Wikipedia:Reference desk/Archives/Mathematics/2015 February 2

= February 2 =

Quiz Question
In a newspaper in the UK there was a quiz question that stumped me, I'd be grateful for any hints.

The question is: In one season of a football (soccer) league, one sixth of the goals were scored by defenders, two and a half times as many strikers, eighteen were own goals, and the remaining quarterby midfielders. If none were scored by goalkeepers, how many were scored by strikers? I added 1/6th to 5/12ths and thought that the total ( including the 18) would represent three quarters of the total and tried to solve for n, but got it wrong. What I can't understand is the own goals. As the could be scored by anyone, I can't see how you can write an equation that allocates them to any category. The answer is 45. Could anyone tell me how to solve it, many thanks.Widneymanor (talk) 10:21, 2 February 2015 (UTC)
 * I think you had the right method, but made an error in your arithmetic somewhere (or ended up looking for the wrong value). Since the soccer team contains only defenders, strikers, midfielders, and goalkeepers, own goals must not be counted in the other totals (possibly because they aren't thought of as the player "scoring").  3n/4 - 1n/6 - (2.5)n/6 = 9n/12 - 2n/12 - 5n/12 = 2n/12 = 18, so n/12=9, so 5n/12=45 (the number scored *by strikers*) MChesterMC (talk) 10:42, 2 February 2015 (

Thank you very muchWidneymanor (talk) 13:34, 2 February 2015 (UTC)

Probability of cashing in
If I place a $1 bet on a six horse accumulator each day, and I always bet on the favourites, given that a favourite horse will win about 1/3 of the time, how many days will I have to bet before I win?

(I know statistics doesnt work like that but I'm not sure how to phrase the question - is it how many days do I need to bet before my chances of winning become 1? I know I could still bet forever and not win but I hope you understand the thrust of the question.) 5.148.151.18 (talk) 10:45, 2 February 2015 (UTC)
 * A good question to ask is what is the expected number of times to bet until you win. The answer is 3.
 * A different question is how many days to bet until your probability of winning at least one is greater than $$p$$. The answer is $$\frac{\log(1-p)}{\log(2/3)}$$, rounding as necessary. -- Meni Rosenfeld (talk) 12:34, 2 February 2015 (UTC)

Thanks - although I think if there was one horse with 3.1 odds I would have to bet 3 times, but this is a six-horse accumulator, meaning each horse needs to win, each with a 33% chance of winning the race. 5.148.151.18 (talk) 13:15, 2 February 2015 (UTC)
 * I forgot to mention that I have no idea what "six horse accumulator" means. I assumed that each day you bet on a horse, each day you have a probability of 1/3 to win, and the different bets are independent of each other. -- Meni Rosenfeld (talk) 14:38, 2 February 2015 (UTC)


 * Until we hear back on what an accumulator is in this sense, here's some WP articles that relate to the problem at hand: Martingale_(betting_system), Martingale_(probability_theory), Gambler's_fallacy, Gambler's ruin, Mathematics_of_bookmaking, Horse_racing. SemanticMantis (talk) 15:20, 2 February 2015 (UTC)

Apologies, you choose 6 horses and place a $1 bet. If the first one wins, the money won rolls to the second etc. If they all win, you get the payout. Assuming each horse has a 1/3 chance of winning, what is the expected number of days I need to bet in order to win? (assuming I have phrased that correctly, see above) 5.148.151.18 (talk) 16:43, 2 February 2015 (UTC)


 * If there is a six horse accumulator bet then the horses selected in each of six races must win for the bet to pay off. If each horse has a probability of winning of 1/3 then the probability that all six will win is (1/3)^6, which is 1/729. That is, this bet will pay off successfully, on average, once every 729 days. Further, if you start following this strategy today the "expected" number of days before your first win will be 729 days. What mathematicians mean by "expected" is "the average number of days you'll have to wait"; you might get lucky and get your first win before this time, or you might be unlucky and have to wait much longer, but "on average" you'll have to wait 729 days. RomanSpa (talk) 16:52, 2 February 2015 (UTC)


 * Also note that there is no number of bets that will guarantee a win. The more bets you make, the better your chance of having a win, but that chance never gets to 100%.  And, by the time you do win, you are likely to have lost more money than you won. StuRat (talk) 16:57, 2 February 2015 (UTC)

Pólya Problem to Prove (Part I Chap 19)
Pólya's hypothetical teacher poses the following problem to prove to the hypothetical student: ''Two angles are in different planes but each side of one is parallel to the corresponding side of the other, and also has the same direction. Prove that such angles are equal. Through some work, he introduces the auxiliary problem to solve, adding an extra line segment, turning the angles into triangles, and going to prove that the triangles are congruent, but it gets to where the "teacher" says "Did you use the hypothesis? What is the hypothesis?" And the "student" says "We suppose that AB || A'B', AC || A'C'. Yes, of course, I must use that." But then the "teacher" says: "Did you use the whole hypothesis? You say that AB || A'B'. Is that all you know about these lines?" and the "student" says: "No; AB is also equal to A'B', by construction. They are parallel and equal to each other. And so are AC and A'C'."''

That last statement by the "student" is what I don't get from the original problem statement. I copied the original problem statement verbatim in full. From that first italicized bit of text at the beginning of my question, where is it to be gotten that any line from one angle is also equal to any other by construction? It seems to me only to say the angles are in different planes, and that corresponding sides are parallel and in the same direction. I don't see anything about construction or how I can assume lengths are equal from that statement. Can anyone explain? 20.137.2.50 (talk) 18:34, 2 February 2015 (UTC)
 * Here's my attempt to reconstruct (in outline) the proof Pólya has in mind; not sure that it explains the statement but it can't hurt. First prove a lemma (*) that if AB = A'B' and AB || A'B' and in the same direction, then AA'B'B is a parallelogram. This should be easy, even if you don't assume the points are in the same plane. Note the lemma is false if AB and A'B' are in opposite directions; you get a self intersecting quadrilateral.


 * We are given two angles A, A' with corresponding sides parallel and in the same direction. Add points B, B', on corresponding sides so AB = A'B' and C, C' on the other sides so AC = A'C'. AB || A'B' and AC || A'C' so AA'B'B and AA'C'C are parallelograms by (*). Then BB' = AA' = CC' and BB' || AA' || CC'. Apply (*) again to get BB'C'C a parallelogram. This gives you BC = B'C' and, with AB = A'B' and AC = A'C', ABC ≝ A'B'C'. --RDBury (talk) 17:19, 3 February 2015 (UTC)

percentage
Is 0.3408 % 0.0408 percent than 0.3 % or 13.6 % greater? — Preceding unsigned comment added by 199.119.235.181 (talk) 18:56, 2 February 2015 (UTC)


 * It depends on the base:

0.3408% is 0.0408% of 100 more than 0.3%

0.3408% is 13.6% of 0.3 more than 0.3%


 * Now, where would you use each ? Let's say the interest rate on your checking account just went up from 0.3% to 0.3408%.  The bank might claim that's a 13.6% increase, but you will only make .0408% more on your money, so I'd use that figure.  In other words, you're still not getting much. StuRat (talk)

So if 0.3 % of population group A has a disease, and 0.3408 % of group b has a disease, group b has 0.0408 percent more people with disease than group a.  — Preceding unsigned comment added by 199.7.157.123 (talk) 23:01, 2 February 2015 (UTC)