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April 10

Odds and evens

I'm reviewing a wagering-proposition where I suspect the house has made a mistake in the odds it is offering. In a twenty-horse race (assuming all of the horses are of equal ability) the result of the first three finishers all being odd- or even-numbered was offered at three-to-one odds. This seems to defy logic, as two of the horses will have to be either even or odd. The proposition seems to boil down to: "Will the third horse (of the remaining eighteen) be the same type of number as the other two?". The true odds of this happening seem to only be slightly less than even, certainly nowhere near three-to-one. I want to rush to put the maximum-bet of ~~$300~~ ($100, my mistake) down on this proposition, but being far from the smartest fellow on the planet, I figured I should double-check this with our learned and perspicacious mathematicians first. Thanks in advance! Joefromrandb (talk) 04:39, 10 April 2014 (UTC)[reply]

It is a little more complicated than that and it is just about 4:1, according to my calculations. For one thing, we are assuming that half of the horses have odd numbers. Also, it depends on the number of horses in the race. For instance if there were only six horses, it would have to be a particular three out of those six. And whatever the odd/even of the first horse is, the second one is a little less than 50% (9/19 with 20 horses). And if those two match, the probability that the third matches the first two is a little less than 50% (8/18 with 20 horses). Bubba73 ^{You talkin' to me?} 04:53, 10 April 2014 (UTC)[reply]

Thanks a lot! Your solution was the way I was originally thinking, before I tripped myself up figuring there was some kind of Monty Hall-paradox going on. With your solution, the true odds seem to be slightly higher than 4–1, giving the house quite a substantial edge by offering the 3–1 odds. Joefromrandb (talk) 05:20, 10 April 2014 (UTC)[reply]

Make the following experiment of thought: Let there be infinitely many horses in the race. (Hey, we just had a huge thread with infinitely many monkeys above, why not horses?) Then there are eight equally possible outcomes; (odd, odd, odd), (odd, odd, even), ..., (even, even, odd), (even, even, even). Only two of these are in your favor. YohanN7 (talk) 05:35, 10 April 2014 (UTC)[reply]

Yes, and it approaches that as the number of horses increases. Bubba73 ^{You talkin' to me?} 05:41, 10 April 2014 (UTC)[reply]

If each horse's number is drawn as even or odd independently, the expected probability is exactly 0.25 if you do not know the actual numbers in the race. Since you know the actual number of each horse before the race, you can calculate the odds. If the number of even numbers equals the numbers of odd numbers, the odds (betting at 3:1) are slightly against you. If they're sufficiently skewed to numbers of one parity, the odds are in your favour. I'd bet that the bookies only offer the bet on races that are sufficiently balanced between even and odd numbers (and for all I know, numbers are allocated consecutively), i.e. when there is a margin in the bookie's favour. Joefromrandb, I think you've misinterpreted the odds: as I understand it, 3:1 means you get back your own bet plus 3 times if you win, which makes the margin very slim. There would be zero margin when the probability of winning is 1/(1+3) = 0.25. —Quondum 15:31, 10 April 2014 (UTC)[reply]

When I do the math shown above (10x9x8/20x19x18), I get 0.105. Multiple that by 2 for both the odd and even case, and I get 0.21. Take the reciprocal to get 4.75:1 odds. So, there's still a substantial margin there. However, if we knew the odds of each particular horse winning, that might change things, say, if most of the even-numbered or odd-numbered horses happen to have better than even odds. StuRat (talk) 15:52, 10 April 2014 (UTC)[reply]

Odds are expressed as a ratio of the respective lose:win probabilities, not as the probability of losing (or winning) as a ratio. Thus, a probability of winning of 0.21 is is expressed as odds of 0.79:0.21, i.e. 3.76:1. Think of it as two people as betting against each other. The odds for one person must simply be reversed to give the odds for the other. 1:1 odds are when each party has a 50% probability of winning. —Quondum 17:18, 10 April 2014 (UTC)[reply]

I agree that the margin is a little larger than the "slim margin" – higher than I'd expected. The bookie's margin is 0.79 ⋅ 1 − 0.21 ⋅ 3 = 0.16, i.e. 16% of every dollar that is bet on this is the bookie's average gain. —Quondum 17:53, 10 April 2014 (UTC)[reply]

Saltire question

I'm making a flag, and I want to make sure I have the proportions right. Let's imagine that this flag is 1000 units wide and 500 units tall. Each arm of the cross has a thickness (A) of 100 units, and is centered on a line running, at a 26.57° angle, from one corner of the flag to the other. What is B, the length of white along the bottom edge on the left? --Lazar Taxon (talk) 11:40, 10 April 2014 (UTC)[reply]

That would be 50√5, or about 111.80. 86.160.87.195 (talk) 12:46, 10 April 2014 (UTC)[reply]

This is an interesting question. I'd be grateful if you could explain how you got that answer. Thanks.

