Three cocktail waitress problem
Poll
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 | 2 votes (13.33%) | ||
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| | 1 vote (6.66%) | ||
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| | 9 votes (60%) | ||
| | 5 votes (33.33%) | ||
| | 5 votes (33.33%) |
15 members have voted
January 10th, 2019 at 3:23:49 PM
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At the Wizard casino there are three cocktail waitresses -- Amy, Betty, and Cathy. It is a small casino, so each makes rounds of the entire place, as opposed to having territories, like most casinos do. Amy passes by any one location every 3 minutes. Betty is a little slower and it takes 5 minutes to make the same round. Cathy is slower still and takes 7 minutes. However, all are like clockwork, moving each at her own speed, never taking breaks, and always taking the same time to circle around to the same place.
All three are at work, but you have no idea when each started or where they are in their rotations. You arrive and plop down at a River Dragons machine, because the large jackpot is over $4,950. What is the probability each one of them arrives to you first to offer a nice libation?
Like last time, I'd like to ask Don to give this a 24-hour waiting period to give someone else a chance. As usual, please put answers and solutions in spoiler tags.
The question for the poll is what is the probability Amy is first?
All three are at work, but you have no idea when each started or where they are in their rotations. You arrive and plop down at a River Dragons machine, because the large jackpot is over $4,950. What is the probability each one of them arrives to you first to offer a nice libation?
Like last time, I'd like to ask Don to give this a 24-hour waiting period to give someone else a chance. As usual, please put answers and solutions in spoiler tags.
The question for the poll is what is the probability Amy is first?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 3:38:25 PM
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This math is too complex for me.
Since I assume that one of the poll choices is the correct answer, using no math, I picked 1/2 simply because it feels right... at least from the options provided.
For what its worth, I used the same type of picking an answer on many of my SAT questions. I did rather well.
Since I assume that one of the poll choices is the correct answer, using no math, I picked 1/2 simply because it feels right... at least from the options provided.
For what its worth, I used the same type of picking an answer on many of my SAT questions. I did rather well.
I invented a few casino games. Info:
http://www.DaveMillerGaming.com/
Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
January 10th, 2019 at 3:40:59 PM
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How are you counting Amy arriving at the same time as one of the others? Is Amy there first, though tied, or does that disqualify that visit/round from her?
The sine waves don't completely repeat until 105 minutes, but there will be co-visits to the same machine at 15 minute, 21 minute, and 35 minute intervals. You could have sat down anywhere within that time frame. So am I looking for times that end with co-visits, or only solo visits? If co-visits count, do I give Amy full credit, or 1/2 credit?
Edit: I am assuming they all clock in at hour 0 and leave the cocktail station at the same time.
The sine waves don't completely repeat until 105 minutes, but there will be co-visits to the same machine at 15 minute, 21 minute, and 35 minute intervals. You could have sat down anywhere within that time frame. So am I looking for times that end with co-visits, or only solo visits? If co-visits count, do I give Amy full credit, or 1/2 credit?
Edit: I am assuming they all clock in at hour 0 and leave the cocktail station at the same time.
If the House lost every hand, they wouldn't deal the game.
January 10th, 2019 at 3:47:50 PM
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Quote: beachbumbabsHow are you counting Amy arriving at the same time as one of the others? Is Amy there first, though tied, or does that disqualify that visit/round from her?
Time is divided into infinitely small units, so ties are not an issue. Much like the probability of picking pi from from the uniform distribution from 0 to 10 -- the probability of a tie is zero -- but it could happen.
p.s. Just so it's clear, let's assume the River Dragons machine is just a point on the floor.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 3:50:22 PM
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Quote: WizardTime is divided into infinitely small units, so ties are not an issue. Much like the probability of picking pi from from the uniform distribution from 0 to 10 -- the probability of a tie is zero -- but it could happen.
Please see the spoiler I added as you were responding. I'm calculating in whole minutes.
If the House lost every hand, they wouldn't deal the game.
January 10th, 2019 at 3:55:06 PM
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Quote: beachbumbabsPlease see the spoiler I added as you were responding. I'm calculating in whole minutes.
Again, there can be no ties, so your answer is to a different question.
