Easy math puzzles
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46 members have voted
October 12th, 2020 at 3:37:48 PM
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Quote: Gialmere
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Find the roulette wheel that has a black instead of green zero.
:strip_icc()/pic5708069.png)
This bonus puzzle was much more difficult!
start from the lower left corner of the image. Move one roulette wheel to the right. Immediately above that roulette wheel is a wheel that is almost totally hidden, except that it has a black zero showing!
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 12th, 2020 at 5:45:27 PM
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Quote: gordonm888This bonus puzzle was much more difficult!
start from the lower left corner of the image. Move one roulette wheel to the right. Immediately above that roulette wheel is a wheel that is almost totally hidden, except that it has a black zero showing!
Correct!
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Russian roulette is an amazing game...
5 out of 6 people recommend it.
Have you tried 22 tonight? I said 22.
October 13th, 2020 at 8:00:50 AM
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And now a toughie for Tuesday...
Two gamblers agree to an unusual wager.
Each of them receives a standard deck of 52 cards which they thoroughly shuffle.The players then deal the cards face up one at a time from the top of their decks (like the card game War), checking for an exact match (such as both players tossing down the queen of spades).
Player A wins if, once the packs are fully dealt, no matches are found. Player B wins if at least one match occurs.
What is the probability that player A wins?
Two gamblers agree to an unusual wager.
Each of them receives a standard deck of 52 cards which they thoroughly shuffle.The players then deal the cards face up one at a time from the top of their decks (like the card game War), checking for an exact match (such as both players tossing down the queen of spades).
Player A wins if, once the packs are fully dealt, no matches are found. Player B wins if at least one match occurs.
What is the probability that player A wins?
Note that the solve for this puzzle is beyond my meager math skills and I can only hope that your solve matches the answer I do have.
Have you tried 22 tonight? I said 22.
October 13th, 2020 at 9:21:52 AM
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To be fair, I had to look this one up, if for nothing else but to see if a solution actually existed - this is a rather famous problem.
Edited - the original "solution" was for Player B winning
It appears to be 1 - (1 - 1/2! + 1/3! - 1/4! + ... + 1/51! - 1/52!), which is very close to 1/e.
In fact, as the number of cards approaches positive infinity, the solution approaches 1/e.
Edited - the original "solution" was for Player B winning
It appears to be 1 - (1 - 1/2! + 1/3! - 1/4! + ... + 1/51! - 1/52!), which is very close to 1/e.
In fact, as the number of cards approaches positive infinity, the solution approaches 1/e.
October 13th, 2020 at 9:27:42 AM
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Quote: ThatDonGuyTo be fair, I had to look this one up, if for nothing else but to see if a solution actually existed - this is a rather famous problem.
It appears to be 1 - 1/2! + 1/3! - 1/4! + ... + 1/51! - 1/52! = about 0.632120558828558
As the number of cards approaches positive infinity, the solution approaches 1 - 1/e.
This is an old bar bet.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 13th, 2020 at 9:55:49 AM
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I knew the answer p (as you say you need 1-p here) but never knew the proof. There's an interesting proof by induction at https://www.dartmouth.edu/~chance/teaching_aids/books_articles/probability_book/Chapter3.pdf p85ffQuote: ThatDonGuyTo be fair, I had to look this one up, if for nothing else but to see if a solution actually existed - this is a rather famous problem.
Edited - the original "solution" was for Player B winning
It appears to be 1 - (1 - 1/2! + 1/3! - 1/4! + ... + 1/51! - 1/52!), which is very close to 1/e.
In fact, as the number of cards approaches positive infinity, the solution approaches 1/e.
October 13th, 2020 at 10:09:50 AM
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Here's the proof that I saw, although I can't remember exactly where I saw it:
The probability that N particular cards are right is (52-N)! / 52!
By the Inclusion/Exclusion Principle, the probability that at least 1 card is right is
P(#1 is right) + P(#2 is right) + ... + P(#52 is right)
- P(#1 and #2 are both right) - P(#1 and #3 are both right) - ... - P(#51 and #52 are both right)
+ P(1,2,3 are right) + P(1,2,4 are right) + P(1,2,5 are right) + ... + P(50,51,52 are right)
- P(1,2,3,4 are right) - ... - P(49,50,51,52) are right
+ ...
