Easy math puzzles

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February 23rd, 2023 at 7:11:55 AM permalink

Only one of these statements is true. Which is it?

A. All of the below
B. None of the below
C. One of the above
D. All of the above
E. None of the above
F. None of the above

Have you tried 22 tonight? I said 22.

chevy

chevy

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February 23rd, 2023 at 8:38:43 AM permalink

Quote: Gialmere

Only one of these statements is true. Which is it?

A. All of the below
B. None of the below
C. One of the above
D. All of the above
E. None of the above
F. None of the above
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I will go with E

if A is true, then B - F are true, but that contradicts problem construction of only one true
if B is true, then C is false, which means neither A nor B are true, contradiction.....note Can't be A&B both true either by problem construction of only one true
if C is true, then A or B are true, which is again ruled out by problem construction of only one true
if D is true....again, A - C are true, again violates problem construction of only one true.
if F is true, Then E would be false, meaning one of A-D is also true, again violating only one true.

E is true....so A-D are false (which agrees with above), F is not affected by E, but F is seen to not be true

Joeman

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February 23rd, 2023 at 8:58:49 AM permalink

E.

"Dealer has 'rock'... Pay 'paper!'"

ThatDonGuy

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February 23rd, 2023 at 5:01:23 PM permalink

It cannot be A, as that would mean 5 are true
If B is true, then C is false, but that is contradicted by the fact that B is true; therefore, B is false
It cannot be C, because A and B are both false
It cannot be D, as that would mean 3 are true
This makes E true; as a result, F is false

The only correct statement is E

Gialmere

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February 23rd, 2023 at 5:27:46 PM permalink

Quote: chevy

I will go with E

if A is true, then B - F are true, but that contradicts problem construction of only one true
if B is true, then C is false, which means neither A nor B are true, contradiction.....note Can't be A&B both true either by problem construction of only one true
if C is true, then A or B are true, which is again ruled out by problem construction of only one true
if D is true....again, A - C are true, again violates problem construction of only one true.
if F is true, Then E would be false, meaning one of A-D is also true, again violating only one true.

E is true....so A-D are false (which agrees with above), F is not affected by E, but F is seen to not be true

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Quote: Joeman
E.

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Quote: ThatDonGuy

It cannot be A, as that would mean 5 are true
If B is true, then C is false, but that is contradicted by the fact that B is true; therefore, B is false
It cannot be C, because A and B are both false
It cannot be D, as that would mean 3 are true
This makes E true; as a result, F is false

The only correct statement is E

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Correct!!

Well done.

Chevy's and ThatDonGuy's responses are as good as the official solve. Joeman's is the most nostalgic.

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

Stop boolean me!

Have you tried 22 tonight? I said 22.

ThatDonGuy

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February 24th, 2023 at 12:32:21 PM permalink

You're playing a carnival game where the object is to throw a quarter onto a flat surface with circles on it; you win if the quarter is entirely within any of the circles.
In the picture, part of the quarter is just outside of the circle, so this is a loss.

For purposes of the problem, assume the surface is infinite, and you are equally likely to land the quarter on any point on the surface.
The circles are placed in a square grid pattern.
The diameter of each circle = 2 times the diameter of the quarter, and the distance between the closest edges of any two "adjacent" circles is also 2 times the diameter of the quarter.

What is the probability of landing a quarter entirely within a circle?

Ace2

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February 24th, 2023 at 1:02:53 PM permalink

pi / 64

It�s all about making that GTA

aceside

aceside

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February 24th, 2023 at 1:53:34 PM permalink

Pi/64

charliepatrick

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February 24th, 2023 at 2:06:03 PM permalink

The trick is to realise where the centre of the coin must land, that's within a circle the same size as the coin.

So let the width of the coin be C = 2R.

Therefore the area the centre of the coin needs to land in is Pi R².

The pattern repeats itself every 4 coin widths, so the height/width can be considered as 4C = 8R. So the pattern's area is 64 R².

Hence the probability of winning is Pi/64.

Ace2

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February 24th, 2023 at 2:19:13 PM permalink

If it was a carnival game, the player would get five tries to get the quarter in the dot. Win pays even money

It�s all about making that GTA

ThatDonGuy

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February 25th, 2023 at 3:10:43 PM permalink

Quote: Ace2

pi / 64

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Quote: aceside

Pi/64

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Quote: charliepatrick
The trick is to realise where the centre of the coin must land, that's within a circle the same size as the coin.

So let the width of the coin be C = 2R.

Therefore the area the centre of the coin needs to land in is Pi R².

The pattern repeats itself every 4 coin widths, so the height/width can be considered as 4C = 8R. So the pattern's area is 64 R².

Hence the probability of winning is Pi/64.

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All correct.
I would post the solution here, but charliepatrick's is the one I used.

