superprismatic
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Everything posted by superprismatic
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I meant "fallible".
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All I have to add is that the reverend Bayes would be proud of this debate! But, ultimately, he agreed with me. That's why the CBSE board exam accepts my answer as correct. I do, however, understand other opinions. After all, Bayes was just an infallible human as all of us are. Basically, prior probabilities are ALWAYS subject to debate. So, be calm......
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I agree with BebopKid.
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The expected number of iterations is the sum of products of the probability of a path with its length. The sum is over all possible paths. The expected length is what we usually call the average length. I hope this helps.
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Include the new ball which was placed into B in the same iteration.
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Hey! Good luck with that!
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You have two bags containing balls. Bag A contains 2 white balls, bag B contains 2 black balls. You randomly pick a ball from bag A and place it into bag B, then you randomly pick a ball from bag B and place it into bag A - this will be called an iteration. Note that after any number of iterations the two bags will each contain 2 balls. What is the expected number of iterations needed to go from the initial configuration of 2 white balls in bag A and 2 black balls in bag B to a final configuration of 2 black balls in bag A and 2 white balls in bag B?
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Sorry if I sounded condescending. I really only meant to point out to marsupialsoup that his analysis was fine except for the fact that the three cases may not be of equal probability. I don't think this is a hot point. There is good reason to disagree with the 0.6 answer. Pointing out where the problem lies is good, though.
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I think we all noticed that that was a typo! For your information, mmiguel1 asserts that the three cases have different probabilities, thus causing the answer to be 0.25 rather than 0.60. I say this so you can appreciate the disagreement.
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Please explain in detail how you came up with 0.6. I would like to see a simpler explanation than an application of Bayes Theorem.
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Plainglazed, It's not clear to me that there are 55 codes. Those switches may be rocker switches in which case there may be 105 codes! I tried to find out which it is on line, but I had no luck. Everything seemed to assume the reader knows what it is. Bummer. It's much harder to find a deBruijn sequence if there are 105 codes, but I have a program to find one.
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mmiguel1's got it for sure with his "Return 42"!
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You make a lot of good points. If you're not a mathematician, you should consider being one! Mathematics needs thinkers that challenge other people's ideas. You have a very admirable quality. I don't know if you are right in this case or not, but you approach the problem with gusto! I'll call a truce. I hope you accept my terms.
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Spring has sprung The grass is riz I wonder where the birdies iz? --- simple and to the point
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Yes, it is an infinite series. But since every permutation of every distribution is included, each term in the average must be equal.
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Actually, I tried to say that, if you were to average over all distributions -- all equally likely in the absence of information -- then you get the uniform distribution. So, in that sense, I didn't just pick it out of the air. After all, that bag may have been filled by someone who has a marked preference for pastel colors and added some white ones for balance. Or, perhaps, it is a bag made by someone who was filling bags with balls of colors in alphabetical order and he started this bag with Ultramarine (U) ,Violet (V), White (W) but, since he couldn't come up with a color starting with X, he got frustrated and just stuck another White (W) just to get four in there! I give each distribution (whether binomial or even one conveying a very personal bias) equal weight.
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Please post the dimensions of figure 1. Otherwise, one has to guess to insure making a square.
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