dgreening
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Posts posted by dgreening
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A friend and I were working on this. Mark came up with an elegant solution.
Close-form (explicit function) solution:
SpoilerDefine as the number of Happy Bugs in Generation g, where initial conditions, . Then,
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Etc
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This is very interesting, but I think there is a more concrete answer.
SpoilerFor each "On - Off" cycle, the light is ON 2/3 of the time and OFF 1/3 of the time.
After not too many cycles the additions make very little difference in the total On Time and Off time.
This would indicate that the light is on for 1:20 and off for 0:40.
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My first thought was like Capt Ed.
BUT thinking about it some more.
I suspect that the person sitting at 4 or 6 will probably have a better chance.
Rationale: after a few moves, the passing is going to tend to go one direction [say CCW] and will tend to go back and forth on that side of the host. By that logic, #5 may often be the next to last guest to get the bottle.
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No dispute here.
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Aha!
I just didn't follow the same process through to get the supper precision devices!
Thanks
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I was working on the strategy that you describe:
- Bet half my money [rounded to an integer]; or
- bet the "ceiling" [twice my original money - Current money].
Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.
Curious about your result. If you average over many trials are your expected winnings positive?
As I mentioned, I only did a small number of runs [about 100].
BTW - I like your title "Retired Expert"!
I think that if you do a large number of runs, you eventually get to about 50%.
Now I am confused.
I think I followed the strategy you described yesterday, when you postulated that it would deliver 100% chance of winning.
But, now your simulation it gives you a 50% chance of winning.
Did I miss something??
I was quoting bonanova solution/answer, not my proposed strategy.
I was working on the strategy that you describe:
- Bet half my money [rounded to an integer]; or
- bet the "ceiling" [twice my original money - Current money].
Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.
Your strategy is not applicable, you can't bet more money than you have, if your current money is 50$, your original money is 100$, so 2*100-50 = 150$, which you don't have.
Now I explain, my proposed strategy.
Hidden Content
yes, if you [and the house] are willing to bet micro, nano or pico dollars, then the game can go on forever. But your money is gone, so it is theoretically correct, but not very practical.
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An answer for part 1
This is actually a common problem with some electrical components.
The solution is that you build a large quantity of Bazfaz and then test them
The 30% defines the limits of the curve, but some of the units will be very close [ less than 1%].
You sell the ones that happen to be more precise for a higher price. Less precise units are sorted by precision and sold at lower prices.
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Perhaps some more description of the task would help.
What is the context for this question??
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Now I am confused.
I think I followed the strategy you described yesterday, when you postulated that it would deliver 100% chance of winning.
But, now your simulation it gives you a 50% chance of winning.
Did I miss something??
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I was working on the strategy that you describe:
- Bet half my money [rounded to an integer]; or
- bet the "ceiling" [twice my original money - Current money].
Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.
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The hint about primary numbers did not help, but I noticed a pattern
Could it be 91, 71, 31 ??
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I would be interested in how you calculated your answer.
I came up with something slightly different
I assumed that the critical position is at 45 degrees.
At that point the ladder would extend:
- 7 foot behind the corner on the far side of the 7 foot hallway; and
- 5 foot beyond the corner on the far side of the 5 foot hallway
Each of these points would be 12 feet from the [outside] corner where the halls meet
Therefore the ladder could be no larger than the distance between those 2 points
D = Sqrt ((12*12)+(12*12)) = Sqrt (144*2) = Sqrt (288) ~ 16.97
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If we assume that all 1000 bottles might be used ... one approach
Hidden Content
While driving home last night I realized that I had made the problem much more complicated than it needed to be
Use the same "Binary Coding " approach that I described earlier .... BUT soldier only need to drink from the bottles with a value of zero [or one] for each power of 2. Therefore:
number each bottle from 0 to 999 - use the binary representation [requires 10 bits]
For every binary position [1,2,4, ... 512] one soldier drinks from every bottle that has a
1 in that position and another drinks from every bottle witha 0 in that position:- for the least significant bit - soldier A drinks from every bottle where that value is 0 [even],
soldier B drinks from every bottle where that value = 1 [odd] - for the next significant bit, soldier C drinks from every bottle where the value = 0
and soldier D drinks from every bottle where the value = 1 - ...