Sorry, I forgot to sign outWidneymanor (talk) 15:19, 10 April 2014 (UTC)[reply]

Draw a perpendicular from the right hand end of arrow "B" to the main diagonal of the flag. This makes a little right-angled triangle which is similar to the triangle formed by the bottom edge, right edge and diagonal of the flag. Calculate the length of the diagonal of the flag using Pythagoras, and then use the fact the ratios of corresponding sides of similar triangles are equal. 86.171.43.111 (talk) 17:43, 10 April 2014 (UTC)[reply]

Thank you very much.Widneymanor (talk) 19:28, 10 April 2014 (UTC)[reply]

Rotation of the 24-cell question

Is it possible to color the vertices of a 24-cell red yellow and blue such that a rotation (or combination of rotations) would take the red vertices to where the yellow were and similarly the yellow to blue and the blue to red? Secondly, is there a coloring of the octahedral cells orange, green and purple so that the *same* rotations in the first question "rotates the colors" of the cells as well?Naraht (talk) 15:48, 10 April 2014 (UTC)[reply]

I don't know the answer but the 24-cell is self dual, so if you can answer the question about the vertices then take the dual to get the answer for the cells, and vice-versa.

Looking at the construction at 24-cell#Constructions you could try colouring the eight vertices at (±1, 0, 0, 0) red. Then colour non-adjacent vertices out of(±⁠12⁠, ±⁠12⁠, ±⁠12⁠, ±⁠12⁠) yellow and blue. To get non-adjacent vertices form sets of vertices with even and odd numbers of minus signs, this means no two adjacent vertices are in the same set (as they would differ only in one coordinate). I strongly suspect this is the colouring you want but I don't know how to prove it by e.g. giving the rotation.--JohnBlackburne^words_deeds 16:52, 10 April 2014 (UTC)[reply]

Yes, the 24-cell is self dual, which I believe means that if a coloring of the vertices has a rotation that works then there is a rotation that will work for the cells, *but* it may not be the same rotation.

And with your coloration every one of the 96 triangles has three different colors at the corners, which would seem to be the most symetrical way to color them. Naraht (talk) 19:00, 10 April 2014 (UTC)[reply]

Here's how I'd construct a rotation. With the above colouring you have every vertex one of three colours, no two adjacent vertices the same colour. The faces are triangles, so each triangle must have each colour for a vertex. So pick a triangular face then do a simple rotation in the plane of this triangle through ±120° or ±2π/3. I would use geometric algebra for this, building the bivector in the plane then a rotor from it, but without doing it and applying it to a few vertices I don't know if it will work. It seems likely but it's a bit of work to check. Note this is only one or two of many rotations - the full rotational symmetry group has order 576.--JohnBlackburne^words_deeds 23:55, 10 April 2014 (UTC)[reply]

I'm currently looking at a rotation on the axis that goes through w=0, x=y=z, I think that may work, it rotates for example, (0,1,1,0), (0,1,0,1), and (0,0,1,1) to each other.Naraht (talk) 16:11, 11 April 2014 (UTC)[reply]

You don't rotate in 4D about an axis. You rotate in a plane. The article I linked to earlier, Rotations in 4-dimensional Euclidean space has details while Plane of rotation has more information on rotating in planes in general. That may be what you are doing though: two non-parallel vectors define a plane (in any dimension) and you have the vectors (1,0, 0, 0) and (0, 1, 1, 1) fixed, so a fixed plane. It's certainly a rotation: it permutes the x, y, z coordinates and acts as so on all coordinates of the dual. The question is what's your colouring scheme? It's not obvious for the vertices you're using.--JohnBlackburne^words_deeds 17:20, 11 April 2014 (UTC)[reply]

The axis goes through x=y=z and as such is a plane, I believe. (ignore the w=0 before) since it has two degrees of freedom (w & x) The points are the commutation of (±1,±1,0,0). If abs(w)=abs(x) then red, if abs(w)=abs(y) then yellow, if abs(w)=abs(z) then blue. So (1,1,0,0) red rotates to (1,0,1,0) yellow which rotates to (1,0,0,1) blue.