Quote: Ace2
If it was a carnival game, the player would get five tries to get the quarter in the dot. Win pays even money
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Actually, this is a carnival game, but usually, the circles aren't quite as big, they're surrounded by a different-color ring which, in turn, is inside of a square/rectangle.
Something like this:

Ace2

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February 25th, 2023 at 3:35:18 PM permalink

I was only trying to make an example of carnival house edges, which tend to be outrageous

It�s all about making that GTA

ThatDonGuy

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February 25th, 2023 at 4:05:16 PM permalink

Quote: Ace2
I was only trying to make an example of carnival house edges, which tend to be outrageous
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If a carnival game's house edge isn't 100%, the carny isn't doing their job properly.

Gialmere

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February 27th, 2023 at 9:28:58 AM permalink

It's easy Monday. Let's play cards...

Each player in a game of cribbage has a hand of four cards. A single further card is turned up and serves as the fifth card in every player�s hand. Part of the game involves scoring your hand. You get points for any combination of cards that adds to 15, like 9 4 2; for two or more of any rank, like 3 3; or for any run of three or more, like Ace 2 3. As in blackjack, the Jack, Queen, and King each have a point value of 10.

If a 5 is turned up as the community card, is it possible to have a hand that doesn't score any points?

Have you tried 22 tonight? I said 22.

charliepatrick

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February 27th, 2023 at 11:21:26 AM permalink

I don't think it's possible to have an non-scoring hand.
(i) You can't have any cards with a value of 10 (as with the 5 they add up to 15), so this immediately rules out K Q J T
(ii) You can't have any pairs, so this rules out 5, and means the remaining cards have to be four different from A 2 3 4 6 7 8 9.
(iii) You can't have two cards that add up to 10 (as with the 5 they add up to 15), so you have to have one for each of these four pairs
(a) 1 or 9 (i.e. if you have 1 you can't have 9, if you have 9 you can't have 1, but you have to have one from each of a b c d)
(b) 2 or 8
(c) 3 or 7
(d) 4 or 6.
(iv) You can't have a straight, so can't have 7 and 6, and also can't have 4 and 3. Hence you have 74 or 63.
(v) You can't have two cards adding up to 15, hence 74 can't have 8 (7+8=15), and 63 can't have 9 (6+9=15); so this means 742 or 369. But you can't have 369 and 6+9-15. Thus leaving 742.
(vi) You can't have three/four cards adding up to 10. With 742 you can't have 1 (7+2+1+5=15), and you can't have 9 (4+2+9=15).

Thus there are no four cards that prevent you from scoring.

You must have a 15, straight or pair somewhere.

ChesterDog

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February 27th, 2023 at 11:33:45 AM permalink

No.

I did an exhaustive (hopefully) search to avoid pairs, straights, and fifteens and couldn't find any zero-point hands with a cut card of 5.

ThatDonGuy

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February 27th, 2023 at 4:40:19 PM permalink

The hand has no 10-point cards, as otherwise they would go with the 5 for a 15
Since pairs score, the four cards are different, and none are 5s
A 2 cannot have 3 (A-2-3 run), 7 (A-2-7-5 15), 8 (2-8-5 15), or 9 (A-9-5 15), leaving 4 and 6, so the only possible hand is A 2 4 6, which has 4-6-5 15
A 3 cannot have 7 (3-7-5 15) or 9 (A-9-5 15), leaving 4, 6, 8
A 3 4 6 has 4-6-5 15; A 3 4 8 has 3-4-8 15; A 4 6 8 has 4-6-5 15
A 4 cannot have 6 (4-6-(5) 15) or 9 (A-9-5 15), leaving 7 and 8, so the only possible hand is A-4-7-8, which is 7-8 15
A 6 7 8 has 7-8 15; A 6 7 9 and A 6 8 9 both have 6-9 15; A 7 8 9 has 7-8 15
Therefore, the hand cannot have an Ace
2 3 cannot have 4 (2-3-4 run), 7 (3-7-5 15), or 8 (2-8-5 15), leaving 6 and 9, so the only possible hand is 2 3 6 9, which has 6-9 15
2 4 cannot have 6 (4-6-5 15) or 8 (2-8-5 15), leaving 7 and 9, so the only possible hand is 2 4 7 9, which has 2-4-9 15
2 6 cannot have 8 (2-8-5 15), leaving 7 and 9, so the only possible hand is 2 6 7 9, which has 6-9 15
2 7 8 9 has 7-8 15
Therefore, the hand cannot have a 2
3 4 is a 3-4-5 run
3 6 cannot have a 7 (3-7-5 15), leaving 8 and 9, so the only possible hand is 3 7 8 9, which is 7-8 15
3 7 8 9 is also 7-8 15
Therefore, the hand cannot have a 3
4 cannot have a 6, so the only possible hand is 4 7 8 9, which is another 7-8 15
The only remaining hand is 6 7 8 9, which is yet still even another 7-8 15 (not to mention a 6-9 15 and, when you include the 5, a run of 5)

Therefore, a 5-card hand with a 5 cannot be scoreless

Gialmere

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February 27th, 2023 at 5:22:04 PM permalink

