k-man
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Spoiler
Here are my thoughts...
How can the Angel win? Since the Angel can only turn the lanterns OFF then the only way for him to win is to get all the lanterns to be OFF. He can never win with all lanterns being ON because that would require help from the Devil to turn ON the final lantern and we can't count on that.
So, the last winning move by the Angel must be to turn OFF the final lantern. How does the angel get to the scenario when all the lanterns are OFF except one? The previous lantern must have been also ON and the Angel must turn it OFF. Because if the previous lantern was already OFF then the Devil would see that he is about to lose and he would turn it ON! The same reasoning follows for the lantern before that and so on... So, for the Angel to win there must be all three conditions:
- a continuous sequence of lit lanterns that he can turn off
- all the other lanterns must be already off
- they must be in front of the first lit lantern in that sequence
So, the Devil's strategy is to prevent this from happening and it's easy to do. All the Devil needs to do is to ensure that there is always at least 2 lit lanterns and not next to each other. If they go through a lit sequence and the Angel turns them all off just turn on the next lantern and continue to the next lit sequence. If the next sequence got turned off by the Angel then again just turn on the following lantern. This can continue for a while, but since the number of states is finite the Devil will eventually win.
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I understood the OP that you can only remove X coins from ONE chest and deduce the contents of all chests by observing removed coins. What's the lowest X?
SpoilerYou need to remove 11 coins from chest C.
If they are all Gold then A = 40S, B = 10G+10S, C = 50G
If they are all Silver then A = 50G, B = 10G+10S, C = 40S
If they are a mix then A = 50G, B = 40S, C = 10G+10S-
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Highlight text below to see my theory
While the average wait time should be 15 minutes, there are longer and shorter wait times during the day. A random placement of a point in time is more likely to land on a longer wait time period. For example, if we look at a 1 hour period that has two 10-minute waiting intervals and two 20-minute waiting intervals, then an average wait time is 15 minutes, but a random point in time selected within that hour is twice more likely to be within the 20-minute interval.
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55 minutes.
Explanation:
Let's say his wife takes X minutes to drive from home to the train station. It also takes her X minutes to drive back. So, every day (including that one day) she leaves home X minutes before 5 o'clock to meet her husband and they arrive home X minutes after 5. That day they arrived home X-10 minutes after five, so the wife drove 5 minutes less in each direction. She drove X-5 minutes until she met with her husband, which puts the time of their meeting at 4:55 pm. He started walking at 4 pm, so he walked for 55 minutes.
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...or more in general...
Let y=f(x)
Solve f(x)x=xf(x) for x. There are unlimited number of y=f(x) for which a real solution exists for f(x)x=xf(x)
For example, let y=2x. Solving (2x)x=x2x results in x=2. Therefore, y=2x=4. 2^4=4^2.
Another example is what CaptainEd found above
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I'm getting the answer of about 47.32%
Here is how I'm getting it.
Standard deviation (SD) of the entire population [0-100] is approximately 29.3. We take a sample of size n=32 from the population and calculate the mean. The standard error will be about SE=SD/sqrt(n) ~ 5.18. This means that the mean of a sample will follow a normal distribution with the mean of 50 and standard deviation of ~5.18. Half of sample means will be above 50 and half will be below. To be in the 40s the sample mean needs to be above the 40 mark. Z score of the 40 mark is -1.93 and the p-value is 0.0268, which means that there is 2.68% chance that the mean of a sample will be below 40 resulting in the answer of .5 - .0268 = .4732
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New York. And they all work as cab drivers.
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I built the probability tree using PowerPoint. It's not freeware, but it's easy to learn and use. If you don't have PowerPoint you can try Google Docs, but I don't know if they have the same widgets that PowerPoint has or not.
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I've been following this thread without comment as there wasn't anything I could contribute (didn't have the aha! moment like harey did). I'm convinced that harey's approach is correct. Probably the easiest way to see it is to draw the probability tree of complementary events (see below).
The green track leads to the complete annihilation with total probability established earlier, the red track is the "no collision" track with total probability of 1/n! and the yellow tracks are all the partial annihilation tracks.