- for the most significant bit [512] soldier S drinks from every bottle where that value = 0 [less that 512]
and Soldier T drinks from every bottle with the value = 1 [512 or greater]
At the end of 31 days some of the testers will die and the bits associated with them can be used to decode the poisonous bottle.
You do need 10 soldiers --- the number that die is purely a function of the binary value of the poisonous bottle.
- for the least significant bit - soldier A drinks from every bottle where that value is 0 [even],
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If we assume that all 1000 bottles might be used ... one approach
number each bottle from 0 to 999 - use the binary representation [requires 10 bits]
For every binary position [1,2,4, ... 512] one soldier drinks from every bottle that has a 1 in that position and another drinks from every bottle with a 0 in that position:
- for the least significant bit - soldier A drinks from every bottle where that value is 0 [even], soldier B drinks from every bottle where that value = 1 [odd]
- for the next significant bit, soldier C drinks from every bottle where the value = 0 and solder D drinks from every bottle where the value = 1
- ...
- for the most significant bit [512] soldier S drinks from every bottle where that value = 0 [less that 512] and Soldier T drinks from every bottle with the value = 1 [512 or greater]
At the end of 31 days half of the tasters [10] will die and the bits associated with them can be used to decode the poisonous bottle.
Minimum number is 10
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Maybe I am looking at this too simplistically
There is one way [out of 36] to make 12 [6,6]. P(12) = 1/36 = ~ 2.77%
There are 6 ways to make 7 [1,6] [2,5] [3,4] [4,3], [5,2] and [6,1]. So P(7) = 6/36 = 1/6
To make 2 consecutive 7's will be P(7) * P(7). So P(7,7) = 1/6 * 1/6 = 1/36 = ~ 2.77%
Thus the odds of making 12 or consecutive 7's are equal.
Therefore the P(A wins) = P(B wins) = 50%
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If we go with the obvious.
4/7 [57.1%] is slightly higher than 5/9 [55.5%].
If both are equally good, then the odds would suggest that it is more likely to win 5 out of 9 [than 4 out of 7]
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I seem to recall, something similar to this.
I assume we can use the following simplifying assumptions:
- Bullets travel to infinity [unless annihilated]
- Bullets follow the same path [or can be assumed to follow a straight line]
As I recall, it doesn't take many bullets before the leaders continue on undamaged.
Obviously if the first bullet is the fastest of the 10, then they will not all annihilate [10%]
similarly if the last bullet is the slowest of the 10, then they will not all annihilate [10%]
There are many variations of these conditions that lead to at least 1 [actually 2] bullets continuing on.
So the probability is certainly less than 80% and I suspect it is closer to 20% ..
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I d not see that coming.
The classic solution is to find numbers that sandwiched between 2 prime numbers.
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Very nice apprach!
I missed that relationship ... good job.
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Question for harey: how did you solve for the quantitites??
Thanks ....
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I have been working on this a few days.
bonanova is right
There must be
- an odd number of the $1501.5 TVs and
- 5n + 4 of the $600.6 models [for n = 0, 1, ...]
interestingly these 2 prices have a 5/2 ratio, so the combinations have the same values
e.g., (4* 600.6) + 3*(1501.5) = (9*600.6)+ (1*1501.5)
If we make the assumption that there is a unique solution, then it is probably safe to work with the only unique value to these two sets
(4*600.6) + (1501.5) = 3903.90
I have not found a good metholoology for finding combination of the other 4 sets to equal 9616.0. -
A card will go into the low pile when it immediately follows a higher card.
card 1000 can never meet this criteria
card 999 can meet this only if it follows card 1000 [P = 0.1%]
card 998 can meet this criteria if it follows cards 1000 or 999 [P= 0.2%] ...
The expected value should be the pioint at which The sum of the probabilities = 50%
By my calculations this occurs between 968 and 969.
So the expected value will be about 968.88
Simple classic puzzle on profit
in New Logic/Math Puzzles
Posted
Pickett makes a good point about not knowing the original price that Bill paid, but based on the data as presented
Bill acquired this for some value X in the past.
Bill sold to [$100] and bought from [$80] Tom for a net profit of $20
Bill bought from Tom [$80] sold to Herman [$90] for a net profit of $10
For these 3 transactions the net profit is $30
Since we don't know X Bill's profit will be affected by the cost of Acquisition
If it cost Bill
Therefore Bill's total profit = X - $100 + $30