Easy math puzzles
Poll
 | 21 votes (45.65%) | ||
| | 14 votes (30.43%) | ||
| | 6 votes (13.04%) | ||
| | 3 votes (6.52%) | ||
| | 12 votes (26.08%) | ||
| | 3 votes (6.52%) | ||
| | 6 votes (13.04%) | ||
| | 5 votes (10.86%) | ||
| | 12 votes (26.08%) | ||
| | 10 votes (21.73%) |
46 members have voted
January 4th, 2023 at 9:36:51 AM
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I think you need to generate all the numbers 2 thru 12 with the fewest number of faces on each die. Once you've done that, then create as many sevens as possible.
If die1 would be ABCccc and die2 EFGeee then you could only create nine different numbers; thus you need at least three different numbers on one die and four on the other. Once you've done this, e.g. ABCDdd EFGggg such that d+g=7 you have at least 12 sevens. ABCD and EFG create twelve permutations, so ideally you'd like them to create 7s via DG but also one other. You can't have three ways to make 7 as you need to create 11 different totals.
Thus the puzzle boils down to finding ABCD and EFG (obviously ABCD are all different, so are EFG, and A=1, E=1, and D+G=12) so that they create all the numbers 2-12 and 7 twice. If you can, the probability of a 7 would be 13/36 rather than the value I found. If not, then I'm happy with my solution, although there may be more than one possibility.
(I feel a spreadsheet is about to be created!!)
If die1 would be ABCccc and die2 EFGeee then you could only create nine different numbers; thus you need at least three different numbers on one die and four on the other. Once you've done this, e.g. ABCDdd EFGggg such that d+g=7 you have at least 12 sevens. ABCD and EFG create twelve permutations, so ideally you'd like them to create 7s via DG but also one other. You can't have three ways to make 7 as you need to create 11 different totals.
Thus the puzzle boils down to finding ABCD and EFG (obviously ABCD are all different, so are EFG, and A=1, E=1, and D+G=12) so that they create all the numbers 2-12 and 7 twice. If you can, the probability of a 7 would be 13/36 rather than the value I found. If not, then I'm happy with my solution, although there may be more than one possibility.
(I feel a spreadsheet is about to be created!!)
This gives two solutions 1234 148 and 1234 158, which create a duplicate 5 and 9 respectively. So I still think 12 is the best I can do.
January 4th, 2023 at 10:38:40 AM
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Quote: GialmereWhat is the largest possible probability of rolling a 7?
I also could not come up with anything better than .3333.
January 4th, 2023 at 10:50:14 AM
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Good stuff so far, but nothing correct.
Have you tried 22 tonight? I said 22.
January 4th, 2023 at 10:58:14 AM
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Quote: GialmereGood stuff so far, but nothing correct.
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Oh, my gosh.
This dice arrangement seems to satisfy all of the criteria.
January 4th, 2023 at 12:30:06 PM
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I agree...Quote: EdCollins...Oh, my gosh...
It's obvious once you see it - well spotted!! Using 123 as EFG it would enable the other numbers of 1 4 7 10 to create all the numbers 2 thru 13. However by using 1469 instead means you can include two sevens.
1 2 3 3 3 3
1 2 3 4 .. .. ..
4 5 6 7 .. ... .
6 7 8 9 .. .. ..
9 10 11 12 .. .. ..
4 .. .. .. .. .. ..
4 .. .. .. .. .. ..
Initial 4x3 has all the numbers 2 thru 12 including two Sevens.
1 2 3 3 3 3
1 2 3 4 .. .. ..
4 5 6 7 +7 +7 +7
6 7 8 9 .. .. ..
9 10 11 12 .. .. ..
4 .. .. +7 +7 +7 +7
4 .. .. +7 +7 +7 +7
Adding more 3s and 4s, create another eleven Sevens.
January 4th, 2023 at 4:08:57 PM
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Quote: EdCollinsOh, my gosh.
This dice arrangement seems to satisfy all of the criteria.
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Quote: charliepatrickI agree...Quote: EdCollins...Oh, my gosh...
It's obvious once you see it - well spotted!! Using 123 as EFG it would enable the other numbers of 1 4 7 10 to create all the numbers 2 thru 13. However by using 1469 instead means you can include two sevens.
1 2 3 3 3 3
1 2 3 4 .. .. ..
4 5 6 7 .. ... .
6 7 8 9 .. .. ..
9 10 11 12 .. .. ..
4 .. .. .. .. .. ..
4 .. .. .. .. .. ..
Initial 4x3 has all the numbers 2 thru 12 including two Sevens.
1 2 3 3 3 3
1 2 3 4 .. .. ..
4 5 6 7 +7 +7 +7
6 7 8 9 .. .. ..
9 10 11 12 .. .. ..
4 .. .. +7 +7 +7 +7
4 .. .. +7 +7 +7 +7
Adding more 3s and 4s, create another eleven Sevens.
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Correct!!!
Excellent. In fact, that set of dice isn't included in the official solve. (We'll call it the EdCollins special "come out roll" dice.)
If the first die has six different numbers, then any number on the second die has at most a 1/6 probability of getting the right number on the first die to get a total of 7. Therefore, the probability of rolling a 7 is at most 1/6.
If the first die has five different numbers, with one appearing twice, the second die must have at least three different numbers. There is only one possible number on the second die which totals a 7 with the duplicated number on the first die. The second die has at most four copies of the number which totals a 7 with the duplicated number and has two ways to roll a 7; all other numbers on the second die have only one way to roll a 7. This is a maximum total of 10 rolls for a probability of 10/36.