To help visualize it, assume the casino is like a big circle and you are at a random point on it. The three waitresses are like clock hands. What is the probability the clock hand that makes a 3-minute revolution reaches your point first?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 4:03:20 PM
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Quote: WizardAgain, there can be no ties, so your answer is to a different question.
To help visualize it, assume the casino is like a big circle and you are at a random point on it. The three waitresses are like clock hands. What is the probability the clock hand that makes a 3-minute revolution reaches your point first?
I think the clock analogy is a good one for demonstrating there CAN be ties. The minute hand and the hour hand will be tied 12 times in 12 hours. The minute hand and second hand will be tied every just-over-a-minute. As will the second hand and the hour hand. And all 3 will be together 12 times in every 12 hours. What if I'm seated at exactly 12 on your clock face?
If the House lost every hand, they wouldn't deal the game.
January 10th, 2019 at 4:08:17 PM
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First stab with simple math... Might come back and check later.
We split into 4 cases. 1st Case - all 3 waitresses are coming in 3 minutes. 2nd Case - 1st and second waitress are coming in 3 minutes. 3rd Case - 1st and 3rd waitress are coming in 3 minutes. 4th Case - only 1st waitress is coming in 3 minutes. Since waitresses coming in 3 minutes all have the same probability of showing up first, this leads to probabilities of 1/3, 1/2, 1/2, and 1. Then, we calculate the probabilities of each case. 1st - (3/5)(3/7) = 9/35, 2nd - (3/5)(4/7) = 12/35, 3rd - (2/5)(3/7) = 6/35, 4th - (2/5)(4/7) = 8/35. Summing up, we get 3/35 + 6/35 + 3/35 + 8/35 = 20/35 = 4/7.
January 10th, 2019 at 4:12:07 PM
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I guess I only did Amy... The same cases can give the probabilities for the other two. Betty can win in case 1 and 2, yielding 3/35+6/35 = 9/35. Cathy can win in case 1 and 3 yielding 3/35 + 3/35 = 6/35. The total probability is 4/7 + 9/35 + 6/35 = 35/35, so that checks out at least.
January 10th, 2019 at 4:12:13 PM
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She works hard for the money.
For every 1 revolution for Amy, Betty only gets 21/35ths of the way around, and Cathy only 15/35ths. At best, the two slowpokes would cover 1 and 1/35ths of the stations every time Amy made a round, so 34 out of 35 times Amy will get there first.
For every 1 revolution for Amy, Betty only gets 21/35ths of the way around, and Cathy only 15/35ths. At best, the two slowpokes would cover 1 and 1/35ths of the stations every time Amy made a round, so 34 out of 35 times Amy will get there first.
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 10th, 2019 at 4:16:18 PM
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Hehe... Then hope it is/isn't a cuckoo? Hey Yoooooooo...Quote: beachbumbabs...What if I'm seated at exactly 12 on your clock face?
Simplicity is the ultimate sophistication - Leonardo da Vinci
January 10th, 2019 at 4:17:54 PM
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Making a big messy graph of 105 minutes (3 x 5 x 7), I end up with probabilities of 53% for Amy, 28% for Betty, and 19% for Cathy.
I may have made some counting / arithmetic mistakes. In that case, I guess 50, 30 and 20 percents, because it feels right
January 10th, 2019 at 4:33:53 PM
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Quote: beachbumbabsI think the clock analogy is a good one for demonstrating there CAN be ties. The minute hand and the hour hand will be tied 12 times in 12 hours. The minute hand and second hand will be tied every just-over-a-minute. As will the second hand and the hour hand. And all 3 will be together 12 times in every 12 hours. What if I'm seated at exactly 12 on your clock face?
If you're at a random point on the clock, the probability of both hands passing you at exactly the same time is zero -- but it could happen. Like my random number from 0 to 10 analogy, which was discussed in a math paradox thread a while ago.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 4:34:46 PM
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Quote: TomG
Making a big messy graph of 105 minutes (3 x 5 x 7), I end up with probabilities of 53% for Amy, 28% for Betty, and 19% for Cathy.
I may have made some counting / arithmetic mistakes. In that case, I guess 50, 30 and 20 percents, because it feels right
The first answer was pretty close, but not quite.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 4:38:33 PM
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Let me reword the problem, to help avoid the confusion about ties.
