## Poll

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46 members have voted

ThatDonGuy
• Posts: 6457
Joined: Jun 22, 2011
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January 24th, 2023 at 4:28:45 PM permalink
Quote: aceside

Chess solution:

Rook#1: place it at H5 so that King must move down;
Rook #2: place it two squares directly underneath King;
Rook 3: place it in the same column above Rook #2 to check.
Rook #4: place it in the same column with Rook #3 to checkmate.

1 - Rook in h5; King to e4
2 - Rook in e2; King to f3
3 - Rook in e3; King to g4
4 - Rook in e4; King captures Rook on h5

I think your solution is correct if you change Rook #4 to always go on h4.

Of course, if the King goes up on the first turn, then the second Rook is two squares above the King, and the fourth Rook is on h6.

aceside
• Posts: 538
Joined: May 14, 2021
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January 24th, 2023 at 5:04:07 PM permalink
Quote: ThatDonGuy

Quote: aceside

Chess solution:

Rook#1: place it at H5 so that King must move down;
Rook #2: place it two squares directly underneath King;
Rook 3: place it in the same column above Rook #2 to check.
Rook #4: place it in the same column with Rook #3 to checkmate.

1 - Rook in h5; King to e4
2 - Rook in e2; King to f3
3 - Rook in e3; King to g4
4 - Rook in e4; King captures Rook on h5

I think your solution is correct if you change Rook #4 to always go on h4.

Of course, if the King goes up on the first turn, then the second Rook is two squares above the King, and the fourth Rook is on h6.

ThatDonGuy is correct! Let me revise my solution:
Rook #1: place it at H5 so that King must move down;
Rook #2: place it two squares directly underneath King;
Rook #3: place it in the same column of Rook #2 to check;
Rook #4: place it in the same column of H5 to checkmate.
Last edited by: aceside on Jan 24, 2023
Gialmere
• Posts: 2994
Joined: Nov 26, 2018
January 24th, 2023 at 6:27:11 PM permalink
Quote: aceside

I guess only two ways. One is this square, and the other one is a triangle with a 4th dot on its weight center.

Quote: unJon

Too lazy to solve but I would have thought there would also be a

trapezoid solution where the diagonal distance equals the length of the longer side. And the three short sides are the same distance.

Quote: ChesterDog

Dart

Correct!!!

Impressive. (As a joint solve I didn't quote every post.)

These six shapes are the only ones possible. For a lengthy proof go here. For a short version...

There are only 2 distances: short and long. Note that each shape has in it a triangle consisting of 2 shorts and a long. (Note also that point B can’t be higher than the dotted line since then the blue lines would no longer be the short distances.) By placing dot D somewhere around this triangle you can then try to connect it.

There are 8 possible ways it could do that. The distances from D to A, B and C could be

short, short, short
short, short, long
short, long, short
short, long, long
long, short, short
long, short, long
long, long, short
long, long, long

For each combo you simply check (admittedly time consuming) to see if the shape is valid.

-----------------------------------------------

Quote: Gialmere

To kill some time before a meeting of chess grandmasters, Burt Hochberg offered this anonymous puzzle from the 15th century...

White must place four white rooks on the board, one at a time, giving check with each one. After each placement the black king can respond with any normal legal move.

How can White plan his moves so that the fourth rook reliably gives checkmate?

Quote: aceside

Chess solution:

Rook#1: place it at H5 so that King must move down;
Rook #2: place it two squares directly underneath King;
Rook 3: place it in the same column above Rook #2 to check.
Rook #4: place it in the same column with Rook #3 to checkmate.

Quote: ThatDonGuy

1 - Rook in h5; King to e4
2 - Rook in e2; King to f3
3 - Rook in e3; King to g4
4 - Rook in e4; King captures Rook on h5

I think your solution is correct if you change Rook #4 to always go on h4.

Of course, if the King goes up on the first turn, then the second Rook is two squares above the King, and the fourth Rook is on h6.

Quote: aceside

ThatDonGuy is correct! Let me revise my solution:

Rook #1: place it at H5 so that King must move down;
Rook #2: place it two squares directly underneath King;
Rook #3: place it in the same column above Rook #2 to check.
Rook #4: place it either at H4 or in the same column above Rook #2 (depending on where King is) to checkmate.

Correct!!

Very good.

The key is placing the second rook so that it traps the black king on two ranks or files with two rooks left to place.

Hochberg says the grandmasters studied the position for several minutes before Paul Keres came up with an answer.

---------------------------------------------

The circle got up and looked round, the triangle suffered acute injuries and the square was alright.

