## Poll

21 votes (46.66%) | |||

14 votes (31.11%) | |||

6 votes (13.33%) | |||

2 votes (4.44%) | |||

12 votes (26.66%) | |||

3 votes (6.66%) | |||

6 votes (13.33%) | |||

5 votes (11.11%) | |||

12 votes (26.66%) | |||

10 votes (22.22%) |

**45 members have voted**

That matches what the poster said. I failed this one. It was on the tip of my brain.Quote:Wizard12/29 =~ 41.38%

A guy walks into a 7-11 store and selects four items to buy. The clerk at the counter informs the gentleman that the total cost of the four items is $7.11. The guy is completely surprised that the cost is the same as the name of the store and says so. The clerk informs the man that he simply multiplied the cost of each item to arrive at the total. The guy calmly informs the clerk that the items should be added and not multiplied. The clerk then adds the items together and informs the astonished customer that the total is still exactly $7.11.

What is the exact cost of each item?

I hope it's okay to give a hint: Don't convert to cents! There's a big difference between 1.1 as a factor ($1.10) and 110 as a factor (110¢).Quote:GialmereWhat is the exact cost of each item? link to original post

With cents, I got close with these numbers, assuming that there were some items they were so desperate to get rid of that they paid you to take them.

-1¢, -1¢, 1¢, $7.11: Product=$7.11 (no pun intended), Sum=$7.10.

Let a, b, c, d be the amounts in cents

a + b + c + d = 711

(a / 100)(b / 100)(c / 100)(d / 100) = 7.11, so abcd = 711,000,000 = 79 x 2^6 x 3^2 x 5^6

Okay, this isn't a "math" answer, but is there an easier way to solve this than brute force?

I did a brute force search, and got one solution: 3.16, 1.50, 1.25, 1.20

It gives the answer as 1.20 1.25 1.50 3.16.

Factor 711 gives 3 3 79. But you also need to have values with cents.

So of the four prices are v.vv x.xx y.yy z.zz then the sum of them is $7.11. However the product would have eight decimal places, whereby the last six are all zeroes (i.e. 7.11000000). So you need to add 10

^{6}which gives six more 2s and 5s.

The way to split up the 3 3 79 amongst the four prices is

(a) 1 3 3 79

(b) 1 1 9 79

(c) 1 1 1 711 (but this gives a total that is too big, so can be ignored)

(d) 1 1 3 237

On these bases prices, you can allocate all the 2s and 5s (this is where the brute force can apply).

There are a number of ways (84) you can allocate six balls into four boxes, so you use this list for the 2s, and also for the 5s.

So look at all permutations of a/b/d 1-84 1-84.

This gives (offsets, base numbers, powers of 2, powers of 5),

1 11 73 1 3 3 79 0 1 3 2 3 2 1 0 1.25 1.50 1.20 3.16

Sum=7.11

Product=7.1148844

Less than 0.5 cent off. If we use rounding, this solution works too. This is the best I could get.

Also, looking at the 1c rather than 10c part, they add up to 1c but when multiplied together mustn't leave any fractions after the second decimal place. So, ignoring the Dollar parts, the cents parts have to multiply together to leave something and 11c. So the last digits of all the prices have to add up to 1 mod 10, but when multiplied together 0 mod 10.

If the price was 79c, then this means the final digits of the other three prices add up to 2 mod 10. So the most zeroes is 0.x9 * 0.y0 * 0.z0 * 0.v2. Similarly n=2 n=3, until you get to 3.16 which allows 0.x5 0.y0 0.z0 and an even number. This looks the best bet for more zeroes at the end of the product. In the end the result is 5/4 * 3/2 * 6/5 * 4/9*$7.11 = 90/40 * 4/9 * $7.11 = $7.11.

Quote:ThatDonGuy

Let a, b, c, d be the amounts in cents

a + b + c + d = 711

(a / 100)(b / 100)(c / 100)(d / 100) = 7.11, so abcd = 711,000,000 = 79 x 2^6 x 3^2 x 5^6

Okay, this isn't a "math" answer, but is there an easier way to solve this than brute force?