Quote: charliepatrick
I don't think it's possible to have an non-scoring hand.
(i) You can't have any cards with a value of 10 (as with the 5 they add up to 15), so this immediately rules out K Q J T
(ii) You can't have any pairs, so this rules out 5, and means the remaining cards have to be four different from A 2 3 4 6 7 8 9.
(iii) You can't have two cards that add up to 10 (as with the 5 they add up to 15), so you have to have one for each of these four pairs
(a) 1 or 9 (i.e. if you have 1 you can't have 9, if you have 9 you can't have 1, but you have to have one from each of a b c d)
(b) 2 or 8
(c) 3 or 7
(d) 4 or 6.
(iv) You can't have a straight, so can't have 7 and 6, and also can't have 4 and 3. Hence you have 74 or 63.
(v) You can't have two cards adding up to 15, hence 74 can't have 8 (7+8=15), and 63 can't have 9 (6+9=15); so this means 742 or 369. But you can't have 369 and 6+9-15. Thus leaving 742.
(vi) You can't have three/four cards adding up to 10. With 742 you can't have 1 (7+2+1+5=15), and you can't have 9 (4+2+9=15).

Thus there are no four cards that prevent you from scoring.

You must have a 15, straight or pair somewhere.

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Quote: ChesterDog

No.

I did an exhaustive (hopefully) search to avoid pairs, straights, and fifteens and couldn't find any zero-point hands with a cut card of 5.

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Quote: ThatDonGuy

The hand has no 10-point cards, as otherwise they would go with the 5 for a 15
Since pairs score, the four cards are different, and none are 5s
A 2 cannot have 3 (A-2-3 run), 7 (A-2-7-5 15), 8 (2-8-5 15), or 9 (A-9-5 15), leaving 4 and 6, so the only possible hand is A 2 4 6, which has 4-6-5 15
A 3 cannot have 7 (3-7-5 15) or 9 (A-9-5 15), leaving 4, 6, 8
A 3 4 6 has 4-6-5 15; A 3 4 8 has 3-4-8 15; A 4 6 8 has 4-6-5 15
A 4 cannot have 6 (4-6-(5) 15) or 9 (A-9-5 15), leaving 7 and 8, so the only possible hand is A-4-7-8, which is 7-8 15
A 6 7 8 has 7-8 15; A 6 7 9 and A 6 8 9 both have 6-9 15; A 7 8 9 has 7-8 15
Therefore, the hand cannot have an Ace
2 3 cannot have 4 (2-3-4 run), 7 (3-7-5 15), or 8 (2-8-5 15), leaving 6 and 9, so the only possible hand is 2 3 6 9, which has 6-9 15
2 4 cannot have 6 (4-6-5 15) or 8 (2-8-5 15), leaving 7 and 9, so the only possible hand is 2 4 7 9, which has 2-4-9 15
2 6 cannot have 8 (2-8-5 15), leaving 7 and 9, so the only possible hand is 2 6 7 9, which has 6-9 15
2 7 8 9 has 7-8 15
Therefore, the hand cannot have a 2
3 4 is a 3-4-5 run
3 6 cannot have a 7 (3-7-5 15), leaving 8 and 9, so the only possible hand is 3 7 8 9, which is 7-8 15
3 7 8 9 is also 7-8 15
Therefore, the hand cannot have a 3
4 cannot have a 6, so the only possible hand is 4 7 8 9, which is another 7-8 15
The only remaining hand is 6 7 8 9, which is yet still even another 7-8 15 (not to mention a 6-9 15 and, when you include the 5, a run of 5)

Therefore, a 5-card hand with a 5 cannot be scoreless

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Correct!!

Good show.

Yes. It's impossible. Hence, when a 5 turns up, a longtime player might declare, "Five's a gift!" or "Everybody's friend!" or "Fives help everyone!"

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

They practice peggin' rituals.

Have you tried 22 tonight? I said 22.

Gialmere

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March 1st, 2023 at 10:00:31 AM permalink

It's winter Wednesday. (By which I mean my work was cancelled due to snow.) Here is a handful of puzzles...

Twenty-five ants are placed randomly on a meter stick. Each faces east or west. At a signal they all start to march at 1 centimeter per second. Whenever two ants collide they reverse directions.

How long must we wait to be sure that all the ants have left the stick?

Two U.S. state capitals have the same last 8 letters in their names.

Which capitals are they?

Only two U.S. state names can be typed with a single hand on a normal keyboard.

What are they?

You�re playing darts and trying to maximize the number of points you earn with each throw. You are deciding which sector to aim for. Your dart has a 50 percent chance of landing in that sector and a 25 percent chance of landing in one of the two neighboring sectors. Reading clockwise, the sectors are worth 20, 1, 18, 4, 13, 6, 10, 15, 2, 17, 3, 19, 7, 16, 8, 11, 14, 9, 12 and 5 points, as shown above. (For the purposes of this puzzle, don�t worry about the bullseye, the outer ring that�s worth double or the inner ring that�s worth triple.)

Which sector should you aim for to maximize your expected score?

Extra credit: How would you �fairly� (by some definition of fair for you to define) assign the point values around a dartboard? Explain your thinking.