The row where N cards are correct has Combin(52,N) entries, each of which has probability (52-N)!,
so the sum of the terms in row N is 52! / (N! (52-N)!) x (52-N)! / 52! = 1 / N!
and the sum of all of the rows is 1 - 1/2! + 1/3! - 1/4! + ... - 1/52!
The probability that none of the cards are correct is 1 minus this value
The probability that N particular cards are right is (52-N)! / 52!
By the Inclusion/Exclusion Principle, the probability that at least 1 card is right is
P(#1 is right) + P(#2 is right) + ... + P(#52 is right)
- P(#1 and #2 are both right) - P(#1 and #3 are both right) - ... - P(#51 and #52 are both right)
+ P(1,2,3 are right) + P(1,2,4 are right) + P(1,2,5 are right) + ... + P(50,51,52 are right)
- P(1,2,3,4 are right) - ... - P(49,50,51,52) are right
+ ...
The row where N cards are correct has Combin(52,N) entries, each of which has probability (52-N)!,
so the sum of the terms in row N is 52! / (N! (52-N)!) x (52-N)! / 52! = 1 / N!
and the sum of all of the rows is 1 - 1/2! + 1/3! - 1/4! + ... - 1/52!
The probability that none of the cards are correct is 1 minus this value
October 13th, 2020 at 2:54:45 PM
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Standard derangement problem, often called the hat check problem. If x number of people check their hat at the door, then the clerk returns them randomly, what is the chance no one gets their hat back. The answer converges quickly to 1/e as x increases
It’s all about making that GTA
October 13th, 2020 at 5:13:57 PM
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At the Wizard of Vegas, the time between suspensions is on average 10 days. The probability of a suspension at any given time is always the same, regardless of historical suspensions. What is the probability of exactly 3 suspensions in 30 days?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 13th, 2020 at 5:27:41 PM
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Quote: WizardAt the Wizard of Vegas, the time between suspensions is on average 10 days. The probability of a suspension at any given time is always the same, regardless of historical suspensions. What is the probability of exactly 3 suspensions in 30 days?
3^3 / (3! * e^3) =~ 22.4%
It’s all about making that GTA
October 13th, 2020 at 5:32:18 PM
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Quote: Ace2
3^3 / (3! * e^3) =~ 22.4%
That was fast. I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 14th, 2020 at 8:03:38 AM
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Back to candles...
/pic5559639.png)
I have two candles, one an inch longer than the other.
I light the longer candle at 3 PM and the shorter one at 5 PM. At 9 PM, they are the same length. The longer candle goes out at 11 PM and the shorter one goes out at 10:30 PM.
How long were they at the start?
I have two candles, one an inch longer than the other.
I light the longer candle at 3 PM and the shorter one at 5 PM. At 9 PM, they are the same length. The longer candle goes out at 11 PM and the shorter one goes out at 10:30 PM.
How long were they at the start?
Have you tried 22 tonight? I said 22.
October 14th, 2020 at 8:29:53 AM
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y/x = 16/11
y - x = 1
x = short = 2.2 inches
y = long = 3.2 inches
It’s all about making that GTA
October 14th, 2020 at 9:00:36 AM
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Quote: GialmereBack to candles...
...I have two candles, one an inch longer than the other.
I light the longer candle at 3 PM and the shorter one at 5 PM. At 9 PM, they are the same length. The longer candle goes out at 11 PM and the shorter one goes out at 10:30 PM.
How long were they at the start?
Let L be the long candle's initial length, and s be the short candle's initial length.
And let K be the rate in inches per hour that the long candle burns, and k be the short candle's rate.
Then:
L = s + 1
L - 6K = s - 4k
L - 8K = 0
s - 5.5k = 0
k = s/5.5 = 2s / 11
K = L/8
L = s + 1
L - 6L/8 = s - 8s/11
L = s + 1
L/4 = 3s/11
L = s + 1
11L = 12s
11s + 11 = 12s
s = 11 in.
L = 12 in.
October 14th, 2020 at 9:14:23 AM
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I get 12" and 11".
After 9pm the longer candle took 2 hours and the shorter candle took 1.5 hours to burn the same length.
The longer candle was lit at 3pm (this is 6 hours prior to 9pm) and took a total of 8 hours to burn. Thus the last 2 hours represent 1/4 of its entire length.
If the "shorter" candle had been the same length (as the longer candle) its last quarter would have taken 1.5 hours, so in total the whole candle would have taken 6 hours. However the actual shorter candle only took 5.5 hours (17h00 to 22h30) and the difference in time, 1/2 hour, represents 1 inch.