I did a brute force search, and got one solution: 3.16, 1.50, 1.25, 1.20

link to original post

Quote:charliepatrickHere's an approach to reduce the number of options.

It gives the answer as 1.20 1.25 1.50 3.16.

Factor 711 gives 3 3 79. But you also need to have values with cents.

So of the four prices are v.vv x.xx y.yy z.zz then the sum of them is $7.11. However the product would have eight decimal places, whereby the last six are all zeroes (i.e. 7.11000000). So you need to add 10^{6}which gives six more 2s and 5s.

The way to split up the 3 3 79 amongst the four prices is

(a) 1 3 3 79

(b) 1 1 9 79

(c) 1 1 1 711 (but this gives a total that is too big, so can be ignored)

(d) 1 1 3 237

On these bases prices, you can allocate all the 2s and 5s (this is where the brute force can apply).

There are a number of ways (84) you can allocate six balls into four boxes, so you use this list for the 2s, and also for the 5s.

So look at all permutations of a/b/d 1-84 1-84.

This gives (offsets, base numbers, powers of 2, powers of 5),1 11 73 1 3 3 79 0 1 3 2 3 2 1 0 1.25 1.50 1.20 3.16

link to original post

Correct!!

Very good.

I don't, however, know who wrote this particular solve...

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

In looking for the 4 numbers that add and multiply to $7.11 I find that it is easy to come up with four numbers whose sum is correct, and it is much more difficult to think about numbers whose product is correct.

Thus, to find numbers that will multiply my first instinct is to factor the number. When I factored 7.11 I found that 3, 3, and 0.79 divide into it evenly. But 0.79 + 3.00 + 3.00 + 1.00 = 7.79. So I will have to manipulate the factors to accommodate for the decimals.

Decimals are nasty to work with, so I will convert everything into pennies. My sum is then multiplied by 100 to give me 711, and the product is multiplied by 100^4 to give me 711,000,000. This can be seen in the equations below:

100a + 100b + 100c + 100d = 100 (a + b + c + d) = 100 (7.11) = 711

100a * 100b * 100c * 100d = (100^4)(a * b * c * d) = (100^4)(7.11) = 711,000,000

Now when I factor 711,000,000 I can see that I am left with 79, 3, 3, 2, 2, 2, 2, 2, 2, 5, 5, 5, 5, 5, and 5. So now it's a matter of rearranging these factors so that their four products sum to 711.

I begin with 79 since it is the largest and will yield the fewest combinations. I can determine that one of my prices is a multiple of 79 cents as seen in the chart. I did not include 79 * 3 * 3 = 7.11 because then the other prices would be zero and the multiplication would fail.

From here forward the method to solve is based on combinatorics (combinations). To narrow my permutations, I can first use some logic.

At least one of the remaining numbers ends in something other than a 0 because none of the above end in 1. Thus, in order for the ones digit of our sum to be a 1, we need the digits to add to 1, 11, 21, 31. This means that at least one number will not contain a factor of 5.

For numbers 79, 158, 237, 474, and 632, this means that the six 5's would be split among two prices which would total at least 125 cents each. So I can rule out 474 and 632 as possible values because 125 + 125 = 250 and 474 +250 > 711 as well as 632 + 250 > 711.

It is at this point that I notice a pattern that gives me a hunch. My guess is that the value will be either 3.16 or 3.95. Why? Because these numbers allow for the six factors of 5 to be spread among three prices instead of two.

I begin with 3.16 so that I can examine the prices who all have factors of 5. So for my bases I have 316; 5*2; 5*2; and 5. This leaves me with the remaining factors of 3, 3, 2, 2, 5, 5, and 5. With addition limiting the maximum and minimum sizes of the prices, I can quickly come to find that the four prices are $3.16, $1.20, $1.50, and $1.25.

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

Welp, I'm on vacation.

Conventional wisdom is to rotate the tires on your car because they'll last longer. (Front tires wear faster than rear tires.) I was skeptical, thinking that rotating just distributes the wear, so over 120k miles, you're still looking at buying the same number of tires overall. But really, rotating does mean going through fewer tires.

Consumer Reports says typical tire life is:

20k for unrotated front tires

60k for unrotated rear tires

40k for tires that are rotated

(And indeed, 40k looks like the average of 20 and 60.)

