Wikipedia:Reference desk/Archives/Mathematics/2009 September 29

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September 29

Pair and more pairs

I was playing poker today, and in one hand, four of us had pocket pairs; and, even more unusually, two of us had the same pair. (In other words, of the nine players at the table, there was an AA, a JJ, a 99, and a 99.) It got me to thinking about it, and I'm asking here because I'm both too damned lazy to figure it out myself, and my math for such things is a bit rusty. Here's my question: Given N sets of pairs, what is the probability that two of the pairs match? --jpgordon^{::==( o )} 05:53, 29 September 2009 (UTC)

Since you assume N pairs this is close to the Birthday problem. Although care must be taken since the pairs are not independent. Taemyr (talk) 12:23, 29 September 2009 (UTC)

I just started to try to work this out and it looks a bit tricky since the probabilities of the hands being a pairs are not independent as you have in the birthday problem. For example the probability of another player having a pair goes up if you have four of a kind. It might be interesting to compute the amount of the change but if you assume it's negligible then you essentially have the birthday problem.--RDBury (talk) 14:19, 29 September 2009 (UTC)

Yeah, I had the birthday problem in mind. To refine: I'm playing Texas Hold'em, and I'm concerned solely with the hole cards ("before the flop"). PS: I had one of the pairs of nines; I got the hell out of there when the AA and JJ started crazy raising. The other 99 was, apparently, overly optimistic. --jpgordon^{::==( o )} 14:22, 29 September 2009 (UTC)

Open for criticism, but here is my take: If you are dealt first, the probability of the second player getting a matching card would be 3/51. As the deal comes back around to you, your probability of getting a matching card, given that the second player has matched, is 2/50. And the probability of the second player getting the fourth matching card, given the preceding, is 1/49. Something like 8E-6 when multiplied together.Alfrodull (talk) 21:34, 30 September 2009 (UTC)

But there are 3 other players who might have had the second pair of 9's.70.90.174.101 (talk) 08:55, 1 October 2009 (UTC)

Right. My original question (before I posted it here) was: how many pairs do there need to be before the probability of two identical pairs is greater than .5? Forget the probability of there being so many pairs in the first place; that's the given. N pairs exist; what's the probability that two of them are identical? --jpgordon^{::==( o )} 15:39, 1 October 2009 (UTC)

Darn. Was hoping someone would find this an interesting puzzle. --jpgordon^{::==( o )} 05:45, 4 October 2009 (UTC)

I think I may have worked out the answer, and it is a slightly modified version of the birthday problem. There are a total of 26 pairs in the deck of cards. If we know there are only two pairs, the chance of them matching is one in 25 (one pair matches, the other 24 don't). If we have two that don't match each other, the chance of a third pair matching either of them is one in 24 - or one in 12 for both. The probability that there are matching pairs in three pairs is the probability the first two match (1/25) plus the probability that the third matches either, given that the first doesn't match (1/25 + 24/25 x 2/24). Similar logic follows for four or more pairs, giving the table:

Total number of pairs	Likelihood of matching pairs
2	4.0%
3	12.0%
4	23.5%
5	37.4%
6	52.3%
7	66.6%
8	78.9%
9	88.3%
10	94.5%
11	97.9%
12	99.4%
13	99.9%
14	100%

The result for 14 is obvious - if there is no matching pair in the first 13, then all possible pairs are covered so the fourteenth must find a match.

So whilst the likelihood of getting 4 pairs in the first place is quite low, once you know it's happened, the likelihood of finding a matching pair is almost one in 4.

And, to answer the initial question, there need to be 6 pairs before the likelihood gets more than 50%.

Please feel free to correct if I'm wrong.Fletch79 (talk) 18:16, 4 October 2009 (UTC)

It feels right, but maybe not precise. Yes, for 2, there are 1 pair that match for the 24 that don't. But the probability of getting that one pair that matches is less than the probability of getting any of the of the other pairs -- there are six ways to make pairs out of the non-paired cards, and only one for the other. And that's where my head got stuck. --jpgordon^{::==( o )} 05:37, 5 October 2009 (UTC)

Hmm, that got me thinking... I think you're right. So what we're saying is that there is only one way of picking the "right" pair, but six ways (rather than the two I assumed above) of picking each of the "wrong" pairs.

