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There is a square of side length 1. Each corner of the square is the center of a circle of radius one. What is the area of the red region, where all four circles overlap?

AC = BC = 1

AD = 1/2

This means CAD is a 30-60-90 right triangle, and CD = sqrt(3)/2

Similarly, BCF is also 30 degrees, so ACB = 90 - 30 - 30 = 30 degrees

This means sector AB is 1/12 of a unit circle, so its area is PI/12

The area of the sector inside of the upper left quadrant = PI/12 - AXH - BXJ - HXJ - HCJ

AG = sqrt(3)/2, so AX = BX = (sqrt(3) - 1)/2

CHE is a 30-60-90 triangle; CE = 1/2, so HE = 1 / 2 sqrt(3) = sqrt(3) / 6, and the area = 1/2 * 1/2 * sqrt(3)/6 = sqrt(3)/24

HXJ + HCJ = 1/4 - CEH - CGJ = 1/4 - 2 CEH = 1/4 - sqrt(3)/12

BXJ = AXH = 1/2 * AX * HX = 1/2 * AX * (AX / sqrt(3)) = (AX)^2 / (2 sqrt(3)) = (sqrt(3)-1)^2 / (8 sqrt(3))

= (2 - sqrt(3)) / (4 sqrt(3)) = (2 sqrt(3) - 3) / 12

Sector area inside the quadrant = PI/12 - (2 sqrt(3) - 3) / 6 - (1/4 - sqrt(3)/12)

= PI/12 - sqrt(3)/3 + 1/2 - 1/4 + sqrt(3)/12

= PI/12 + 1/4 - sqrt(3)/4

The total area = 4x the area inside the quadrant = PI/3 + 1 - sqrt(3) = about 0.31515

Quote:Wizard

There is a square of side length 1. Each corner of the square is the center of a circle of radius one. What is the area of the red region, where all four circles overlap?

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Quote:ChesterDog

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I agree! I see you used calculus in your solution. Mine was brute force geometry.

Wizard solution (PDF).

Quote:WizardQuote:ChesterDog

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I agree! I see you used calculus in your solution. Mine was brute force geometry.

Wizard solution (PDF).

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I would also have used a geometric method as did Don and you if I had thought of it.

Quote:ThatDonGuyHere's my solution, ...

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Well done Don! I overlooked this post before. I think this one was beer worthy.

While killing some time at your desk one afternoon, you fire up a new game of Solitaire on your computer (specifically the version where you deal out three cards from the deck at a time). But your boredom quickly turns to rage because your game is unplayable — there's nothing you can do on the board and you can flip through your deck, but you never have any legal moves!

Rounded to the nearest quarter percent, what is the probability of being dealt such a nightmare scenario?

Quote:WikipediaThe probability of being able to win a game of Klondike with best-possible play is not known, although Hoyle's Rules of Games suggests the chances of winning as being 1 in 30 games. The inability of theoreticians to precisely calculate these odds has been referred to by mathematician Persi Diaconis as "one of the embarrassments of applied probability".

The game enjoyed a surge in popularity when it was included as part of the Windows operating package. The idea was to get users comfortable using a mouse by dragging and dropping red 6s onto black 7s. The result was workers wasting endless hours playing the game. But hey, it was worth all that losing to see those cards fly around the screen when you finally won right? According to Microsoft, Solitaire was its most popular program for many years.

Klondike was also called Canfield in America, perhaps because it was once a casino game at the Canfield Casino in Saratoga Springs, New York. I've heard of Solitaire in casinos but have never seen it myself. I do think that the reason "Las Vegas Scoring" is so popular on computer versions is that you can lose the game but still claim a minor victory if you got enough cards up top to end with a positive cash amount.

Anyways, although we may never know the true odds of winning Klondike, the above puzzle (from the Riddler) is solvable (but not a cakewalk).

Quote:Ace2For a bet that follows the normal distribution exactly, the risk of ruin for a given bankroll and playing time is exactly double the probability of ending the session with a busted bankroll. For example, if you have a 63 unit bankroll and bet on 1,000 fair coin flips with an even money payout, your RoR is 4.62912 %. If you have unlimited bankroll, the probability of ending the session down more than 62 units is 2.31456%. Incidentally, a z-score calculator approximates the latter as 2.31726%

Can someone prove why the RoR is double?

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I think the RoR is double ONLY when μ = 0 and σ = 1, am I correct ?

The first thing is to remember the pack can be considered as two of 26 different cards, as red Kings are different from black Kings. So consider the deck as 26x2. For convenience alternate the colours and use

1+2 Red/Black Kings

3+4 Black/Red/Queens

...

25+26 Aces (the colour of these doesn't really matter as they can go to the top)

To enable a card to be played either it is an Ace (25/26) or the card two greater than it is visible, e.g. 5 - Red Queen, 7 = Black Jack.

Therefore stage one is to look at the original seven visible cards, these are all permutations of seven of the numbers 1 thru 26 (allowing for duplicates) where there are no possible movements (you probably do this with distributions of 2221 22111 211111 1111111).

The next thing to notice is after dealing out the board, this leaves 24 cards for the "pack", so if there are no possible movements only eight of these will be looked at.

For each combination in phase 1, work out how many possible cards left could cause a movement (i.e. be an Ace or two more than a visible card) and how many don't. Work out how often the eight are solely in the don't category.

(In practice the eight will be a random selection from the remaining pack of 45. The important feature of 24 is only eight cards can ever be seen if there are no movements.)

Quote:charliepatrickI haven't managed to work it out but here is a thought and possible approach (which would eventually give an exact answer).

The first thing is to remember the pack can be considered as two of 26 different cards, as red Kings are different from black Kings. So consider the deck as 26x2. For convenience alternate the colours and use

1+2 Red/Black Kings

3+4 Black/Red/Queens

...

25+26 Aces (the colour of these doesn't really matter as they can go to the top)

To enable a card to be played either it is an Ace (25/26) or the card two greater than it is visible, e.g. 5 - Red Queen, 7 = Black Jack.

Therefore stage one is to look at the original seven visible cards, these are all permutations of seven of the numbers 1 thru 26 (allowing for duplicates) where there are no possible movements (you probably do this with distributions of 2221 22111 211111 1111111).

The next thing to notice is after dealing out the board, this leaves 24 cards for the "pack", so if there are no possible movements only eight of these will be looked at.

For each combination in phase 1, work out how many possible cards left could cause a movement (i.e. be an Ace or two more than a visible card) and how many don't. Work out how often the eight are solely in the don't category.

(In practice the eight will be a random selection from the remaining pack of 45. The important feature of 24 is only eight cards can ever be seen if there are no movements.)

Keep in mind that there are only 15 cards you need to worry about; the 7 you see in the initial deal, and the 8 you will see in the 24 cards in the deck.

It is a "bad deal" only if all three of these conditions are true:

1. None of the 15 are Aces

2. None of the 7 visible cards are one rank lower than, and the opposite color of, another visible card

3. None of the 8 cards in the deck that you will see are one rank lower than, and the opposite color of, any of the 7 initially visible cards.