Have you tried 22 tonight? I said 22.
Gialmere
• Posts: 2994
Joined: Nov 26, 2018
January 25th, 2023 at 7:10:19 AM permalink

Able, Baker, and Charlie are playing tag. Able is faster than Baker, who’s faster than Charlie. All three of them start at point P, and Able is “it.” At time -T, Baker runs north and Charlie runs south. After a count that takes time T, Able starts chasing one of the two quarries. The game ends when Able has tagged both Baker and Charlie.

If Baker and Charlie maintain their speeds and directions, who should Able chase first to minimize the time required to make the second tag?
Have you tried 22 tonight? I said 22.
chevy
• Posts: 148
Joined: Apr 15, 2011
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January 25th, 2023 at 9:23:09 AM permalink

-Whomever Able chases first, "1", will be at a location of v1*T when Able starts to run
-To catch "1", given the differences in speed takes distance/(rel. vel.) = v1*T/(vA-v1)
-So total time elapsed so far is t1 = T + T* v1/(vA-v1) = T*(1+v1/(vA-v1)) = T * vA/(vA-v1)

-In time t1, since Baker and Charlie run in opposite directions, they are a distance vB*t1 + vC*t1 = (vB+vC)*t1 apart
-To catch "2" given differences in speed takes distance/(rel. vel.) = (vB+vC)*t1 / (vA-v2)

= (vB+vC) * T * vA / (vA-v1) / (vA-v2)
=(vB+vC) * T * vA / ( (vA-v1) * (vA-v2) )

***EDIT : This is the time from catching 1 to 2 and is symmetric thus seems to not matter which is chased first.

Total time is then
t1 +t2

t2 is indecent of order, but t1=T * vA/(vA-v1) is smallest when the difference (vA-v1) is largest....

so 1 should be Charlie??

Wizard
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Joined: Oct 14, 2009
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January 25th, 2023 at 9:26:35 AM permalink
Quote: Gialmere

If Baker and Charlie maintain their speeds and directions, who should Able chase first to minimize the time required to make the second tag?

I show A should go after C first.

I tried this under the following conditions:

A, B, and C run at rates 3, 2, and 1 respectively.
The countdown is one unit of time.

I show chasing B first will take 7.5 units of time, while chasing C first takes 6 units of time.

To be honest, my initial instinct was to go after the faster person first.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
chevy
• Posts: 148
Joined: Apr 15, 2011
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January 25th, 2023 at 9:36:04 AM permalink
Quote: Wizard

Quote: Gialmere

If Baker and Charlie maintain their speeds and directions, who should Able chase first to minimize the time required to make the second tag?

I show A should go after C first.

I tried this under the following conditions:

A, B, and C run at rates 3, 2, and 1 respectively.
The countdown is one unit of time.

I show chasing B first will take 7.5 units of time, while chasing C first takes 6 units of time.

To be honest, my initial instinct was to go after the faster person first.

I was editing my answer as Wiz posted his. My edit agree with Wiz's result....I originally listed just the time from first catch to second catch...edit shows full time

rackuun
• Posts: 6
Joined: Jan 24, 2023
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January 25th, 2023 at 10:02:54 AM permalink
The 4-points-2-distances puzzle is a good one Gialmere. I'm going to file that one away for the mathletes. The trapezoid solution is constructed by removing one vertex from a regular pentagon. And it turns out that the 5-points-2-distances problem has only one solution: the regular pentagon.
Wizard
• Posts: 26679
Joined: Oct 14, 2009
January 25th, 2023 at 5:10:46 PM permalink
Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2994
Joined: Nov 26, 2018
January 25th, 2023 at 6:50:38 PM permalink
Quote: chevy

-Whomever Able chases first, "1", will be at a location of v1*T when Able starts to run
-To catch "1", given the differences in speed takes distance/(rel. vel.) = v1*T/(vA-v1)
-So total time elapsed so far is t1 = T + T* v1/(vA-v1) = T*(1+v1/(vA-v1)) = T * vA/(vA-v1)

-In time t1, since Baker and Charlie run in opposite directions, they are a distance vB*t1 + vC*t1 = (vB+vC)*t1 apart
-To catch "2" given differences in speed takes distance/(rel. vel.) = (vB+vC)*t1 / (vA-v2)

= (vB+vC) * T * vA / (vA-v1) / (vA-v2)
=(vB+vC) * T * vA / ( (vA-v1) * (vA-v2) )

***EDIT : This is the time from catching 1 to 2 and is symmetric thus seems to not matter which is chased first.

Total time is then
t1 +t2

t2 is indecent of order, but t1=T * vA/(vA-v1) is smallest when the difference (vA-v1) is largest....

so 1 should be Charlie??

Quote: Wizard

I show A should go after C first.

