:11:53 PM
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Quote: GialmereQuote: ThatDonGuy
5D: Forever and a day (24 "ours")
Correct
1a: Back to square one
1b: Deep rooted
1c: Paint by numbers
1d: Against all odds
2a: Wait on tables
2b: Times Square
2c:
2d: A finger in every pie
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a:
4b: Y intercept
4c:
4d: Close quarters
5a: Positive attitude
5b: Money is the root of all evil
5c: Apple pie
5d: Forever and a day
Still Needed
2c:
3b:
4a:
4c:
Hints posted later if needed.
2c: Bald headed/ Bald on top?
3b: For being out of line . . . ?
4a: $1000 in the hole?
4c: The force is with us?
OR
4c: Overall force?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 10th, 2020 at 3:14:51 PM
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Quote: unJonthink for 4a
4a: grandstand
Correct
Quote: gordonm888
2c: Bald headed/ Bald on top?
3b: For being out of line . . . ?
4a: $1000 in the hole?
4c: The force is with us?
OR
4c: Overall force?
Good guesses but incorrect.
1a: Back to square one
1b: Deep rooted
1c: Paint by numbers
1d: Against all odds
2a: Wait on tables
2b: Times Square
2c:
2d: A finger in every pie
3a: Too close to call
3b:
3c: Custard pie
3d: Task Force
4a: Grandstand
4b: Y intercept
4c:
4d: Close quarters
5a: Positive attitude
5b: Money is the root of all evil
5c: Apple pie
5d: Forever and a day
Still Needed
2c:
3b:
4c:
Hints posted later if needed.
Have you tried 22 tonight? I said 22.
September 10th, 2020 at 4:00:24 PM
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Cry of innocence
Urich after Vega$
Like a baby (or dog)
Have you tried 22 tonight? I said 22.
September 10th, 2020 at 4:19:09 PM
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4c - On all fours
September 10th, 2020 at 7:47:25 PM
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Nothing to do with me
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 10th, 2020 at 7:53:01 PM
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For hire
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 10th, 2020 at 10:17:53 PM
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Quote: charliepatrick4c - On all fours
Quote: gordonm888
Nothing to do with me
Quote: gordonm888
For hire
DING! DING! DING!
CORRECT!
Joint solve by...
gordonm888
Joeman
ThatDonGuy
unJon
charliepatrick
1a: Back to square one
1b: Deep rooted
1c: Paint by numbers
1d: Against all odds
2a: Wait on tables
2b: Times Square
2c: Nothing to do with me
2d: A finger in every pie
3a: Too close to call
3b: For hire
3c: Custard pie
3d: Task Force
4a: Grandstand
4b: Y intercept
4c: On all fours
4d: Close quarters
5a: Positive attitude
5b: Money is the root of all evil
5c: Apple pie
5d: Forever and a day
----------------------------
Have you tried 22 tonight? I said 22.
September 11th, 2020 at 9:41:44 AM
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Here's a classic puzzle...
Using each of the digits 1, 2, 3, and 4 twice, write out an eight-digit number in which there is one digit between the 1s, two digits between the 2s, three digits between between the 3s and four digits between the 4s.
Have you tried 22 tonight? I said 22.
September 11th, 2020 at 9:55:14 AM
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Quote: Gialmere
Here's a classic puzzle...
Using each of the digits 1, 2, 3, and 4 twice, write out an eight-digit number in which there is one digit between the 1s, two digits between the 2s, three digits between between the 3s and four digits between the 4s.
41312432
Seems like there are others. At least the reverse is one.
Seems like there are others. At least the reverse is one.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
September 11th, 2020 at 11:34:39 AM
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I haven't looked at it yet but there was a similar puzzle in last week's Sunday Times.
( https://www.thetimes.co.uk/article/teaser-3024-csbvz2lxl or https://brg.a2hosted.com/ ).
( https://www.thetimes.co.uk/article/teaser-3024-csbvz2lxl or https://brg.a2hosted.com/ ).