If each die has four different numbers, they can be split 2,2,1,1 or 3,1,1,1. Each number on each die can pair with at most one number on the other die to make a 7. The largest possible sum of products, which gives the number of ways to roll a 7, is 3 × 3 + 1 × 1 + 1 × 1 + 1 × 1 = 12. The maximum probability is thus 12/36.
If the first die has four different numbers and the second has three, there are only 12 different pairs of values, and thus there must be at most two different value pairs which total 7 in order to have 11 possible totals.
Therefore, the largest possible sum of products is 3 × 4 + 1 × 1 = 13, not 3 × 4 + 1 × 1 + 1 × 1 = 14. And 13 is possible, with one die having 1,2,6,6,6,7 and the other having 1,1,1,1,3,5 (or 1,2,2,2,6,7 and 1,3,5,5,5,5); this gives a probability of 13/36.
If the first die has five different numbers, with one appearing twice, the second die must have at least three different numbers. There is only one possible number on the second die which totals a 7 with the duplicated number on the first die. The second die has at most four copies of the number which totals a 7 with the duplicated number and has two ways to roll a 7; all other numbers on the second die have only one way to roll a 7. This is a maximum total of 10 rolls for a probability of 10/36.
If each die has four different numbers, they can be split 2,2,1,1 or 3,1,1,1. Each number on each die can pair with at most one number on the other die to make a 7. The largest possible sum of products, which gives the number of ways to roll a 7, is 3 × 3 + 1 × 1 + 1 × 1 + 1 × 1 = 12. The maximum probability is thus 12/36.
If the first die has four different numbers and the second has three, there are only 12 different pairs of values, and thus there must be at most two different value pairs which total 7 in order to have 11 possible totals.
Therefore, the largest possible sum of products is 3 × 4 + 1 × 1 = 13, not 3 × 4 + 1 × 1 + 1 × 1 = 14. And 13 is possible, with one die having 1,2,6,6,6,7 and the other having 1,1,1,1,3,5 (or 1,2,2,2,6,7 and 1,3,5,5,5,5); this gives a probability of 13/36.
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Have you tried 22 tonight? I said 22.
January 4th, 2023 at 4:50:26 PM
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I've continued the spreadsheet idea, by hand, to come up with the two 13-way solutions and four 12-way solutions. One thing I noticed with the 13-way was a symmetry of the numbers.
1267 135 have 1+7=2+6, and 2-1=7-6=1, 6-2=4, 5-3=3-1=2 (i.e. not 1 nor 4). Also the average of the first die is 4 and the second 3, 4+3=7.
1469 123 have 1+9=4+6, 4-1=9-6=3, 6-4=2, 3-2=2-1=1 (i.e. not 2 nor 3). The average of the first die is 5 and the second 2, 5+2=7.
Is it obvious or a coincidence that the average of ABCD and average of EFG add up to 7, that A+D must equal B+C, and EFG have equal spacing? Not at the time!!
1267 135 have 1+7=2+6, and 2-1=7-6=1, 6-2=4, 5-3=3-1=2 (i.e. not 1 nor 4). Also the average of the first die is 4 and the second 3, 4+3=7.
1469 123 have 1+9=4+6, 4-1=9-6=3, 6-4=2, 3-2=2-1=1 (i.e. not 2 nor 3). The average of the first die is 5 and the second 2, 5+2=7.
Is it obvious or a coincidence that the average of ABCD and average of EFG add up to 7, that A+D must equal B+C, and EFG have equal spacing? Not at the time!!
January 5th, 2023 at 7:15:14 AM
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Paint the outside of an n × n × n cube red, then chop it into n^3 unit cubes. Put the unit cubes in a box, mix them up thoroughly, withdraw one at random and throw it across a table.
What’s the probability that it comes to rest with a red face on top?
Have you tried 22 tonight? I said 22.
January 5th, 2023 at 7:53:54 AM
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Quote: GialmereWhat’s the probability that it comes to rest with a red face on top?
I'm probably not expressing this properly, but I'll go out on a limb and say the probability is ( n / 100) / 100
January 5th, 2023 at 8:00:48 AM
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Having brute-forced n=1 through n=4 (including taking apart my Rubik's Cube at n=3*), it appears that the probability of a red face is 1/n.
* - j/k!
* - j/k!
"Dealer has 'rock'... Pay 'paper!'"
January 5th, 2023 at 8:05:43 AM
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Quote: JoemanHaving brute-forced n=1 through n=4 (including taking apart my Rubik's Cube at n=3*), it appears that the probability of a red face is 1/n.
* - j/k!
link to original post
The number of red squares is 6n2.
The number of squares is 6n3.
So, the probability of a red square is 6n2 / (6n3) = 1/n.
January 5th, 2023 at 8:28:29 AM
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Quote: JoemanHaving brute-forced n=1 through n=4 (including taking apart my Rubik's Cube at n=3*), it appears that the probability of a red face is 1/n.
* - j/k!
link to original post
Yep, I knew my answer wasn't the best way to express it. Thanks.
January 5th, 2023 at 8:34:35 AM
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I hope nobody minds me saying the best I can do with the two dice problem is eight sevens. I haven't looked at any spoilers on this. G, can you tell me if it's possible to do better than 8?
| 1 | 2 | 3 | 3 | 3 | 4 | |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 4 | 4 | 5 |
| 3 | 4 | 5 | 6 | 6 | 6 | 7 |
| 4 | 5 | 6 | 7 | 7 | 7 | 8 |
| 4 | 5 | 6 | 7 | 7 | 7 | 8 |
| 5 | 6 | 7 | 8 | 8 | 8 | 9 |
| 9 | 10 | 11 | 12 | 12 | 12 | 13 |
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 5th, 2023 at 8:49:12 AM
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Quote: WizardI hope nobody minds me saying the best I can do with the two dice problem is eight sevens. I haven't looked at any spoilers on this. G, can you tell me if it's possible to do better than 8?
