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| Author | Topic: Dicing With Probability |
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Brad D Byrne Member |
David, from a math point I am very inclined to agree w/ yayet, hehe, I think our math must be a little incomplete, because, "...we cannot know when that balance will occur..." true, well it should a dilemna I know ------------------ IP: Logged |
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David Roberts Member |
There is no dilemma, Brad. We have '50:50' on the first toss of a coin. On subsequent tosses the average will oscillate about '50:50'. As the number of tosses tends to infinity the average will tend to '50:50'. Finally, to use a mathematics term, 'in the limit' the average will be '50:50'. That is, at the point of infinity which, admittedly, is beyond imagination but used in the same way as a convergent series summation at the point of infinity determined without the need for an infinite number of steps. In effect we have a vibrating string fixed at 1 and infinity with the vibrations increasingly dampened as we approach infinity. Sounds like quantum mechanics. Come to think of it we haven't got an average on the first toss. I may be able to scrape through by saying we have '50:50' whilst the coin is in the air spinning. On landing the probability density function collapses, à la Schrodinger's wave function collapsing, and we have either a head or a tail. Added: I'm not out of the soup until the second lands. OK, during the second spin we have both 0.75 and 0.25 coexisting given that the first is a head or not; the average being 0.5. On collapsing we have either 1, 0 or two chances at 0.5; the average again being 0.5. Spooky, isn't it? I think I'll fix the string at zero whilst I'm at it giving exactly 50:50 at both ends. Of course, we could have 50:50 between but we'll never know a priori. [This message has been edited by David Roberts (edited January 21, 2007).] IP: Logged |
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Charles Pegge Member |
Caution needed here. While it is true that as you increase the number of flips, your average For instance: probabilities of getting exactly equal heads and tails: 2 flips 0.5 4 flips 0.375 6 flips 0.318 8 flips 0.273 10 flips 0.246 Your standard deviation from the mean only decreases by the square root of the number of flips. ------------------ IP: Logged |
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Brad D Byrne Member |
Cool!! David, Charles, this is getting Fun!! wish we new the length of the string need to go work right now, but looking forward to where this goes B ------------------ IP: Logged |
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Michael Mattias Member |
Simplified Rules of Probability For Dice Games: "Ya pays yer money and takes yer chances." IP: Logged |
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David Roberts Member |
quote: We could map the number of tosses into the width of our monitor with 50:50 at the half height. Eventually the oscillations will be barely discernible and, as Charles points out, the likelihood of crossing the line will almost vanish leaving us with an asymptotic curve heading either north or south to the point of getting under our monitor resolution appearing then as a flat line ie visually fully dampened. To use Charles' word paradoxically we could have the difference between heads and tails in the billions but the relative value compared with unity may be infinitesimal. This is typical of a determinate process where the outcome is known but the path to it is not as opposed to an indeterminate process such as a chaotic one where the path in small increments is almost known but the outcome, like whether it will be raining on the first Sunday in 2009, is not. IP: Logged |
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Ian Cairns Member |
Flipping coins with a computer is not the same as flipping coins in practice. We humans build bias into everything we do. I am sure that a practiced coin tosser could flip a coin a calculated number of turns and have it land (mostly but not always) with the desired side up. regards, Ian Cairns ------------------ IP: Logged |
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Dave Stanton Member |
quote: Not and live long with both knee caps. ------------------ IP: Logged |
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Charles Pegge Member |
Okay guys I give up. These were more tricky than I originally anticipated, so I'll just get it over with. Here are the answers: Throwing a dice 5 times: principle idea:
Possible combinations that satisfy: 3*(1/6)^3*(5/6)^2 + .011 B: getting a ONE 3 times only. 10*(1/6)^3*(5/6)^2 C. getting a ONE 3 times or more. 1*(1/6)^5 + 5*(1/6)^4*(5/6) + 10*(1/6)^3*(5/6)^2 D. getting a single ONE, a single TWO and a single THREE in any order. 10*(3/6)*(2/6)*(1/6)*(3/6)^2 E. getting a single ONE, a single TWO and a single THREE in ascending order. 10*(1/6)*(1/6)*(1/6)*(3/6)^2 May your kneecaps live long. ------------------ IP: Logged |
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Ian Cairns Member |
Whenever you are dealing with anything that a human being can influence (even if you think they can't), especially if the outcome will cost you something, be suspicious. Memorable Quotes from Guys and Dolls (1955) by Sky Masterson (Marlon Brando) quote: source: http://us.imdb.com/title/tt0048140/quotes regards, Ian Cairns ------------------ IP: Logged |
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Donald Darden Member |
Let;s look at coin flipping another way. Say that this is az coin that all factors are balanced so that it truly should come up either heads ot tails equally. As you flip the coin, there will always be a tendancy for the Since there is no memory associated with coin flipping, what is Probability says that the odds are exactly the same, but in actual ------------------ IP: Logged |
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Brad D Byrne Member |
Yeah, but the pendulum does swing! ![]() not to distract from the original topic too much and Charles' fine math! Thanks! Charles ADDED: and btw, your paintings are GREAT! Charles! http://www.pegge.net/paintings/PEGGE.HTM ------------------ [This message has been edited by Brad D Byrne (edited January 23, 2007).] IP: Logged |
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Emil Menzel Member |
>>Probability says that the odds are exactly the same, but in actual practice, there would be no point where you could predict that the quantity of heads and tails would exactly match. I agree, as would (I believe) any statistician. To elaborate: Of course I might possibly lose my bet -- especially (say) if we The probabilities that Charles is talking about are exact, but More could be said on this score, both mathematically & Another thing I have found with computer simulations is that if ------------------ IP: Logged |
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Emil Menzel Member |
Another puzzle: What is the probability of getting exactly 50% heads & 50% tails if ------------------ IP: Logged |
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David Roberts Member |
The question can be put: What is the probability of N/2-combinations from N. The number of r-combinations from n is given by nCr = n!/(r!(n-r)!) So, the number of N/2-combinations from N is given by N!/((N/2)!^2) The population is simply the summation of the coefficients of Pascal's pyramid ( see Charles' first post) which is 2^N So, the answer to Emil's question is N!/((N/2)!^2/(2^N)) As N tends to infinity this probability tends to zero confirming Charles' cautionary note earlier. Added: The trend will be inverse exponential. Added further: The humdinger of a paradox is that as N tends to infinity the ratio of heads vs tails tends to 50:50 with a probability tending to zero. In other words the theoretical expected value ain't going to happen. A bit more: An analogy would be a glass half full of beer. We then half fill the empty part. We then half fill the new empty part. The theoretical limit is a full glass but we never get there because we keep dividing the empty part. In practice we hit the Planck length when the empty part is smaller than anything which is not space so we either leave it as falling short or overflow. Such is life. [This message has been edited by David Roberts (edited January 23, 2007).] IP: Logged |
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