Quote: the puzzleTeaser 3024
Triple Jump
From a set of playing cards, Tessa took 24 cards consisting of three each of the aces, twos, threes, and so on up to the eights. She placed the cards face up in single row and decided to arrange them such that the three twos were each separated by two cards, the threes were separated by three cards and so forth up to and including the eights, duly separated by eight cards. The three aces were numbered with a one and were each separated by nine cards. Counting from the left, the seventh card in the row was a seven.
In left to right order, what were the numbers on the first six cards?
September 11th, 2020 at 1:13:33 PM
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After a bit of trial and error I get...Quote: charliepatrickI haven't looked at it yet...
A268257246A584736543A873
September 11th, 2020 at 2:39:14 PM
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Quote: unJon41312432
Seems like there are others. At least the reverse is one.
Correct!
It and its mirror are the only two solves.
------------------------------
I got into a fight with 1, 3, 5, 7 and 9.
The odds were against me.
Have you tried 22 tonight? I said 22.
September 11th, 2020 at 2:48:23 PM
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Here's another simple one...
:strip_icc()/pic5539870.png)
What percentage of the above combined shape is blue?
What percentage of the above combined shape is blue?
Have you tried 22 tonight? I said 22.
September 11th, 2020 at 4:26:34 PM
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Quote: Gialmere...What percentage of the above combined shape is blue?
Let each equilateral triangle have a side of 1. By sliding the triangles against each other, each side of the hexagon can be reduced to 0 or expanded to 1. When each side of the hexagon is 1, then the area of the hexagon is equal to the area of the six triangles. Therefore, the percentage blue of the figure can be adjusted anywhere from 0 to 50. In other words, the figure lacks sufficient information to determine an answer.
But for this particular figure, the blue looks to be about 12% of the total.
September 11th, 2020 at 4:34:51 PM
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Quote: ChesterDog
Let each equilateral triangle have a side of 1. By sliding the triangles against each other, each side of the hexagon can be reduced to 0 or expanded to 1. When each side of the hexagon is 1, then the area of the hexagon is equal to the area of the six triangles. Therefore, the percentage blue of the figure can be adjusted anywhere from 0 to 50. In other words, the figure lacks sufficient information to determine an answer.
But for this particular figure, the blue looks to be about 12% of the total.
You have a point.
The sides of the triangles are three times longer than the sides of the regular hexagon.
Have you tried 22 tonight? I said 22.
September 11th, 2020 at 5:53:45 PM
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Quote: GialmereQuote: ChesterDog
Let each equilateral triangle have a side of 1. By sliding the triangles against each other, each side of the hexagon can be reduced to 0 or expanded to 1. When each side of the hexagon is 1, then the area of the hexagon is equal to the area of the six triangles. Therefore, the percentage blue of the figure can be adjusted anywhere from 0 to 50. In other words, the figure lacks sufficient information to determine an answer.
But for this particular figure, the blue looks to be about 12% of the total.
You have a point.
The sides of the triangles are three times longer than the sides of the regular hexagon.
Thanks--your information does make it an easy one.
The regular hexagon can be decomposed into 6 little triangles, and each big red triangle can be decomposed into 9 little triangles. So, blue/whole = 6 / [6+6(9)] = 1/10 = 10%.
September 11th, 2020 at 11:18:58 PM
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Quote: ChesterDog
Thanks--your information does make it an easy one.
The regular hexagon can be decomposed into 6 little triangles, and each big red triangle can be decomposed into 9 little triangles. So, blue/whole = 6 / [6+6(9)] = 1/10 = 10%.
Correct!
:strip_icc()/pic5539871.png)
--------------------------
Have you tried 22 tonight? I said 22.
September 12th, 2020 at 9:47:51 AM
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Gialmere, I love your puzzles but I almost love your jokes and comics more. The last one about Oregon was priceless.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 12th, 2020 at 1:30:23 PM
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Here's a toughie for the weekend...
You are a gaming mathematician and you're being consulted by an eccentric game designer who needs your help concerning his new creation called "Casino 52 Pick-Up".
The game is played by throwing a standard 52-card deck into the air in such a way that each card, independently, is equally likely to land face up or face down. The total value of the cards which land face up is then calculated. Card values are assigned as follows: Ace=1, 2=2, ... , 10=10, Jack=11, Queen=12, King=13.
