aarthrj3811
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- Joined
- Apr 1, 2004
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- Location
- Northern Nevada
- Detector(s) used
- Dowsing Rods and a Ranger Tell Examiner
- #1
Thread Owner
OK now ....I am a Dowser..I am going to take one of those Scentific Double-Blind Test. There will be one coin placed under one of 10 cups. I will make my run and be scored. I will make a total of 10 runs. How many coins will I find?. Now you have to remember that all Dowsing is the same as guessing.The only question here is HOW MANY COINS WILL I FIND?
When I ask this simple Question these are some of the answers I recieved..
Odds of finding 1 is 38.7%.
Odds of finding 0 is 34.9%.
Odds of finding 2 is 19.4%.
Odds of finding 3 is 5.7%.
Odds of finding 4 is 1.1%.
Odds of finding 5 is 0.15%.
So, you have a 98.7% chance of finding between 0 and 3 coins.
Art it's already been answered, you sum the probabilities from the binomial distribution just as Carl has already shown. Here I'll break it down for you even more because you obviously have some kind of patronizing manner as if you think this question is too hard to figure out (when in reality I first learned all the basic discrete probability distributions in my VERY FIRST INTRO PROB STATS CLASS). If you're looking for a number, if you ran this test an infinite number of times it will average out 1 correct guess. That is the expected number of coins on average you'll guess correctly.
Sorry Art you have not beat the chances for guessing at random, you have beaten the expected AVERAGE in one single instance.
Sure it won't be exact for a single instance. The fact is a dowser would still have to have it occur at that exact moment for it to matter. The odds of him dowsing say 5 or 6 is still so stacked against him that it probably WON'T happen in that instance.
.....you are free to label probability and its application, by any name you choose. However, as a tool for predicting the expected outcome of a process (test) having a finite number of possible outcomes;
If it's done correctly, double-blind, ect, I think you'd be lucky to get one in ten.
I fully "expect" Art to find on average 1 coin in his attempts. On average, the more replications of his test he runs, the average of all replications will converge asymptotically to the expected average of the binomial distribution, in this case 1.
As you can see, roughly 93% of the time you will experience Zero, One or Two Hits