This tree also explains jasen's finding that the most likely outcome is that 2 bullets remain. Half the leaf nodes of this tree will have that outcome and the total probability will be the sum of those leaf node's probabilities.
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cheating...
1011010000 = 720 in binary
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My thoughts...
Maybe.
But I think I know what BMAD is growing...
Cannabis
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I would agree with jasen's answer. I think you're reading too much into details that are merely artifacts of the quality of the test. In this second puzzle there are 9 ways to combine 2 sticks of 3 types. 8 of them are given and the 9th is missing.
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Well, I don't have the context that you have as I don't see the rest of the questions, but looking at this question alone I wouldn't call it particularly complex and my solution/answer seems reasonable to me. BTW, I've never seen a test where two of the possible answers were exactly the same as in this test.
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I think the answer is...
the leftmost image in the second row - the triangle.
Here are my reasons:
1) The first two rows contain each one square, one rectangle and one triangle, so to complete the pattern the third row should have some sort of a triangle. This is consistent with triangles dominating the answer space. (As a side note, two images in the middle of the top row seem to be identical. Also first and third image from the left in the bottom row are almost identical, but the leftmost image seems to match the size of the other triangles in the puzzle, so would be preferred to the other one.)
2) The square, rectangle and triangle can either
- have a circle around them
- have circle arcs within them
- have no circle arcs
The only triangle that fits that pattern is the leftmost in the bottom row.
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It depends on how good you are with your eyes open.
Let P be the probability of scoring with your eyes open, then P/5 is the probability of scoring with your eyes closed.
1-P = probability of missing one shot with eyes open. (1-P/5)6 = probability of missing all 6 shots with eyes closed.
These probabilities are equal when P is approximately 0.367403, so if you're scoring better than 36.7% with your eyes open, go for a single shot. If you're worse than that then 6 shots with your eyes closed is a better choice.
P.S. Calculation results are courtesy of WolframAlpha http://www.wolframalpha.com/input/?i=1-p%3D(1-p%2F5)^6%2C+0<p<1.
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My attempt...
The answer is M+1 choose N.
The way to explain this logically seems to be easier if we reverse the question (at least it's easier for me). What is the difference (M+1 choose N) - (M choose N)? In other words the question is: How many combinations are introduced by choosing the same number of objects from a set that's larger by 1 object. The answer to this question is (M choose N-1) and here is why.
Say that we start with M white numbered balls and we choose N balls randomly. There are (M choose N) different results that we can have. Let's add a black ball and see how many new results we can have if we choose N balls from the new set that now includes the black ball. We have all the original results that don't include the black ball (M choose N) plus new results that include the black ball. Those that include the black ball have N-1 white balls and the number of those results is (M choose N-1).
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Since no one has attempted to answer this yet...
The ladder can be of any length up to approx. 16.89 feet.
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Given that John and Julie play on the same level and have 50/50 chance of winning any particular game, I agree with Bonanova that Julie has 50% chance of winning whatever series they choose to play. However, I think the OP is asking about the probability of Julie winning with the score 4:3 in the 7 game series vs. the probability of winning 5:4 in the 9 game series. Those probabilities are different...
The probability that the outcome of a 9 game series will be 5:4 is 63/256, but the probability that the outcome of a 7 game series will be 4:3 is 35/128 and is higher, so Julie is more likely to win 4 games out of 7 than 5 games out of 9.
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Assuming that all heavy halves are identical (and therefore all light halves are identical too), so we can pair any heavy half with any light half to create a sphere.
Using the balance scale it can be done with 5 tries. You need to distinguish from 8C4=70 possible distributions (HHHHLLLL, HLHLHLHL, etc.), but we don't always have to identify every H and L. As soon as we found a HL pair we can make a sphere, so it may be possible to do it in 4 tries, but I seriously doubt it can be done in 3.
Using the digital scale, it can be done in 3 tries. Here is how...
Take any 6 halves and weigh them on the digital scale. Based on the result you can distinguish from 3 possible heavy to light (H:L) ratios - 4:2, 3:3, and 2:4. Note that 4:2 and 2:4 cases are symmetrical, so I will only cover 4:2 case as the other can be resolved the same way.
