Probability/Chance up if applying more Ivies schools?

Parents Forum
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<p>That’s incorrect. H and Y and P decisions are independent, in every sense of that word. Probabilities of those decisions can be multiplied. This means that the “wrong” calculations tokenadult has so repetitively FAQ-ed about, were correct all along. </p>

<p>This is all a very different matter from the one you’re talking about, which is whether, given Harvard and Yale admission results (for one applicant or for all of them) we can draw conclusions about the Princeton results. The subject of the totally wrong FAQ was a specific type of calculation that does not proceed from any aggregated outcomes. </p>

<p>The only thing that is sometimes wrong in the argument is the set of probabilities that a particular applicant inserts into the calculations when using them to make strategic decisions as to whether more applications will make a difference. But the inability to know or, in many cases, to estimate, an individual’s probabilities at the time of application is another argument for the correctness of the strategic conclusions.</p>

<p>For example, I have argued here that a majority or near-majority of the 30000+ applications to Harvard and Stanford have a probability of literally zero (the schools are fleecing the applicants) and similar but only slightly less extreme fractions hold at the other top ten schools. The real selection is of 2-4000 from a pool of 6000 to 15000 remotely plausible candidates. However, if you don’t think that is necessarily correct, and are willing to send in a 60-100 dollar donation to the admissions office as a lottery ticket, the same logic leads to buying more tickets at the other schools, too. For applicants in all other probability ranges, chances that sending eight applications instead of two will make a difference, that is, the chance that one or more of the added attempts will come up “ADMIT!”, can be substantial. </p>

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<p>There are no underlying variables, and nothing to vary. The variables in question are constant; we are considering a single applicant. The only random variation implicit in this problem is random fluctuations in the outcome for a given applicant were the entire process to be re-run (on the same set of applications) over and over. i.e., a fixed admissions probability for each school. This probability is applicant-dependent but we are not varying the applicant. Rather, each applicant guesses his own probabilities, or considers several scenarios for what these probabilities might be, and makes his own calculation, using the correct method that tokenadult has FAQed as being “wrong”.</p>

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<p>A definition of independent events, from Charles M. Grinstead and J. Laurie Snell, Introduction to Probability, book published by the American Mathematical Society:
“It often happens that the knowledge that a certain event E has occurred has no effect on the probability that some other event has occurred, that is, that P(F|E) = P(F),” or in words, the conditional probability of F given that E has occurred is equal to the unconditional probability of F. In this case, also P(E|F) = P(E).
(Technicall, Grinstead and Snell also require the probabilities to be positive.)
Now let E = Student X is admitted to Harvard.
And let F = Student X is admitted to Yale.
Even though Harvard and Yale make their decisions completely independently (at least one assumes so), if one knows that event E has occurred, then student X has better than the raw odds of also being admitted to Yale. This is due to the features that (probably) caused X’s admission to Harvard.
Now, X has a personal probability of being admitted to Harvard, which is pH for X. This is not the raw probability–however, I do believe that the decision has an element of probability in it for an individual student, for reasons explained above. X can compute his/her personal probability of being rejected by both as (1 - pH) (1 - pY), if X knows pH and pY. This doesn’t mean that E and F are independent events.</p>

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<p>siserune and QuantMech</p>

<p>I really think you guys are just having a semantic argument. I think what siserune is saying is that for a given student, there is a probability of admission to Harvard that is pH. We assume first that you could possibly know this number, which you really can’t. We assume this probability will not change, regardless of whether the student is accepted to every other top 10 school, or doesn’t even apply to any other top 10 school. There is no official enhancing bonus point a person gets at one school if they are accepted at another school, nor is there some sort of demerit. The probability is what it is, and assuming you could possibly know this number, and a similar number for all the other schools where the student applies, you could use the basic rules of probability to compute the student’s chances at this school or any combination of schools.</p>

<p>What might change is your estimate of these number based on a student’s acceptance at another similar school. </p>

<p>At least that’s how I see it. I sort of agree with both arguments.</p>

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<p>Reasonableness like that can spoil a good argument.</p>

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<p>Oh, heh, I can reduce my “semantics” to the form of equations and numerical demonstrations. I thought that tokenadult and I were having a semantic disagreement.</p>

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<p>One needs to actually understand what the book says, not just who published it. </p>