No rotation for 120k miles

Buy 3 sets of tires (at 40k, 80k, 120k miles) = 12#

Rotation for 120k miles

Buy 6 pairs of front tires (at 20k, 40k, 60k, 80k, 100k, 120k miles) = 12#

Buy 2 pairs of rear tires (at 60k and 120k miles) = 4#

16# total

Skipping the calculations on the suggestion of aceside, the average for not rotating is 30k miles/tire, vs. 40k for rotated.

I understand that's the calculated result. I just can't wrap my head around why the calculated result isn't the same figure for both rotated and unrotated. I need a conceptual explanation to understand it, not a formula, since I can already calculate it, the result just doesn't make sense.

Quote:MichaelBluejay

Conventional wisdom is to rotate the tires on your car because they'll last longer. (Front tires wear faster than rear tires.) I was skeptical, thinking that rotating just distributes the wear, so over 120k miles, you're still looking at buying the same number of tires overall. But really, rotating does mean going through fewer tires.

link to original post

(clipped)

This overlooks the recommendation that new tires should go on the back and that tires have an expiration date in spite of remaining tread.

Tire age probably doesn't come into play unless you mostly don't use the car.

You also may not qualify for the usual buy 3 get 1 free pricing for only two tires.

Quote:MichaelBluejayI can't understand the concept / reasoning behind this and I'm hoping someone can give a dumbed-down explanation.

Conventional wisdom is to rotate the tires on your car because they'll last longer. (Front tires wear faster than rear tires.) I was skeptical, thinking that rotating just distributes the wear, so over 120k miles, you're still looking at buying the same number of tires overall. But really, rotating does mean going through fewer tires.

Consumer Reports says typical tire life is:

20k for unrotated front tires

60k for unrotated rear tires

40k for tires that are rotated

(And indeed, 40k looks like the average of 20 and 60.)

No rotation for 120k miles

Buy 3 sets of tires (at 40k, 80k, 120k miles) = 12#

Rotation for 120k miles

Buy 6 pairs of front tires (at 20k, 40k, 60k, 80k, 100k, 120k miles) = 12#

Buy 2 pairs of rear tires (at 60k and 120k miles) = 4#

16# total

Skipping the calculations on the suggestion of aceside, the average for not rotating is 30k miles/tire, vs. 40k for rotated.

I understand that's the calculated result. I just can't wrap my head around why the calculated result isn't the same figure for both rotated and unrotated. I need a conceptual explanation to understand it, not a formula, since I can already calculate it, the result just doesn't make sense.

link to original post

If we accept the premise that rotated tires last 40K miles, then I agree with your conclusion that over 120,000 miles you would buy 12 tires if you rotate and 16 if you don't.

However, I don't buy the 40K for rotated tires, assuming the 20K and 60K are correct. Following is my math on how long tires will last by rotating. To keep things simple, let's consider a motorcycle with the same tire longevity given.

Let's say tires start as 12 mm thick.

After 15K miles, the front tire will have 3mm left and the rear 9mm.

You then switch them, so the front has 9mm and the rear 3mm.

After another 15K, they both wear out completely at the same time.

So, rotated tires last on average 30K.

The general formula for how long rotated tires last is 2R/(1+R/F), where:

R = How long rear tires last

F = How long front tires last

where R>=F

I hope that helps.

Quote:WizardThe general formula for how long rotated tires last is 2R/(1+R/F), where:

R = How long rear tires last

F = How long front tires last

where R>=F

I hope that helps.

link to original post

I got to thinking an easier way to calculate the lifespan for rotated tires.

1/average(1/F,1/R)

In other words, the inverse of the average of the inverses.

In this case

1/[average(1/20000,1/60000)] = 1/(1/30000) = 30000

However, I think I figured it out: All the articles talk about "the" tread on the tires, like how to measure "the" tread depth, but there are actually FOUR treads on each tire. Some articles do point out that the sides wear faster than the middle, but none of them really explore the ramifications of that, with numbers. Once I take that into account, I do find a benefit for rotating, but it's quite modest (3.5%).