4:2 caseThe remaining 2 halves that we didn't weigh are light, so put them aside. Take any 4 of the initial 6 and weigh them on the digital scale. Possible results are 4:0, 3:1, and 2:2.
-- 4:0 case is trivial as we've identified all heavy and all light halves and can put them together.
-- In 3:1 case we know that the 2 halves we didn't weigh the second time make up a sphere, so put them together. Weigh any 2 from 3:1 group to complete the remaining 3 spheres.
-- In 2:2 case we've already identified 2 heavy and 2 light halves, so we can put together 2 spheres. Weigh any 2 of the remaining 4 to complete the remaining 2 spheres.
3:3 caseThe remaining 2 halves that we didn't weigh make up a sphere, so put them together. Take any 4 of the 6 and weigh them on the digital scale. Possible results are 3:1, 2:2 and 1:3. Again, 3:1 and 1:3 are symmetrical, so we don't need to review both of them.
-- In 3:1 case we've identified 2 light halves, so one more weighing of 2 will identify the third.
-- In 2:2 case we can construct another sphere, so it's the same as the 2:2 case above.Solved in 3 weighings.
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The first image
The hex converted to ASCII translates into "Sjö sjö tveir einn fimm sjö núll", which in turn means "7 7 2 1 5 7 0" in Islandic
No idea what the other 2 images mean.
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Looks like after the site update I can now upload Excel files, so here you go...
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Welcome back, araver!
to produce a single closed-form expression as it will be very long and complex, but I was intrigued by this problem enough to attempt to solve it in Excel. I think that if you solve it in Excel using only math operations, then there is a closed-form expression solution. I was able to solve it in Excel without writing a single line of code and I can share it with you. Brainden doesn't allow me to upload an Excel spreadsheet, but you can PM me if you are interested. Maybe it will help you to arrive to the closed-form expression you're looking for.
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I would say that for distribution to be fair the average cost per slice (CPS) paid should be as close as possible. For example if the cost of pizza was $7 the solution would be easy and they would split the cost 3+3+1 for $.50 per slice paid by all.
For $8 pizza, the extra dollar has to come from someone and it should be either A or B, but not C. If it came from C then A and B would still be paying $.50/slice while C would be paying $1/slice or twice as much as A and B. So 4+3+1 or 3+4+1 distribution for $8 pizza seems fair and one of the three pays $.67/slice while others are still paying $.50/slice
For $9 pizza, it should be 4+4+1 for the same reason.
Now, for $10 pizza the options are
- 5+4+1 with CPS at $.83 / $.67 / $.50
- 4+4+2 with CPS at $.67 / $.67 / $1
The absolute difference between the highest and the lowest CPS is the same in both cases and is $.33. However, by taking the lowest CPS as the base and measuring the premium paid by others as a percentage compared to the base, option 2 becomes preferred - Charlie pays 50% premium ($1 over $.67) compared to option 1 when either Alice or Bob have to pay 66% premium ($.83 over $50).
For $11 pizza, it's either
- 5+4+2 with CPS at $.83 / $.67 / $1, or
- 5+5+1 with CPS at $.83 / $.83 / $.50
- 4+4+3 with CPS at $.67 /$.67 /$1.5
Again, using the analysis above option 1 is preferred.
So, basically I arrived at the same result as dgreening, but using a slightly different "measure of fairness". I think there isn't an absolute best answer here. It all depends on what criteria is used to establish what's fair.
Lanterns of Eden.
in New Logic/Math Puzzles
Posted
Since the Angel cannot turn ON a lantern that is currently OFF (only the Devil can do that), as long as there is at least one lantern in the OFF state from the beginning, the Angle cannot win by having all lanterns ON. This would require the Devil to turn ON the final lantern that will deliver the win to the Angel. Assuming the Devil wants to win, this will never happen, so the only way the Angel can win in practice is when all the lanterns are OFF.
These are some trivial and uninteresting examples and I'm not sure how they help.