Easy math puzzles
Poll
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| | 12 votes (26.08%) | ||
| | 10 votes (21.73%) |
46 members have voted
June 12th, 2023 at 9:34:38 PM
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Sounded vaguely familiar but wasn't sure it was here or not. Used the search function and bingo. Some thing I remember sometimes I can't remember what I had for breakfast
June 12th, 2023 at 10:27:14 PM
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Quote: avianrandySounded vaguely familiar but wasn't sure it was here or not. Used the search function and bingo. Some thing I remember sometimes I can't remember what I had for breakfast
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avianrandy,
Don't feel bad... I can't remember what you had for breakfast either.
;-)
Dog Hand
June 15th, 2023 at 7:17:38 AM
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From the Riddler comes this variation of the Monty Hall problem...
In a game show, there are three identical doors arranged in a row from left to right. The host of the show, “Monty,” chooses one of the doors and places a prize of $100 dollars behind it. There is no prize behind the other two doors. You are not present when Monty chooses the door and places the money behind it, so you cannot say for certain which door the prize is behind.
You are then brought to the stage and must select one of the three doors to open. If the prize money is behind it, then you win! But if you guess incorrectly, all is not lost. You can pay $80 to pick a second door. Before you make that second selection, however, Monty will give you a hint, telling you whether the prize is behind a door that’s to the left or to the right of your first choice. (Note that this hint is only helpful when you previously selected the middle door.) If the prize isn’t behind that second door, you can pay another $80 to try a third time.
Assume that both you and Monty play with optimal strategies — you to maximize your expected net earnings (prize winnings minus payments for hints), and Monty to minimize the same.
How much net earnings can you expect to make on average?
In a game show, there are three identical doors arranged in a row from left to right. The host of the show, “Monty,” chooses one of the doors and places a prize of $100 dollars behind it. There is no prize behind the other two doors. You are not present when Monty chooses the door and places the money behind it, so you cannot say for certain which door the prize is behind.
You are then brought to the stage and must select one of the three doors to open. If the prize money is behind it, then you win! But if you guess incorrectly, all is not lost. You can pay $80 to pick a second door. Before you make that second selection, however, Monty will give you a hint, telling you whether the prize is behind a door that’s to the left or to the right of your first choice. (Note that this hint is only helpful when you previously selected the middle door.) If the prize isn’t behind that second door, you can pay another $80 to try a third time.
Assume that both you and Monty play with optimal strategies — you to maximize your expected net earnings (prize winnings minus payments for hints), and Monty to minimize the same.
How much net earnings can you expect to make on average?
Have you tried 22 tonight? I said 22.
June 15th, 2023 at 10:14:10 AM
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I originally over thought it but then realised something simple. Let's call the outside boxes A and C, the middle one B.
Clearly the host should pick A or C with indifference, otherwise you would have an advantage in picking between A and C. Similarly the probabilities should be such that your chances are the same whether you pick box A (or Box C) versus box B.
Let's assume the chances of the host picking the middle box is q and each outside box p/2. NOTE { p/2 , q , p/2 }
If you pick the middle box you will win $100 with Pr=q, and (after paying $80) otherwise will be able to pick the correct box have a net win of $20 - this will happen p or 1-q. So you will always make $20 and with pr=q will $80 extra; this is equivalent to making $100 but making $80 less when it's an outside box, so is also 100-80p.
if you pick one of the outside boxes, say LHS, then your chances of being correct is p/2. This part is worth 100*p/2 = 50p.
If not you have to guess whether to pick the middle box or the other outside box. If you pick the outside box then the chances of this can be calculated as p/2 (you can use long hand to say it's (1-p/2)*(p/2)/(1-p/2)). So this part is worth 20*p/2 = 10p. If this doesn't find the winner the you lost $60 with Pr=q or -60*(1-p) = 60p-60.
The total of these is 50p + 10p + (60p-60) = 120p-60.