1 2 3 3 3 4 1 2 3 4 4 4 5 3 4 5 6 6 6 7 4 5 6 7 7 7 8 4 5 6 7 7 7 8 5 6 7 8 8 8 9 9 10 11 12 12 12 13
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Yes it is.
Have you tried 22 tonight? I said 22.
January 5th, 2023 at 9:28:17 AM
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Deletep
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 5th, 2023 at 10:42:49 AM
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Quote: Gialmere
Yes it is.
link to original post
Thanks. How about ten sevens?
| 1 | 2 | 3 | 3 | 3 | 4 | |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 4 | 4 | 5 |
| 4 | 5 | 6 | 7 | 7 | 7 | 8 |
| 4 | 5 | 6 | 7 | 7 | 7 | 8 |
| 4 | 5 | 6 | 7 | 7 | 7 | 8 |
| 5 | 6 | 7 | 8 | 8 | 8 | 9 |
| 8 | 9 | 10 | 11 | 11 | 11 | 12 |
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 5th, 2023 at 11:00:29 AM
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Quote: WizardQuote: Gialmere
Yes it is.
link to original post
Thanks. How about ten sevens?
1 2 3 3 3 4 1 2 3 4 4 4 5 4 5 6 7 7 7 8 4 5 6 7 7 7 8 4 5 6 7 7 7 8 5 6 7 8 8 8 9 8 9 10 11 11 11 12
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Getting warmer.
Have you tried 22 tonight? I said 22.
January 5th, 2023 at 7:09:34 PM
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Of the n^3 cubes, the (n-2)^3 cubes in the interior have zero red faces, the 6 (n-2)^2 cubes on a face but not an edge/corner each have one, the 12 (n-2) on an edge each have 2, and the 8 on the corners each have 3.
The probability of a red face = ((n-2)^3 x 0/6 + 6 (n-2)^2 x 1/6 + 12 (n-2) x 2/6 + 8 x 3/6) / n^3
= (6n^2 - 24n + 24 + 24n - 48 + 24) / (6 n^3)
= 6n^2 / (6 n^3)
= 1 / n
January 5th, 2023 at 7:24:55 PM
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Quote: JoemanHaving brute-forced n=1 through n=4 (including taking apart my Rubik's Cube at n=3*), it appears that the probability of a red face is 1/n.
* - j/k!
link to original post
Quote: ChesterDog
The number of red squares is 6n2.
The number of squares is 6n3.
So, the probability of a red square is 6n2 / (6n3) = 1/n.
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Quote: ThatDonGuy
Of the n^3 cubes, the (n-2)^3 cubes in the interior have zero red faces, the 6 (n-2)^2 cubes on a face but not an edge/corner each have one, the 12 (n-2) on an edge each have 2, and the 8 on the corners each have 3.
The probability of a red face = ((n-2)^3 x 0/6 + 6 (n-2)^2 x 1/6 + 12 (n-2) x 2/6 + 8 x 3/6) / n^3
= (6n^2 - 24n + 24 + 24n - 48 + 24) / (6 n^3)
= 6n^2 / (6 n^3)
= 1 / n
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Correct!!
Very good.
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He died on the cross.
Have you tried 22 tonight? I said 22.
January 6th, 2023 at 7:02:17 AM
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Alice has a dozen cartons, arranged in a 3×4 grid, which for convenience we'll labeled A through L:
:strip_icc()/pic7266039.png)
She has randomly chosen two of the cartons and hidden an Easter egg inside each of them, leaving the remaining ten cartons empty. She gives the dozen cartons to Bob, who opens them in the order A, B, C, D, E, F, G, H, I, J, K, L until he finds one of the Easter eggs, whereupon he stops. The number of cartons that he opens is his score. Alice then reseals the cartons, keeping the eggs where they are, and presents the cartons to Chris, who opens the cartons in the order A, E, I, B, F, J, C, G, K, D, H, L, again stopping as soon as one of the Easter eggs is found, and scoring the number of opened cartons. Whoever scores lower wins the game; if they score the same then it’s a tie.
For example, suppose Alice hides the Easter eggs in cartons H and K. Then Bob will stop after reaching the egg in carton H and will score 8, while Chris will stop after reaching the egg in carton K and will score 9. So Bob wins in this case.
Who is more likely to win this game, Bob or Chris? Or are they equally likely to win?
She has randomly chosen two of the cartons and hidden an Easter egg inside each of them, leaving the remaining ten cartons empty. She gives the dozen cartons to Bob, who opens them in the order A, B, C, D, E, F, G, H, I, J, K, L until he finds one of the Easter eggs, whereupon he stops. The number of cartons that he opens is his score. Alice then reseals the cartons, keeping the eggs where they are, and presents the cartons to Chris, who opens the cartons in the order A, E, I, B, F, J, C, G, K, D, H, L, again stopping as soon as one of the Easter eggs is found, and scoring the number of opened cartons. Whoever scores lower wins the game; if they score the same then it’s a tie.
For example, suppose Alice hides the Easter eggs in cartons H and K. Then Bob will stop after reaching the egg in carton H and will score 8, while Chris will stop after reaching the egg in carton K and will score 9. So Bob wins in this case.
Who is more likely to win this game, Bob or Chris? Or are they equally likely to win?
Have you tried 22 tonight? I said 22.
January 6th, 2023 at 7:29:32 AM
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I guess Bob is more likely to win because there is a better chance for him to have two eggs in the same horizontal line so that the two eggs are separated more apart horizontally, statistically.
Last edited by: aceside on Jan 6, 2023
January 6th, 2023 at 7:53:50 AM
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Quote: GialmereWho is more likely to win this game, Bob or Chris? Or are they equally likely to win?