So the shorter candle burns at 2 inches/hour and was originally 11 inches tall. The longer candle burns at 1.5 inches/hour and was originally 12 inches tall.
After 9pm the longer candle took 2 hours and the shorter candle took 1.5 hours to burn the same length.
The longer candle was lit at 3pm (this is 6 hours prior to 9pm) and took a total of 8 hours to burn. Thus the last 2 hours represent 1/4 of its entire length.
If the "shorter" candle had been the same length (as the longer candle) its last quarter would have taken 1.5 hours, so in total the whole candle would have taken 6 hours. However the actual shorter candle only took 5.5 hours (17h00 to 22h30) and the difference in time, 1/2 hour, represents 1 inch.
So the shorter candle burns at 2 inches/hour and was originally 11 inches tall. The longer candle burns at 1.5 inches/hour and was originally 12 inches tall.
October 14th, 2020 at 9:47:58 AM
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11" and 12".
Height of candle one at time t = 16.5 - 3t/2.
Height of candle two at time t = 21 - 2t.
Where t is hours since noon. These equations only work once the candle is lit (so t=3 for candle 1 and t=5 for candle 2).
Last edited by: Wizard on Oct 14, 2020
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 14th, 2020 at 9:52:04 AM
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Let L be the shorter length in inches; the longer length is L + 1
Let C be the "common" length at 9 PM
Shorter candle:
(L - C) / 4 = C / 1.5
1.5 L = 5.5 C
C = 3/11 L
Longer candle:
((L + 1) - C) / 6 = C / 2
L + 1 = 4C
C = 1/4 L + 1/4
3/11 L = 1/4 L + 1/4
12 L = 11 L + 11
L = 11 (and C = 3), so the shorter candle is 11 inches and the longer candle is 12.
October 14th, 2020 at 11:09:28 PM
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Quote: ChesterDog
Let L be the long candle's initial length, and s be the short candle's initial length.
And let K be the rate in inches per hour that the long candle burns, and k be the short candle's rate.
Then:
L = s + 1
L - 6K = s - 4k
L - 8K = 0
s - 5.5k = 0
k = s/5.5 = 2s / 11
K = L/8
L = s + 1
L - 6L/8 = s - 8s/11
L = s + 1
L/4 = 3s/11
L = s + 1
11L = 12s
11s + 11 = 12s
s = 11 in.
L = 12 in.
Quote: charliepatrickI get 12" and 11".
After 9pm the longer candle took 2 hours and the shorter candle took 1.5 hours to burn the same length.
The longer candle was lit at 3pm (this is 6 hours prior to 9pm) and took a total of 8 hours to burn. Thus the last 2 hours represent 1/4 of its entire length.
If the "shorter" candle had been the same length (as the longer candle) its last quarter would have taken 1.5 hours, so in total the whole candle would have taken 6 hours. However the actual shorter candle only took 5.5 hours (17h00 to 22h30) and the difference in time, 1/2 hour, represents 1 inch.
So the shorter candle burns at 2 inches/hour and was originally 11 inches tall. The longer candle burns at 1.5 inches/hour and was originally 12 inches tall.
Quote: Wizard
11" and 12".
Height of candle one at time t = 16.5 - 3t/2.
Height of candle two at time t = 21 - 2t.
Where t is hours since noon. These equations only work once the candle is lit (so t=3 for candle 1 and t=5 for candle 2).
Quote: ThatDonGuy
Let L be the shorter length in inches; the longer length is L + 1
Let C be the "common" length at 9 PM
Shorter candle:
(L - C) / 4 = C / 1.5
1.5 L = 5.5 C
C = 3/11 L
Longer candle:
((L + 1) - C) / 6 = C / 2
L + 1 = 4C
C = 1/4 L + 1/4
3/11 L = 1/4 L + 1/4
12 L = 11 L + 11
L = 11 (and C = 3), so the shorter candle is 11 inches and the longer candle is 12.
Correct!
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When it gets cold in Ireland, we all sit round a candle.
And when it gets REALLY cold, we light it.
Have you tried 22 tonight? I said 22.
October 15th, 2020 at 2:08:04 PM
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My daughter's homework asked this question: What is the least possible sum of a 4-digit number?
I answered 1
1,000 is broken into these 4 digits 1+0+0+0 = 1
The answer is Wrong. Any ideas what I'm missing?