If the total of these are the same then 120p-60=100-80p; so 200p=160, so p=4/5. Thus the chances of each outside box is 40% and the inside 20%. (It also means you were correct to pick the other outside box at Step 2.)
This gives an EV of $36.
Clearly the host should pick A or C with indifference, otherwise you would have an advantage in picking between A and C. Similarly the probabilities should be such that your chances are the same whether you pick box A (or Box C) versus box B.
Let's assume the chances of the host picking the middle box is q and each outside box p/2. NOTE { p/2 , q , p/2 }
If you pick the middle box you will win $100 with Pr=q, and (after paying $80) otherwise will be able to pick the correct box have a net win of $20 - this will happen p or 1-q. So you will always make $20 and with pr=q will $80 extra; this is equivalent to making $100 but making $80 less when it's an outside box, so is also 100-80p.
if you pick one of the outside boxes, say LHS, then your chances of being correct is p/2. This part is worth 100*p/2 = 50p.
If not you have to guess whether to pick the middle box or the other outside box. If you pick the outside box then the chances of this can be calculated as p/2 (you can use long hand to say it's (1-p/2)*(p/2)/(1-p/2)). So this part is worth 20*p/2 = 10p. If this doesn't find the winner the you lost $60 with Pr=q or -60*(1-p) = 60p-60.
The total of these is 50p + 10p + (60p-60) = 120p-60.
If the total of these are the same then 120p-60=100-80p; so 200p=160, so p=4/5. Thus the chances of each outside box is 40% and the inside 20%. (It also means you were correct to pick the other outside box at Step 2.)
This gives an EV of $36.
June 15th, 2023 at 10:53:51 AM
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Quote: Gialmere
How much net earnings can you expect to make on average?
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This is probably wrong, but I get $38.46.
I show the player should pick door 2 with 5/13 probability and Monty should put the money there with 3/13 probability. Otherwise, for both, pick between 1 and 3 with equal probability.
If the player picks 1 or 3 and is wrong, he should decline to pay the $80 for another pick. I could be wrong about that.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 15th, 2023 at 11:13:08 AM
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^ I agree
I hadn't checked whether it was better for the player to continue playing having picked a losing outside box, the EV for standing in my original answer was 40, but it went down to 36 because the player played on. Using the same logic, but the player gives in if they're wrong on an outside box means if the host picks the boxes 5/13 3/13 5/13 then the player is indifferent whether to chose the middle box or outside box. Thus the EV is $100*5/13 for an outside choice, and $100*3/13+$20*10/13, both which come to 500/13 = $38.46.
If the payment was $75 rather than $80 I get the player would be indifferent whether to play on and the EV would be $40 (using the 40%/20%/40%).
100-75 player doesn't play again if outside is wrong 75-0 player does play again.
Cost to pay vs hosts probabilities vs EV
100 (5 5 5) 33.333
90 (5 4 5) 35.714
80 (5 3 5) 38.462
75 (5 2.5 5) 40
50 (same) 60
25 (same) 80
0 (same) 100.
100-75 player doesn't play again if outside is wrong 75-0 player does play again.
Cost to pay vs hosts probabilities vs EV
100 (5 5 5) 33.333
90 (5 4 5) 35.714
80 (5 3 5) 38.462
75 (5 2.5 5) 40
50 (same) 60
25 (same) 80
0 (same) 100.
Last edited by: charliepatrick on Jun 15, 2023
June 15th, 2023 at 4:38:28 PM
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Quote: WizardQuote: Gialmere
How much net earnings can you expect to make on average?
link to original post
This is probably wrong, but I get $38.46.
I show the player should pick door 2 with 5/13 probability and Monty should put the money there with 3/13 probability. Otherwise, for both, pick between 1 and 3 with equal probability.