First, let me cite the sources I consulted before I ran my own numbers.

Sources:

• Consumer Reports. 20k front, 60k rear, 40k rotated

• Car and Driver. "Rotating your tires evens out the wear and makes them last longer." (No figures offered.)

• Tire Grades. "Rotating your tires can easily increase tire life 25%....some may get only 10%, though others could get 100%."

• How Stuff Works. "Will rotating make your tire last longer? Yes...A 60,000-mile tire run without rotation may last, say, 50,000 miles. But with rotation there is good chance that same tire could reach 60,000 miles."

Data from my own car

Tread depth (32nds of an inch) after 40k miles

Front: left • right 5.25 , 5.75, 6.25, 5.25 • 4.75 , 5.50, 5.50, 5.25

Rear left • right: 5.75 , 6.12, 6.00, 5.25 • 5.25 , 6.00, 5.75, 6.00

Tread decrease from new (=10)

Front: left • right -4.75 , -4.25, -3.75, -4.75 • -5.25 , -4.50, -4.50, -4.75

Rear left • right: -4.25 , -3.88, -4.00, -4.75 • -4.75 , -4.00, -4.25, -4.00

Miles per tread until tread=3 (new=10)

Front: left • right 59 , 66, 75, 59 • 53 , 62, 62, 59

Rear left • right: 66 , 72, 70, 59 • 59 , 70, 66, 70

Tire life, unrotated (lowest value for each tire from above)

Front: left • right 59 • 53

Rear left • right: 59 • 59

Tire life, rotating FrontRight & RearLeft

Front: left • right 59 • 59.5

Rear left • right: 59.5 • 59

So, I could get 3.5% more life from rotation. At $686 for a set of four tires including installation, tax, and fees, that's a savings of $24. There's some environmental benefit too, but not very much.

Does that seem right?

To five decimal places, the number of days in a year is 365.24217.

The current method to determine if a year is a leap year is that if a year is evenly divisible by 4, it is a leap year, with an exception for years evenly divisible by 100 are not, with an exception to the exception that years evenly divisible by 400 are. That comes to 97 leap years every 400 years, for 356.2425 days per year. Pretty close, off by 0.00033.

However, there is a way to be more accurate with a shorter cycle. Your task is to find an N and D, where N/D is as close as possible to 0.24217 and D < 400.

For extra credit, device a rule and an exception for whether a year is a leap year, to achieve the optimal fraction N/D above, using as small numbers as possible.

I used an online calculator to find this 3027/12500=0.24216

97/400=0.24250

61/250=0.244

The closest fraction I get is 89 / 367, which is off by 1 / 146,800.

A closer one is impossible - proof:

Let N / D be the closest solution.

If N / D < 97 / 400, then the difference is (97 D - 400 N) / 146,800

If N / D > 97 / 400, then the difference is (400 N - 97 D) / 146,800

400 x 89 - 97 x 367 = 1, so any closer fraction must be exact, but there cannot be any other exact fractions as 97 and 400 are relatively prime.

That said, I think Don's answer is right, but to the wrong goal.

Quote:WizardI think I had a typo in my question. The object is to get as close as possible to 0.24217.

That said, I think Don's answer is right, but to the wrong goal.

link to original post

Assuming 0.24217 is "exact," the closest fraction I get is 85 / 351

Quote:ThatDonGuy

Assuming 0.24217 is "exact," the closest fraction I get is 85 / 351

link to original post

I agree!

If a year is evenly divisible by 4, but not by 31, then it's a leap year.

If you bet $1 per flip on one million fair coin flips, what is the probability you win exactly $1000?

Closed-form approximations are acceptable, but they should get at least the first four digits correctly

To win $1000, there have to be exactly 500,500 heads and 499,500 tails

Since the probability of a head and the probability of a tail are both 1/2, this is C(1,000,000, 499,500) / 2^1,000,000.

This overflows pretty much everything I have, but rewriting it as:

C(1,000,000, 499,500) / 2^1,000,000

= (1,000,000 x 999,999 x ... x 500,501) / (499,500 x 499,499 x ... x 1 x 2^1,000,000)

= 10^(log 1,000,000 + log 999,999 + ... + log 500,501 - log 499,500 - log 499,499 - ... - log 1 - 1,000,000 log 2)

which is 0.00048394, or about 1 / 2066.3653.