There are three ants on a circle, both going around the circumference in the same direction, and constant but different rates. It takes ant A three minutes to make a revolution. It takes ant B five minutes and ant C seven minutes to do the same. All three ants started at random points on the circle before you arrived.
You pick a random point on the circumference of the circle too and stay there. What is the probability each ant reaches you first?
Extra credit: Does the answer change if the any given ant goes the opposite direction as the other two?
There are three ants on a circle, both going around the circumference in the same direction, and constant but different rates. It takes ant A three minutes to make a revolution. It takes ant B five minutes and ant C seven minutes to do the same. All three ants started at random points on the circle before you arrived.
You pick a random point on the circumference of the circle too and stay there. What is the probability each ant reaches you first?
Extra credit: Does the answer change if the any given ant goes the opposite direction as the other two?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 4:41:46 PM
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Quote: WizardIf you're at a random point on the clock, the probability of both hands passing you at exactly the same time is zero -- but it could happen. Like my random number from 0 to 10 analogy, which was discussed in a math paradox thread a while ago.
Ok, thanks. Sorry to belabor the point.
If the House lost every hand, they wouldn't deal the game.
January 10th, 2019 at 4:44:30 PM
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Yes, congratulations, you're right! I like your solution too, much more intuitive than mine, which involved a triple integral.
I owe you a beer for being first.
I owe you a beer for being first.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 4:46:08 PM
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Quote: Wizard
The first answer was pretty close, but not quite.
Hopefully I was on the right track, just getting too sloppy on a first try. Took some extra time, made it neater. Amy = 74/138 = 54%; Betty = 38/138 = 28%; and Kathy = 26/138 = 19%
edit: see someone else got it better using an actual math method.
edit: see someone else got it better using an actual math method.
January 10th, 2019 at 4:52:34 PM
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Quote: WizardYes, congratulations, you're right! I like your solution too, much more intuitive than mine, which involved a triple integral.
I owe you a beer for being first.
Cool, thanks! My first thought was the triple integral, but I definitely wanted to avoid it :p.
January 10th, 2019 at 4:53:23 PM
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Quote: TomGHopefully I was on the right track, just getting too sloppy on a first try. Took some extra time, made it neater. Amy = 74/138 = 54%; Betty = 38/138 = 28%; and Kathy = 26/138 = 19%
Again, close but no cigar
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 5:05:31 PM
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Extra Credit- Does not matter which direction the ants are going
Beware, I work for the dark side.... We have cookies
January 10th, 2019 at 5:32:51 PM
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Wrong solution, but I don't know why.
In a 105 minute period Amy visits the River Dragon machine 35 times, Betty 21, and Cathy 15. Service is offered 71 times for the period. The probability of who is offering service for a visit is Amy 35/71, Betty 21/71. Cathy 15/71.
You dont bring a bone saw to a negotiation. - Robert Jordan, former U.S. ambassador to Saudi Arabia
January 10th, 2019 at 6:01:19 PM
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Going to give this a shot before reading the other answers...
The way I did this was come up with a lowest common denominator of minutes... so let A = 3, B=5, and C=7. The LCD is 105. Meaning that if they all started at exactly the same time, after 105 minutes, they all would have come up with a finished round. Anna would have done 35 rounds, Betty would have done 21 rounds, and Cathy would have done 15 rounds. Adding those up, we get 71 rounds total. And the chances that any particular waitress would be visiting you next on her round is her percentage of that 71 total rounds.
SO... A = 35/71, B = 21/71, and C=15/71.
The chances would be as follows:
Anna: .49295774 or ~ 49%
Betty: .29577464 or ~ 30%
Cathy: .21126760 or ~ 21%
I'm probably oversimplifying this, and I look forward to seeing the other answers. But that's what I'm submitting.
SO... A = 35/71, B = 21/71, and C=15/71.
The chances would be as follows:
Anna: .49295774 or ~ 49%
Betty: .29577464 or ~ 30%
Cathy: .21126760 or ~ 21%
I'm probably oversimplifying this, and I look forward to seeing the other answers. But that's what I'm submitting.
I want to stay as close to the edge as I can without going over. Out on the edge you see all kinds of things you can't see from the center.
