## 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**

Quote:MichaelBluejaySave any time versus what, versus both of you walking? Seems like it:

walk 3mph = 20m/mile

bike 12mph = 5m/mile

A starts on bike

B starts on foot

00:00 A departs on bike, B departs on foot

00:05 A finishes Mile 1 on bike and starts walking

20:00 B picks up bike at Mile 1

25:00 A finishes Mile 2 on foot at same time B finishes on bike

So after 2 miles, each person took 25 minutes (a 5-minute bike mile and a 20-minute walk mile). That's 12.5 minutes per mile, and beats walking.

This helped me wrap my mind around why it works that way: Imagine just a single person taking the trip, alternating walking and biking each mile. After every mile s/he walks, there happens to be a bicycle there. So, s/he's walking and biking half the time, which is gonna be faster than just walking. The presence of the other person is just a distraction, it doesn't change the time taken.

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Quote:charliepatrickAs has been said regardless of the two relative speeds, assuming the cycling is quicker than walking, one person cycles the first mile and then walks. The other person has to walk a mile before they reach the bicycle. By the time the two meet again the first person has cycled a mile and walked for some time, while the second person walked a mile and cycled for some distance. When they meet the time elapsed and distance covered is the same. If you were to plot the distance against time on a graph, the slopes of the walking would be the same, and the slopes of the cycling would be the same. Hence the shape of the two lines form a rhombus (is that the correct name for a rectangle where all the sides are equal). Hence each person, in whichever order, walked one mile and cycled one mile; and they meet every two miles.

Thus the combination of cycling and walking is clearly quicker than just walking.

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Quote:ThatDonGuy

He is. Your claim that "every inch of the distance is traversed by someone on foot" does not take into account that there are times when both of you are walking simultaneously, so it is faster than a single person walking the entire distance.

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Correct!!

Well done.

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

I tried.

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

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

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

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:JoemanE.

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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.

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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?

pi / 64

Pi/64

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

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

^{2}.

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 R

^{2}.

Hence the probability of winning is Pi/64.

Quote:Ace2

pi / 64

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Quote:aceside

Pi/64

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Quote:charliepatrickThe 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 R^{2}.

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 R^{2}.

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:Ace2If 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:

Quote:Ace2I 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.

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?

(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.

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.

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

Quote:charliepatrickI 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.

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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.

Carson City and Jefferson City

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.

Quote:ChesterDog

Carson City and Jefferson City

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Quote:JoemanOhio (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!

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

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

Estimates are acceptable.

Quote:Ace2A 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.

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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?

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

Quote:Ace2I don’t think you need to choose a basis for the bet.

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

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

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.

Quote:Ace2Estimates acceptable

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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

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

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 lossesQuote:ThatDonGuyQuote:Ace2Estimates acceptable

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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

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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

Quote:Ace2I assume you “Markoved” it for 500 iterations and backed out half the value of all the losses

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You assume correctly.

I accept your answer as well.Quote:charliepatrickTo the first four significant digts I agree with both ThatDonGuy's figures

based on a simple spreadsheet using its builtin function NORM.DIST .1666.76, 0.14893

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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.

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.

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.

seed

+iced

------

spice

1663

+9263

---------

10926

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.

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.

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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.

Quote:ThatDonGuy

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.

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

Quote:MichaelBluejayWhen 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?"

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

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

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

Quote:WizardQuote:ThatDonGuyPlease tell me I am not the only person who sees the problem with the question as stated.

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

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"Who left?" can be interpreted as "Who's left in the room?" Change the end of the question to, "Who left the room?"

It's funnier as just "Who left?", and that's the point: it's a joke. No way would this be actually implemented, so there are no actual problems with a customer misinterpreting it.Quote:ThatDonGuy"Who left?" can be interpreted as "Who's left in the room?" Change the end of the question to, "Who left the room?"

Quote:avianrandyNot really a math problem but involves numbers. How many times does the word one appear on the current one dollar bill?

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It appears to be six times on the back of the one dollar bill and another, 7th, time on the front of the bill.

One might claim the word 'ONE' actually appears a 2nd time on the front of the bill. On the top of the front of the bill are the words:

which does contain the word ONE displayed backwards.