I think Bob is more likely to win this interesting game.
Bob should win the game approximately 42.36% of the time.
Chris should win the game approximately 37.49% of the time.
Approximately 20.15% of the time the game should end in a tie.
Last edited by: EdCollins on Jan 6, 2023
January 6th, 2023 at 8:36:56 AM
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Whoops!!! I'd like to revise my initial answer. Now I come up with...
Bob wins with a probability of .42424
Chris wins with a probability of .37879
The probability of a tie is .19697
Bob wins with a probability of .42424
Chris wins with a probability of .37879
The probability of a tie is .19697
Last edited by: EdCollins on Jan 6, 2023
January 6th, 2023 at 10:28:59 AM
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Quote: GialmereAlice has a dozen cartons, arranged in a 3×4 grid, which for convenience we'll labeled A through L:
She has randomly chosen two of the cartons and hidden an Easter egg inside each of them, leaving the remaining ten cartons empty. She gives the dozen cartons to Bob, who opens them in the order A, B, C, D, E, F, G, H, I, J, K, L until he finds one of the Easter eggs, whereupon he stops. The number of cartons that he opens is his score. Alice then reseals the cartons, keeping the eggs where they are, and presents the cartons to Chris, who opens the cartons in the order A, E, I, B, F, J, C, G, K, D, H, L, again stopping as soon as one of the Easter eggs is found, and scoring the number of opened cartons. Whoever scores lower wins the game; if they score the same then it’s a tie.
For example, suppose Alice hides the Easter eggs in cartons H and K. Then Bob will stop after reaching the egg in carton H and will score 8, while Chris will stop after reaching the egg in carton K and will score 9. So Bob wins in this case.
Who is more likely to win this game, Bob or Chris? Or are they equally likely to win?
link to original post
I couldn’t think of a clever way so used excel for Brute force.
I get a tie 26/132 (19.70%) of the time, Bob wins 54/132 (40.91%) of the time and Chris wins 52/132 (39.39%) of the time.
Extra credit, if Chris gets to know Bob’s choosing order and then decide how to open the cartons, how often should Chris win.
Chris should skip Bob’s first carton and then do Bob’s order of opening beginning with the second carton she opened. Chris would win unless the first crate Bob opened had an Easter egg. So Chris would win 110/132 (83.33%) of the time, and tie 2/132 (1.52%) of the time.
ETA: fixed names
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 6th, 2023 at 10:37:25 AM
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Unjon, you have Bob winning one less time (and Chris winning one more time) than I do... which means I likely made a mistake. Darn for me.
Edit: Yep. Found my mistake. I agree with your answer.
Edit: Yep. Found my mistake. I agree with your answer.
January 6th, 2023 at 11:36:51 AM
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My calculation numbers agree with unJon.
January 6th, 2023 at 4:53:26 PM
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Quote: unJonI couldn’t think of a clever way so used excel for Brute force.
I get a tie 26/132 (19.70%) of the time, Bob wins 54/132 (40.91%) of the time and Chris wins 52/132 (39.39%) of the time.
Extra credit, if Chris gets to know Bob’s choosing order and then decide how to open the cartons, how often should Chris win.Chris should skip Bob’s first carton and then do Bob’s order of opening beginning with the second carton she opened. Chris would win unless the first crate Bob opened had an Easter egg. So Chris would win 110/132 (83.33%) of the time, and tie 2/132 (1.52%) of the time.
ETA: fixed names
link to original post
Quote: EdCollinsUnjon, you have Bob winning one less time (and Chris winning one more time) than I do... which means I likely made a mistake. Darn for me.
Edit: Yep. Found my mistake. I agree with your answer.
link to original post
Quote: acesideMy calculation numbers agree with unJon.
link to original post
Correct!!
Well done. (I wasn't sure what order to place you guys, so I settled on the last three posts.)
Label a carton with a “b” (respectively, a “c”) if Bob (respectively, Chris) reaches that carton more quickly, and also record Bob’s (respectively, Chris’s) score upon reaching that carton. Label cartons A and L with “xx” since both players reach those cartons simultaneously.
We obtain:
xx b2 b3 b4
c2 c5 b7 b8
c3 c6 c9 xx
Note that there are five b cartons and five c cartons. So the cases in which Alice selects carton A or carton L are equally split between Bob and Chris. Similarly if Alice selects two b cartons then Bob necessarily wins, but these are balanced out by an equal number of cases in which Alice selects two c cartons and C necessarily wins.
The crucial cases occur when Alice selects one b carton and one c carton. Bob wins if the b carton has a lower score than the c carton:
b2 and (c3 or c5 or c6 or c9)
b3 and (c5 or c6 or c9)
b4 and (c5 or c6 or c9)
b7 and c9
b8 and c9
Chris wins if the c carton has a lower score than the b carton:
c2 and (b3 or b4 or b7 or b8)
c3 and (b4 or b7 or b8)
c5 and (b7 or b8)
c6 and (b7 or b8)
Since this yields 12 cases in Bob’s favor and only 11 cases in Chris’s favor, Bob has the advantage.
We obtain:
xx b2 b3 b4
c2 c5 b7 b8
c3 c6 c9 xx
Note that there are five b cartons and five c cartons. So the cases in which Alice selects carton A or carton L are equally split between Bob and Chris. Similarly if Alice selects two b cartons then Bob necessarily wins, but these are balanced out by an equal number of cases in which Alice selects two c cartons and C necessarily wins.