(It reminds you of a lunatic's Westernized version of Fan Tan complete with plastic dome.)
What the designer wants to know is...
What is the probability that the sum of the face up cards will be evenly divisible by 13?
You are a gaming mathematician and you're being consulted by an eccentric game designer who needs your help concerning his new creation called "Casino 52 Pick-Up".
The game is played by throwing a standard 52-card deck into the air in such a way that each card, independently, is equally likely to land face up or face down. The total value of the cards which land face up is then calculated. Card values are assigned as follows: Ace=1, 2=2, ... , 10=10, Jack=11, Queen=12, King=13.
(It reminds you of a lunatic's Westernized version of Fan Tan complete with plastic dome.)
What the designer wants to know is...
What is the probability that the sum of the face up cards will be evenly divisible by 13?
Have you tried 22 tonight? I said 22.
September 13th, 2020 at 9:54:54 AM
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To a very close approximation, I think the answer is 1/13 +2^51.
First of, we apply clock arithmetic. We consider the sum of every face-up card as ( kmod13).
Now consider any scenario where N cards are face-up, such that N= 1 to 51. If ΣN = 0mod13 then Σ(52-N) must also be congruent to 0mod13 and more generally if ΣN=Kmod13 then Σ(52-N)=(13-K)mod13.
So, another words, we are seeking a case in which both the face-up cards and the face down cards are evenly divisible by 13. We can validly consider either ΣN or Σ(52-N) depending upon whether N or 52-N is smaller. Thus, we can can say that we will consider N as ranging from 1 to 25.
Now for any given scenario in which N cards are face up, and N=1 to 25: after (N-1) cards are face-up the probability of k in (N-1)= kmod13 is evenly distributed between 0,1,2,3,4,5,6,7,8,9,10,11 and 12. Thus, the chance of the ranks of the Nth card bringing the sum up to 0mod13 is, on average, 1 in 13. (admittedly, this is an approximation)
Now let's consider the case when all 52 cards are either face-up or face down. In both those cases, the sum of the face-up cards will be evenly divisible by 13. and the combined liklihood of those 2 cases is 2^51.
So my answer is: Approximately 1/13 + (1/2^51)= 0.0769230769230774.
Last edited by: gordonm888 on Sep 13, 2020
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 13th, 2020 at 11:06:33 AM
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Quote: GialmereWhat is the probability that the sum of the face up cards will be evenly divisible by 13?
I couldn't figure out a way to calculate it other than with brute force.
5,412,980,321,359 / 70,368,744,177,664
Note that 1/13 = 5,412,980,321,359 / 70,368,744,177,667
Last edited by: ThatDonGuy on Sep 13, 2020
September 13th, 2020 at 7:22:35 PM
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Quote: gordonm888
To a very close approximation, I think the answer is 1/13 +2^51.
First of, we apply clock arithmetic. We consider the sum of every face-up card as ( kmod13).
Now consider any scenario where N cards are face-up, such that N= 1 to 51. If ΣN = 0mod13 then Σ(52-N) must also be congruent to 0mod13 and more generally if ΣN=Kmod13 then Σ(52-N)=(13-K)mod13.
So, another words, we are seeking a case in which both the face-up cards and the face down cards are evenly divisible by 13. We can validly consider either ΣN or Σ(52-N) depending upon whether N or 52-N is smaller. Thus, we can can say that we will consider N as ranging from 1 to 25.
Now for any given scenario in which N cards are face up, and N=1 to 25: after (N-1) cards are face-up the probability of k in (N-1)= kmod13 is evenly distributed between 0,1,2,3,4,5,6,7,8,9,10,11 and 12. Thus, the chance of the ranks of the Nth card bringing the sum up to 0mod13 is, on average, 1 in 13. (admittedly, this is an approximation)
Now let's consider the case when all 52 cards are either face-up or face down. In both those cases, the sum of the face-up cards will be evenly divisible by 13. and the combined liklihood of those 2 cases is 2^51.
So my answer is: Approximately 1/13 + (1/2^51)= 0.0769230769230774.
Correct!
--------------------------------
Have you tried 22 tonight? I said 22.