Note this does *not* rotate the octahedra in the same way since the octahedron made up from the 6 vertices where w=1 simply rotates in place.Naraht (talk) 18:03, 11 April 2014 (UTC)[reply]

It seems what you're doing works. It may be just a matter of terminology but I think it's an important one: there are no axes of rotation in 4D, only planes of rotation. It's a frequent source of misunderstanding as people familiar with 3D rotations, which can described in a coordinate free way as 'about an axis', look for similar axes in 4D. But they don't exist. While a axis of rotation exists only in 3D, a plane of rotation exists in all dimensions ≥ 2, and the rotation can be entirely described in terms of these planes (in four or more dimensions you need more than one plane for a general rotation). So planes are a much better tool. Unless you're working only in 3D you should forget about axes of rotation.--JohnBlackburne^words_deeds 12:27, 12 April 2014 (UTC)[reply]

Agreed. Dealing with multiple planes of rotation in 4-D can be fun.Naraht (talk) 22:59, 12 April 2014 (UTC)[reply]

Does the D₄-symmetric tricolouring work for the cells, constructing the 24-cell as ? Double sharp (talk) 14:27, 13 April 2014 (UTC)[reply]

I don't know, can you give me more information on that tricoloring either at a Wikipedia page or elsewhere?Naraht (talk) 14:25, 14 April 2014 (UTC)[reply]

A net of the 24-cell tricoloured that way is the leftmost third of this image:

. Double sharp (talk) 15:24, 14 April 2014 (UTC)[reply]

Oh that is the natural tricoloring for the cells. And is the dual of the coloring for the verticies that I had.Naraht (talk) 15:23, 15 April 2014 (UTC)[reply]

Commutative exponentiation

We know that addition and multiplication are commutative, but exponentiation is not. However, I also know this inconsistency:

Let's look at 2 number lines; the additive number line and the multiplicative number line. The additive line converts to the multiplicative line by replacing all $a$ with $2^{a}$ .

Every addition problem using numbers on the additive number line $a+b=c$ has a corresponding multiplication problem $2^{a}*2^{b}=2^{c}$ . That is, each number on the additive line becomes a number on the multiplicative number line. For example, $2+3=5$ becomes $4*8=32$ .

Every multiplication problem using numbers on the additive number line $a*b=c$ has a corresponding exponentiation problem $(2^{a})^{b}=2^{c}$ . That is, $a$ and $c$ are replaced with $2^{a}$ and $2^{c}$ , but $b$ is retained, not replaced by $2^{b}$ . For example, $2*3=6$ becomes $4^{3}=64$ , not $4^{8}=64$ .

Suppose we had used a commutative version of exponentiation. This form of exponentiation can be notated $commexp(a,b)$ . It can be defined as follows: for any $a$ and $b$ , $a^{b}=commexp(a,2^{b})$ . For example, $4^{3}$ becomes $commexp(4,8)$ .

Do you notice that this "commutative exponentiation" shares several properties with addition and multiplication?? These are the commutative property, the associative property, the identity property (2 is the exponentiative identity for commutative exponentiation,) the special-value property (for addition; this number is $-\infty$ ; for multiplication it is $0$ ; for commutative exponentiation it is $1$ ); and the inverse property (the exponentiative inverse of $a$ is the number $b$ for which $commexp(a,b)=2$ .

Any thoughts on whether this can go in any Wikipedia article?? Georgia guy (talk) 17:43, 10 April 2014 (UTC)[reply]

I forget the article's name, but it is probably already there: certain fields (such as real numbers) have an infinite sequence of endomorphisms via exponentiation. —Quondum 17:58, 10 April 2014 (UTC)[reply]

There exists commutative versions of hyperoperations, for example in Hyper operator#Commutative hyperoperations. --Mark viking (talk) 20:00, 10 April 2014 (UTC)[reply]

Nice find. I've also found exponential field, which is probably what I had seen. —Quondum 22:00, 10 April 2014 (UTC)[reply]