If the player picks 1 or 3 and is wrong, he should decline to pay the $80 for another pick. I could be wrong about that.
link to original post
Quote: charliepatrick^ I agree
I hadn't checked whether it was better for the player to continue playing having picked a losing outside box, the EV for standing in my original answer was 40, but it went down to 36 because the player played on. Using the same logic, but the player gives in if they're wrong on an outside box means if the host picks the boxes 5/13 3/13 5/13 then the player is indifferent whether to chose the middle box or outside box. Thus the EV is $100*5/13 for an outside choice, and $100*3/13+$20*10/13, both which come to 500/13 = $38.46.If the payment was $75 rather than $80 I get the player would be indifferent whether to play on and the EV would be $40 (using the 40%/20%/40%).
100-75 player doesn't play again if outside is wrong 75-0 player does play again.
Cost to pay vs hosts probabilities vs EV
100 (5 5 5) 33.333
90 (5 4 5) 35.714
80 (5 3 5) 38.462
75 (5 2.5 5) 40
50 (same) 60
25 (same) 80
0 (same) 100.
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Correct!!!
Excellent.
This one was game theory with a unique Nash equilibrium.
-------------------------------------------------
Today's Monty Hall joke is simply a puchline without a setup...
Then the goat behind door #3 drove off in the car.
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Puzzle news...
It's announced that Friday, June 30 will mark the final column for The Riddler at FiveThirtyEight. He does, however, hint at a future project.
Have you tried 22 tonight? I said 22.
June 17th, 2023 at 3:41:31 PM
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There is a train with 20 passenger carriages, numbered 1 to 20 from the engine. The entire train holds 2001 passengers. Any three consecutive carriages hold 300 passengers. How many passengers are in carriage #18?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 17th, 2023 at 4:03:15 PM
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Cars 1, 4, 7, 10, 13, 16, and 19 each have the same number (otherwise, for exampe, the number in cars 1, 2, and 3 cannot equal the number in cars 2, 3, and 4); let this number be A
Similarly, cars 2, 5, 8, 11, 14, 17, and 20 each have the same number; let this number be B
Cars 1 through 18 have a total of 6 x 300 = 1800 passengers, so cars 19 and 20 have 201; since 19 has A and 20 has B, A + B = 201
The number in 18 = 300 - the number in cars 19 and 20 = 300 - 201 = 99.
June 18th, 2023 at 2:07:06 AM
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C1+C2+C3=300, C4+C5+C6=300,...,C16+C17+C18=300; also note C18+C19+C20=300.
so C1+...+C18=6*300=1800
but C1+...+C20=2001
Hence C19+C20=201
Hence C18=99.
While it's true C1=C4 etc, I don't think you need that to solve the problem.
so C1+...+C18=6*300=1800
but C1+...+C20=2001
Hence C19+C20=201
Hence C18=99.
While it's true C1=C4 etc, I don't think you need that to solve the problem.
June 20th, 2023 at 2:31:32 PM
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I know we recently did a problem similar to this, but I plan to make an Ask the Wizard question out of it so want to make sure my answer is right.
On a game show the host presents two doors. One is labeled car and the other is labeled jewelry. The host explains that one door contains what is on the label and the other has a goat.
The player may pick either door, with the goal of maximizing the win. The host may choose either door to hide the prize, with the goal of minimizing cost to the show. Both are logicians.
Due to advertising considerations, the car has a cost to the show of $20,000 and a value to the player of $25,000.
The jewelry has a cost to the show of $5,000 and a value to the player of $10,000.
The goat has a value of $0 both ways.
What should the strategy of both participants be? What is the player's expected win, assuming both follow optimal strategy?
On a game show the host presents two doors. One is labeled car and the other is labeled jewelry. The host explains that one door contains what is on the label and the other has a goat.
The player may pick either door, with the goal of maximizing the win. The host may choose either door to hide the prize, with the goal of minimizing cost to the show. Both are logicians.
Due to advertising considerations, the car has a cost to the show of $20,000 and a value to the player of $25,000.
The jewelry has a cost to the show of $5,000 and a value to the player of $10,000.