I tried this under the following conditions:

A, B, and C run at rates 3, 2, and 1 respectively.
The countdown is one unit of time.

I show chasing B first will take 7.5 units of time, while chasing C first takes 6 units of time.

To be honest, my initial instinct was to go after the faster person first.

Correct!!

Very good.

Let a, b, and c be the respective speeds of Able, Baker, and Charlie, and let x and y be the times required for Able to catch the first and second person. If Able chases Baker first, then

If Able chases Charlie first, then

Since both y’s are equal, the x’s show that the least time will elapse when Able chases Charlie first.

-------------------------------------------

When the clown is it.

--------------------------------------------
Quote: rackuun

The 4-points-2-distances puzzle is a good one Gialmere. I'm going to file that one away for the mathletes. The trapezoid solution is constructed by removing one vertex from a regular pentagon. And it turns out that the 5-points-2-distances problem has only one solution: the regular pentagon.

Yeah, that's a fun one. I think young mathletes would enjoy it.
--------------------------------------------

Currently in play...
Quote: Wizard

Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?

Have you tried 22 tonight? I said 22.
ThatDonGuy
• Posts: 6457
Joined: Jun 22, 2011
January 25th, 2023 at 6:53:06 PM permalink
Quote: Wizard

Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?

Let P(n) be the probability of two consecutive 7s being rolled first given that n 7s have been rolled

P(0) = 1/6 P(1) + 29/36 P(0)
7/36 P(0) = 1/6 P(1)
P(1) = 7/6 P(0)

P(1) = 1/6 + 29/36 P(0)

1/6 + 29/36 P(0) = 7/6 P(0)
1/6 = 13/36 P(0)
P(0) = 1/6 * 36/13 = 6/13

Thus, a 12 is more likely to happen first (7/13 of the time)
Wizard
• Posts: 26679
Joined: Oct 14, 2009
January 25th, 2023 at 7:22:26 PM permalink
Quote: ThatDonGuy

Let P(n) be the probability of two consecutive 7s being rolled first given that n 7s have been rolled

P(0) = 1/6 P(1) + 29/36 P(0)
7/36 P(0) = 1/6 P(1)
P(1) = 7/6 P(0)

P(1) = 1/6 + 29/36 P(0)

1/6 + 29/36 P(0) = 7/6 P(0)
1/6 = 13/36 P(0)
P(0) = 1/6 * 36/13 = 6/13

Thus, a 12 is more likely to happen first (7/13 of the time)

I agree!
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
• Posts: 2994
Joined: Nov 26, 2018
January 26th, 2023 at 7:01:20 AM permalink

A boy and a girl played rock paper scissors 10 times. Altogether the boy played rock three times, scissors six times, and paper once, and the girl played rock twice, scissors four times, and paper four times (though, in each case, the order of these plays is unknown). There were no ties.

Who won?
Have you tried 22 tonight? I said 22.
Joeman
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January 26th, 2023 at 7:20:03 AM permalink
The boy, whom I'll call Brian, with a score of 7-3.

Because boys rule!

If Brian played 6 scissors and Gigi played 4 scissors, and there were no ties, then all four of Gigi's papers would have lost to Brian's scissors. Conversely, all three of Brian's rocks were pitted against Gigi's scissors, giving him 3 more victories, for a total of 7.
"Dealer has 'rock'... Pay 'paper!'"
Wizard
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January 26th, 2023 at 8:12:52 AM permalink

The boy won 7, the girl 3.

Since there were no ties, the boy's 6 scissors must have gone against the girls rock (2x) and paper (4x). Those six games would have gone four for the boy, two for the girl.

Likewise, the girl's four scissors must have gone against the boy's rock 3x and paper 1x. Those four games would have gone 3 for the boy, 1 for the girl.

Final score:
Boy 4+3 = 7
Girl 2+1=3
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 8:56:48 AM permalink
Quote: ThatDonGuy

Quote: Wizard

Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?

Let P(n) be the probability of two consecutive 7s being rolled first given that n 7s have been rolled

P(0) = 1/6 P(1) + 29/36 P(0)
7/36 P(0) = 1/6 P(1)
P(1) = 7/6 P(0)

P(1) = 1/6 + 29/36 P(0)

1/6 + 29/36 P(0) = 7/6 P(0)
1/6 = 13/36 P(0)
P(0) = 1/6 * 36/13 = 6/13

Thus, a 12 is more likely to happen first (7/13 of the time)

Have a question about the math problem itself. The problem states “Which event is more likely to happen first?”, but your solution solved this “Which event is more likely to happen?”