The crucial cases occur when Alice selects one b carton and one c carton. Bob wins if the b carton has a lower score than the c carton:
b2 and (c3 or c5 or c6 or c9)
b3 and (c5 or c6 or c9)
b4 and (c5 or c6 or c9)
b7 and c9
b8 and c9
Chris wins if the c carton has a lower score than the b carton:
c2 and (b3 or b4 or b7 or b8)
c3 and (b4 or b7 or b8)
c5 and (b7 or b8)
c6 and (b7 or b8)
Since this yields 12 cases in Bob’s favor and only 11 cases in Chris’s favor, Bob has the advantage.
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Box Jokes...
...but it folded
A box that was SUPPOSED to be full of snakes.
"Who's thinking outside the box now Gary?"
Have you tried 22 tonight? I said 22.
January 6th, 2023 at 5:18:47 PM
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There are 66 combinations of two letters. Then I saw that if there was an "A" in the pair, both players would score 1. Since there are 11 other letters, this is 11 combinations. Then I looked at the letters where players could score 2. If there was both a "B" and "E", then each player scores 2 and it's a tie (1 way). If there was "B" but not an "E" (and also not an "A") then Bob would score 2 and win (9 ways). Similarly if there was an "E" but not a "B" (and also not an "A") then Chris would win. You continue this idea looking for who could win in the nth round and what combinations of letters work given which letters would have gone earlier.
A - 1st round
B or E - 2nd round
C or I - 3rd round (so CI is a tie, C but not ABEI is a win, I but not ABCE is a win).
D - 4th round (Chris can't win in the 4th round as A would have won if there was a B)
etc.
So the totals are 27/66 for Bob, 13/66 for Tie, 26/66 for Chris.
A - 1st round
B or E - 2nd round
C or I - 3rd round (so CI is a tie, C but not ABEI is a win, I but not ABCE is a win).
D - 4th round (Chris can't win in the 4th round as A would have won if there was a B)
etc.
XY NN Bb Ti Ch
Ax 11 .. 11 ..
BE 1 .. .1 ..
B2 9 9 .. ..
E2 9 .. .. 9
CI 1 .. 1 ..
C4 7 7 .. ..
I4 7 .. .. 7
D5 6 6 .. ..
F6 5 .. .. 5
J7 4 .. .. 4
G8 3 3 .. ..
H9 2 2 .. ..
KL 1 .. .. 1
Total 27 13 26
January 6th, 2023 at 5:27:01 PM
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If Bob picks ABCD EFGH IJKL then Chris should pick BCDE FGHI JKLA. Bob wins on any combinations with an "A" (11 of them), and Chris wins all the others. So Chris wins 5/6 of the time.
January 7th, 2023 at 1:47:30 PM
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That last puzzle reminded me of one I posted on a puzzle thread here a few years back. So, if you remember this post, please abstain from responding...
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/pic5045763.png)
Chip Chase
Well, viva Las Vegas. You've made it to Cutting Edge and are attending a meet-and-greet for designers. To your surprise, your name is called out as a possible door prize winner. You are taken to a table that has five boxes (numbered 1-5) placed in a line and told that one of the boxes contains a $5000 casino chip.
/pic5045745.png)
Also on the table is a bag containing ten keno balls (numbered 1-10). You need to draw a ball from the bag to indicate how many guesses you get at which box contains the five grand cheque. The catch is, after each incorrect guess, you must turn your back and the chip is moved to an adjacent box before you can guess again.
Fortunately, being a game designer, you've seen this sort of mechanic before and realize that the game is beatable. You just need to get lucky with the draw.
What is the lowest numbered keno ball you can draw that guarantees you'll find the $5000 chip?
/pic5045762.png)
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Chip Chase
Well, viva Las Vegas. You've made it to Cutting Edge and are attending a meet-and-greet for designers. To your surprise, your name is called out as a possible door prize winner. You are taken to a table that has five boxes (numbered 1-5) placed in a line and told that one of the boxes contains a $5000 casino chip.
Also on the table is a bag containing ten keno balls (numbered 1-10). You need to draw a ball from the bag to indicate how many guesses you get at which box contains the five grand cheque. The catch is, after each incorrect guess, you must turn your back and the chip is moved to an adjacent box before you can guess again.
Fortunately, being a game designer, you've seen this sort of mechanic before and realize that the game is beatable. You just need to get lucky with the draw.
What is the lowest numbered keno ball you can draw that guarantees you'll find the $5000 chip?
Have you tried 22 tonight? I said 22.
January 7th, 2023 at 4:59:31 PM
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Can you pick the same box twice?
Can the host move the money to same box it has been before?
Can the host move the money to a box that has been picked before?
Can the host move the money to same box it has been before?
Can the host move the money to a box that has been picked before?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 7th, 2023 at 5:05:08 PM
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Yes, yes and yes.
Have you tried 22 tonight? I said 22.
January 7th, 2023 at 5:07:04 PM
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Quote: GialmereYes, yes and yes.
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Is box 1 adjacent to box 5?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 7th, 2023 at 5:09:56 PM
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No. A chip at the end has only one legal move.
Have you tried 22 tonight? I said 22.
January 7th, 2023 at 5:12:28 PM
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unJon, you were in on the solve a few years ago.
Have you tried 22 tonight? I said 22.
January 7th, 2023 at 5:14:46 PM
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Quote: GialmereNo. A chip at the end has only one legal move.
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In that case you can win with:
6 guesses.
Round 1: box 2
Round 2: box 2
You will have won if the chip started in either box 1 or 2. So you now know the chip is either in box 3, 4 or 5.
Round 3: box 4
Round 4: box 4
You will have won if the chip was either in box 4 or 5 at the start of round 3. That means the chip must have started round 3 in box 3 and then moved to box 2 at the start of round 4. So at the start of round 5 the chip is either in box 1 or 3.