The goat has a value of $0 both ways.
What should the strategy of both participants be? What is the player's expected win, assuming both follow optimal strategy?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 21st, 2023 at 8:22:39 AM
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Scientific American had an article yesterday about the circle around a circle problem that DRich posted back in March, noting that the SAT test writers failed to include the correct answer as one of the multiple choices because the answer isn't intuitive.
Presidential Election polls and odds: https://2605.me/p
June 21st, 2023 at 9:53:44 AM
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Quote: WizardWhat should the strategy of both participants be? What is the player's expected win, assuming both follow optimal strategy?
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I am learning about this. Here is what I got.
The strategy for the Show is to put a car in its door 28.6% of the time. The strategy for the Player is to choose the car consistently all the time. With this strategy, the player's EV is $7143.
The strategy for the Show is to put a car in its door 28.6% of the time. The strategy for the Player is to choose the car consistently all the time. With this strategy, the player's EV is $7143.
June 21st, 2023 at 10:40:26 AM
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Quote: acesideI am learning about this. Here is what I got.
The strategy for the Show is to put a car in its door 28.6% of the time. The strategy for the Player is to choose the car consistently all the time. With this strategy, the player's EV is $7143.
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I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
June 22nd, 2023 at 7:00:05 AM
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Here's another confusing collection of rattling rebus puzzles. Although they all have a mathematical angle, word skills are helpful too...
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1A=?
1B=?
1C=?
1D=?
2A=?
2B=?
2C=?
2D=?
3A=?
3B=?
3C=?
3D=?
4A=?
4B=?
4C=?
4D=?
5A=?
5B=?
5C=?
5D=?
(Puzzle by David Pleacher)
1A=?
1B=?
1C=?
1D=?
2A=?
2B=?
2C=?
2D=?
3A=?
3B=?
3C=?
3D=?
4A=?
4B=?
4C=?
4D=?
5A=?
5B=?
5C=?
5D=?
(Puzzle by David Pleacher)
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 8:44:29 AM
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1A=There's safety in numbers
1B=Something for nothing
1C=Tackle box
1D=Daily mass?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=?
3C=The whole is greater than its parts
3D=Over cloud 9
4A=?
4B=?
4C=?
4D=?
5A=?
5B=Much ado about nothing
5C=Upset over nothing
5D=?
1B=Something for nothing
1C=Tackle box
1D=Daily mass?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=?
3C=The whole is greater than its parts
3D=Over cloud 9
4A=?
4B=?
4C=?
4D=?
5A=?
5B=Much ado about nothing
5C=Upset over nothing
5D=?
"Dealer has 'rock'... Pay 'paper!'"
June 22nd, 2023 at 9:07:11 AM
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Quote: Joeman1A=There's safety in numbers
1B=Something for nothing
1C=Tackle box
1D=Daily mass?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=?
3C=The whole is greater than its parts
3D=Over cloud 9
4A=?
4B=?
4C=?
4D=?
5A=?
5B=Much ado about nothing
5C=Upset over nothing
5D=?
link to original post
1C=Good guess but no.
1D=No
1A=Safety in numbers
1B=Something for nothing
1C=?
1D=?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=?
3C=The whole is greater than the sum of its parts.
3D=On cloud 9
4A=?
4B=?
4C=?
4D=?
5A=?
5B=Much ado about nothing
5C=Upset over nothing
5D=?
1B=Something for nothing
1C=?
1D=?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=?
3C=The whole is greater than the sum of its parts.
3D=On cloud 9
4A=?
4B=?
4C=?
4D=?
5A=?
5B=Much ado about nothing
5C=Upset over nothing
5D=?
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 9:37:18 AM
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1C = Tentacle
5D = St. Peters Square
May the cards fall in your favor.
June 22nd, 2023 at 9:53:35 AM
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Quote: Dieter
1C = Tentacle
5D = St. Peters Square
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Both correct.
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=?
3C=The whole is greater than the sum of its parts.
