dgreening
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Posts posted by dgreening
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Dig along the green lines
1 + 1 + √(0,52 + 0,52) = 2,7071
I think maybe Sam could dig even less, following the pink lines BMD + MC + AO (green), but I don't know, how to find point M.
It would be minimum for M chosen at a distance of (3 -root(3)) / 6 = 0.211 from both sides
This gives total length of digging as 2.64
very elegant solution(s)! My hat is off to both of you!
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I think you are both on the right track.
either approach will reveal one of the "constant" scales [normal or reverse] as the scale that is different from the others. This will take 1 weighing of either combination described previously [1 coin vs no coins or 1 coins vs 7 coins]. [at this point you have weighed the items once on each scale]
You no longer have to do any more weighing on the scale that is different on the first weighing.
It will take at least 1 more weighing to determine which of the remaining scales is random.
you could simply reverse the loads [or re-weigh them on the same sides] and weigh them again and again.
when the results diverge, you will have determined which one is which.
The hard part is to try to estimate how long the random scale and the remaining "constant" scale track each other.
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Hi
I have been reading this site for sometime, but only recently started actually posting anything.
I enjoy puzzles [crossword, KenKen and logic are my current faves], world travel and good novels audio or visual.
Looking forward to more great puzzlers!
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After all of those calculations, I decided to run some simulations on the 3 strategies put forward:
- 9 9 9 [ stay on 9 or better for each of the first 3 cards
- 9 8 7 [stay for 9 on the first card, 8 on the second and 7 on the third]; and
- 11 10 7 [described above]
When I ginned this up on the spread sheet, I entered the limits incorrectly for " 11 10 7" and actually entered "10 9 6".
Imagine my surprise when I found that "10 9 6" out performed all 3 of the other strategies.
I ran 100k trials and captured the cumulative results at 50k and 100k.
The column below the strategy name is the average pay off for that strategy
"Best" is the number of trials in which this strategy out performed the other.
Results
"9 9 9 " Best "9 8 7" Best "10 9 6" Best "11 10 7 " Best Trials 0.058 1,205 0.054 655 0.097 509 0.071 3,082 50k 0.069 2,422 0.070 1,435 0.103 1,015 0.080 6,153 100kIt is interesting that "11 10 7" had a lot more bests that any others, but did not have the highest payoff.
I will be glad to share the spread sheet with you, if you want to look it over or dabble with it.
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I went back and looked at the numbers a little differently.
If you have 1 card, on average what card [or higher] should you expect if you draw 3 more cards?
One way to look at this is to choose a card and see if it's probability is more than 0.5 for 3 more card draws.
Since we are looking at a series of cards, this is similar to the problem hitting a target with a dart. If the P [hit] = 0.6, then P [miss] = 0.4. If you throw 2 darts, the P [hit 2 darts] = 1 - P[miss (1 dart)]*P[miss (1 dart)] = 1- [0.4 * 0.4] = 1 - 0.16 = 0.84
For the shady game a similar calculation can be used.
For the first card
What is the likelihood that 10 or better will come up in the next 3 cards dealt?
P[ 10 thru 13 (1 card)] = 4/13 = 1 - P[1 thru 9 (1 card)]
thus P[1 thru 9 (1 card)] = 9/13
P[10 thru 13 (3 cards)] = 1 - {P[1 thru 9 (1 card)]}3 = 1 - [9/13]3 = 1- (0.69)3 = 1-0.332 = 0.668
So well more than half the time, you should expect to get a 10 or higher in the next 3 cards.
What about 11 or higher?
P[11 thru 13 (3 cards)] = 1 - {P[1 thru 10 (1 card)]}3 = 1 - [10/13]3 = 1- (0.77)3 = 1-0.455 = 0.545
So more than half the time, you should expect to get a 11 or higher in the next 3 cards [indicating that you should draw on 10]
The same calculation will yield P[12 thru 13 (3 cards)] = .0394 well below .5
Indicating that you should draw on 10 and hold on 11 for the first card.