January 10th, 2019 at 6:09:13 PM
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If it makes you feel better, BleedingChipsSlowly, I came up with the same wrong answer. :)
I want to stay as close to the edge as I can without going over. Out on the edge you see all kinds of things you can't see from the center.
January 10th, 2019 at 6:09:22 PM
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Nice to see I have company. Hope, perhaps, that we are not wrong?Quote: vonnegutGoing to give this a shot before reading the other answers...
You dont bring a bone saw to a negotiation. - Robert Jordan, former U.S. ambassador to Saudi Arabia
January 10th, 2019 at 6:49:25 PM
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Quote: BleedingChipsSlowlyWrong solution, but I don't know why.
In a 105 minute period Amy visits the River Dragon machine 35 times, Betty 21, and Cathy 15. Service is offered 71 times for the period. The probability of who is offering service for a visit is Amy 35/71, Betty 21/71. Cathy 15/71.
You're answering the wrong question. If the question was who is the waitress for a random visit, you would be right. Why is that different that what I'm asking? Hard to explain. It's because the regular cycles. I'm still struggling with how to explain it.
Let's say you are playing in a contest with Tiger Woods to see who hits the first hole in one. Let's say that your chances per shot are 1 in 1,000,000 and Tigers are 1 in 100. Your best hope of winning is you have very good luck and Tiger has very bad luck. However, if you both hit them at exact regular cycles, you tease out some of that luck, and improve Tiger's chance of winning.
Let's say you are playing in a contest with Tiger Woods to see who hits the first hole in one. Let's say that your chances per shot are 1 in 1,000,000 and Tigers are 1 in 100. Your best hope of winning is you have very good luck and Tiger has very bad luck. However, if you both hit them at exact regular cycles, you tease out some of that luck, and improve Tiger's chance of winning.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 6:52:07 PM
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Quote: vonnegutGoing to give this a shot before reading the other answers...
The way I did this was come up with a lowest common denominator of minutes... so let A = 3, B=5, and C=7. The LCD is 105. Meaning that if they all started at exactly the same time, after 105 minutes, they all would have come up with a finished round. Anna would have done 35 rounds, Betty would have done 21 rounds, and Cathy would have done 15 rounds. Adding those up, we get 71 rounds total. And the chances that any particular waitress would be visiting you next on her round is her percentage of that 71 total rounds.
SO... A = 35/71, B = 21/71, and C=15/71.
The chances would be as follows:
Anna: .49295774 or ~ 49%
Betty: .29577464 or ~ 30%
Cathy: .21126760 or ~ 21%
I'm probably oversimplifying this, and I look forward to seeing the other answers. But that's what I'm submitting.
Nice try. You had the same answer as BleedingChipsSlowly. See my response to him for my attempt to explain why it's wrong, but it probably isn't the best explanation. Maybe Don can put it better.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 7:26:12 PM
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Quote: BleedingChipsSlowlyWrong solution, but I don't know why.
In a 105 minute period Amy visits the River Dragon machine 35 times, Betty 21, and Cathy 15. Service is offered 71 times for the period. The probability of who is offering service for a visit is Amy 35/71, Betty 21/71. Cathy 15/71.
I thought this at one point as well, but...
Suppose instead that all three are going the same speed. By your reasoning, each has an equal chance of being first. However, if Amy is just ahead of Betty, who is just ahead of Cathy, then Amy will be the first to reach the machine far more often than the other two.
I also know I'm supposed to wait 24 hours, but since somebody already posted the correct answer, I figure it's safe now...and I just realized this is pretty much just like a problem I solved some time ago
Let A, B, and C be the number of minutes before Amy, Betty, and Cathy will get to the machine. A is in the interval [0,3], B is in [0,5], and C is in [0,7].
Divide the interval [0,7] into three pieces; 0-3, 3-5, and 5-7.