3D=On cloud 9
4A=?
4B=?
4C=?
4D=?
5A=?
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 10:02:11 AM
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1D Out of hours service
3B Hole In One
4B Calculated Risk
5A A Score To Settle
3B Hole In One
4B Calculated Risk
5A A Score To Settle
June 22nd, 2023 at 10:15:55 AM
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Quote: charliepatrick1D Out of hours service
3B Hole In One
4B Calculated Risk
5A A Score To Settle
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1D=Good guess but no.
The rest are correct.
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=?
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=?
4B=Calculated risk
4C=?
4D=?
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 10:59:11 AM
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1D = 24 hour mass ??? maybe 24 hr is a day so ....Day Mass???
Neither of those make sense ... but definitely seems to be 24 hr something
EDITE: I see Joeman already guessed "daily Mass" which makes more sense
June 22nd, 2023 at 11:06:09 AM
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Quote: chevy
1D = 24 hour mass ??? maybe 24 hr is a day so ....Day Mass???
Neither of those make sense ... but definitely seems to be 24 hr something
EDITE: I see Joeman already guessed "daily Mass" which makes more sense
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You're right on top of it.
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 12:50:13 PM
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...try 24 hour service
June 22nd, 2023 at 1:13:16 PM
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Quote: charliepatrick...try 24 hour service
link to original post
Correct
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=?
4B=Calculated risk
4C=?
4D=?
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
Hints (if needed) coming later.
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 1:14:19 PM
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4C ... Sandbox?
(Sandy blonde?)
May the cards fall in your favor.
June 22nd, 2023 at 1:15:15 PM
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Quote: Dieter
4C ... Sandbox?
(Sandy blonde?)
link to original post
Sorry, no
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 1:18:09 PM
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Quote: GialmereQuote: Dieter
4C ... Sandbox?
(Sandy blonde?)
link to original post
Sorry, no
link to original post
I can't guess again, because I forgot the no-searching rule.
Foiled again!
May the cards fall in your favor.
June 22nd, 2023 at 1:32:03 PM
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^ Dieter I can't think why you can't guess again, on one of my guesses I had accidentally been very close and combining this with someone else's guess, got the correct answer. I might not be correct but ....
Light Box.
I did think of White Box but doubt everyone knows what White Box Testing and Black Box Testing are. The term is applied to software, usually a subroutine but can be a program.
White Box means you know the code inside and are trying to test it works, especially any decisions/boundary conditions you know are made inside. Black Box means you don't know the code so just have to do tests that represent the things you might think happen.
I did think of White Box but doubt everyone knows what White Box Testing and Black Box Testing are. The term is applied to software, usually a subroutine but can be a program.
White Box means you know the code inside and are trying to test it works, especially any decisions/boundary conditions you know are made inside. Black Box means you don't know the code so just have to do tests that represent the things you might think happen.
June 22nd, 2023 at 1:41:39 PM
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Quote: charliepatrick^ Dieter I can't think why you can't guess again, on one of my guesses I had accidentally been very close and combining this with someone else's guess, got the correct answer. I might not be correct but ....
Light Box.
I did think of White Box but doubt everyone knows what White Box Testing and Black Box Testing are. The term is applied to software, usually a subroutine but can be a program.
White Box means you know the code inside and are trying to test it works, especially any decisions/boundary conditions you know are made inside. Black Box means you don't know the code so just have to do tests that represent the things you might think happen.
link to original post
Sorry no.
(I assume Dieter searched and found the puzzle answers. Consider his honorable response to be a clue for 4C.)
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 2:04:05 PM
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Quote: GialmereQuote: charliepatrick^ Dieter I can't think why you can't guess again, on one of my guesses I had accidentally been very close and combining this with someone else's guess, got the correct answer. I might not be correct but ....
Light Box.
I did think of White Box but doubt everyone knows what White Box Testing and Black Box Testing are. The term is applied to software, usually a subroutine but can be a program.