September 13th, 2020 at 8:29:37 PM
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Deleted post.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 13th, 2020 at 8:56:38 PM
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Having just finished watching the American Football it's much too late here (4:50am) to work this out.
But here's an idea.
For each of the ranks you can have 0,1,2,3 or 4 cards showing, that's a probability of 1,4,6,4,1 / 16.
For the one which is 0 mod 13 it doesnt matter how many there are, so ignore those.
So let's look at 1s and 12s, then 2s and 11s. etc.
If there are the same number (of say 1s and 12s) then the total is 0 mod 13.
If not then you can work out the net total (mod 13). If you have more 1s then it's +1 or +2 etc, if fewer then it's -1, -2 etc. So you can work out the chances of being -4 thru +4.
Doing this for each of the six pairs means you only have to work out a few combinations (96 rather than over 1bn) results.
But here's an idea.
For each of the ranks you can have 0,1,2,3 or 4 cards showing, that's a probability of 1,4,6,4,1 / 16.
For the one which is 0 mod 13 it doesnt matter how many there are, so ignore those.
So let's look at 1s and 12s, then 2s and 11s. etc.
If there are the same number (of say 1s and 12s) then the total is 0 mod 13.
If not then you can work out the net total (mod 13). If you have more 1s then it's +1 or +2 etc, if fewer then it's -1, -2 etc. So you can work out the chances of being -4 thru +4.
Doing this for each of the six pairs means you only have to work out a few combinations (96 rather than over 1bn) results.
September 14th, 2020 at 10:51:23 AM
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The 15 winners of a contest have been invited to a congratulatory party.
Although the average age of the group is exactly 25, some of the winners are quite young (15, 16, and 17).
The host asks the whole group to wear name tags with their names and ages.
Shortly after the party begins, three gatecrashers sneak in, fill out blank name tags with their own names and ages, and mingle with the legitimate guests.
Below are the 18 name tags.
Can you identify the 3 gatecrashers?
Al, 33
Bev, 15
Carl, 22
David, 26
Ed, 23
Fran, 20
Guy, 19
Heidi, 25
Irv, 32
John, 24
Kate, 18
Les, 21
Mark, 35
Nancy, 31
Oscar, 16
Patty, 17
Quincy, 28
Richard, 27
Have you tried 22 tonight? I said 22.
September 14th, 2020 at 12:20:18 PM
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Quote: Gialmere
The 15 winners of a contest have been invited to a congratulatory party.
Although the average age of the group is exactly 25, some of the winners are quite young (15, 16, and 17).
The host asks the whole group to wear name tags with their names and ages.
Shortly after the party begins, three gatecrashers sneak in, fill out blank name tags with their own names and ages, and mingle with the legitimate guests.
Below are the 18 name tags.
Can you identify the 3 gatecrashers?
Al, 33
Bev, 15
Carl, 22
David, 26
Ed, 23
Fran, 20
Guy, 19
Heidi, 25
Irv, 32
John, 24
Kate, 18
Les, 21
Mark, 35
Nancy, 31
Oscar, 16
Patty, 17
Quincy, 28
Richard, 27
Yes! The 3 gatecrashers are Fran, Guy, and Kate.
The sum of all 18 ages is 432. The sum of the 15 invitees' ages is 15*25 = 375. Therefore, the sum of the ages of the crashers is 432 - 375 = 57.
Taken three-at-a-time, many triples have sums of ages of 57. To narrow the choices, we can use the additional bit of information that three of the invitees have ages of 15, 16, and 17. The only triple not eliminated is 18-19-20 which are the ages of Kate, Guy, and Fran.
September 14th, 2020 at 12:43:15 PM
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Quote: charliepatrickHaving just finished watching the American Football it's much too late here (4:50am) to work this out.
But here's an idea.
For each of the ranks you can have 0,1,2,3 or 4 cards showing, that's a probability of 1,4,6,4,1 / 16.
For the one which is 0 mod 13 it doesnt matter how many there are, so ignore those.
So let's look at 1s and 12s, then 2s and 11s. etc.
If there are the same number (of say 1s and 12s) then the total is 0 mod 13.