Wizard
• Posts: 26679
Joined: Oct 14, 2009
January 26th, 2023 at 11:46:52 AM permalink
Quote: aceside

Have a question about the math problem itself. The problem states “Which event is more likely to happen first?”, but your solution solved this “Which event is more likely to happen?”

The question asks if you rolled the dice until either event happened, which event would it more likely be, and what is that exact probability.

I didn't see any flaw in Don's solution.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 12:02:17 PM permalink
Quote: Wizard

Quote: aceside

Have a question about the math problem itself. The problem states “Which event is more likely to happen first?”, but your solution solved this “Which event is more likely to happen?”

The question asks if you rolled the dice until either event happened, which event would it more likely be, and what is that exact probability.

I didn't see any flaw in Don's solution.

Thank you! But if you ask “Which event is more likely to happen first?” you mean which event takes fewer rolling turns to happen. Is this understanding correct? Or, they are the same thing?
Gialmere
• Posts: 2994
Joined: Nov 26, 2018
January 26th, 2023 at 4:08:24 PM permalink
Quote: Joeman

The boy, whom I'll call Brian, with a score of 7-3.

Because boys rule!

If Brian played 6 scissors and Gigi played 4 scissors, and there were no ties, then all four of Gigi's papers would have lost to Brian's scissors. Conversely, all three of Brian's rocks were pitted against Gigi's scissors, giving him 3 more victories, for a total of 7.

Quote: Wizard

The boy won 7, the girl 3.

Since there were no ties, the boy's 6 scissors must have gone against the girls rock (2x) and paper (4x). Those six games would have gone four for the boy, two for the girl.

Likewise, the girl's four scissors must have gone against the boy's rock 3x and paper 1x. Those four games would have gone 3 for the boy, 1 for the girl.

Final score:
Boy 4+3 = 7
Girl 2+1=3

Correct!!

Well done.
---------------------------------

I can’t believe I lost the Rock’s Paper Scissors.
Have you tried 22 tonight? I said 22.
ThatDonGuy
• Posts: 6457
Joined: Jun 22, 2011
January 26th, 2023 at 4:33:43 PM permalink
Quote: aceside

Quote: ThatDonGuy

Quote: Wizard

Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?

Let P(n) be the probability of two consecutive 7s being rolled first given that n 7s have been rolled

P(0) = 1/6 P(1) + 29/36 P(0)
7/36 P(0) = 1/6 P(1)
P(1) = 7/6 P(0)

P(1) = 1/6 + 29/36 P(0)

1/6 + 29/36 P(0) = 7/6 P(0)
1/6 = 13/36 P(0)
P(0) = 1/6 * 36/13 = 6/13

Thus, a 12 is more likely to happen first (7/13 of the time)

Have a question about the math problem itself. The problem states “Which event is more likely to happen first?”, but your solution solved this “Which event is more likely to happen?”

No, my solution solved the probability of two consecutive 7s being rolled before a 12.

Since that probability < 1/2, the only other result - a 12 being rolled first - is more likely to happen.

aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 7:34:56 PM permalink
Quote: ThatDonGuy

Quote: aceside

Quote: ThatDonGuy

Quote: Wizard

Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?

Let P(n) be the probability of two consecutive 7s being rolled first given that n 7s have been rolled

P(0) = 1/6 P(1) + 29/36 P(0)
7/36 P(0) = 1/6 P(1)
P(1) = 7/6 P(0)

P(1) = 1/6 + 29/36 P(0)

1/6 + 29/36 P(0) = 7/6 P(0)
1/6 = 13/36 P(0)
P(0) = 1/6 * 36/13 = 6/13

Thus, a 12 is more likely to happen first (7/13 of the time)

Have a question about the math problem itself. The problem states “Which event is more likely to happen first?”, but your solution solved this “Which event is more likely to happen?”

No, my solution solved the probability of two consecutive 7s being rolled before a 12.

Since that probability < 1/2, the only other result - a 12 being rolled first - is more likely to happen.

If we do an experiment to make either of the two endings happen 130 times, 70 of these will be a #12 ending and 60 of these will be a #7-#7 ending. The result says the first case is more likely to happen, but it does not say that a fewer rolling number is required to roll a #12. Is this right?
unJon
• Posts: 4661
Joined: Jul 1, 2018
January 26th, 2023 at 7:45:19 PM permalink
Quote: aceside

Quote: ThatDonGuy

Quote: aceside

Quote: ThatDonGuy

Quote: Wizard

Sorry if this has been asked before.

Two six-sided dice are rolled until one of the following events occurs:

1. Two consecutive totals of seven.
2. One total of 12.

Which event is more likely to happen first?

