Wikipedia:Reference desk/Archives/Mathematics/2008 October 25

= October 25 =

Roulette Odds
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 that there is no flaws or manipulation going on then yes, the odds is 50/50 that the next will be red. That assumption is getting stretched though, so you might want to lean the odds in favor of next spin ending as black. Taemyr (talk) 04:23, 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" :-)
 * This is one of those anomalous propositions which, seemingly, appealing to thought altogether apart from the mathematical, is yet one which only the mathematician can fully entertain. Nothing, for example, is more difficult than to convince the merely general reader that the fact of sixes having been thrown twice in succession by a player at dice, is sufficient cause for betting the largest odds that sixes will not be thrown in the third attempt. A suggestion to this effect is usually rejected by the intellect at once. It does not appear that the two throws which have been completed, and which lie now absolutely in the Past, can have influence upon the throw which exists only in the Future. The chance for throwing sixes seems to be precisely as it was at any ordinary time -- that is to say, subject only to the influence of the various other throws which may be made by the dice. And this is a reflection which appears so exceedingly obvious that attempts to controvert it are received more frequently with a derisive smile than with anything like respectful attention. The error here involved -- a gross error redolent of mischief -- I cannot pretend to expose within the limits assigned me at present; and with the philosophical it needs no exposure. It may be sufficient here to say that it forms one of an infinite series of mistakes which arise in the path of Reason through her propensity for seeking truth in detail.

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)
 * Well, it is true that (say with probability one) at a certain moment a number of reds will appear, sufficient to bring the surplus to 0 again. But in fact more is true: in some future time arbitrarily large deviations from 0 will be reached, in positive and negative, and as you see, if you keep this in mind you don't need to recall what was the zero point to conclude that it will be reached again. Any number in fact will be reached again and again (almost surely). To say it in another way, suppose that another person started the count in another moment, so they have different counts of the surplus...--PMajer (talk) 12:08, 25 October 2008 (UTC)


 * On the original question Given the number of times each roulette wheel is spun every day in a casino, how often between spins? five minutes? that would be 400 a day for 365 days a year assuming they are ungodly comes to over 100 thousand, whereas 15 blacks in a row the chances are a measly 1 in about 30 thousand. So it will probably occur a few times every year on every wheel just by chance. However where it could be fixed I'd put it on the opposite colour of a high roller the next time. On the other hand they might be in league with the house so you never know - just stay away. If it really would make a difference if you doubled your money and having what you have is no good then by all means bet once to double it but I can't see any other good reason. Dmcq (talk) 11:35, 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.


 * On the second question see Random walk, or what's more commonly called a drunkard's walk. There is no memory, the distribution of possible values gets wider and wider, though in your particular case rather surprisingly it is almost certain the value will go back over the 'base point' - it will almost certainly reach every value eventually. Dmcq (talk) 12:02, 25 October 2008 (UTC)
 * Apart from the minor detail that you're going to run out of money eventually! That's way the Martingale better system does work despite the fact that you will almost surely win. --Tango (talk) 11:53, 25 October 2008 (UTC)
 * There is no reason why it should go back to zero. As a percentage of the number of spins, it would be expected to get smaller (see Law of large numbers) but that doesn't mean it will necessarily get to zero. The expectation is always that the net number of blacks in the *future* will be zero (so the net will stay the same as an absolute number), the ones in the past aren't random since you already know what they are. They can have no effect on the table (assuming it really is fair), so you can forget about them. If you spin it 10 times and get 10 blacks, that gives you net of 10 or 100%. Over the next 10 spins I would expect to get 5 black and 5 reds, leaving the net at 10, which is now 50%. As you can see, it has reduced proportionally speaking. If you do 1000s of spins the chance of the net difference between blacks and whites being more than say 1% becomes every more and more remote, but there is always a good chance that it will be quite high during the first few spins. --Tango (talk) 11:53, 25 October 2008 (UTC)


 * I believe you meant 'less than 1%' but here's a quote from the Random walk article: "The following, perhaps surprising theorem is the answer: simple random walk on \mathbb Z will almost surely cross every point an infinite number of times" Dmcq (talk) 12:02, 25 October 2008 (UTC)
 * No, more than. "Remote" means unlikely - the chance of it being more than 1% is unlikely. What you say about random walks is, of course, true, but that's for an infinitely long walk. No-one plays infinitely long games of roulette! --Tango (talk) 14:14, 25 October 2008 (UTC)


 * It seems to me, 94.27.170.127, that the very reason why you were puzzled relies in the unclear expression the average count has to be 0. If you try and give a precise mathematical meaning to average count any contradiction should disappear. If N(k) is the number of blacks exits (that we assume indipendend to each other etc) at time k, we can prove for instance that a.s. the average number of blacks at time k, that is N(k)/k, tends to 1/2 as k goes to infinity, and this is in accordance with the count of the other persons (if I enter the room at time=100, my count will be N'(k)=N(100+k)-N(100), so in any case my N'(k)/k has the same limit as your N(k)/k). But for instance we cannot infer that N(k)-k/2 is bounded: the behaviour should look like the one pictured in the random walk article quoted above; in fact your argument proves that it has to be unbounded, otherwise there should be a preferred observer (or a preferred starting time).--PMajer (talk) 13:08, 25 October 2008 (UTC)

The article gambler's fallacy discusses these questions. —Preceding unsigned comment added by Aenar (talk • contribs) 13:56, 25 October 2008 (UTC)


 * I see no one's mentioned statistical hypothesis testing here. The difference between a probabilist and a statistician is that when the coin turns up "heads" 20 times in a row, the statistician suspects the coin is biased. Michael Hardy (talk) 21:30, 25 October 2008 (UTC)
 * That was mentioned right at the top (in the 1st 2 replies), although not by name. --Tango (talk) 21:34, 25 October 2008 (UTC)

Solving diophantine equations
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


 * Are you looking for all the solutions or just one? And presumably you don't want trivial solutions like (A,B,C,X)=(0,z,p,1)? Also, is p prime? --Tango (talk) 20:58, 25 October 2008 (UTC)
 * Isn't this completely trivial as stated? Take A and X arbitrary integers; then there are unique B and C that make (A,B,C,X) a solution. Algebraist 21:29, 25 October 2008 (UTC)
 * Well spotted! --Tango (talk) 21:39, 25 October 2008 (UTC)
 * Let us make the problem (marginally) more interesting by specifying that one or more of A, B, C, X are given. If two or more are given, then we have only two variables, and two equations, so we can immediately solve the problem in the complex numbers and check if the solutions are integer.  Therefore for an interesting problem exactly one of those four variables are given.  With Algebraists' argument, either B or C must be given, else the problem is trivial.  Therefore there are only two potentially interesting problems:  B is given, with A, C, X free to vary, and C is given, with A, B, X free to vary.  In the former case the problem is equivalent to factoring z - B;  in the latter, with slightly more work, equivalent to factoring p - C.  Eric.  131.215.45.193 (talk) 00:28, 28 October 2008 (UTC)

NNOs
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)
 * For a really boring example, there is the category with only one object and only the identity morphism. The unique object trivially satisfies the requirements in that article. Aenar (talk) 16:56, 26 October 2008 (UTC)
 * Here's an example a bit more interesting: Let the category be the category of modules over a commutative ring with 1-element. Choose N to be the free module with basis the natural numbers. Let the zero element be the basis element 0 and let the successor function be defined as the homomorphism mapping any basis element i to i+1. This works since giving a homomorphism from N to a module M is the same as giving a function (in Set) from the the natural numbers to the underlying set of M. Aenar (talk) 17:21, 26 October 2008 (UTC)