Round 5: box 3
If you haven’t won they means the chip was in box 1 so is now in box 2
Round 6: box 2
Winner for sure. I don’t think you can do better than that, but maybe I haven’t thought of a more efficient sorting method.
Round 1: box 2
Round 2: box 2
You will have won if the chip started in either box 1 or 2. So you now know the chip is either in box 3, 4 or 5.
Round 3: box 4
Round 4: box 4
You will have won if the chip was either in box 4 or 5 at the start of round 3. That means the chip must have started round 3 in box 3 and then moved to box 2 at the start of round 4. So at the start of round 5 the chip is either in box 1 or 3.
Round 5: box 3
If you haven’t won they means the chip was in box 1 so is now in box 2
Round 6: box 2
Winner for sure. I don’t think you can do better than that, but maybe I haven’t thought of a more efficient sorting method.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
January 7th, 2023 at 5:34:49 PM
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Quote: unJonQuote: GialmereNo. A chip at the end has only one legal move.
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In that case you can win with:6 guesses.
Round 1: box 2
Round 2: box 2
You will have won if the chip started in either box 1 or 2. So you now know the chip is either in box 3, 4 or 5.
Round 3: box 4
Round 4: box 4
You will have won if the chip was either in box 4 or 5 at the start of round 3. That means the chip must have started round 3 in box 3 and then moved to box 2 at the start of round 4. So at the start of round 5 the chip is either in box 1 or 3.
Round 5: box 3
If you haven’t won they means the chip was in box 1 so is now in box 2
Round 6: box 2
Winner for sure. I don’t think you can do better than that, but maybe I haven’t thought of a more efficient sorting method.
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You open an empty box.
1) you 2, chip 4
2) you 2, chip 3
3) you 4, chip 2
4) you 4, chip 3
5) you 3, chip 4
6) you 2, chip 3 or 5
(I guess you don't remember this one.)
Have you tried 22 tonight? I said 22.
January 7th, 2023 at 9:44:46 PM
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Quote: GialmereWhat is the lowest numbered keno ball you can draw that guarantees you'll find the $5000 chip?
Keno Ball 6.
The trick is to narrow down where the chip is as much as possible with each guess.
1st guess: Box 4
2nd guess: Box 3
If I haven't found it, I know the chip cannot be in Box 5 right now. (Why? Because if it IS in 5, it had in 4 on the prior turn, but that was my first guess which obviously came up empty.)
3rd guess: Box 2
Right now we know the chip isn't in Box 4 because a turn earlier we eliminated it being in 5. (It can only move from 5 to 4.) However, I think I can make a case that right now it COULD be in all of the others. So my 3rd guess was Box 2 and if not right...
4th guess: Box 2
...I'll chose Box 2 again! If it's not here, I think I can now eliminate the chip is not in Boxes 1, 2, 3, or 5! (Not in 1 because it wasn't in 2 a turn earlier. Not in Box 5 because it wasn't in Box 4 a turn earlier. Not in Box 3 because it wasn't in Box 2 or Box 4 earlier.) So if Box 2 isn't correct, I found the chip... it's in Box 4 right now! However, we know it could move adjacently to 3 or 5 next time. So, I will then guess...
5th guess: Box 3!
... and if that's not correct it MUST be in Box 5 right and can only move adjacently to Box 4. (This is why we don't guess Box 5 here, because if we are wrong, sure it was in Box 3 but now COULD move to 2 OR 4.)
6th guess: Box 4. The chip has to be here.
So I think it's possible I might need as many as six guesses to select a box that I know for sure the chip is in, and thus I hope I draw Keno Ball #6 or higher. It does seem if I guess in this specific order each time (4-3-2-2-3-4) I will have tracked it down by Guess #6 for sure.
(I haven't figured out a way yet, to guarantee I will always get it in 5 turns.)
January 8th, 2023 at 8:32:46 AM
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Quote: EdCollinsQuote: GialmereWhat is the lowest numbered keno ball you can draw that guarantees you'll find the $5000 chip?
Keno Ball 6.
The trick is to narrow down where the chip is as much as possible with each guess.
1st guess: Box 4
2nd guess: Box 3
If I haven't found it, I know the chip cannot be in Box 5 right now. (Why? Because if it IS in 5, it had in 4 on the prior turn, but that was my first guess which obviously came up empty.)
3rd guess: Box 2
Right now we know the chip isn't in Box 4 because a turn earlier we eliminated it being in 5. (It can only move from 5 to 4.) However, I think I can make a case that right now it COULD be in all of the others. So my 3rd guess was Box 2 and if not right...
4th guess: Box 2
...I'll chose Box 2 again! If it's not here, I think I can now eliminate the chip is not in Boxes 1, 2, 3, or 5! (Not in 1 because it wasn't in 2 a turn earlier. Not in Box 5 because it wasn't in Box 4 a turn earlier. Not in Box 3 because it wasn't in Box 2 or Box 4 earlier.) So if Box 2 isn't correct, I found the chip... it's in Box 4 right now! However, we know it could move adjacently to 3 or 5 next time. So, I will then guess...
5th guess: Box 3!
... and if that's not correct it MUST be in Box 5 right and can only move adjacently to Box 4. (This is why we don't guess Box 5 here, because if we are wrong, sure it was in Box 3 but now COULD move to 2 OR 4.)
6th guess: Box 4. The chip has to be here.
So I think it's possible I might need as many as six guesses to select a box that I know for sure the chip is in, and thus I hope I draw Keno Ball #6 or higher. It does seem if I guess in this specific order each time (4-3-2-2-3-4) I will have tracked it down by Guess #6 for sure.
(I haven't figured out a way yet, to guarantee I will always get it in 5 turns.)