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<p>This refers, as it always does in every orthodox textbook of probability, to “knowledge that event E has occurred” together with complete knowledge of which random process has produced E as an outcome — a single, specific, unambiguous, non-hidden probability distribution. It is never the case in probability calculations (such as the ones that tokenadult’s FAQ claims are “wrong”), and very commonly the case in statistical inference problems (an entirely different context that you are raising and is unrelated to the bogus-ness of the assertions in the FAQ), that the underlying random processes are unknown.</p>

<p>For instance, if you observe many tosses of the same coin, the results are clearly “independent” of each other. However, knowing the result of the first 999 tosses does provide information on the 1000th toss <em>if the coin’s probability of Heads isn’t considered as given</em>, because it provides an observational estimate of that probability. This does not change the fact that each toss is independent of all the others. Exactly the same is true if we toss three similar coins labelled “Harvard”, “Yale”, and “Princeton” (or “Joe Gets Into Princeton”). The coins are indeed independent, in every sense including that of Grinstead and Snell’s textbook quoted above. </p>

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<p>The formula correctly contradicts tokenadult’s claims in the “FAQ”. It also contradicts the second, incorrect, sentence about E and F not being independent. The problem is that you are confusing the clear, well-defined and entirely standard notion of “independent events” with a hazy undefined notion of “independent probabilities” (i.e., the ability or inability to presume relationships between the parameters defining one’s chances at different schools).</p>

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<p>I’ll just point out that I have Grinstead and Snell on my bookshelf because I’ve read it and worked the problems, a few years ago. But I’m off the thread at this point.</p>

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<p>A simple mathematical explanation is that your chance will always increase because your chance to get acceptance from any school is absolute 0% if you don’t apply.</p>

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<p>Disputing whether a formula is logically correct or produces numbers that are a valid basis for action in the real world, is the very opposite of a semantic argument.</p>

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<p>^^^
My point was that I believed, and still do,that you two actually agree but are operating from different definitions of “probability of admission”. Frankly, the mathematics of the discussion is not that complicated. I think you two are talking around each other.</p>

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<p>Again, there is a very concrete issue that cannot be reduced to terminological differences and communication problems: do the calculations that tokenadult’s FAQ claims to discredit actually work? That is, are they correct mathematically (can one multiply the assumed admission probabilities? does averaging the per-school probabilities cause a problem, or not?); and do the numbers that come out correctly suggest a many-application strategy such as the one advocated by cellardweller, or do they lead to what tokenadult calls “suboptimal application behavior”?</p>

<p>The nice thing here is that the rubber meets the road, numerically. The “semantic dodge” is not available. It’s a math problem, and a fairly specific one, and we are dealing with rather precise claims about the math problem.</p>

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<p>Okay, it’s an incurable addiction–I’m back. siserune, in Post #13 I suggested the calculations that tokenadult was trying to discredit, but with individual probabilities for a given applicant. I later tried to explain how a table of the outcomes at the two schools could make it look as if they were not statistically independent, even though the multiplicative probability calculation was valid. We agree on that.</p>

<p>Admittedly, there are some knotty difficulties in probability calculations. It’s claimed that Erdos never really got his head around the Monty Hall problem. And I do make mistakes from time to time (like R2D2). However, I don’t see a reason to retract anything I’ve written, based on the comments on this thread so far.</p>

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<p>Well yes siserune, I would say I agree with you and cellardweller (at least I think you two agree - I didn’t read that post in detail). I guess I just am not picking up on the major bone of contention betwen you and QM. I guess I’ll have to go back and read all the posts. Maybe tomorrow.</p>

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<p>The OP asked, back on 7 February 2010: </p>

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<p>Since many other issues have come up meanwhile, which I am finding quite interesting, allow me to say that, yes, as long as the applicant prepares the applications well, more applications is better. My general advice to applicants, “apply boldly to a range of colleges of varying selectivity,” is one way that I say that an applicant should apply to the colleges that the applicant would genuinely like to attend, and has some reasonable chance of being admitted to. It’s too bad when high school counselors discourage applicants from applying–one has to apply to have any reasonable chance at all. And of course I always advise applicants to build their application lists on the foundation of a [safety</a> college](<a href=“http://talk.collegeconfidential.com/1061209889-post28.html]safety”>http://talk.collegeconfidential.com/1061209889-post28.html) and add other colleges to the list according to well considered personal preference, based on whatever criteria are meaningful to the applicants. </p>