If not then you can work out the net total (mod 13). If you have more 1s then it's +1 or +2 etc, if fewer then it's -1, -2 etc. So you can work out the chances of being -4 thru +4.
Doing this for each of the six pairs means you only have to work out a few combinations (96 rather than over 1bn) results.
I agree. I started out this way and tried different things. But eventually I realized that splitting a deck into n and 52-n cards does not change the expected values of the sum of the ranks in either batch to be different than the entire deck.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 14th, 2020 at 4:18:58 PM
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Quote: ChesterDog
Yes! The 3 gatecrashers are Fran, Guy, and Kate.
The sum of all 18 ages is 432. The sum of the 15 invitees' ages is 15*25 = 375. Therefore, the sum of the ages of the crashers is 432 - 375 = 57.
Taken three-at-a-time, many triples have sums of ages of 57. To narrow the choices, we can use the additional bit of information that three of the invitees have ages of 15, 16, and 17. The only triple not eliminated is 18-19-20 which are the ages of Kate, Guy, and Fran.
Correct!
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Have you tried 22 tonight? I said 22.
September 14th, 2020 at 6:04:24 PM
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I ignored the 0-value cards so there's 48 cards in the deck. The spreadsheet was easy to write but took a while to calculate, old version of excel!Quote: gordonm888I agree....
Thus there are 248 ( X=281 474 976 710 656 ) options to be considered and I get (X-1)/13 for all the Nmod13 outcomes other than 0mod13 which is one more.
So using algebra where X=248
pr(0 mod 13) = { (X-1)/13+1 }/X = {X-1+13)/13 X = 1/13 + 12/13X. I'm guessing the other possibilities are slightly less than 1/13 (1/13 - 1/13X). So the extra 1 option for no cards face up already has 1/13 of it in the 0mod13 total of 1/13, hence the 12/13th.
So using algebra where X=248
pr(0 mod 13) = { (X-1)/13+1 }/X = {X-1+13)/13 X = 1/13 + 12/13X. I'm guessing the other possibilities are slightly less than 1/13 (1/13 - 1/13X). So the extra 1 option for no cards face up already has 1/13 of it in the 0mod13 total of 1/13, hence the 12/13th.
September 15th, 2020 at 8:01:01 AM
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A sort of sequel to the famous birthday problem...
What is the probability that in a group of 4 people, at least 2 of them are born in the same month?
Assume a 365 day year (i.e. 28 days in February) but, apart from this caveat, use the correct probabilities for each of the months.
What is the probability that in a group of 4 people, at least 2 of them are born in the same month?
Assume a 365 day year (i.e. 28 days in February) but, apart from this caveat, use the correct probabilities for each of the months.
Have you tried 22 tonight? I said 22.
September 15th, 2020 at 8:39:59 AM
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This problem involves a corner of combination theory that I am trying to become more familiar with, so I look forward to any discussion of your answers. Of course, there is more than one approach to solving this.
(1) How many bridge deals (13 cards out of 52) on average will it take before a player has held all 52 cards (in his own hand) at least once?
(2) How many Pai Gow Poker hands (7 cards out of 53) on average will it take before a player has held all 53 cards at least once?
(1) How many bridge deals (13 cards out of 52) on average will it take before a player has held all 52 cards (in his own hand) at least once?
(2) How many Pai Gow Poker hands (7 cards out of 53) on average will it take before a player has held all 53 cards at least once?
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 15th, 2020 at 8:52:51 AM
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Quote: Gialmere...What is the probability that in a group of 4 people, at least 2 of them are born in the same month?...use the correct probabilities for each of the months.
I couldn't see an easy way other than working out all the combinations of 31-day,30-day,28-day months could happen, then the number of days in a year that satisfy each person being in a different month from previous people. The answer in this table is where all people are in different months. So the answer to the puzzle is 1-X = 41.781 374%.