Here's another attempt (the answer doesn't feel intuitively as right as the previous one but that doesn't necessarily mean it's wrong!) n = 2: number of ways of picking matching pair = 1, number of ways of picking non-match = 12 x 6 = 72. Likelihood of matching pair is 1/73 = 1.4%. n = 3: number of ways of picking matching pair = 2, number of ways of picking non-match = 11 x 6 = 66. Likelihood of matching pair is 2/68 = 4.3%. Total likelihood of a match in 3 pairs is (1/73 + 72/73 x 2/68) = 4.3%.

In a table (horizontally this time):

Total number of pairs	2	3	4	5	6	7	8	9	10	11	12	13	14
Likelihood of matching pairs	1.4%	4.3%	8.8%	15.1%	23.1%	32.7%	43.7%	55.5%	67.7%	79.2%	89.2%	96.4%	100.0%

Interesting problem - still not totally sure that's the right answer (9 seems too high) Fletch79 (talk) 22:05, 5 October 2009 (UTC)

Ellipse theorem

In Proposition XI, Problem VI of the Principia (the inverse square law for ellipse focus), Newton refers to the 'writers on Conics' for proof of this property of the ellipse : all 'tangentially circumscribing' parallelograms have the same area.

I've tried Appolonius and Archimedes for this theorem but cannot find it there. Geometric proof anyone?

Dunloskinbeg (talk) 06:48, 29 September 2009 (UTC)

That doesn't sound right at all to me, are you sure you're quoting him right? A circle is an ellipse and you can get as large an area as you like with a rhombus round it. Dmcq (talk) 09:55, 29 September 2009 (UTC)

It sounds plausible to me. Fix a circle with radius r and then any circumscribing parallelogram that's tangent to the circle in four places must have some fixed area, namely 4r²? Think of the unit circle and the square ∂{(x,y) : |x| ≤ 1 ∧ |y| ≤ 1}. Any other circumscribing parallelogram must also be a square, obtained from ∂{(x,y) : |x| ≤ 1 ∧ |y| ≤ 1} by rotation about the origin. Taking special affine transformations (SL(2,R) $\ltimes$ R²) of this we get parallelograms (of area 4) circumscribing ellipses of (area π)~~ Dr Dec (Talk) ~~ 10:16, 29 September 2009 (UTC)

This:

Think of the unit circle and the square (...). Any other circumscribing parallelogram must also be a square (...)

is false. Think of a unit circle $x^{2}+y^{2}=1$ and a rhombus with vertices $(\pm 1/\sin \alpha ,0)$ and $(0,\pm 1/\cos \alpha )$ for some small alpha. --CiaPan (talk) 11:45, 29 September 2009 (UTC)

Yeah, quite right. Silly me. I didn't think it through properly. ~~ Dr Dec (Talk) ~~ 16:10, 29 September 2009 (UTC)

I agree with Dmcq. For a circle of radius 1 the tangential parallelogram of smallest area (4) is a square, but by bringing an opposite pair of vertices closer the other pair expand. and the area increases. In the limit a pair of parallel lines enclose infinite area.→217.43.212.252 (talk) 10:27, 29 September 2009 (UTC)

Also agree with Dmcq. Four tangents to $x^{2}+y^{2}=1$ at $(\pm \cos \theta ,\pm \sin \theta )$ enclose a rhombus with area $4/|\sin(2\theta )|$. This area is minimised when $\theta =\pi /4$, which is the square case, but can be made as large as we like by a suitable choice of θ. Gandalf61 (talk) 10:43, 29 September 2009 (UTC)

Hah! The explanation is in De motu corporum in gyrum article, first section 'Contents of "De Motu"', part 2 Lemmas:

All parallelograms touching a given ellipse (to be understood: at the end-points of conjugate diameters) are equal in area.