Have you tried 22 tonight? I said 22.

ChesterDog

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March 1st, 2023 at 10:05:26 AM permalink

Carson City and Jefferson City

Joeman

Joeman

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March 1st, 2023 at 10:17:50 AM permalink

Ohio (right hand) & Texas (left hand)

You should aim for the 7, which would earn you 12.25 points per throw.

"Dealer has 'rock'... Pay 'paper!'"

ChesterDog

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March 1st, 2023 at 7:02:44 PM permalink

My guess is 100 seconds.

If only one ant was placed facing right on the left end of the meter stick, it would have to walk 100 centimeters.

And if there were two ants placed facing each other on the meter stick with at least one on an end, then at least one of them would have to walk 100 centimeters.

I'm hoping that if 25 ants are put randomly on the meter stick, then 100 centimeters is the most any of them would have to walk.

Gialmere

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March 1st, 2023 at 8:21:33 PM permalink

Quote: ChesterDog

Carson City and Jefferson City

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Quote: Joeman
Ohio (right hand) & Texas (left hand)

You should aim for the 7, which would earn you 12.25 points per throw.

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Quote: ChesterDog

My guess is 100 seconds.

If only one ant was placed facing right on the left end of the meter stick, it would have to walk 100 centimeters.

And if there were two ants placed facing each other on the meter stick with at least one on an end, then at least one of them would have to walk 100 centimeters.

I'm hoping that if 25 ants are put randomly on the meter stick, then 100 centimeters is the most any of them would have to walk.

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Correct!!

Very good.

Ants: The key is to realize that two ants colliding and switching directions is equivalent to two ants passing through one another. So the answer is 100 seconds � the time it takes one ant to travel the full length of the stick.

Riddler Darts: Now the average value of all the sectors was the sum of the numbers from 1 to 20 divided by 20, or 1/20*20(21)/2, or 10.5. Surely it was possible to do better than that.

Aiming for the 16-point sector did fairly well. Half of 16 plus a quarter of 7 plus a quarter of 8 resulted in an expected score of 11.75 points. Meanwhile, both the 19-point and 14-point sectors were better targets. Half of 19 plus a quarter of 3 plus a quarter of 7 resulted in an expected score of 12 points. And half of 14 plus a quarter of 9 plus a quarter of 11 similarly resulted in 12 points.

But the best sector to aim for was the 7-point sector. Half of 7 plus a quarter of 16 plus a quarter of 19 resulted in an expected score of 12.25 points. While the individual sectors ranged from 1 to 20 points, the expected scores formed a much tighter distribution around the mean of 10.5, ranging only from 8.75 to 12.25 points.

For extra credit, you had to �fairly� (by some definition of fair for you to define) assign the point values around a dartboard in some other way. Solver Michael Greenberg worked under the assumption that you still had a 50 percent chance of hitting your target sector and a 25 percent chance of hitting either neighbor. Michael then sought to have all 20 expected values be as close as possible to the average of 10.5.

He immediately realized they couldn�t all equal 10.5, since targeting the 20-point sector (with minimal neighbors worth 1 and 2 points) resulted in an expected score of 10.75 points. But Michael still came up with the following ordering of sectors: 20, 1, 19, 3, 17, 5, 15, 7, 13, 9, 11, 10, 12, 8, 14, 6, 16, 4, 18 and 2. The 20-point and 10-point sectors both had an expected score of 10.75 points, while the 11-point and 1-point sectors both had an expected score of 10.25 points. The remaining 16 sectors were right on target, with an expected score of 10.5 points. As noted by solver Emily Kelly, this arrangement decreased the standard deviation of the expected scores by an order of magnitude. Not bad!

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

Unfortunately, the correct answer is Frankfort.

Have you tried 22 tonight? I said 22.

Ace2

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March 3rd, 2023 at 1:31:30 PM permalink

A casino offers you a 50% loss rebate if you make at least 500 bets. You accept the offer and decide to make exactly 500 passline bets (no more, no less) with 3-4-5 odds. All losses are valid for the rebate including odds.

What�s the house edge or player advantage for this scenario?

Estimates are acceptable.

It�s all about making that GTA

ThatDonGuy

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March 3rd, 2023 at 7:45:30 PM permalink

Quote: Ace2
A casino offers you a 50% loss rebate if you make at least 500 bets. You accept the offer and decide to make exactly 500 passline bets (no more, no less) with 3-4-5 odds. All losses are valid for the rebate including odds.

What�s the house edge or player advantage for this scenario?