| No of | No of | No of | Ways | Days | Days | Days | Days | Contribution |
|---|---|---|---|---|---|---|---|---|
| 31-day | 30-day | 28-day | first | first | first | first | ||
| months | months | months | person | person | person | person | ||
4 | 1 | 217 | 186 | 155 | 124 | 775 757 640 | ||
3 | 1 | 4 | 217 | 186 | 155 | 120 | 3 002 932 800 | |
3 | 1 | 4 | 217 | 186 | 155 | 28 | 700 684 320 | |
2 | 2 | 6 | 217 | 186 | 120 | 90 | 2 615 457 600 | |
2 | 1 | 1 | 12 | 217 | 186 | 120 | 28 | 1 627 395 840 |
1 | 3 | 4 | 217 | 120 | 90 | 60 | 562 464 000 | |
1 | 2 | 1 | 12 | 217 | 120 | 90 | 28 | 787 449 600 |
4 | 1 | 120 | 90 | 60 | 30 | 19 440 000 | ||
3 | 1 | 4 | 120 | 90 | 60 | 28 | 72 576 000 | |
Total | 10 164 157 800 | |||||||
365 | 364 | 363 | 362 | 17 458 601 160 | ||||
Prob | 58.218 626% |
September 15th, 2020 at 10:05:00 AM
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Quote: gordonm888This problem involves a corner of combination theory that I am trying to become more familiar with, so I look forward to any discussion of your answers. Of course, there is more than one approach to solving this.
(1) How many bridge deals (13 cards out of 52) on average will it take before a player has held all 52 cards (in his own hand) at least once?
(2) How many Pai Gow Poker hands (7 cards out of 53) on average will it take before a player has held all 53 cards at least once?
I'll use the Bridge problem as an example.
Let E(n) be the expected number of deals needed to get all 52 cards at least once, if you have already had at least n cards dealt at least once.
If n <= 39:
E(n) = 1 + (C(52-n, 13) x E(n+13) + C(n, 1) x C(52-n, 12) x E(n + 12) + C(n, 2) x C(52-n, 11) x E(n + 11) + ... + C(n, 12) x C(52-n, 1) x E(n + 1) + C(n, 13) x E(n)) / C(52, 13)
E(40) = 1 + (C(40, 1) x C(12, 12) x E(52) + C(40, 2) x C(12, 11) x E(51) + C(40, 3) x C(12, 10) x E(50) + ... + C(40, 12) x C(12, 1) x E(41) + C(40, 13) x E(40)) / C(52, 13)
E(41) = 1 + (C(41, 1) x C(11, 11) x E(52) + C(41, 2) x C(11, 10) x E(51) + C(41, 3) x C(11, 9) x E(50) + ... + C(41, 12) x C(11, 1) x E(42) + C(41, 13) x E(41)) / C(52, 13)
and so on through E(51). Note that E(52) = 0.
This is 52 equations in 52 unknowns; solve for E(0).
It appears to be easiest to solve by working backwards from E(51).
September 15th, 2020 at 12:42:38 PM
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Quote: charliepatrickI couldn't see an easy way other than working out all the combinations of 31-day,30-day,28-day months could happen, then the number of days in a year that satisfy each person being in a different month from previous people. The answer in this table is where all people are in different months. So the answer to the puzzle is 1-X = 41.781 374%.
No of No of No of Ways Days Days Days Days Contribution 31-day 30-day 28-day first first first first months months months person person person person
Sorry. Incorrect.
Have you tried 22 tonight? I said 22.
September 15th, 2020 at 3:35:56 PM
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Quote: GialmereQuote: charliepatrickI couldn't see an easy way other than working out all the combinations of 31-day,30-day,28-day months could happen, then the number of days in a year that satisfy each person being in a different month from previous people. The answer in this table is where all people are in different months. So the answer to the puzzle is 1-X = 41.781 374%.
No of No of No of Ways Days Days Days Days Contribution 31-day 30-day 28-day first first first first months months months person person person person
Sorry. Incorrect.
0.427335923
I agree with Charlie's approach and calculations except that the total number of combinations should be 365^4 because the four people can be born on the same day.
I agree with Charlie's approach and calculations except that the total number of combinations should be 365^4 because the four people can be born on the same day.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
September 15th, 2020 at 4:12:50 PM
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Quote: gordonm8880.427335923
I agree with Charlie's approach and calculations except that the total number of combinations should be 365^4 because the four people can be born on the same day.
Close, but no cigar.
Have you tried 22 tonight? I said 22.