White Box means you know the code inside and are trying to test it works, especially any decisions/boundary conditions you know are made inside. Black Box means you don't know the code so just have to do tests that represent the things you might think happen.
link to original post
Sorry no.
(I assume Dieter searched and found the puzzle answers. Consider his honorable response to be a clue for 4C.)
link to original post
I searched for a list of idioms, and it became facepalmingly obvious what the answer was.
I had it all typed in and everything before I had to honourably withdraw.
May the cards fall in your favor.
June 22nd, 2023 at 2:41:37 PM
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2D = duodecimal
May the cards fall in your favor.
June 22nd, 2023 at 2:57:01 PM
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Quote: Dieter
2D = duodecimal
link to original post
Good guess, but no.
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 3:04:16 PM
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4A. ones (once) upon a time
June 22nd, 2023 at 3:51:48 PM
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Quote: ChesterDog
4A. ones (once) upon a time
link to original post
Correct
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=Once upon a time
4B=Calculated risk
4C=?
4D=?
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 5:09:42 PM
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1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=Once upon a time
4B=Calculated risk
4C=?
4D=?
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=Once upon a time
4B=Calculated risk
4C=?
4D=?
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
2D=Like one third
4C=Like an honest game
4D=Like a bad car wreck
4C=Like an honest game
4D=Like a bad car wreck
I'll buzz this out later tonight if unsolved.
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 5:14:27 PM
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4D = total loss (I needed the hint.)
June 22nd, 2023 at 5:16:42 PM
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4C = fair and square (needed the hint)
June 22nd, 2023 at 5:34:40 PM
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Quote: ChesterDog
4D = total loss (I needed the hint.)
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Quote: ChesterDog
4C = fair and square (needed the hint)
link to original post
Both correct
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=Once upon a time
4B=Calculated risk
4C=Fair and square
4D=Total loss
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
2D=Like one third on a calculator
Have you tried 22 tonight? I said 22.
June 22nd, 2023 at 5:54:45 PM
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Quote: GialmereQuote: ChesterDog
4D = total loss (I needed the hint.)
link to original postQuote: ChesterDog
4C = fair and square (needed the hint)
link to original post
Both correct
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=?
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=Once upon a time
4B=Calculated risk
4C=Fair and square
4D=Total loss
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square2D=Like one third on a calculator
link to original post
Repeating decimal
Extended hint required.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
June 22nd, 2023 at 6:34:24 PM
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Quote: unJonRepeating decimal
Extended hint required.
link to original post
Correct!!!
Excellent
There's too many posts to reference so here's a list of the joint solvers...
Joeman
Dieter
charliepatrick
ChesterDog
unJon
And an assist from: chevy
1A=Safety in numbers
1B=Something for nothing
1C=Tentacle
1D=Twenty-four hour service
2A=Lump sum
2B=Lean-to
2C=Good for nothing
2D=Repeating decimal
3A=Gross injustice
3B=Hole in one
3C=The whole is greater than the sum of its parts
3D=On cloud 9
4A=Once upon a time
4B=Calculated risk
4C=Fair and square
4D=Total loss
5A=A score to settle
5B=Much ado about nothing
5C=Upset over nothing
5D=St Peter's Square
-------------------------------------------------------
Have you tried 22 tonight? I said 22.
June 23rd, 2023 at 5:33:00 AM
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Quote: Gialmere
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Yeah, I don't got it. ;)
Thanks for the fun puzzle.
Thanks for the fun puzzle.
May the cards fall in your favor.
June 23rd, 2023 at 7:20:35 AM
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In a World Cup soccer group round, four teams in a section play a round robin series of games. Winning a game is worth 3 points, a tie is worth 1 point, and a loss scores zero points.
In a certain section, the competing teams were Australia, India, Turkey, and the U.S.A.
From the following clues, can you determine each team's point total?
1. Only the U.S.A. had more than one tie.
2. Turkey had one more point than the U.S.A. and one less point than Australia.
Have you tried 22 tonight? I said 22.