Thanks
I answered 1
1,000 is broken into these 4 digits 1+0+0+0 = 1
The answer is Wrong. Any ideas what I'm missing?
Thanks
October 15th, 2020 at 2:51:17 PM
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Quote: gerback123My daughter's homework asked this question: What is the least possible sum of a 4-digit number?
That depends on what the definition of the sum of one number is.
If it's the number itself (i.e. in the same way that the sum of the set of all even prime numbers is 2), then the answer is 1000...unless the teacher is one of those people who thinks something like -1234 is "a four-digit number," in which case it's -9999. (This brings back memories - apparently, the only reason I got an 800 on my math SAT back in 1979 instead of a 790 was, when a question asked, "Which is always larger - a three-digit number, or the sum of its digits," I used the interpretation that "n-digit number" is positive and does not start with zero.)
October 15th, 2020 at 4:16:33 PM
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what about (-9),(-9),(-9),(-9)? Its sum is -36.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 15th, 2020 at 5:00:19 PM
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Quote: gordonm888what about (-9),(-9),(-9),(-9)? Its sum is -36.
That's four 1-digit numbers (and again, assuming that a "one-digit number" can be negative), not one 4-digit number
October 15th, 2020 at 6:42:48 PM
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Quote: ThatDonGuyThat's four 1-digit numbers (and again, assuming that a "one-digit number" can be negative), not one 4-digit number
I generally agree, but it does depend what radix/number system you are in, A system such as ternary does allow negative numbers as digits. You're assuming a standard decimal representation of numbers (which is probably appropriate for this particular context.)
Here's an example of a very low four digit number: VIII.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 15th, 2020 at 8:08:29 PM
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Quote: ThatDonGuyThat's four 1-digit numbers (and again, assuming that a "one-digit number" can be negative), not one 4-digit number
In that case -9,000 would be -9 and the minimum. I’m not sure that the question is asking for the minimum sum of the digits though. But I’m not really sure what the question is otherwise asking for either.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
October 15th, 2020 at 8:40:11 PM
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Quote: gerback123My daughter's homework asked this question: What is the least possible sum of a 4-digit number?
I answered 1
1,000 is broken into these 4 digits 1+0+0+0 = 1
The answer is Wrong. Any ideas what I'm missing?
Thanks
I agree with your answer if the exact question was: What is the least possible sum of the digits of a 4-digit number?
Perhaps some words of the question are missing. Maybe, the question was: What is the least possible sum of the factors of the prime factorization of a 4-digit number? As an example, 1024 = 2*2*2*2*2*2*2*2*2*2, and the sum of the factors is 20.
By the way, for what year of school was this homework question?
October 16th, 2020 at 2:17:53 AM
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Nice idea. Some clocks actually have IIII, so 4 might be the answer!Quote: gordonm888...Here's an example of a very low four digit number: VIII.
It's also interesting whether something like 999i, a halfword x0000, 1E-999, etc. count.
October 16th, 2020 at 7:54:57 AM
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From 538:
On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.
If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?
Extra credit: Suppose there are N candidates instead of three. What is the probability of a runoff?
Not sure how to proceed. Thoughts?
On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.
If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?
Extra credit: Suppose there are N candidates instead of three. What is the probability of a runoff?
Not sure how to proceed. Thoughts?
October 16th, 2020 at 8:17:59 AM
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Quote: rsactuaryNot sure how to proceed. Thoughts?
Let P0 through PN be points on the line 0 <= x <= 1, with P0 = 0 and Pk - Pk-1 be the fraction of the vote that candidate k received. Note that PN = 1.
What is the probability that none of Pk - Pk-1 > 1/2 for k = 1, 2, ..., N?
Let x be the fraction of votes Candidate A got (0 < x < 1). There is a runoff if Candidate B got between (0.5 - x) and 0.5; if B gets less than 0.5, then C has a majority.