Since you could not (feasibly) post the entire equation, I don’t consider that a closed-form solutionQuote:ThatDonGuy

To win $1000, there have to be exactly 500,500 heads and 499,500 tails

Since the probability of a head and the probability of a tail are both 1/2, this is C(1,000,000, 499,500) / 2^1,000,000.

This overflows pretty much everything I have, but rewriting it as:

C(1,000,000, 499,500) / 2^1,000,000

= (1,000,000 x 999,999 x ... x 500,501) / (499,500 x 499,499 x ... x 1 x 2^1,000,000)

= 10^(log 1,000,000 + log 999,999 + ... + log 500,501 - log 499,500 - log 499,499 - ... - log 1 - 1,000,000 log 2)

which is 0.00048394, or about 1 / 2066.3653.

link to original post

Quote:Ace2Since you could not (feasibly) post the entire equation, I don’t consider that a closed-form solution

link to original post

I was going to say the same thing Don did. My WizCalc can handle the numbers and gives an answer to ...

I'll have to work on getting a tighter form of an expression.

Quote:Ace2It’s “Too Easy Tuesday”

If you bet $1 per flip on one million fair coin flips, what is the probability you win exactly $1000?

Closed-form approximations are acceptable, but they should get at least the first four digits correctly

link to original post

Building on Don's work and using Stirling's approximation for the factorial function, I get a closed form approximation of:

1 / [ sqrt( 499999.5 * pi * e) ]

This is about 0.00048394, which matches Don's and Wizard's answers to five places.

You have an unopened 3-liter bag of win.

Wine drains out of the bag at a rate in proportion to the wine still in the bag.

Based on your last (identical) bag, wine poured at a rate of one centiliter per second when there was one liter left in the bag.

How many seconds will it take for your 3-liter bag of wine to drain to one milliliter of wine?

Quote:WizardNot to interrupt the discussion on the last problem, but here is the next one in line.

You have an unopened 3-liter bag of win.

Wine drains out of the bag at a rate in proportion to the wine still in the bag.

Based on your last (identical) bag, wine poured at a rate of one centiliter per second when there was one liter left in the bag.

How many seconds will it take for your 3-liter bag of wine to drain to one milliliter of wine?

link to original post

Wiz,

I get

Since the rate of loss is proportional to the wine in the bag, we know:

dw/dt = -k*w

From the data, when w = 1000 ml, the rate of loss is 10 ml/s, so -10 ml/s = -k*(1000 ml), which gives a value of k = 0.01 1/s.

Integrating the ODE gives

ln(w/w0) = -k*t

So t = ln(w/w0)/(-k)

Plugging in the numbers with the boundary condition that w0 = 3000 ml at t = 0 gives

t = ln((1 ml)/(3000 ml))/(-0.01 1/s) = 800.6367... seconds.

Dog Hand

Quote:DogHandWiz,

I getLet w = wine in bag in milliliters, and let k = rate constant for drainage in 1/s.

Since the rate of loss is proportional to the wine in the bag, we know:

dw/dt = -k*w

From the data, when w = 1000 ml, the rate of loss is 10 ml/s, so -10 ml/s = -k*(1000 ml), which gives a value of k = 0.01 1/s.

Integrating the ODE gives

ln(w/w0) = -k*t

So t = ln(w/w0)/(-k)

Plugging in the numbers with the boundary condition that w0 = 3000 ml at t = 0 gives

t = ln((1 ml)/(3000 ml))/(-0.01 1/s) = 800.6367... seconds.

Dog Hand

link to original post

I agree! This one is worthy of a glass of cheap boxed wine.

I’ll accept that answerQuote:ChesterDogQuote:Ace2It’s “Too Easy Tuesday”

If you bet $1 per flip on one million fair coin flips, what is the probability you win exactly $1000?