Estimates are acceptable.
link to original post

Assume your bet is 2, with 6/8/10 odds
4/36 x 155/155 of the time, you roll craps; result is -1
8/36 x 155/155 of the time, you roll 7 or 11; result is +2
6/36 of the time, you roll a 4 or 10; that means 6/36 x 55/165 of the time, you make a point for +14, and 6/36 x 110/165 of the time, you seven out for -4
8/36 of the time, you roll a 5 or 9; that means 8/36 x 66/165 of the time, you make a point for +14, and 8/36 x 99/165 of the time, you seven out for -6
10/36 of the time, you roll a 6 or 8; that means 10/36 x 75/165 of the time, you make a point for +14, and 10/36 x 90/165 of the time, you seven out for -8

Wins = ((8 x 155 x 2) + (6 x 55 x 14) + (8 x 66 x 14) + (10 x 75 x 14)) / (36 x 165) = 24,920 / 5940
Losses: ((4 x 155 x 1) + (6 x 110 x 4) + (8 x 99 x 5) + (10 x 90 x 6)) / (36 x 165) = 12,620 / 5940
Expected Win = 12,300 / 5940 = 205 / 99

"House Edge / Player Advantage" depends on what you use as the basis for the bet. If you use the expected average bet:

The average bet = 12/36 x 2 + 6/36 x 8 + 8/36 x 10 + 10/36 x 12 = 68/9
PA = (205 / 99) / (68 / 9) = 205 / 748, or about 27.4%

Of course, this assumes that I understand how the loss rebate works in this case. I am assuming that every lost bet is refunded 50%. Is this correct, or is it only that if the result after 500 bets is a loss, then that amount is cut in half?

Ace2

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March 3rd, 2023 at 8:45:07 PM permalink

I don�t think you need to choose a basis for the bet.

I�ll restate the problem as:

A casino offers you a 50% loss rebate (all loses including odds) if you make at least 500 bets. You decide to make exactly 500 passline bets for $100 each, adding 3-4-5 odds whenever possible. The loss rebate only comes into effect after 500 bets are resolved and, obviously, only if you had a loss. What is your expected result, expressed in dollars, net of any rebate, after 500 bets?

Estimates acceptable

It�s all about making that GTA

ThatDonGuy

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March 4th, 2023 at 7:32:10 AM permalink

Quote: Ace2
I don�t think you need to choose a basis for the bet.
link to original post

You don't, if you want the answer in terms of dollars, as opposed to a house edge percentage of some sort.

Ace2

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March 4th, 2023 at 7:57:01 AM permalink

My goodness!!

You may express the answer in:

1) dollars
2) edge/advantage as a percentage of passline bets: divide #1 by (500 * $100)
3) edge/advantage as a percentage of total money wagered: multiply #2 by 9/34
4) British pounds. Divide #1 by 1.2

It�s all about making that GTA

charliepatrick

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March 4th, 2023 at 9:20:06 AM permalink

I'm not sure I've got the maths right but I think I have an aproach. I'm going to assume each wager is $1 and you just continue to make Pass line bets (i.e. they are independent).

If you play with or without odds, on average you lose 1.4c per bet. Thus for 500 of these, you would lose $7.07.

I get the SD without odds for a wager to be 1, so 500 bets would be 22.36 (SQRT(500)). This means that you come out ahead if you can get (roughtly) 7/22 SDs above average, which is about 37.6% (using https://www.mathsisfun.com/data/standard-normal-distribution-table.html ).

I get the SD when you play with odds as being 4.91, so 500 bets would be 109.92. This means the average loss would be $7.07 but the chances of coming out ahead are about 47.44%.

Without integrating what I've done is put it into a spreadsheet and look at losses for each $1 (e..g "-1000" means somewhere between -1000.99 and -1000) and looking at the chances of being in that range (excel can work out the normal curve figures).

For the 500 bets...
No odds - average loss $7.07, with rebate loss $0.62.
With odds - average loss $7.07, with rebate profit $16.67.

What makes the difference is being able to make an odds bet which is odds against but gets you half back if you lose. If you were able to do this for several days, your best tactic would be to make one large bet of $500 each day; even better would be to find someone else who does Don't Pass at the same time and share the spoils!

There may be a way of comparing the areas of profit (i.e. 0 and more) with losses (-14.14 and less) as they will have the same area under the curve; without rebate they would offset each other to give the expected loss (e.g. {0 and -14.14}, {10,-24.14}, {20,-34.14}/etc to get an average loss of -7.07), with rebate these would be {0,-7.07} {10,-12.07} {20,-17.14} so you see that eventually they become a profit. Similarly look the the bits between which happen about 5% of the time.

ThatDonGuy

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March 4th, 2023 at 11:51:32 AM permalink

Quote: Ace2
Estimates acceptable
link to original post

Estimates? The very idea!