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...if it was in Box 5 originally, it moves to Box 4 after the first incorrect guess (by the way, now it cannot be in box 5), so if the second guess is 3, it can move from 4 to 3 or 5. It cannot be in 4, as it was not in 5, and if it was in 3, then you would have found it, but it can be in 1, 2, 3, or 5.
Apologies if I solved this one the first time, but I don't remember it
...6 notes
Guess 1: #4 - 1 moves to 2; 2 moves to 1 or 3; 3 moves to 2 or 4; 5 moves to 4 - the chip can be moved to 1, 2, 3, or 4
Guess 2: #3 - 1 moves to 2; 2 moves to 1 or 3; 4 moves to 3 or 5; it cannot be in 5 - the chip can be moved to 1, 2, 3, or 5
Guess 3: #4 - 1 moves to 2; cannot be in 2; 3 moves to 2 or 4; 5 moves to 4 - the chip can be moved to 2 or 4
Guess 4: #2 - if the chip is in 4, it can be moved to 3 or 5
Guess 5: #3 - if the chip is in 5, it has to be moved to 4
Guess 6: #4
January 8th, 2023 at 8:56:34 AM
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Don, thanks for responding.
Hmmm.... Not sure I'm following you. I'll check again, but if it was in 5 originally, I still think I find it no matter where it ends up.
If it always alternates between 4 and 5, I find it on Guess #5.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 5
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 5
6th guess: Box 4 / It's in 4 Found
If it stops at Box 3 for a moment, but still ends up in 5 at my 5th guess, I'll again find it on Guess #6.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 5
6th guess: Box 4 / It's in 4 Found
If it stops at Box 2 from Box 3, I'll find it on Guess #4.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 2 Found
If it stops at Box 3 from Box 4, I'll find it on Guess #5.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 3 Found
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 5
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 3 Found
Which travel path by the chip am I missing, where I don't find it with 4-3-2-2-3-4?
Hmmm.... Not sure I'm following you. I'll check again, but if it was in 5 originally, I still think I find it no matter where it ends up.
If it always alternates between 4 and 5, I find it on Guess #5.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 5
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 5
6th guess: Box 4 / It's in 4 Found
If it stops at Box 3 for a moment, but still ends up in 5 at my 5th guess, I'll again find it on Guess #6.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 5
6th guess: Box 4 / It's in 4 Found
If it stops at Box 2 from Box 3, I'll find it on Guess #4.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 2 Found
If it stops at Box 3 from Box 4, I'll find it on Guess #5.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 3 Found
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 5
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 3 Found
Which travel path by the chip am I missing, where I don't find it with 4-3-2-2-3-4?
January 8th, 2023 at 9:00:56 AM
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Don, does this chip path refute your solution?
Don's Guess 1: #4 / Chip is in #2
Don's Guess 2: #3 / Chip is in #1
Don's Guess 3: #4 / Chip is in #2
Don's Guess 4: #2 / Chip is in #3
Don's Guess 5: #3 / Chip is in #4
Don's Guess 6: #4 / Chip is in #3 or #5
Don's Guess 1: #4 / Chip is in #2
Don's Guess 2: #3 / Chip is in #1
Don's Guess 3: #4 / Chip is in #2
Don's Guess 4: #2 / Chip is in #3
Don's Guess 5: #3 / Chip is in #4
Don's Guess 6: #4 / Chip is in #3 or #5
January 8th, 2023 at 10:08:54 AM
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I pick Keno Ball 7 as the lowest try number because all above solutions do not work.
January 8th, 2023 at 12:31:49 PM
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Quote: EdCollinsDon, thanks for responding.
Hmmm.... Not sure I'm following you. I'll check again, but if it was in 5 originally, I still think I find it no matter where it ends up.
If it always alternates between 4 and 5, I find it on Guess #5.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 5
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 5
6th guess: Box 4 / It's in 4 Found
If it stops at Box 3 for a moment, but still ends up in 5 at my 5th guess, I'll again find it on Guess #6.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 5
6th guess: Box 4 / It's in 4 Found
If it stops at Box 2 from Box 3, I'll find it on Guess #4.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 2 Found
If it stops at Box 3 from Box 4, I'll find it on Guess #5.
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 3
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 3 Found
1st guess: Box 4 / It's in 5
2nd guess: Box 3 / It's in 4
3rd guess: Box 2 / It's in 5
4th guess: Box 2 / It's in 4
5th guess: Box 3 / It's in 3 Found
Which travel path by the chip am I missing, where I don't find it with 4-3-2-2-3-4?
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I checked all six possible first moves where the box is not in 4; all six can be solved with 4,3,2,2,3,4 in six or fewer guesses
January 8th, 2023 at 12:33:27 PM
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Quote: EdCollinsDon, does this chip path refute your solution?
Don's Guess 1: #4 / Chip is in #2
Don's Guess 2: #3 / Chip is in #1
Don's Guess 3: #4 / Chip is in #2
Don's Guess 4: #2 / Chip is in #3
Don's Guess 5: #3 / Chip is in #4
Don's Guess 6: #4 / Chip is in #3 or #5
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I'll have to check my answer again...
January 8th, 2023 at 1:37:30 PM
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I think I now found four possible solutions which also guarantee a find in six moves.
The first one I listed earlier... 4-3-2-2-3-4
I just realized there's the mirror of that, of course, which should also work: 2-3-4-4-3-2
However, I believe 2-3-4-2-3-4 and its mirror, 4-3-2-4-3-2 also work.
January 8th, 2023 at 1:57:45 PM
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Quote: EdCollinsQuote: GialmereWhat is the lowest numbered keno ball you can draw that guarantees you'll find the $5000 chip?
Keno Ball 6.
The trick is to narrow down where the chip is as much as possible with each guess.
