dgreening

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A friend and I were working on this. Mark came up with an elegant solution.
Closeform (explicit function) solution:
SpoilerDefine as the number of Happy Bugs in Generation g, where initial conditions, . Then,
For þ
For þ
For þ
For þ
For þ
For þ
For þ
Etc
.

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.

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.

No dispute here.

Aha!
I just didn't follow the same process through to get the supper precision devices!
Thanks

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*10050 = 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.

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.
 1

Perhaps some more description of the task would help.
What is the context for this question??

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 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.

The hint about primary numbers did not help, but I noticed a pattern
Could it be 91, 71, 31 ??


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

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],

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
 1

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%

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]

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% ..

I d not see that coming.
The classic solution is to find numbers that sandwiched between 2 prime numbers.
 1

Very nice apprach!
I missed that relationship ... good job.

Question for harey: how did you solve for the quantitites??
Thanks ....

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