Let P(n) be the probability of two consecutive 7s being rolled first given that n 7s have been rolled

P(0) = 1/6 P(1) + 29/36 P(0)
7/36 P(0) = 1/6 P(1)
P(1) = 7/6 P(0)

P(1) = 1/6 + 29/36 P(0)

1/6 + 29/36 P(0) = 7/6 P(0)
1/6 = 13/36 P(0)
P(0) = 1/6 * 36/13 = 6/13

Thus, a 12 is more likely to happen first (7/13 of the time)

Have a question about the math problem itself. The problem states “Which event is more likely to happen first?”, but your solution solved this “Which event is more likely to happen?”

No, my solution solved the probability of two consecutive 7s being rolled before a 12.

Since that probability < 1/2, the only other result - a 12 being rolled first - is more likely to happen.

If we do an experiment to make either of the two endings happen 130 times, 70 of these will be a #12 ending and 60 of these will be a #7-#7 ending. The result says the first case is more likely to happen, but it does not say that a fewer rolling number is required to roll a #12. Is this right?

It does say that. They are equivalent statements, mathematically.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 8:09:03 PM permalink
In that case, we don’t need any calculation for this conclusion . Let’s just say we roll the dice once; there is a chance to get a #12 ending but absolutely no chance to get a #7-#7 ending.
unJon
• Posts: 4661
Joined: Jul 1, 2018
January 26th, 2023 at 9:06:21 PM permalink
Quote: aceside

In that case, we don’t need any calculation for this conclusion . Let’s just say we roll the dice once; there is a chance to get a #12 ending but absolutely no chance to get a #7-#7 ending.

That doesn’t follow. Three 7s are more likely than 12 twelves.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
• Posts: 26679
Joined: Oct 14, 2009
January 26th, 2023 at 9:06:58 PM permalink
I'm getting confused about the point of departure, but if the original question was badly worded, I'm open to correction. I plan to make this an "ask the wizard" question in the next column and would like to word the question as well as possible.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
unJon
• Posts: 4661
Joined: Jul 1, 2018
January 26th, 2023 at 9:08:12 PM permalink
Quote: Wizard

I'm getting confused about the point of departure, but if the original question was badly worded, I'm open to correction. I plan to make this an "ask the wizard" question in the next column and would like to word the question as well as possible.

I think aceside just wanted to see the answers in terms of average number of rolls to roll a 12 and average number of rolls to roll consecutive 7s.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 9:30:18 PM permalink
Quote: unJon

Quote: Wizard

I'm getting confused about the point of departure, but if the original question was badly worded, I'm open to correction. I plan to make this an "ask the wizard" question in the next column and would like to word the question as well as possible.

I think aceside just wanted to see the answers in terms of average number of rolls to roll a 12 and average number of rolls to roll consecutive 7s.

Very nice of you! I am learning. Let me refine my question a little.
What is the average number of rolls to roll a 12 before rolling two consecutives 7s?
What is the average number of rolls to roll two consecutive 7s before rolling a 12?
Last edited by: aceside on Jan 26, 2023
charliepatrick
• Posts: 2972
Joined: Jun 17, 2011
January 26th, 2023 at 9:56:29 PM permalink
Coming to this party late, so I just ran a quick simulation.
`Parms: sh:1000000 Time:5:44:51:27177 2 2770677 3 2253877 4 2194677 5 20792...77 222 177 241 177 256 177 304 112 1 2777812 2 2684312 3 2531012 4 23817...12 220 112 227 112 230 112 235 1Total: 77s : 462100 (19.85129409218784) 12s : 537900 (18.972372188139058)Parms: sh:1000000 Time:5:44:54:62`
aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 10:07:36 PM permalink
It looks like your simulation says 18.97 rolls for a 12 ending but 19.85 rolls for a 7-7 ending.
Ace2
• Posts: 2706
Joined: Oct 2, 2017
January 26th, 2023 at 10:24:52 PM permalink
Avg number of rolls to get 77 and twelve is 42 and 36 respectively. Coincides with the 7:6 ratio of which one is more likely to be rolled first.
It’s all about making that GTA
charliepatrick
• Posts: 2972
Joined: Jun 17, 2011
January 26th, 2023 at 10:25:49 PM permalink
A longer run - Total: 77s : 46157503 (19.850333801635674) 12s : 53842497 (18.98714786574627) , so perhaps 18.99 or 18.98
aceside
• Posts: 538
Joined: May 14, 2021
January 26th, 2023 at 10:33:37 PM permalink
Quote: Ace2

Avg number of rolls to get 77 and twelve is 42 and 36 respectively. Coincides with the 7:6 ratio of which one is more likely to be rolled first.