The profit after 500 bets is:
73,757,475,375,627,846,307,350,078
436,536,886,123,015,558,692,237,218,084
692,158,059,537,793,457,579,981,781,280
182,993,521,187,538,472,164,180,680,104
217,264,621,453,665,842,270,932,365,204
942,756,160,837,394,641,602,376,922,656
629,794,518,893,012,361,700,069,097,416
708,674,157,480,315,734,489,616,273,805
841,861,659,192,166,453,194,723,121,455
402,733,639,736,117,928,058,574,923,471
985,905,474,167,372,420,726,357,160,161
714,830,811,049,311,991,163,790,306,775
217,342,204,513,295,346,053,543,267,155
091,722,965,213,587,322,592,091,184,261
888,503,257,423,955,271,125,909,335,321
085,870,653,981,415,068,197,230,888,165
194,549,612,640,658,289,777,497,642,374
530,061,655,801,041,398,886,370,606,794
078,624,886,187,002,018,927,625,285,361
252,209,596,541,827,484,009,843,207,339
376,072,806,708,086,053,131,731,384,222
419,601,413,874,419,308,397,355,071,283
711,920,733,625,187,423,386,013,113,845
397,688,991,700,058,688,933,903,173,922
957,188,360,650,204,798,031,091,450,228
517,324,439,397,176,579,838,520,595,787
348,987,299,858,068,169,644,388,697,047
798,520,683,822,967,546,322,989,483,881
578,884,410,987,254,394,717,299,161,910
568,910,618,715,338,590,480,118,154,645
973,621,103,368,869,246,254,710,034,831
349,743,888,352,326,236,336,073,655,374
472,227,348,791,547,146,536,842,860,495
433,652,597,587,169,315,756,780,910,907
731,574,923,761,737,606,578,220,402,431
121,446,885,304,781,886,395,603,759,804
433,288,175,819,835,762,580,462,804,643
982,068,455,329,633,522,755,627,202,911
420,301,976,650,964,536,429,078,444,685
337,566,843,945,290,394,622,153,941,570
979,710,497,050,636,449,149,466,379,610
505,427,600,358,589,952,417,606,956,897
505,093,547,354,024,938,297,228,697,200
377,334,691,775,391,981,743,970,269,332
382,613,429,384,835,428,021,547,149,800
/ 44,236,710,902,545,405,050,796
404,815,190,850,027,930,221,278,714,599
195,789,778,277,649,763,799,461,902,327
371,231,286,828,943,268,385,811,111,488
120,198,541,892,140,231,717,967,876,724
046,239,601,920,203,992,964,896,389,039
344,997,195,963,829,255,017,438,251,051
754,687,994,692,111,708,115,758,815,593
374,976,010,576,333,330,473,099,097,386
454,801,684,422,038,823,391,543,541,041
974,980,155,343,119,980,886,850,504,008
708,396,711,569,534,389,188,261,872,140
309,272,246,723,195,686,848,883,432,521
876,598,048,902,585,547,405,783,719,488
237,187,245,460,774,470,553,661,980,941
080,554,457,783,304,438,494,999,144,847
322,711,341,848,960,050,536,218,365,730
907,349,167,327,826,236,961,756,150,046
484,582,719,322,552,647,275,077,585,443
173,085,181,490,593,579,344,698,250,759
360,244,575,561,141,916,614,363,644,631
261,006,968,541,000,662,429,458,926,774
155,659,104,576,141,463,795,527,596,798
877,045,312,333,133,695,491,022,986,069
805,704,995,012,490,575,045,371,655,154
524,286,569,340,473,172,471,151,346,992
741,189,267,831,995,511,097,951,317,464
254,747,255,117,610,040,877,913,431,478
528,615,907,773,963,290,997,106,502,615
258,949,547,301,399,014,303,393,153,089
052,314,018,843,818,959,342,660,924,164
585,256,646,606,155,013,500,856,276,966
677,928,371,220,664,440,358,750,978,933
266,728,430,731,930,155,416,690,876,804
265,188,146,586,505,318,494,814,346,253
667,638,709,388,362,157,471,936,553,875
607,111,494,063,131,520,726,726,320,942
341,418,907,899,052,315,127,527,695,927
499,313,792,634,507,998,219,095,975,038
767,886,526,317,550,161,438,802,951,888
106,932,499,250,644,356,243,419,380,219
377,367,834,340,577,780,226,639,835,644
698,559,792,815,488,239,473,434,482,025
398,819,256,435,458,692,998,385,910,044
589,763,856,492,936,611,175,537,109,375

or about 1667.3363338

ThatDonGuy

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March 5th, 2023 at 12:26:22 PM permalink

I did also calculate, albeit with an approximation this time

The rebate needed for the house edge to be zero is about 14.8894165%

Ace2

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March 5th, 2023 at 3:17:29 PM permalink

Quote: ThatDonGuy
Quote: Ace2
Estimates acceptable
link to original post

Estimates? The very idea!