The probability of a runoff is Integral(0, 1/2) {x / (1 - x) dx}
= - Integral(0, 1/2) {x / (x - 1) dx}
= - Integral(0, 1/2) {(x - 1 + 1) / (x - 1) dx}
= - Integral(0, 1/2) {1 + 1 / (x - 1) dx}
= - Integral(0, 1/2) {1 dx) + Integral(0, 1/2) {1 / (1 - x) dx}
= - 1/2 - Integral(0, 1/2) {1 / (1 - x) d(1-x)}
= - 1/2 - Integral(0, 1/2) {ln |1 - x|}
= - 1/2 - ln 1/2 + ln 1 = 1/2 - ln (-ln 2)
= ln 2 - 1/2 = about 0.193147
This took me a while to figure out, until I realized that I was trying to take the integral {x / (1 - x) dx} by calculating its derivative instead...that's one reason I prefer discrete methods to integrating
Last edited by: ThatDonGuy on Oct 16, 2020
October 16th, 2020 at 9:16:25 AM
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Quote: rsactuaryFrom 538:
On Nov. 3, the residents of Riddler City will elect a mayor from among three candidates. The winner will be the candidate who receives an outright majority (i.e., more than 50 percent of the vote). But if no one achieves this outright majority, there will be a runoff election among the top two candidates.
If the voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course), then what is the probability of a runoff?
Extra credit: Suppose there are N candidates instead of three. What is the probability of a runoff?
Not sure how to proceed. Thoughts?
Thought 1: A simulation will quickly find a close number with which to compare your later calculations.
Thought 2: The 3-dimensional problem can be visualized as the interior of a cube with points (0,0,0), (1,0,0), (0,1,0), (0,0,1), (1,1,0), (1,0,1), (0,1,1), and (1,1,1).
The values of x, y, and z that would cause a runoff satisfy these inequalities:
x / (x+y+z) < 1/2
y / (x+y+z) < 1/2
z / (x+y+z) < 1/2
along with 0<x<1, 0<y<1, and 0<z<1
Then the probability of a runoff would be the volume of the solid characterized by these inequalities.
October 16th, 2020 at 11:03:04 AM
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By the way, you can type an integral sign: ∫ The Html code is &_in; with the underlined space removed.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 16th, 2020 at 11:03:16 AM
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By the way, you can type an integral sign: ∫ The Html code is &_in; with the underlined space removed.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 16th, 2020 at 11:03:26 AM
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By the way, you can type an integral sign: ∫ The Html code is &_in; with the underlined space removed.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 16th, 2020 at 12:05:50 PM
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Quote: ChesterDog
The values of x, y, and z that would cause a runoff satisfy these inequalities:
x / (x+y+z) < 1/2
y / (x+y+z) < 1/2
z / (x+y+z) < 1/2
along with 0<x<1, 0<y<1, and 0<z<1
Shouldn't the boundary conditions be x > 0, y > 0, z > 0, and x + y + z < 1?
October 16th, 2020 at 2:11:26 PM
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I'm not sure I understand how the percentage each candidate gets is determined. It's easy to pick three random numbers from 0 to 1, but here we require them to add up to 1. The way I did is I picked two random points from 0 to 1, which divides the segment between 0 and 1 into three pieces, which I then gave to each candidate as their share of the vote. For example, if the random points were 0.25 and 0.9, then A got 25%, B got 65%, and C got 10%.
If that is an acceptable way to interpret the problem, then my answer is....
If that is an acceptable way to interpret the problem, then my answer is....
50%
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 16th, 2020 at 2:34:11 PM
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Quote: ThatDonGuyQuote: ChesterDog
The values of x, y, and z that would cause a runoff satisfy these inequalities:
x / (x+y+z) < 1/2
y / (x+y+z) < 1/2
z / (x+y+z) < 1/2
along with 0<x<1, 0<y<1, and 0<z<1
Shouldn't the boundary conditions be x > 0, y > 0, z > 0, and x + y + z < 1?
Let the fractions of the vote that candidates A, B, and C each get are represented by a, b, and c respectively. If a is randomly distributed from 0 and 1, and b is randomly distributed from 0 and 1 - a, then c = 1- a - b. A quick simulation shows that the probability of a runoff is approximately 19%, which agrees with your answer.
I'm not sure that the above distribution is the same as stated in the problem, "...voting shares of each candidate are uniformly distributed between 0 percent and 100 percent (subject to the constraint that they add up to 100 percent, of course)...", but it might be.
I am not confident in my method. By the way, my method produces an answer of 50%, which matches a simulation.
October 16th, 2020 at 6:25:18 PM
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Quote: WizardI'm not sure I understand how the percentage each candidate gets is determined. It's easy to pick three random numbers from 0 to 1, but here we require them to add up to 1. The way I did is I picked two random points from 0 to 1, which divides the segment between 0 and 1 into three pieces, which I then gave to each candidate as their share of the vote. For example, if the random points were 0.25 and 0.9, then A got 25%, B got 65%, and C got 10%.