This is convincing, just like what unJon said.
Gialmere
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January 27th, 2023 at 7:02:04 AM permalink
For your weekend perusal, Easy Math Puzzles proudly presents...

Bar Bets at the Poker Table

During a break at a home poker game a player thoroughly shuffles a standard deck of cards, has you cut them, and then deals the top three cards face down onto the table. He says he'll bet you \$20 that at least one of the three cards is a jack, queen or king.

Is this a fair bet?

During the next break, the player places a quarter, a half dollar, and a silver dollar on the table. He tells you to pick one of them and he'll get the other two. You'll then both place money into a pot equal to 100 times the cash amount of the coin(s). (For example, if you pick the silver dollar you'd place \$100 into the pot. The other player would add his quarter and half dollar together and place \$75 into the pot.)

The three fair coins are then simultaneously flipped and allowed to land on the table. A coin that comes up heads is worth its monetary value. A coin that comes up tails is worth zero. Whichever player has the highest value wins the money in the pot.

(Using the example above, if all three coins come up heads you would have a value of \$1 to the player's 75¢ and would win the pot. If only, say, the quarter comes up heads, the player would have a value of 25¢ to your goose egg and he would win the pot. If all three coins are tails, they are flipped again.)

Is this a fair game?

As the poker game ends for the evening the player offers you one last chance to make money. He again thoroughly shuffles the deck and has you cut them.

He says he'll turn up the cards in pairs. If both cards are red, you'll keep them. If both cards are black, he'll keep them. If one is red and one black, they'll be set aside belonging to no one.

You'll pay him a dollar for the privilege of playing the game, and then you'll go through the whole deck. If you end up with more red cards than he has black cards he'll pay you \$3 for each one you have that exceeds his total. However, if he has more black cards, you must pay him \$2 for each that he has over.

Is this a fair game?
Have you tried 22 tonight? I said 22.
aceside
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January 27th, 2023 at 7:39:58 AM permalink
Jack, Queen, King; Bet #1
This is not a fair game. Player has a better chance of winning, about 0.553, which is greater than 0.5. This result is very useful for gambling!
charliepatrick
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January 27th, 2023 at 7:48:24 AM permalink
For you to win he has to miss the twelve picture cards, the chances of this are 40/52 then 39/51 then 38/50. This is about 44.706%. Hence it is not a fair bet

I like this as it looks as if it's a fair game regardless of which coin you pick. Let's call the 100c=A, 50c=B, 25c=C.
If you pick the 100c then you win with the four times A comes up, lose the three times A doesn't and one other coin does, no coins reflips. Hence your chances of winning are 4 to 3, which corresponds to the wagers you made.
If you pick the 50c then you win two times, if B comes up with C or without C. You lose on the other five outcomes. Thus 2 to 5 corresponds to the wagers you made.
Similarly for 25c you only win once --C, and lose the other six.
charliepatrick
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January 27th, 2023 at 7:57:15 AM permalink
My feeling is no-one ever gets more cards than the other player, so he just takes \$1 off you every hand.

The deck starts with 26 Red and 26 Black, they have been arranged into 26 Pairs. Firstly get rid of all the Pairs with one of each. While it's possible this leaves no other cards, normally it will leave some cards over. Since we have removed the same number of Red and Black cards, it follows that there are the same number of Red and Black cards left. By construction these are now all in coloured Pairs. Therefore there are the same number of Red Pairs as Black Pairs. So no-one can ever obtain a "win".
ThatDonGuy
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January 27th, 2023 at 8:24:32 AM permalink

You win if none of the first three cards are a Jack, Queen, or King.
The probability that none of the first three cards is a Jack, Queen, or King is 40/52 x 39/51 x 38/50 = 114/255 < 1/2, so this is a bad bet

Generalize the problem: let the three coin values be A, B, C, with A < B < C and, to make things easier, A + B < C
There are eight equally likely possible results of the coin flips.
One result is all tails; this is always a push.
If you have coin A, you win (B + C) if only A is heads, and lose A if any of the other six besides all tails occurs; the EV = (B + C - 6A) / 8.
If you have coin B, you win (A + C) if either just B or just A and B are heads, and lose B if any of the other five besides all tails occurs; the EV = (2A + 2C - 5B) / 8.
If you have coin C, you win (A + B) if C is heads (four of the results have this happen), and lose C if C is tails (and A and/or B are heads) the EV = (4A + 4B - 3C) / 8.

In this case, B = 2A and C = 4A, so the EV of choosing A is (2A + 4A - 6A) / 8 = 0, the EV of choosing B is (2A + 8A - 10A) / 8 = 0, and the EV of choosing C is (4A + 8A - 12A) / 8 = 0. No matter what coin you choose, the game is fair.