Closed-form approximations are acceptable, but they should get at least the first four digits correctly

link to original post

Building on Don's work and using Stirling's approximation for the factorial function, I get a closed form approximation of:

1 / [ sqrt( 499999.5 * pi * e) ]

This is about 0.00048394, which matches Don's and Wizard's answers to five places.

link to original post

For one million flips, the expectation is to win 500,000 with a SD of 500. To get the probability of exactly 500 wins, take the height of the normal curve at 1 SD then multiply by the width of 1/500. That’s:

(2πe)^-.5 / 500 =~ 0.00048394145

You can also apply Stirling’s formula to the combination c(1000000,500500) and use logs to evaluate:

1000000^1000000.5 * 499500^-499500.5 * 500500^-500500.5 * (2π)^-.5 * 2^-1000000

which yields 0.00048394165. If you used Stirling, I don’t understand how there is an e in your final answer since the e’s cancel out

Quote:Ace2Since you could not (feasibly) post the entire equation, I don’t consider that a closed-form solution

In that case, the exact answer is...

...too big to fit - even the denominator itself is too large for the board to handle; it's 301,026 digits.

Express 1,000,000 as the product of two numbers, neither of which contains any zeroes.

If I roll three dice and multiply the three resulting numbers together, what is the probability that the product will be odd?

White mates in three.

If either number has both a 2 and a 5 as a prime factor, then it is a multiple of 10, so it has a zero.

The solution is 2^6 x 5^6, or 64 x 15,625.

The product is odd if and only if all three dice have odd numbers. Since the probability that a particular die has an odd number is 1/2, the solution is 1/2 x 1/2 x 1/2 = 1/8.

1.Kc6 / Ka7

2.c8=Q / Ka6

3. Qa8#

Quote:Wizard2^6 * 5^6 = 15625 * 6450%. Contrary to what Don said, two evens and an odd produce an odd.

link to original post

...it was asking for the sum instead of the product

If it was the sum, it would be 50% regardless of how many dice. No matter what the even/odd "parity" of N-1 dice are, the Nth die changes the parity if it is odd, and keeps it the same if it is even.

^{6}=64 and 5

^{6}=15625.

(1) K=c6 k=a7 (the Black king cannot drift into column b)

(2) P=c8(R) k=a6 (note if the pawn became a Queen, then it would be stalemate as per below)

- - Q (what happens if the pawn was promoted to Queen, Black King cannot move to a6)

k - -

- - K

(3) R=a8##

R - -

- - -

k - K

Note - there may be a "feature" as I tried [ SPOILER = 10 [ sup ] 6 [ /sup ] ] and it didn't want brackets round the "/sup". So hope it comes out fine!

Quote:ThatDonGuy

...it was asking for the sum instead of the product

link to original post

D'oh! You're right. What is my punishment?

You have 100 secretarial candidates. You interview them one at a time. After each interview, you must either offer the candidate the job immediately or dismiss them, with no option to offer them the job later. Your goal is to hire the best secretary of the 100. What should be your strategy?

It's well known the strategy is to interview the first 100/e candidates and then hire the next one to come along better than the best of the first 100/e. The chances of picking the best is approaches 1/e as the candidate pool approaches infinity.

However, what if your strategy is to maximize the quality of whoever you hire? You still are strained by the rule about hiring someone on the spot.

To make it easy, assume every candidate has a number of points uniformly distributed from 0 to 100. What is the optimal strategy and how many expected points of the hired secretary does it produce?

When you're looking at the 99th choice you can either pick it "stand" or go on "hit" and then you have to choose the 100th. If you "hit" then your expectation is 50 (i.e. average of 0 and 100). Thus you "stand" if the one you're looking at is greater than 50. The average of this is 75 (average of 50 and 100).

When you're looking at the 98th choice, similar logic applies. If you "hit" then half the time you'll have to "hit" again (Pr =.5, EV=50) and half the time you'll "stand" (Pr=.5, EV=75); so the EV of moving onto the 99th choice is 62.5. Thus in this case you only "Stand" if the 98th is greater than 62.5.

You repeat this logic for the 97th choice using "hit" (Pr=.625, EV=62.5) "stand" (Pr=.375, EV=81.25).

Continuing this gives the 98.121 figure.