Conjugate diameters do not have their own article on Wikipedia nor in http://mathworld.wolfram.com/ but they are two such chords of an ellipse that each of them halves all chords parallel to the other. --CiaPan (talk) 12:41, 29 September 2009 (UTC)

Now they have, thanks to Jim.belk — see conjugate diameters --CiaPan (talk) 06:31, 1 October 2009 (UTC)

Now the proof is quite simple: an ellipse is an image of a circle in some affine transformation; a pair of conjugate diams is an image of a pair of perpendicular diams of the circle, and the parallelogram is an image of a square tangent to the circle. All such squares have equal area, and area ratios are preserved by affine transformation (as long as it's not degenerate, i.e. detA≠0), so all parallelograms considered have equal areas, too. --CiaPan (talk) 12:52, 29 September 2009 (UTC)

I've seen the phrase "Writers on X say..." or "Books on X say..." before, I think always in books written before 1900. Nowadays of course we use the much more self explanatory "It is well known that...".--RDBury (talk) 14:43, 29 September 2009 (UTC)

Sorry, I'm in fact not as fluent in English as may seem, and do not get your point. Do you mean I missed to give some references? AFAIR I did not use 'it is well known' and can't guess where the problem is. Which point, term, implication needs explanation? --CiaPan (talk) 15:08, 29 September 2009 (UTC)

Relax. RDBury's misindented post was apparently not meant to you, it comments on the OP's quote of Newton referring to the "writers on Conics". — Emil J. 15:50, 29 September 2009 (UTC)

Oh, yes, my English is really weak sometimes... You wrote 'misindented' but I've read 'misintended'. That still was making some sense, however a bit different from what you meant, I suppose... LOL at me.
Thank you for explaining the RDBury's indentation/intention. --CiaPan (talk) 07:12, 30 September 2009 (UTC)

Special thanks to User:Jim.belk for creating an article on Conjugate diameters. --CiaPan (talk) 06:31, 1 October 2009 (UTC)

I'm happy to help. It's just a stub for now, so please add any other information or references that you might know. Jim (talk) 16:00, 1 October 2009 (UTC)

Interest on Savings Account

My bank offered 3.75 compounded monthly on my account? Is there a big difference on 20,000 being compounded monthly and 20,000 being compounded daily? My other banks compound daily? Please advise. tks —Preceding unsigned comment added by 74.136.110.175 (talk) 14:27, 29 September 2009 (UTC)

3.75% compounded monthly gives you an annual percentage yield of about 3.815%. 3.75% compounded daily gives you an annual percentage yield of about 3.821%. The difference in annual interest on a principal of $20,000 is about $1.20. Gandalf61 (talk) 14:44, 29 September 2009 (UTC)

On the other hand, if you deposit on the 1st of the month and withdraw on the 30th of the same month then your interest from the first bank is 0. Anyway, aren't banks supposed to give you APR's so you don't have to do this kind of computation?--RDBury (talk) 14:50, 29 September 2009 (UTC)

When a percentage is quoted, it's often an APR - even if the compounding interval is some other amount. So, the next question would be whether the OP changes his balance on a day-to-day basis. Two accounts with equivalent APRs will earn the same amount over a year, unless the balance changes inside a compounding period (deposits and withdrawals - not just accruing interest). Depending on the type and frequency of such transactions, this can be either in favor of the bank or the customer. Nimur (talk) 15:04, 29 September 2009 (UTC)

I very much doubt that any bank would accrue interest on a month-end balance - that is open to obvious abuse, if a customer deposits a large amount just before month end and withdraws it a few days later. Interest will be accrued daily even if it is only compounded monthly. Any, key point here is that difference between daily and monthly compounding is minimal, amounting to less than one day's interest over a year. Gandalf61 (talk) 15:28, 29 September 2009 (UTC)