What value of X will reach 50%?
0.5 = X3/133
X3 = 0.5*[133] = 2197/2 = 1098.5 thus X = 10.318
This would imply that for the first card you draw if the card showing is less than 10.3 and hold if the card is greater than 10.3. For the first card hold 11, draw 10
For the second card
Y2 = 0.5*[132] = 169/2 = 84.5 thus Y = 9.19
Thus for the second card: hold 10 draw 9
For the 3rd card
Z = 13/5 = 6.5
For the 3rd card hold 7 draw 6
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Winning strategy is to keep the card out of the first 3 as soon as you see a 9 or higher card.
Surely your strategy is EV+, but clearly it is not optimal.
For example: after discarding two Deuces your third card turns out to be an Eight. EV of discarding it approximately equals (13+1)/2 = 7, while keeping gives 8, so it's better to keep it, but your strategy says to discard. You've picked a global threshold for each decision, but to make strategy optimal three separate thresholds should be found (for each out of 3 possible decisions of keeping/discarding).
Full disclosure: I read this a few nights ago and started thinking about it [eventually falling asleep], I have not done any detailed calculations,, but I did have some thoughts:
I think there may be a strategy that changes your decision criteria as you go along in the game; a series of thresholds that diminish as the game goes along
If [for example] you have drawn 3 cards:
- if the 3rd card is 8 or higher - stay [your probability of drawing a lower card is more than 50%]
- if the 3rd card is 6 or lower - draw [your probability of drawing a higher card is more than 50%]
- if the 3rd card is a 7 - do you feel lucky?
If you draw the first card, there is some value X [guessing around 9 or 10] that you are probably better off to draw another card [because out of 3 new cards, you should be able to improve your holding. So the strategy for the first card may be something like this:
- If your 1st card is J, Q or K - definitely stay
- If your 1st card is below X - draw [with 3 more cards, you can probably do better]
- If you value is above X - stay
There will be a similar value Y [guessing 7 or 8] that you should use a threshold for the 2nd card.
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I think witzar is close
if the boy has an even chance of arriving at any minute between 7:55 and 8:02, then every minute has a 1/8 probability
if the bus arrival is also evenly distributed between 7:58 and 8:02, then every minute has a 1/5 probability
If the boy arrives at
7:55 the probability of catching the bus is 1.0
7:56 the probability of catching the bus is 1.0
7:57 the probability of catching the bus is 1.0
7:58 the probability of catching the bus is 1.0
7:59 the probability of catching the bus is 4/5
8:00 the probability of catching the bus is 3/5
8:01 the probability of catching the bus is 2/5
8:02 the probability of catching the bus is 1/5
so the probability that he catches the bus is
1/8* [1+1+1+1+ 0.8 + 0.6 + 0.4 + 0.2] = 1/8 * 6 = 3/4 = 75%
The OP does not constrain the boy's arrival to be on the minute.
Between 7:58 and 7:59, for example, p decreases linearly from 1 to 0.8; it does not drop abruptly to 0.8 at 7:58.
Similarly for the next two minute-intervals.
The boy does not arrive later than 8:01.
Make a graph of p vs boy's arrival time and find its average value.
good point!
Now that I re-read the problem I see your point.
In fact the boy does not necessarily arrive on the minute and the bus does not necessarily arrive on the minute either.
I think your approach is better.
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Not sure if it is the most optimum: Dig along the diagonals
The total length dug is 2*root(2)
I agree, we were working in parallel.
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One approach is to dig an "X" diagonally across the square [e.g, a line from the "Top Left" corner to "Bottom Right" corner and another line from the "Top Right" corner to Bottom Left" corner ].
Note: arbitrary reference system just for explanation
The length of each line would be ~1.414 units long.
The total length of the digging would be ~ 2.828 units long [shorter than digging 3 sides]
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I asked some others and after talking it over, I realized that we cannot mix m [distance] with m3 of volume.