The profit after 500 bets is:
73,757,475,375,627,846,307,350,078
436,536,886,123,015,558,692,237,218,084
692,158,059,537,793,457,579,981,781,280
182,993,521,187,538,472,164,180,680,104
217,264,621,453,665,842,270,932,365,204
942,756,160,837,394,641,602,376,922,656
629,794,518,893,012,361,700,069,097,416
708,674,157,480,315,734,489,616,273,805
841,861,659,192,166,453,194,723,121,455
402,733,639,736,117,928,058,574,923,471
985,905,474,167,372,420,726,357,160,161
714,830,811,049,311,991,163,790,306,775
217,342,204,513,295,346,053,543,267,155
091,722,965,213,587,322,592,091,184,261
888,503,257,423,955,271,125,909,335,321
085,870,653,981,415,068,197,230,888,165
194,549,612,640,658,289,777,497,642,374
530,061,655,801,041,398,886,370,606,794
078,624,886,187,002,018,927,625,285,361
252,209,596,541,827,484,009,843,207,339
376,072,806,708,086,053,131,731,384,222
419,601,413,874,419,308,397,355,071,283
711,920,733,625,187,423,386,013,113,845
397,688,991,700,058,688,933,903,173,922
957,188,360,650,204,798,031,091,450,228
517,324,439,397,176,579,838,520,595,787
348,987,299,858,068,169,644,388,697,047
798,520,683,822,967,546,322,989,483,881
578,884,410,987,254,394,717,299,161,910
568,910,618,715,338,590,480,118,154,645
973,621,103,368,869,246,254,710,034,831
349,743,888,352,326,236,336,073,655,374
472,227,348,791,547,146,536,842,860,495
433,652,597,587,169,315,756,780,910,907
731,574,923,761,737,606,578,220,402,431
121,446,885,304,781,886,395,603,759,804
433,288,175,819,835,762,580,462,804,643
982,068,455,329,633,522,755,627,202,911
420,301,976,650,964,536,429,078,444,685
337,566,843,945,290,394,622,153,941,570
979,710,497,050,636,449,149,466,379,610
505,427,600,358,589,952,417,606,956,897
505,093,547,354,024,938,297,228,697,200
377,334,691,775,391,981,743,970,269,332
382,613,429,384,835,428,021,547,149,800
/ 44,236,710,902,545,405,050,796
404,815,190,850,027,930,221,278,714,599
195,789,778,277,649,763,799,461,902,327
371,231,286,828,943,268,385,811,111,488
120,198,541,892,140,231,717,967,876,724
046,239,601,920,203,992,964,896,389,039
344,997,195,963,829,255,017,438,251,051
754,687,994,692,111,708,115,758,815,593
374,976,010,576,333,330,473,099,097,386
454,801,684,422,038,823,391,543,541,041
974,980,155,343,119,980,886,850,504,008
708,396,711,569,534,389,188,261,872,140
309,272,246,723,195,686,848,883,432,521
876,598,048,902,585,547,405,783,719,488
237,187,245,460,774,470,553,661,980,941
080,554,457,783,304,438,494,999,144,847
322,711,341,848,960,050,536,218,365,730
907,349,167,327,826,236,961,756,150,046
484,582,719,322,552,647,275,077,585,443
173,085,181,490,593,579,344,698,250,759
360,244,575,561,141,916,614,363,644,631
261,006,968,541,000,662,429,458,926,774
155,659,104,576,141,463,795,527,596,798
877,045,312,333,133,695,491,022,986,069
805,704,995,012,490,575,045,371,655,154
524,286,569,340,473,172,471,151,346,992
741,189,267,831,995,511,097,951,317,464
254,747,255,117,610,040,877,913,431,478
528,615,907,773,963,290,997,106,502,615
258,949,547,301,399,014,303,393,153,089
052,314,018,843,818,959,342,660,924,164
585,256,646,606,155,013,500,856,276,966
677,928,371,220,664,440,358,750,978,933
266,728,430,731,930,155,416,690,876,804
265,188,146,586,505,318,494,814,346,253
667,638,709,388,362,157,471,936,553,875
607,111,494,063,131,520,726,726,320,942
341,418,907,899,052,315,127,527,695,927
499,313,792,634,507,998,219,095,975,038
767,886,526,317,550,161,438,802,951,888
106,932,499,250,644,356,243,419,380,219
377,367,834,340,577,780,226,639,835,644
698,559,792,815,488,239,473,434,482,025
398,819,256,435,458,692,998,385,910,044
589,763,856,492,936,611,175,537,109,375

or about 1667.3363338

link to original post

I�ll accept that answer. 16.7 unit gain. I assume you �Markoved� it for 500 iterations and backed out half the value of all the losses

I used ((500^.5) * 4.92 * (2pi)^-.5 + 7.07) / 2 - 7.07 = 18.39 unit gain where 4.92 is the standard deviation and 7.07 the expected loss.

This number would be more accurate with more iterations

It�s all about making that GTA

ThatDonGuy

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Posts: 6272

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March 5th, 2023 at 4:42:45 PM permalink

Quote: Ace2
I assume you �Markoved� it for 500 iterations and backed out half the value of all the losses
link to original post

You assume correctly.

charliepatrick

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March 5th, 2023 at 4:54:26 PM permalink

To the first four significant digts I agree with both ThatDonGuy's figures

1666.76, 0.14893

based on a simple spreadsheet using its builtin function NORM.DIST .

Ace2

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March 5th, 2023 at 5:07:06 PM permalink

Quote: charliepatrick
To the first four significant digts I agree with both ThatDonGuy's figures
1666.76, 0.14893
based on a simple spreadsheet using its builtin function NORM.DIST .
link to original post

I accept your answer as well.

It�s all about making that GTA

DRich

DRich

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March 16th, 2023 at 6:58:30 AM permalink

This question was on the SAT exam many years ago. I am sure those of you that have studied math will get this easily but I chose what I thought was the obvious answer and was wrong.