Since the number of red cards in the deck = the number of black cards, and the number of red cards in the discarded red/black pairs = the number of black cards in those pairs, then the number of red cards you have = the number of black cards he has. Since you paid \$1 to play the game, this is a bad bet.

Ace2
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January 27th, 2023 at 10:49:32 AM permalink
I’d counter bet #1 with:

How about I pay you \$90 a \$100 bill?
It’s all about making that GTA
Wizard
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January 27th, 2023 at 11:05:36 AM permalink

The probability of no face card is combin(40,3)/combin(52,3) = 9880/22100 = 44.71%.

So, the probability of at least one face card is 55.29%.

Thus, I would bet on at least one face card.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Gialmere
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January 28th, 2023 at 10:18:14 AM permalink
Quote: aceside

Jack, Queen, King; Bet #1

This is not a fair game. Player has a better chance of winning, about 0.553, which is greater than 0.5. This result is very useful for gambling!

Quote: charliepatrick

For you to win he has to miss the twelve picture cards, the chances of this are 40/52 then 39/51 then 38/50. This is about 44.706%. Hence it is not a fair bet

I like this as it looks as if it's a fair game regardless of which coin you pick. Let's call the 100c=A, 50c=B, 25c=C.
If you pick the 100c then you win with the four times A comes up, lose the three times A doesn't and one other coin does, no coins reflips. Hence your chances of winning are 4 to 3, which corresponds to the wagers you made.
If you pick the 50c then you win two times, if B comes up with C or without C. You lose on the other five outcomes. Thus 2 to 5 corresponds to the wagers you made.
Similarly for 25c you only win once --C, and lose the other six.

Quote: charliepatrick

My feeling is no-one ever gets more cards than the other player, so he just takes \$1 off you every hand.

The deck starts with 26 Red and 26 Black, they have been arranged into 26 Pairs. Firstly get rid of all the Pairs with one of each. While it's possible this leaves no other cards, normally it will leave some cards over. Since we have removed the same number of Red and Black cards, it follows that there are the same number of Red and Black cards left. By construction these are now all in coloured Pairs. Therefore there are the same number of Red Pairs as Black Pairs. So no-one can ever obtain a "win".

Quote: ThatDonGuy

You win if none of the first three cards are a Jack, Queen, or King.
The probability that none of the first three cards is a Jack, Queen, or King is 40/52 x 39/51 x 38/50 = 114/255 < 1/2, so this is a bad bet

Generalize the problem: let the three coin values be A, B, C, with A < B < C and, to make things easier, A + B < C
There are eight equally likely possible results of the coin flips.
One result is all tails; this is always a push.
If you have coin A, you win (B + C) if only A is heads, and lose A if any of the other six besides all tails occurs; the EV = (B + C - 6A) / 8.
If you have coin B, you win (A + C) if either just B or just A and B are heads, and lose B if any of the other five besides all tails occurs; the EV = (2A + 2C - 5B) / 8.
If you have coin C, you win (A + B) if C is heads (four of the results have this happen), and lose C if C is tails (and A and/or B are heads) the EV = (4A + 4B - 3C) / 8.

In this case, B = 2A and C = 4A, so the EV of choosing A is (2A + 4A - 6A) / 8 = 0, the EV of choosing B is (2A + 8A - 10A) / 8 = 0, and the EV of choosing C is (4A + 8A - 12A) / 8 = 0. No matter what coin you choose, the game is fair.

Since the number of red cards in the deck = the number of black cards, and the number of red cards in the discarded red/black pairs = the number of black cards in those pairs, then the number of red cards you have = the number of black cards he has. Since you paid \$1 to play the game, this is a bad bet.

Quote: Ace2

I’d counter bet #1 with:

How about I pay you \$90 a \$100 bill?

Quote: Wizard

The probability of no face card is combin(40,3)/combin(52,3) = 9880/22100 = 44.71%.

So, the probability of at least one face card is 55.29%.

Thus, I would bet on at least one face card.

Correct!!

Good show.

This one is sort of a card game with an RTP of a slot machine. You MIGHT actually win, but you'll DEFINITELY get the Wizard wagging his finger in your face.

Yes. It's a completely fair game regardless of which coin you select. In fact, this game is fair with any set of coins, no matter how they’re divided, provided that their values show a successive doubling (e.g., 25, 50, 100). If you added a 5% commission on wins, it might make for a fun online casino game.

There was a small controversy at the site I found this one at over the wording of the puzzle. In that version, you and the other player find the three coins and play the game to see who gets to keep all three. Needless to say, that does indeed change things.

(Heh, heh. Fortunately we never have issues over the wording of a puzzle on this thread.)