June 23rd, 2023 at 7:52:23 AM
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You put USA at the bottom of the group, which is won by #101 in the world, India (or #60, if we are talking about the women's side)?!
| Team | Win | Loss | Draw | Points |
|---|---|---|---|---|
| India | 2 | 0 | 1 | 7 |
| Austrailia | 1 | 1 | 1 | 4 |
| Turkey | 1 | 2 | 0 | 3 |
| USA | 0 | 1 | 2 | 2 |
ETA: Whoops! Fixed that 'typo!' Thanks, G!
Last edited by: Joeman on Jun 23, 2023
"Dealer has 'rock'... Pay 'paper!'"
June 23rd, 2023 at 8:17:12 AM
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Quote: JoemanYou put USA at the bottom of the group, which is won by #101 in the world, India (or #60, if we are talking about the women's side)?!
Team Win Loss Draw Points India 2 0 1 6 Austrailia 1 1 1 4 Turkey 1 2 0 3 USA 0 1 2 2
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Heh heh. Don't blame me for breaking your bracket. It's a Games Magazine puzzle. Besides, the bookies are happy.
P.S. I spy a typo.
Have you tried 22 tonight? I said 22.
June 23rd, 2023 at 9:04:43 AM
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Fixed, I think! ;)Quote: GialmereHeh heh. Don't blame me for breaking your bracket. It's a Games Magazine puzzle. Besides, the bookies are happy.
P.S. I spy a typo.
Is Games Magazine still around in some form or fashion? I had subscribed to it for a few years in the 80's when I was a kid. Mine are all long gone, but I think Mrs. Joeman still has a stash somewhere.
"Dealer has 'rock'... Pay 'paper!'"
June 23rd, 2023 at 9:19:55 AM
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It merged with another mag and I lost track of it.
If you can find that stash you could post some oldies here.
If you can find that stash you could post some oldies here.
Have you tried 22 tonight? I said 22.
June 23rd, 2023 at 4:04:43 PM
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Quote: JoemanYou put USA at the bottom of the group, which is won by #101 in the world, India (or #60, if we are talking about the women's side)?!
Team Win Loss Draw Points India 2 0 1 7 Austrailia 1 1 1 4 Turkey 1 2 0 3 USA 0 1 2 2
Correct!!
Very good.
The answer for this group is:
India 7 points
Australia 4 points
Turkey 3 points
U.S.A. 2 points
Turkey defeated U.S.A. and lost its other two games for a total of 3 points.
U.S.A. lost to Turkey and tied both India and Australia for 2 points.
India defeated Turkey and Australia and tied U.S.A. for 7 points.
Australia defeated Turkey, tied U.S.A., and lost to India, for 4 points.
Six games were played in all. From clue 1, at least two were ties. If there were a third tie, from clue 1, it went to U.S.A. and India. In that case, the other three games were decisive. There were either three or four decisive games, and so the total number of points earned by all teams was either 15 or 16.
From clue 2, three teams had consecutive point totals. Totals of 0-1-2 are impossible, because that would have required six wins by the fourth team. Similarly, totals of 1-2-3 are impossible, because that would have required five wins by the fourth team (as well as one more tie by Turkey). Totals of 3-4-5 would mean that the fourth team had 3 or 4 points; but scores of 3-3-4-5 or 3-4-4-5 would mean that the U.S.A., the only team with two ties, had the highest point total, contrary to clue 2.
Totals of 4-5-6 would mean that the fourth team scored 0 points and therefore had three losses. The teams with 4 and 6 points each had one loss to account for their scores, but this adds up to five losses, which added to the U.S.A.'s two ties exceeds the number of games played. Totals of 5-6-7 or higher add up to more than the possible point totals of 15 or 16.
By elimination, teams had consecutive point totals of 2-3-4, which means the fourth team (India) had 7 points. The U.S.A. earned 2 points from ties with India and Australia, losing to Turkey. Turkey lost its other two games. India defeated Australia.