The m/ m3 is actually shorthand for m / m3 of fuel.
Dividing through would leave the expression 1/ m2 of fuel.
Bottom line: I think this was a trick question.
I am going to forget about this one.
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Good one
12 000
+1 200
+ 12
= 13,212
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I think witzar is close
if the boy has an even chance of arriving at any minute between 7:55 and 8:02, then every minute has a 1/8 probability
if the bus arrival is also evenly distributed between 7:58 and 8:02, then every minute has a 1/5 probability
If the boy arrives at
7:55 the probability of catching the bus is 1.0
7:56 the probability of catching the bus is 1.0
7:57 the probability of catching the bus is 1.0
7:58 the probability of catching the bus is 1.0
7:59 the probability of catching the bus is 4/5
8:00 the probability of catching the bus is 3/5
8:01 the probability of catching the bus is 2/5
8:02 the probability of catching the bus is 1/5
so the probability that he catches the bus is
1/8* [1+1+1+1+ 0.8 + 0.6 + 0.4 + 0.2] = 1/8 * 6 = 3/4 = 75%
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I think DeGe's strategy is sound.
The term "toggle" may be adding confusion.
Prisoner 1 always turns the light OFF [or leaves it off]
Every other prisoner is allowed to turn the light ON only once [after they turn it on once, the leave it unchanged]. Other prisoners are not allowed to ever turn the light off.
after day 1 prisoner 1 counts the times he turns it off and after 99 he can safely assume that all have been in the room
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not sure what to make of
"Well, I think that you have to add units to the numerator (as well as the denominator) to make any sense of things.
in your example, 25 miles/gal = 118,400,000/m2=118,400,000 parsecs/parsecs-m2. Now that makes physical sense!"
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when I first started mulling this over, I realized that miles/ gal or km/ liter are skipping several steps.
We really mean that gal of gas is sufficient to propel this vehicle so many miles.
I took a typical value for a US car [25 mpg] and started converting it.
Values are approx
25 miles ~ 32 km
1 Gal ~ 3.7 Liters ~ 3.7/1000 Meters 3
25 Mile/ gal = 32 km/ 0.0037 meters 3 ~ 32,000/ 0.0037 /meters 2
therefore 25 miles/ gal ~ 118,400,000 /meter 2
I am at a loss for how this related to the parallel piped, though I cannot find anything else that makes sense.
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I think the smallest square would be a 3 x 3.
If we start with the most compact area to contain 4 unit squares, it would be a 2 x 2 square. The addition of another unit square will force it to add at least one row or column - thus forcing you to a 2x3 configuration. To meet the criteria of the problem, we must expand that to a 3 x 3
There are multiple configurations [many are just reflections of others
# # #
# # -
- - -
or
# # #
# - -
# - -
or
# - #
- # -
# - #
or ...
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We also do not want the Dr to come into contact with the blood from any of the 3 patients
One approach
- Use pair 1 on patient A [turn the gloves inside out as you remove them]
- Use pair 2 on patient B
- Place the "used side" or pair 1 over pair 2 [used side to used side] - and operate on patient C
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If an even number of pages are removed the sum of the missing page numbers [9808] divided by the number of pages missing would have a fractional portion of 0.5
If an odd number of pages are removed the sum of the missing page numbers [9808] divided by the number of pages missing would have a fractional portion of 0.
A trivial solution exist for the removal of a single page [number 9808]. When divided by 1, the result is 9808.
If we ignore this [the problem states that "pages" are missing].