At my age, a "Life In Prison" sentence is not much of a deterrent.

ChesterDog

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March 16th, 2023 at 7:51:42 AM permalink

In school, I learned that that the earth revolves around the sun whereas the earth rotates about the earth's axis.

Probably this SAT question should have worded the question as, "How many times will Circle A rotate in total?"

My answer is 4 rotations, but that is not one of the choices. Maybe, the questions should have included: (f) none of the above.

charliepatrick

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March 16th, 2023 at 8:04:23 AM permalink

Assume the small circle is like a tyre (tire) so sticks to the other circle as it rotates. At the start put a chalk mark at the base of the small circle (i.e. at 6 o'clock). Now push the tyre along the large circle.

The small circle's circumference is 1/3 pf the large circle so as it reaches 120 degrees (4 o'clock on the big circle) the chalk mark will be at the surface of the large circle. Thus the chalk mark will be at 10 o'clock. So at this stage the chalk mark/small circle has rotated 1 1/3 revolutions.

As this process continues, the other revolutions will also be 1 1/3, giving a total of 4 revolutions.

Another way of looking at this is when the small circle is a 3 o'clock, the chalk mark will be back at 6 o'clock, so rotated once.

MichaelBluejay

MichaelBluejay

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March 18th, 2023 at 10:32:55 PM permalink

This was the math problem this week in my local alternative weekly:

 seed
+iced
------
spice

 1663
+9263
---------
10926

Presidential Election polls and odds: https://2605.me/p

Wizard
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Wizard

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March 29th, 2023 at 5:16:11 PM permalink

I get four as well.

The formula I use is (radius of A)/(radius of A) + 1.

If circle A were inside circle B, then you would subtract one, instead of add.

To be honest, I can't explain why, in plain simple English, why you add one.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy

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March 29th, 2023 at 7:09:20 PM permalink

Quote: Wizard

I get four as well.

The formula I use is (radius of A)/(radius of A) + 1.

If circle A were inside circle B, then you would subtract one, instead of add.

To be honest, I can't explain why, in plain simple English, why you add one.

link to original post

Mark the point on circle A where it is initially tangent to circle B.
Now, stretch circle B into a line whose length is the circle's circumference.

Each time you rotate circle A one circumference, the mark will once again be tangent to the line. The number of rotations = B's circumference / A's circumference = B's radius / A's radius.
However, when this is done while B is a circle, each time the mark is tangent to the line, you have to add the amount of rotation to A added by B's curvature. When you are done, the rotation is 1 complete revolution, which is why it is B / A + 1.
If the circle is on the inside, then the marked point's rotation is in the opposite direction, so you are subtracting B's curvature, which is why it is B / A - 1.

Wizard
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Wizard

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March 29th, 2023 at 7:16:38 PM permalink

Quote: ThatDonGuy
link to original post

Mark the point on circle A where it is initially tangent to circle B.
Now, stretch circle B into a line whose length is the circle's circumference.

Each time you rotate circle A one circumference, the mark will once again be tangent to the line. The number of rotations = B's circumference / A's circumference = B's radius / A's radius.
However, when this is done while B is a circle, each time the mark is tangent to the line, you have to add the amount of rotation to A added by B's curvature. When you are done, the rotation is 1 complete revolution, which is why it is B / A + 1.
If the circle is on the inside, then the marked point's rotation is in the opposite direction, so you are subtracting B's curvature, which is why it is B / A - 1.

link to original post

That's good, thanks. I'm ready for a fresh problem to wake up this thread.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

MichaelBluejay

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March 29th, 2023 at 7:17:41 PM permalink

When I did tech support at Apple in the 90s, a small percentage of customers were so literally unbelievably dumb that we all wondered how they were able to even conduct the transaction in the store to buy the computer and then get the computer home, much less have any kind of job where they could earn money to buy a computer. I suggested we put an intelligence test on the phone tree that customers would have to pass before they would could speak to a tech. The question I suggested was: "John, Mary, and Bob are in a room. Bob leaves. Who left?"

Presidential Election polls and odds: https://2605.me/p

ThatDonGuy

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March 29th, 2023 at 7:25:22 PM permalink

Quote: MichaelBluejay
When I did tech support at Apple in the 90s, a small percentage of customers were so literally unbelievably dumb that we all wondered how they were able to even conduct the transaction in the store to buy the computer and then get the computer home, much less have any kind of job where they could earn money to buy a computer. I suggested we put an intelligence test on the phone tree that customers would have to pass before they would could speak to a tech. The question I suggested was: "John, Mary, and Bob are in a room. Bob leaves. Who left?"
link to original post

Please tell me I am not the only person who sees the problem with the question as stated.

Wizard
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Wizard

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March 29th, 2023 at 7:28:31 PM permalink

Quote: ThatDonGuy
Please tell me I am not the only person who sees the problem with the question as stated.
link to original post

I don't see the problem with it, other than the question itself offending people.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ThatDonGuy