Amy is always in 0-3
Betty is in 0-3 3/5 of the time and 3-5 2/5 of the time
Cathy is in 0-3 3/7 of the time, 3-5 2/7 of the time, and 5-7 2/7 of the time
3/5 x 3/7 = 9/35 of the time, all three are in 0-3; each of the three is equally likely to get to the machine first
3/5 x 4/7 = 12/35 of the time, A and B are in 0-3, and C is in either 3-5 or 5-7; Amy and Betty are equally likely to be first
2/5 x 3/7 = 6/35 of the time, A and C are in 0-3, and B is in 3-5; Amy and Cathy are equally likely to be first
2/5 x 4/7 = 8/35 of the time, A is in 0-3, B is in 3-5, and C is in 3-5 or 5-7; Amy will get there first
P(Amy getting there first) = 9/35 x 1/3 + 12/35 x 1/2 + 6/35 x 1/2 + 8/35 = (18 + 36 + 18 + 48) / 210 = 4 / 7
P(Betty getting there first) = 9/35 x 1/3 + 12/35 x 1/2 = (18 + 36) / 210 = 9 / 35
P(Cathy getting there first) = 9/35 x 1/3 + 6/35 x 1/2 = (18 + 18) / 210 = 6 / 35
The earlier question involved A being in the interval [0,1], B in [0,2], and C in [0,3]
Last edited by: ThatDonGuy on Jan 10, 2019
January 10th, 2019 at 7:38:32 PM
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I'd be interested in knowing what the house edge is set at for various games at the Wizard Casino.
Have you tried 22 tonight? I said 22.
January 10th, 2019 at 8:20:05 PM
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Quote: ThatDonGuyI also know I'm supposed to wait 24 hours, but since somebody already posted the correct answer, here's how I get it (at least, the probability that Amy gets there first):
No triple integral needed, although I did use one (where, as it turns out, one wasn't necessary)
Let A, B, and C be the points in the interval (0,1) where they are located, with 0 being where the machine is, and they are moving from 1 towards 0.
A's speed is 1/3, B's is 1/5, and C's is 1/7
A will get to 0 before B if 3A < 5B -> 3/5 A < B
A will get to 0 before C if 3A < 7C -> 3/7 A < C
At any given point A, the probability that A will get to 0 before B and C is (1 - 3/5 A)(1 - 3/7 A) = 1 - 36/35 A + 9/35 A^2
The integral over A from 0 to 1 of (9/35 A^2 - 36/35 A + 1) dA = 3/35 - 18/35 + 1 = 4/7.
That's how I did it too.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 8:20:24 PM
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Quote: GialmereI'd be interested in knowing what the house edge is set at for various games at the Wizard Casino.
All are under 1%.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 10th, 2019 at 8:24:57 PM
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Quote: WizardThat's how I did it too.
Actually, I found a non-calculus solution, and edited my earlier post.
January 10th, 2019 at 8:25:05 PM
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Bonus question: Youve waited one minute at the river dragon slots and no waitress has shown up. Now whats the probability that Amy shows up first?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 10th, 2019 at 9:40:06 PM
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Quote: unJonBonus question: Youve waited one minute at the river dragon slots and no waitress has shown up. Now whats the probability that Amy shows up first?
Amy: 63.89%
Betty: 22.22%
Cathy: 13.89%
Same problem, except Amy shows up every 2 minutes, Betty every 4, and Cathy every 6.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 11th, 2019 at 8:00:51 AM
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Quote: Wizard
Amy: 63.89%
Betty: 22.22%
Cathy: 13.89%
Same problem, except Amy shows up every 2 minutes, Betty every 4, and Cathy every 6.
Cant fool the Wiz . . .
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 11th, 2019 at 8:02:39 AM
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Quote: unJonCant fool the Wiz . . .
Trivia time!
Who said, "Fool me once, shame on me. Foot me twice....we won't get fooled again. "
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 11th, 2019 at 8:09:52 AM
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Quote: WizardTrivia time!
Who said, "Fool me once, shame on me. Foot me twice....we won't get fooled again. "
GWB (Bush 43)
If the House lost every hand, they wouldn't deal the game.
January 11th, 2019 at 11:14:06 AM
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Quote: beachbumbabsGWB (Bush 43)
Correct! Tough to fool BBB ... twice.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 11th, 2019 at 12:17:23 PM
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Quote: WizardTrivia time!
Who said, "Fool me once, shame on me. Foot me twice....we won't get fooled again. "
"Foot me twice"??? It's like you doubled up and did a shoe joke. He did have 2 shoes thrown at him?
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