1st guess: Box 4
2nd guess: Box 3
If I haven't found it, I know the chip cannot be in Box 5 right now. (Why? Because if it IS in 5, it had in 4 on the prior turn, but that was my first guess which obviously came up empty.)
3rd guess: Box 2
Right now we know the chip isn't in Box 4 because a turn earlier we eliminated it being in 5. (It can only move from 5 to 4.) However, I think I can make a case that right now it COULD be in all of the others. So my 3rd guess was Box 2 and if not right...
4th guess: Box 2
...I'll chose Box 2 again! If it's not here, I think I can now eliminate the chip is not in Boxes 1, 2, 3, or 5! (Not in 1 because it wasn't in 2 a turn earlier. Not in Box 5 because it wasn't in Box 4 a turn earlier. Not in Box 3 because it wasn't in Box 2 or Box 4 earlier.) So if Box 2 isn't correct, I found the chip... it's in Box 4 right now! However, we know it could move adjacently to 3 or 5 next time. So, I will then guess...
5th guess: Box 3!
... and if that's not correct it MUST be in Box 5 right and can only move adjacently to Box 4. (This is why we don't guess Box 5 here, because if we are wrong, sure it was in Box 3 but now COULD move to 2 OR 4.)
6th guess: Box 4. The chip has to be here.
So I think it's possible I might need as many as six guesses to select a box that I know for sure the chip is in, and thus I hope I draw Keno Ball #6 or higher. It does seem if I guess in this specific order each time (4-3-2-2-3-4) I will have tracked it down by Guess #6 for sure.
(I haven't figured out a way yet, to guarantee I will always get it in 5 turns.)
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Correct!!
Good show.
Six is the best you can do to guarantee a win. Both 4, 3, 2, 2, 3, 4 and its mirror 2, 3, 4, 4, 3, 2 work. Another solve is 1, 2, 3, 4, 5, 1, 2, 3, 4 but you'd need keno ball 9 to do it.
Although I gave this puzzle a Vegas theme, it comes in various guises. Sometimes it's a gopher popping up from holes in the ground. The last time I saw it, the puzzle was to find a princess in (no less than) 17 rooms with doors opening to the same hallway! To me that seems like overkill.
Although I gave this puzzle a Vegas theme, it comes in various guises. Sometimes it's a gopher popping up from holes in the ground. The last time I saw it, the puzzle was to find a princess in (no less than) 17 rooms with doors opening to the same hallway! To me that seems like overkill.
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...every fifth caller was a winner.
Have you tried 22 tonight? I said 22.
January 8th, 2023 at 4:42:47 PM
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Here's a new one:
The rectangle is 8 x 20, and the circles each have diameter 10.
To make it clear, the top circle is tangent to the top and left sides of the rectangle, and the bottom circle goes through both of the rectangle's bottom corners.
What is the combined area of the green sections?
The rectangle is 8 x 20, and the circles each have diameter 10.
To make it clear, the top circle is tangent to the top and left sides of the rectangle, and the bottom circle goes through both of the rectangle's bottom corners.
What is the combined area of the green sections?
January 8th, 2023 at 7:02:05 PM
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I got this one.
Total area = 25 Pi -48 ~ 30.54.
If you draw straight lines from each circle center to its intersection points, and then draw lines toward the arc midpoint, all of the angles are formed by right triangles of sides (3, 4, 5). Each shaded area is just the area of the big fan formed by 2x the angle subtracted by 2x the triangle area. I just add all of them up.
Total area = 25 Pi -48 ~ 30.54.
If you draw straight lines from each circle center to its intersection points, and then draw lines toward the arc midpoint, all of the angles are formed by right triangles of sides (3, 4, 5). Each shaded area is just the area of the big fan formed by 2x the angle subtracted by 2x the triangle area. I just add all of them up.
January 9th, 2023 at 8:39:42 AM
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I hope I'm not stepping on the last puzzle (which is beyond my modest skills) but...
It's easy Monday. Here's another trilogy of tiny teasers...
You're shopping for a doohickey online. You find four similar doohickeys that have the same price, the same functions, the same look, but different ratings.
Based on the ratings, which doohickey should you buy?
A. 177 positive ratings (198 total ratings)
B. 90 positive ratings (98 total ratings)
C. 45 positive ratings (48 total ratings)
D. 8 positive ratings (8 total ratings)
The Red Sox beat the Orioles 9 to 4 in 17 innings.
Where was the game played?
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Who moved last?
It's easy Monday. Here's another trilogy of tiny teasers...
You're shopping for a doohickey online. You find four similar doohickeys that have the same price, the same functions, the same look, but different ratings.
Based on the ratings, which doohickey should you buy?
A. 177 positive ratings (198 total ratings)
B. 90 positive ratings (98 total ratings)
C. 45 positive ratings (48 total ratings)
D. 8 positive ratings (8 total ratings)
The Red Sox beat the Orioles 9 to 4 in 17 innings.
Where was the game played?
Who moved last?
Have you tried 22 tonight? I said 22.
January 9th, 2023 at 9:12:43 AM
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Quote: Gialmere
The Red Sox beat the Orioles 9 to 4 in 17 innings.
Where was the game played?
Camden Yards
Iowa -- if it was the Filed of Dreams game and the Orioles were the home team.
Only the visiting team can win an extra innings game by 5 or more runs. If the score is tied in the bottom half of the inning, the first run the home team scores will end the game. If a home run is hit, all the runs would count, limiting the maximum margin of an extra innings home team victory to 4 runs.
Oddly enough, if the home team batter hits a bases-loaded ground-rule double, only the first run counts.
Oddly enough, if the home team batter hits a bases-loaded ground-rule double, only the first run counts.
"Dealer has 'rock'... Pay 'paper!'"