Although a little lengthy, this one is a true bar bet. I can see some smooth talker making a few shekels on the side with it (provided there's no mathematicians around).

---------------------------------------

… A bittersweet victory.
Have you tried 22 tonight? I said 22.
ThatDonGuy
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January 28th, 2023 at 12:23:23 PM permalink
Time for a toughie...hopefully I didn't post this already years ago:

You download a free calculator, but you notice a small problem: while the keys [0] through [9], [ + ], [ - ], [ = ], [ ( ], [ ) ], and [ 1/x ] work, whenever you press the multiplication and division keys, a message pops up asking if you want to upgrade to the premium version that includes those keys for a small fee.
There's one other feature; you can't use any particular key more than 50 times without the app shutting down.

You're better than that - using the keys you are given, multiply 256 by 4096 in a single equation. None of this "4096 + 4096 + ... + 4096 =, then write down the answer, then reload the app, then do it again until you have added 4096 256 times" nonsense, either.
Dieter
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Joined: Jul 23, 2014
January 28th, 2023 at 4:53:59 PM permalink

... to restate the terms as 2^8 * 2^12 = 2^20?
May the cards fall in your favor.
charliepatrick
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Joined: Jun 17, 2011
January 28th, 2023 at 7:14:24 PM permalink
Quote: ThatDonGuy

Time for a toughie..multiply 256 by 4096 in a single equation.

I am hoping that you allowed to realise that it's the same as 1024 * 1024.

Since you have access to the "1/x" key it's useful to notice that 1/x-1/(x+1) = 1/(x*(x+1)) Then using the "1/x" key this gives you x*(x+1). For example 1/3-1/4 = 4/12-3/12 = 1/12. 1/(1/12)=12.

Thus if you use this method to calculate 1024*1025 and then subtract 1024.

If you're not allowed to do this then calculate 255*256 and 256*257, do this eight times and add them all up.
aceside
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January 30th, 2023 at 11:19:02 PM permalink
Quote: charliepatrick

A longer run - Total: 77s : 46157503 (19.850333801635674) 12s : 53842497 (18.98714786574627) , so perhaps 18.99 or 18.98

Wizard
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January 31st, 2023 at 10:46:07 AM permalink
I was studying a McDonald's cup this morning. Here are the dimensions:

Radius of top of cup = 1.75"
Radius of bottom of cup = 1.25"
Height of cup = 6"

After doing the math (hopefully correctly), here are the key figures:
volume = 42.804200 cubic inches (this equals 19.0 ounces)
surface area = 61.653416 square inches

This got me to wondering that if the goal were to hold 33 cubic inches of beverage, what would be the ideal dimensions of the cup to minimize it's surface area? The cup must be in cup shape, including the extreme cases of a cone and cylinder, as well as everything in between.

I'm looking for the top radius, bottom radius, and height to two decimal places each.
Last edited by: Wizard on Jan 31, 2023
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Ace2
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January 31st, 2023 at 11:47:10 AM permalink
I’d suggest using metric…makes conversions so easy.

1 cm^3 = 1 ml = 1 gr (of water. close enough especially for McD coffee)
It’s all about making that GTA
Wizard
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January 31st, 2023 at 12:05:39 PM permalink
Quote: Ace2

I’d suggest using metric…makes conversions so easy.

1 cm^3 = 1 ml = 1 gr (of water. close enough especially for McD coffee)

I only mentioned the ounces to illustrate the actual size of the cup. Nobody thinks of them in cubic inches. However, I'll consider that if I make an Ask the Wizard question out of this.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
ChesterDog
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January 31st, 2023 at 1:19:32 PM permalink
Quote: Wizard

I was studying a McDonald's cup this morning. Here are the dimensions:

Radius of top of cup = 1.75"
Radius of bottom of cup = 1.25"
Height of cup = 6"

After doing the math (hopefully correctly), here are the key figures:
volume = 32.986723 cubic inches (this equals 19.0 ounces)
surface area = 57.383775 square inches

This got me to wondering that if the goal were to hold 33 cubic inches of beverage, what would be the ideal dimensions of the cup to minimize it's surface area? The cup must be in cup shape, including the extreme cases of a cone and cylinder, as well as everything in between.

I'm looking for the top radius, bottom radius, and height to two decimal places each.

I get a top radius of 2.67 inches, a bottom radius of 1.47 inches, and a height of 2.38 inches.

However, I don't the same answers you did for the real McDonald's cup.
unJon
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January 31st, 2023 at 1:29:48 PM permalink
Does the surface area include a lid on top?
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
Wizard
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January 31st, 2023 at 3:02:44 PM permalink
Quote: unJon

Does the surface area include a lid on top?