If we define N as the number of pages missing then we must test the possible number of missing pages using the following:
Frac [9808/ N] =
- 0.5 where N is an even number; or
- 0.0 where N is an off number
testing for up to 100 pages missing there is only 1 value of N that satisfies this criteria
For N = 32
9808/ 32 = 306.5
This tells us that 32 pages are missing and the average value of these page numbers is 306.5
Taking 16 integers values below and 16 integer values above 306.5 gives us a range of
pages 291 thru 322
The sum of these pages = 9808
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X = miles at 55 mph
X+20 = miles driven at 40 mph
and
55 mi/ hr = 0.917 mi/ min
40 mi/ hr = 0.667 mi/ min
D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph)
T = Total Time = T1 (at 55 mph) + D2 (at 40 mph)
D1 = T1 ( 0.917 mi/ min)
D2 = T2 ( 0.667 mi/ min)
Therefore
D = D1 + D2 = T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min)
We also know that D1 +20 = D2
These can be rearranged into 2 simultaneous equations in the form
20
I hit the wrong key
20 = - T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min) [Eqn 1]
100 = T1 +T2 [Eqn 2]
Multiplying all terms in Eqn 1 by {- 1/0.667} yeilds
-30 = 1.375 T1 - T2
Adding this to Eqn2 results in
70 = 2.375 T1 Therefore
T1 = 70 /2.375 = 29.47 min, so
D1 = 29.47 min * ( 0.917 mi/ min) = 27.02
D2 = D1 + 20 = 47.02
Therefore Total distance = 74.04 Miles
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X = miles at 55 mph
X+20 = miles driven at 40 mph
and
55 mi/ hr = 0.917 mi/ min
40 mi/ hr = 0.667 mi/ min
D = Total Distance = D1 (at 55 mph) + D2 (at 40 mph)
T = Total Time = T1 (at 55 mph) + D2 (at 40 mph)
D1 = T1 ( 0.917 mi/ min)
D2 = T2 ( 0.667 mi/ min)
Therefore
D = D1 + D2 = T1 ( 0.917 mi/ min) + T2 ( 0.667 mi/ min)
We also know that D1 +20 = D2
These can be rearranged into 2 simultaneous equations in the form
20
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each person rank orders his choice.
These translate into votes [4 for my first choice, 3 for my second and 1 for my third]
The votes are tallied and the one with the most votes is the choice.
There is still a chance of getting a tie.
So the person ______________ [random variable] will have 3.1 votes for his second choice.
This random variable could be any of the following:
- first to reply
- look at the seconds on the time that the e-mail is posted and the one that is closest to the average of the 3 e-mails,
- the one who's email posting time [last digit only] is closest to the closing DJIE for the day. or
- .....
This approach encourages everyone to think seriously about there 2nd choice.
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The problem statement says
- "His truck can only hold 1000 apples."
the update says that the store is 1000 miles away
If the driver starts with a full load [1000 apples] and drives 1000 miles, he will have 0 apples left upon arrival.
Thus, unless he can ferry apples to some point(s) in between and then move these forward, there is no way to every deliver even 1 apple.
If we assume that there is a point halfway [500 miles] then he could do the following:
- leave the warehouse with 1000 apples
- drive 500 miles [now has 500 apples]
- leave the apples at the halfway point [now has 500 apples]
- drive back to the ware house
- leave the warehouse with 1000 apples
- drive 500 miles [now has 500 apples]
- Picks up 500 the apples at the halfway point [now has 1000 apples in the truck and 0 apples at the halfway point]]
- Drives the remaining 500 miles to the store and delivers 500 apples
Thus for 2 trips[500 and 1000 miles one way (3000 miles total round trip)] 500 apples can be delivered
to deliver 3000 apples would require:
- an interim storage area at 500 miles
- 6 "short" round trips between the warehouse and the interim storage area [1000 miles R/T each]
- 6 "long" round trips from the warehouse [stopping at the interim storage area to reload] to the store [2000 miles R/T each]
Notes:
- you can break this up many ways
- it is not clear what the optimal interim storage area distance [from the ware house] is [and if more than 1 helps]
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jhawk has it right
Weighing Champ
in New Logic/Math Puzzles
Posted
I am not sure if this satisfies the requirements about not impacting the eco-system.