Quote:WizardTo make it easy, assume every candidate has a number of points uniformly distributed from 0 to 100. What is the optimal strategy and how many expected points of the hired secretary does it produce?

link to original post

Questions:

1. By "uniformly distributed," does this mean that each one's points is a uniform random number from 0 to 100, or that 100 of the numbers {0, 1, 2, ..., 99, 100} are distributed equally randomly among the 100 secretaries?

2. Are points necessarily integers?

Quote:charliepatrickThe only way I can see to solve the problem is via a spreadsheet to get an answer about 98.121. But here is the logic I used.

When you're looking at the 99th choice you can either pick it "stand" or go on "hit" and then you have to choose the 100th. If you "hit" then your expectation is 50 (i.e. average of 0 and 100). Thus you "stand" if the one you're looking at is greater than 50. The average of this is 75 (average of 50 and 100).

When you're looking at the 98th choice, similar logic applies. If you "hit" then half the time you'll have to "hit" again (Pr =.5, EV=50) and half the time you'll "stand" (Pr=.5, EV=75); so the EV of moving onto the 99th choice is 62.5. Thus in this case you only "Stand" if the 98th is greater than 62.5.

You repeat this logic for the 97th choice using "hit" (Pr=.625, EV=62.5) "stand" (Pr=.375, EV=81.25).

Continuing this gives the 98.121 figure.

link to original post

I agree.

Quote:GialmereIt's the first easy Monday off the new year. Here is a trio of tiny teasers to contemplate through your hangover haze...

Express 1,000,000 as the product of two numbers, neither of which contains any zeroes.

Also 142857.142857142857... x 7

:-)

Dog Hand

Quote:ThatDonGuy

If either number has both a 2 and a 5 as a prime factor, then it is a multiple of 10, so it has a zero.

The solution is 2^6 x 5^6, or 64 x 15,625.

The product is odd if and only if all three dice have odd numbers. Since the probability that a particular die has an odd number is 1/2, the solution is 1/2 x 1/2 x 1/2 = 1/8.

link to original post

Quote:Wizard2^6 * 5^6 = 15625 * 64

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Quote:charliepatrickAny power of 10 has factors of 2 and 5. So the two numbers share those powers. If any of the 2s were with the 5s then the product would end in a zero. Hence the two numbers are ine is a power of 2 and the other the same power but of 5. Thus the numbers are 2^{6}=64 and 5^{6}=15625.Similar to previous puzzle, none of the three numbers can have a factor of 2. Hence they all have to be odd. The chances of this is 1/2*1/2*1/2=1/8.The important thing here is to recognise that a Rook can create a mate. Thus

(1) K=c6 k=a7 (the Black king cannot drift into column b)

(2) P=c8(R) k=a6 (note if the pawn became a Queen, then it would be stalemate as per below)- - Q (what happens if the pawn was promoted to Queen, Black King cannot move to a6)

k - -

- - K

(3) R=a8##R - -

- - -

k - K

Note - there may be a "feature" as I tried [ SPOILER = 10 [ sup ] 6 [ /sup ] ] and it didn't want brackets round the "/sup". So hope it comes out fine!

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Correct!!

Good show.

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Quote:DogHand333,333.333... x 3

Also 142857.142857142857... x 7

:-)

Dog Hand

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* Be accepting of paradoxes

* Use the binary number system more often

The Tumble Dryer

What fraction of the big square is shaded?

(x+2x)^2 + x^2=1^2.

Solving this we get x=1/sqrt(10).

The area of the four small squares is

4x (2x)^2;

The area of the big square is

2^2=4.

The ratio of the shaded area is 0.4.

Quote:acesideSet the circle radius=1; set a half of the small square side length=x, then we have this equation,

(x+2x)^2 + x^2=1^2.

Solving this we get x=1/sqrt(10).

The area of the four small squares is

4x (2x)^2;

The area of the big square is

2^2=4.

The ratio of the shaded area is 0.4.

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Correct!!

Very good.

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Each die of a pair of non-identical dice has six faces, but some numbers are missing, others are duplicated, and some faces may have more than six pips. No face is blank.

The dice can roll every number from 2 to 12.

What is the largest possible probability of rolling a 7?