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Imagine a roulette table (with no zeros for argument's sake, and no flaws, manipulation etc) where the spin has come up on one colour, say black, for a surprisingly large number of times in a row, say 14 times. Are the odds still 50/50 that the 15th spin will fall on red, or does the possibility somehow start leaning towards red being the next number at say 55/45 ? Fundamentally I would guess that there is still a one in two chance, but then again ... ? -- 196.207.47.60 ( talk) 03:47, 25 October 2008 (UTC)
Assuming 50/50 means that observing black for a large number of times in a row is surprising or improbable. Observing black for a large number of times in a row means that the assumption 50/50 is unlikely or incredible. The probability has a binomial distribution. The likelihood has a beta distribution. Bo Jacoby ( talk) 07:21, 25 October 2008 (UTC).
I cannot help reporting the opinion of E.A.Poe in a concluding remark of his short story " The Mystery of Marie Rogêt" :-)
It is also worth noting that the very high opinion that Poe shows here about mathematics and mathematicians disappears suddenly few time later, most likely, I guess, after an argument with a mathematician about his beliefs about Probability... -- PMajer ( talk) 11:27, 25 October 2008 (UTC)
I have a related question. Suppose someone has been counting net black surplus. At the start it should be 0, and it should quickly cross 0 again. But at some point there will be 13 more blacks that have come up than reds that have come up. When you are at that point, isn't it a new "zero", since there is not, in fact, a history that probability tracks? So how does the probability know that it needs to go down by 13 to get to the "real" zero?? —Preceding
unsigned comment added by
94.27.170.127 (
talk) 11:31, 25 October 2008 (UTC)
Here is another way to look at my new question: If I come into the room and start counting net black surplus, I will start at zero -- but it could be just at the point that another person's count is at 13! So by the time the other person's count is at zero, where it should, on average, stay, my count will be at -13. So isn't this a contradiction? That on the one hand, the count should stay at -13, on average (the first person's zero) and on the other hand it should stay at 0, on average (my zero, the refernce frame I'm using). It doesn't make sense to me that the average net black surplus can remain at once 0 and -13.
In fact, since there's no history, what if we play for a long time and wait until the count is at -25 and then invite a new person in to start keeping count. Then when it is at +25 we can invite a fourth person. So as the four people are watching the game, the average count has to be at -13 (by my framework), 0, -25 (the third person's zero), +25 (the fourth person's zero). This just doesn't make sense to me.
The article gambler's fallacy discusses these questions. —Preceding unsigned comment added by Aenar ( talk • contribs) 13:56, 25 October 2008 (UTC)
Is there a method of solving a problem of this form that does not involve brute force search?
z = A * X + B
p = A ^ 2 * X + C
where z and p are fixed integers and the other variables are adjustable integers.
70.171.7.209 ( talk) 20:30, 25 October 2008 (UTC)op
Can you give me an example of natural number object in any other category than Set? Does it always correspond to natural numbers? 212.87.13.70 ( talk) 21:22, 25 October 2008 (UTC)
Mathematics desk | ||
---|---|---|
< October 24 | << Sep | October | Nov >> | October 26 > |
Welcome to the Wikipedia Mathematics Reference Desk Archives |
---|
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages. |
Imagine a roulette table (with no zeros for argument's sake, and no flaws, manipulation etc) where the spin has come up on one colour, say black, for a surprisingly large number of times in a row, say 14 times. Are the odds still 50/50 that the 15th spin will fall on red, or does the possibility somehow start leaning towards red being the next number at say 55/45 ? Fundamentally I would guess that there is still a one in two chance, but then again ... ? -- 196.207.47.60 ( talk) 03:47, 25 October 2008 (UTC)
Assuming 50/50 means that observing black for a large number of times in a row is surprising or improbable. Observing black for a large number of times in a row means that the assumption 50/50 is unlikely or incredible. The probability has a binomial distribution. The likelihood has a beta distribution. Bo Jacoby ( talk) 07:21, 25 October 2008 (UTC).
I cannot help reporting the opinion of E.A.Poe in a concluding remark of his short story " The Mystery of Marie Rogêt" :-)
It is also worth noting that the very high opinion that Poe shows here about mathematics and mathematicians disappears suddenly few time later, most likely, I guess, after an argument with a mathematician about his beliefs about Probability... -- PMajer ( talk) 11:27, 25 October 2008 (UTC)
I have a related question. Suppose someone has been counting net black surplus. At the start it should be 0, and it should quickly cross 0 again. But at some point there will be 13 more blacks that have come up than reds that have come up. When you are at that point, isn't it a new "zero", since there is not, in fact, a history that probability tracks? So how does the probability know that it needs to go down by 13 to get to the "real" zero?? —Preceding
unsigned comment added by
94.27.170.127 (
talk) 11:31, 25 October 2008 (UTC)
Here is another way to look at my new question: If I come into the room and start counting net black surplus, I will start at zero -- but it could be just at the point that another person's count is at 13! So by the time the other person's count is at zero, where it should, on average, stay, my count will be at -13. So isn't this a contradiction? That on the one hand, the count should stay at -13, on average (the first person's zero) and on the other hand it should stay at 0, on average (my zero, the refernce frame I'm using). It doesn't make sense to me that the average net black surplus can remain at once 0 and -13.
In fact, since there's no history, what if we play for a long time and wait until the count is at -25 and then invite a new person in to start keeping count. Then when it is at +25 we can invite a fourth person. So as the four people are watching the game, the average count has to be at -13 (by my framework), 0, -25 (the third person's zero), +25 (the fourth person's zero). This just doesn't make sense to me.
The article gambler's fallacy discusses these questions. —Preceding unsigned comment added by Aenar ( talk • contribs) 13:56, 25 October 2008 (UTC)
Is there a method of solving a problem of this form that does not involve brute force search?
z = A * X + B
p = A ^ 2 * X + C
where z and p are fixed integers and the other variables are adjustable integers.
70.171.7.209 ( talk) 20:30, 25 October 2008 (UTC)op
Can you give me an example of natural number object in any other category than Set? Does it always correspond to natural numbers? 212.87.13.70 ( talk) 21:22, 25 October 2008 (UTC)