To summarize, here are the results of the six matches:
India defeated Australia.
India defeated Turkey.
India tied the U.S.A.
Australia defeated Turkey.
Australia tied the U.S.A.
Turkey defeated the U.S.A.
India 7 points
Australia 4 points
Turkey 3 points
U.S.A. 2 points
Turkey defeated U.S.A. and lost its other two games for a total of 3 points.
U.S.A. lost to Turkey and tied both India and Australia for 2 points.
India defeated Turkey and Australia and tied U.S.A. for 7 points.
Australia defeated Turkey, tied U.S.A., and lost to India, for 4 points.
Six games were played in all. From clue 1, at least two were ties. If there were a third tie, from clue 1, it went to U.S.A. and India. In that case, the other three games were decisive. There were either three or four decisive games, and so the total number of points earned by all teams was either 15 or 16.
From clue 2, three teams had consecutive point totals. Totals of 0-1-2 are impossible, because that would have required six wins by the fourth team. Similarly, totals of 1-2-3 are impossible, because that would have required five wins by the fourth team (as well as one more tie by Turkey). Totals of 3-4-5 would mean that the fourth team had 3 or 4 points; but scores of 3-3-4-5 or 3-4-4-5 would mean that the U.S.A., the only team with two ties, had the highest point total, contrary to clue 2.
Totals of 4-5-6 would mean that the fourth team scored 0 points and therefore had three losses. The teams with 4 and 6 points each had one loss to account for their scores, but this adds up to five losses, which added to the U.S.A.'s two ties exceeds the number of games played. Totals of 5-6-7 or higher add up to more than the possible point totals of 15 or 16.
By elimination, teams had consecutive point totals of 2-3-4, which means the fourth team (India) had 7 points. The U.S.A. earned 2 points from ties with India and Australia, losing to Turkey. Turkey lost its other two games. India defeated Australia.
To summarize, here are the results of the six matches:
India defeated Australia.
India defeated Turkey.
India tied the U.S.A.
Australia defeated Turkey.
Australia tied the U.S.A.
Turkey defeated the U.S.A.
---------------------------------
Damn, this game is boring.
Have you tried 22 tonight? I said 22.
June 24th, 2023 at 1:59:35 AM
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Quote: Wizard...Due to advertising considerations, the car has a cost to the show of $20,000 and a value to the player of $25,000.
The jewelry has a cost to the show of $5,000 and a value to the player of $10,000....
The player will pick the box with the highest EV. If the chances of a car was 2/7 then the Player's EV by picking the car would be $25k*2/7 = 50/7; similarly the EV by picking the jewelry is $10k*5/7 = 50/7 = $7142.86. So the player would be indifferent. However the TV company would prefer the Player picked the jewelry as its cost to them is less.
Thus the TV company should pick the jewelry with the probability just a tad above 5/7. This makes their cost slightly over $5k*5/7 ($3571.43) and the Player's best option is to pick the jewelry. So the Player's EV is slightly over $7142.86.
Academically there is a point (20%/80%) where the cost to the TV for both selections is equal, $4k. But, as we have seen, their best strategy is offer the player a smaller chance of winning the jewelry, but not too small!
Thus the TV company should pick the jewelry with the probability just a tad above 5/7. This makes their cost slightly over $5k*5/7 ($3571.43) and the Player's best option is to pick the jewelry. So the Player's EV is slightly over $7142.86.
Academically there is a point (20%/80%) where the cost to the TV for both selections is equal, $4k. But, as we have seen, their best strategy is offer the player a smaller chance of winning the jewelry, but not too small!
June 24th, 2023 at 3:10:30 AM
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I’ve been thinking about this too. There is also a difference in variance for the player to pick the car or the jewelry. I guess for a less amount of player’s variance, the strategy should be to pick the jewelry, rather than the car. Is this right?