If that is an acceptable way to interpret the problem, then my answer is....
I like the way you did that. I think that's the best way to do it.
But would you please do a quick simulation to see if it matches your answer?
Last edited by: ChesterDog on Oct 16, 2020
October 16th, 2020 at 8:45:06 PM
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Quote: ChesterDogI like the way you did that. I think that's the best way to do it.
But would you please do a quick simulation to see if it matches your answer?
Thank you. You're right, a simulation does disprove my answer. I still think its a fair way to set up the problem, but can see I made an error in getting to 50%. Mean/median confusion I think.
My new answer is 25%.
It is obvious the chances are 50% that both random numbers are on the same side of over/under 0.5. If that happens, somebody gets over 50% and there is no runoff.
If one is above and one below 0.5, then there is a 50% chance they are within 0.5 of each other. That step is simple geometry.
So, the chances of a runoff are 50% * 50% = 25%.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 16th, 2020 at 9:11:19 PM
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Quote: WizardThank you. You're right, a simulation does disprove my answer. I still think its a fair way to set up the problem, but can see I made an error in getting to 50%. Mean/median confusion I think.
My new answer is 25%.
It is obvious the chances are 50% that both random numbers are on the same side of over/under 0.5. If that happens, somebody gets over 50% and there is no runoff.
If one is above and one below 0.5, then there is a 50% chance they are within 0.5 of each other. That step is simple geometry.
So, the chances of a runoff are 50% * 50% = 25%.
You're welcome!
Your method and answer show that a good way to image the problem is with a "ternary plot."
Here's one from Wikipedia regarding soils:
Suppose the names of the three candidates are Clay, Silt, and Sand. The big triangle can be cut into four congruent triangles with the lines: Clay=50%, Silt=50%, and Sand=50%. Within the center triangle, neither Clay, Silt, nor Sand has a majority, and its area is 1/4 of the big triangle, so the probability of a runoff is 25%.
October 17th, 2020 at 1:25:31 AM
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I get a different number using this method. I also picked two random numbers but then ordered them (lo hi).Quote: WizardI'm not sure I understand how the percentage each candidate gets is determined. It's easy to pick three random numbers from 0 to 1, but here we require them to add up to 1. The way I did is I picked two random points from 0 to 1, which divides the segment between 0 and 1 into three pieces, which I then gave to each candidate as their share of the vote. For example, if the random points were 0.25 and 0.9, then A got 25%, B got 65%, and C got 10%.
If that is an acceptable way to interpret the problem, then my answer is....50%
A gets the votes from 0% to lo % - i.e. a total of lo%.
B gets the votes from lo& to hi% - i.e. a total of (hi-lo)%.
C gets the votes from hi% to 1 - i.e. a total of (1-hi%).
Another way to look at this is start with the first random number chosen (whether highest or lowest). Consider the interval [0,1] divided by this random number. For simplicity only consider those where the number (X) is in the range 0 thru .5. (The ones .5 thru 1 will have the same result, so can be ignored or just use 1-X rather than X.)
The largest section is to the right and is between 1 and .5 wide.
Now consider the position of the second random number and the range of places on [0,1] which result in someone winning with 50% or more.
If X=0 then the remaining sector is 1, so however you cut it one of the parts will be >.5. Pr =1.
If X=0.49999... then if the second number is less than .5 you win, or if it's greater than .99999... Pr = .5.
If X=.25 then if the second number is less than .5 you win, but you also win if you choose a number greater than .75.
It transpires that if the second number is less than .5 you win (as the right hand side left >.5); but also if the second number is sufficently large (X+.5) or more, you win (the middle part is >.5). The chances of this is 1-(X+.5) = .5-X.
Finnaly you take the average of ( .5 + .5-X) = (1-X) from 0 to .5. Since this is a straight line it's the average of 1 and .5 = .75.
Simulation gets the same result.
Probability of someone having a clear win is 75%.
The largest section is to the right and is between 1 and .5 wide.
Now consider the position of the second random number and the range of places on [0,1] which result in someone winning with 50% or more.
If X=0 then the remaining sector is 1, so however you cut it one of the parts will be >.5. Pr =1.
If X=0.49999... then if the second number is less than .5 you win, or if it's greater than .99999... Pr = .5.
