Wikipedia:Reference desk/Archives/Mathematics/2015 December 27

= December 27 =

Point inside a convex polygon
Hi. If x' is the average of the x-coordinates of the vertices of an arbitrary convex 2D polygon, and y' is the average of the y-coordinates, then is the point (x', y') guaranteed to lie within the polygon? 109.151.59.86 (talk) 12:38, 27 December 2015 (UTC)


 * Yes, because (x',y') is a convex combination of the vertices of the polygon.  S ławomir Biały  14:06, 27 December 2015 (UTC)
 * Thank you. 109.151.59.86 (talk) 14:59, 27 December 2015 (UTC)


 * Link: convex combination. -- ToE 16:32, 27 December 2015 (UTC)

Criteria for physical quantity to be vector valued
What is the necessary and sufficient condition to be satisfied to classify a physical quantity as a vector in that can be expressed in math terminology not just an intuitive assertion that the quantity has applications in a directional sense.150.242.150.175 (talk) 13:03, 27 December 2015 (UTC)


 * It has to be a list of numbers that transforms in a certain way under a change of coordinates. For example, the dimensions of a box are the same in all coordinate systems, so would not be a vector.  On the other hand, the components of force depend on what coordinates you use to measure them.  If you measure the components of force in a different coordinate system, they will have transformed in a compensating way.  The mathematically precise way this happens is detailed at  covariance and contravariance of vectors.   S ławomir  Biały  14:05, 27 December 2015 (UTC)

Odds of getting an equal number of heads and tails after X coin flips
Hello. How can I calculate the odds of getting the exact same number of heads and tails after flipping a coin X times (where X is an even number)? Is 50% the correct answer? Thanks.--Leptictidium (mt) 20:40, 27 December 2015 (UTC)


 * The formula for the probability is X! divided by (X/2)! divided by (X/2)! again and finally divided by 2 to the power of X. (The ! means Factorial.)  The probabilities can be found from Binomial coefficients, and the probabilities for 2, 4, 6, 8 flips are 2/4, 6/16, 20/64 and 70/256 respectively.  I've left the fractions uncancelled so that you can see where I'm getting them from.  Your 50% probability (odds of 1 to 1 or "evens") only works for four two flips.  As the number of flips increases, the probability decreases, so the odds against equal numbers increase.  To turn these probabilities into odds, for example, the 8 flips has 70 chances out of 256 of having equal heads and tails, so (256 minus 70) which is 186 chances of different numbers.  Thus the odds are 186 to 70 against (or 93 to 35 against).    D b f i r s   21:13, 27 December 2015 (UTC)
 * Gee, thanks! --Leptictidium (mt) 21:42, 27 December 2015 (UTC)
 * I have replaced "four flips" by " four two flips" in your post. Hope you don't mind. PrimeHunter (talk) 22:29, 27 December 2015 (UTC)
 * Thanks, I was converting n/2 to X, but I should have spotted that mistake!   D b f i r s   22:42, 27 December 2015 (UTC)
 * You might be interested to know that the probability of equal numbers gradually reduces as X increases. A formula that estimates the probability is the square root of (2 divided by pi divided by X) (related to Stirling's approximation for factorials).  The approximation is not very good for low values of X, but improves as X increases.  It's within about 2% for n=12 upwards and within about 1% for X = 24 upwards.  As an example, taking X = 36, the probability is 13.3% using the pi formula (2 / pi / 36 is 0.017684 and the square root of this is approximately 0.132981).  Compare this with the exact value of 9075135300 divided by (2 to the power of 36) which is about 0.132061, so the formula overestimates by about 0.7%
 * The probability of equal numbers drops to about 10% for X = 62 and to about 6% for X = 176. It continues to reduce towards zero, but more and more slowly as X increases.   D b f i r s   22:42, 27 December 2015 (UTC)
 * Am I correct in assuming that, if instead of flipping coins I were rolling six-faced dice, the correct formula to estimate the probability would be the square root of (6 divided by pi divided by X)? --Leptictidium (mt) 11:13, 28 December 2015 (UTC)
 * No, you are less likely to get equal numbers on dice. I'm struggling to apply Stirling's formula to this case.  Perhaps someone else can simplify the expression for X rolls?  If you intended a slightly different question: "What is the probability that all rolls show the same face?" then the answer is just one divided by [six to the power of (X minus 1)].   D b f i r s   13:09, 28 December 2015 (UTC)
 * It's $$(\sqrt{2\pi x})^{1-6}6^{6/2}$$. For dice with a different number of faces, simply put the number of faces instead of every appearance of 6. If you put 2 instead of 6 you get back the coin formula. -- Meni Rosenfeld (talk) 15:24, 28 December 2015 (UTC)
 * Thanks, Meni, I thought of you when I wrote "someone" above.   D b f i r s   15:41, 28 December 2015 (UTC)