If X=.25 then if the second number is less than .5 you win, but you also win if you choose a number greater than .75.
It transpires that if the second number is less than .5 you win (as the right hand side left >.5); but also if the second number is sufficently large (X+.5) or more, you win (the middle part is >.5). The chances of this is 1-(X+.5) = .5-X.
Finnaly you take the average of ( .5 + .5-X) = (1-X) from 0 to .5. Since this is a straight line it's the average of 1 and .5 = .75.
Simulation gets the same result.
Probability of someone having a clear win is 75%.
October 17th, 2020 at 11:52:15 AM
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Here's a classic puzzle...
Two poker players agree to play a game using a standard 52-card deck which is spread face up on the table.
Player A selects any five of the face up cards to form his hand. Player B then selects any five of the remaining cards. Player A may then discard some, none or all of his cards and draw new cards from those still available. Player B then does the same.
The hands are then compared and the player with the highest value poker hand wins. Since player A has the advantage of selecting first, Player B has the advantage of winning all ties.
Which player will win the game and what strategy will he use?
Two poker players agree to play a game using a standard 52-card deck which is spread face up on the table.
Player A selects any five of the face up cards to form his hand. Player B then selects any five of the remaining cards. Player A may then discard some, none or all of his cards and draw new cards from those still available. Player B then does the same.
The hands are then compared and the player with the highest value poker hand wins. Since player A has the advantage of selecting first, Player B has the advantage of winning all ties.
Which player will win the game and what strategy will he use?
Have you tried 22 tonight? I said 22.
October 17th, 2020 at 12:30:09 PM
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Not sure about this but it seems to work
A's first move has to stop B making Royal Flush. If A picks cards all higher than T's, then B can pick Four Tens and cannot be stopped from making 10-high Str Flush or winning with the four 10s. So A should pick four 10s and any other card.
If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.
If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.
October 17th, 2020 at 2:38:12 PM
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Quote: charliepatrickNot sure about this but it seems to work
A's first move has to stop B making Royal Flush. If A picks cards all higher than T's, then B can pick Four Tens and cannot be stopped from making 10-high Str Flush or winning with the four 10s. So A should pick four 10s and any other card.
If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.
I believe you have provided the correct answer, Charlie.
If Pai Gow Poker hand rankings (in which 5432A is the second highest straight hand) are used, then B always wins by getting a tie. I think.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
October 17th, 2020 at 2:54:49 PM
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^ I had actually also thought about the Pai Gow rules and think...
A can ensure a win by choosing four Aces. B can block, say, the Royal Flushes but not all the A2345's. And as there aren't any Aces left B can't tie. So A still wins.
October 17th, 2020 at 3:47:20 PM
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What about the joker in PGP?
Have you tried 22 tonight? I said 22.
October 17th, 2020 at 4:18:18 PM
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Quote: GialmereWhat about the joker in PGP?
Ok he grabs four Aces and Joker! I can't remember if Five Aces beats a Straight Flush! Anyway he's plays as described.
October 17th, 2020 at 4:46:07 PM
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In my previous answer I was actually answering the question about "no runoff," but made an error in calling it a runoff. Stupid mistake.
Since this problem is getting stale, I'll dispense with the spoiler tags.
Let's rephrase the question to:
"Two numbers are chosen between 0 and 1 along the uniform distribution. That will divide the segment into three pieces. What is the probability the longest piece is less than than 0.5?"
The answer would be the same if the first number were drawn between 0 and 0.5, since it must be in one half or the other. May as well be the half between 0 and 0.5. Let's call that number x.
Let the second number be y. The only range for y, where the longest segment is less than 0.5, is from 0.5 to 0.5+x. This distance equals x.
The range of values of x and y where the longest segment is under 0.5 = ...
Integral from 0 to 0.5 of x dx =
x^2/2 from 0 to 0.5
= 0.5^2/2 = 1/8.
However, we must double this, because this only counts the values of x<0.5.
So, the answer is 2/8 = 1/4.
Since this problem is getting stale, I'll dispense with the spoiler tags.
Let's rephrase the question to:
"Two numbers are chosen between 0 and 1 along the uniform distribution. That will divide the segment into three pieces. What is the probability the longest piece is less than than 0.5?"
The answer would be the same if the first number were drawn between 0 and 0.5, since it must be in one half or the other. May as well be the half between 0 and 0.5. Let's call that number x.
Let the second number be y. The only range for y, where the longest segment is less than 0.5, is from 0.5 to 0.5+x. This distance equals x.
The range of values of x and y where the longest segment is under 0.5 = ...
Integral from 0 to 0.5 of x dx =
x^2/2 from 0 to 0.5
= 0.5^2/2 = 1/8.
However, we must double this, because this only counts the values of x<0.5.
So, the answer is 2/8 = 1/4.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 17th, 2020 at 7:35:29 PM
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Quote: charliepatrickNot sure about this but it seems to work
A's first move has to stop B making Royal Flush. If A picks cards all higher than T's, then B can pick Four Tens and cannot be stopped from making 10-high Str Flush or winning with the four 10s. So A should pick four 10s and any other card.
If B doesn't stop A making RF (because B can't) then A wins, so B has to pick something like Four Aces thru Jacks. Then A makes 10-high Str Flush which can't be beaten.
Correct!
Note that a straight in poker MUST contain either a 5 or a 10.
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My wife says she is going to divorce me because I love poker more then her.
I think she’s bluffing.
Have you tried 22 tonight? I said 22.
October 17th, 2020 at 11:08:08 PM
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I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...
Quote: The RiddlerThis week, we return to the brilliant and ageless game show, “The Price is Right.” In a modified version of the bidding round, you and two (not three) other contestants must guess the price of an item, one at a time.
Assume the true price of this item is a randomly selected value between 0 and 100. (Note: The value is a real number and does not have to be an integer.) Among the three contestants, the winner is whoever guesses the closest price without going over. For example, if the true price is 29 and I guess 30, while another contestant guesses 20, then they would be the winner even though my guess was technically closer.
In the event all three guesses exceed the actual price, the contestant who made the lowest (and therefore closest) guess is declared the winner. I mean, someone has to win, right?
If you are the first to guess, and all contestants play optimally (taking full advantage of the guesses of those who went before them), what are your chances of winning?
Officials answers to be released next week.
Have you tried 22 tonight? I said 22.
October 18th, 2020 at 5:23:32 AM
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I would pick two hairs over 200/3 = let's say 66.666668. The second player would pick a hair over 100/3, say 33.333334. The last player would pick 0.
This way you have essentially a 1/3 chance of winning, despite the positional disadvantage.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 18th, 2020 at 4:25:18 PM
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Quote: Gialmere
I see that this week's FiveThirtyEight column containing the election puzzle also has a similar Price is Right puzzle. Note the slight rule changes...
Officials answers to be released next week.
The item is "a randomly selected value between 0 and 100." Thus if we limit ourselves to two digits beyond the decimal point, there are a total of 99.99 distinct possible choices.
If the contestant A picks 33.33 and the contestant B picks 66.67 then the contestant C has these choices:
1. pick 0.00 or 0.01 and win 33.32 percent of the time; A wins 33.34% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time A wins 33.34, B wins 33.33
3. pick 33.34 and win 33.33 percent of the time: A wins 33.33, B wins 33.33 (optimal for C)
4. pick 66.68 and win 33.32 percent of the time A wins 66.66; B wins 0.001
If the 1st contestant picks 33.33 and the 2nd contestant picks 66.66 then the third contestant has these choices:
1. pick 0.00 or 0.01 and win 33.32 percent of the time (because the price cannot be zero); A wins 33.33% of time, B wins 33.33
2. pick 33.32 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
3. pick 33.34 and win 33.32 percent of the time: A wins 33.33; B wins 33.34
4. pick 66.67 and win 33.33 percent of the time; A wins 66.5 and B wins 0.01 (optimal for C)
If A chooses 33.32 and B chooses 66.65 then C chooses 66.66 and A wins 66.64; B wins 0.01; C wins 33.34
If A chooses 33.32 and B chooses 66.66 then C chooses 33.33 and A wins 33.32; B wins 33.34; C wins 33.33
if A chooses 33.34 and B chooses 66.67 then C would choose either 0 or 33.33: A wins 33.33, B wins 33.33, C wins 33.33
if A chooses 33.34 and B chooses 33.33 then C would choose 33.35: A wins 0.01, B wins 33.33, C wins 66.65
IT is OPTIMAL for A to chooses 33.33; it is OPTIMAL for B to then choose either 66.67 or 66.66. In both these cases, A will win 33.33/99.99 or 1/3 of the time.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