<p>Looking back on the thread that first really grabbed my attention on this issue years ago </p>

<p><a href=“http://talk.collegeconfidential.com/college-admissions/417447-if-you-score-2300-you-have-99-admit-chance-hypm-3.html[/url]”>http://talk.collegeconfidential.com/college-admissions/417447-if-you-score-2300-you-have-99-admit-chance-hypm-3.html&lt;/a&gt; </p>

<p>is interesting in light of the very detailed discussion here. When I wrote later on 7 February 2010 in [post</a> # 32](<a href=“http://talk.collegeconfidential.com/1064071808-post32.html]post”>http://talk.collegeconfidential.com/1064071808-post32.html) of this thread, </p>

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<p>I was writing in earnest, because I see there were already several people much more learned in statistics than I posting to the thread as I encountered it. I am convinced that the FAQ needs a rewrite. What I am still trying to wrap my mind around now :slight_smile: is how to express the core advice-to-the-applicant ideas about what to do in terms that neither assume too much nor too little about what an applicant actually knows about chances when applying to colleges. Applying to colleges is an exercise in judgment under uncertainty, and I have heard of more than a few cases of young people who are SURELY more mathematically astute than I am who have not gained admission to colleges where they (and I) thought they were shoo-ins. </p>

<p>I take it the OP’s son is a current junior or student in a later graduation year. Good luck to him finding a good list of colleges to apply to, and cooperation from the high school counselor sufficient to support all the applications. And good luck on the admissions results.</p>

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<p><a href=“QuantMech:”>quote</a></p>

<p>… in Post #13 I suggested the calculations that tokenadult was trying to discredit, but with individual probabilities for a given applicant.

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<p>That “discredited” method is indeed the correct one. The probability that additional applications are effective (i.e., that they lead to one or more acceptances beyond the original set of applications) is simply one minus the product of the rejection probabilities at each of the additional schools. </p>

<p>The orthodox description of this calculation in the standard language of probability and statistics is to say that it reflects the fact that a given applicant’s outcomes at different schools are “independent”. That is a well-defined technical term and it means, for current purposes, that the probability of any combined outcome such as Prob[Joe is admitted to Stanford, rejected by MIT, waitlisted by Harvard] is equal to a product of the per-school probabilities, in this case Prob[Joe admitted to S] x Prob[Joe rejected by M] x Prob[Joe waitlisted by H].</p>

<p>The positions most extensively discussed here are as follows:</p>

<p>A. (I say) The product calculations are correct and the term “independent” applies here.</p>

<p>B. (tokenadult, in FAQ and elsewhere, said) The product calculations are wrong because the concept of “independence” does not apply here.</p>

<p>C. (QuantMech apparently says) The product calculations are correct but the term “independent” does not apply.</p>

<p>QM, because you agree algebraically but disagree terminologically, either:

  • you mis-apply the ordinary terminology (e.g., by omitting a crucial piece of the standard definition of independence, as I mentioned in #146); or,
  • you know of some source corroborating your use of the term “independent”; or,
  • you apply the term correctly to something other than the ground disputed between positions (A) and (B), for example, to interesting but irrelevant examples of 2x2 tables.</p>

<p>I vote for the first and third possibilities. Also, when you write things like “the probabilities [of admission at H and Y and P] are not independent” there is no standard mathematical definition of what that could mean, and certainly no definition that has any implications for the current question of optimizing the number of applications. </p>

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<p>In post #3, you wrote: “Applying to multiple Ivies does not increase your chances as much as a mathematical model based on independent, random selection of students would predict.”. That looks like a succinct restatement of position (B), and if so, it is totally wrong. Independent random selection is precisely the correct mathematical model for what would happen under different application strategies.</p>

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<p>On the last point raised by siserune, let me try to clarify: Applying to multiple Ivies does not increase the chances of most applicants as much as a model based on independent, random selection of students would predict.</p>

<p>I mean the following: if random student X takes the raw odds of admission to each of multiple Ivies–call those rH, rY, and rP–and computes the odds of being rejected by all as (1 - rH) (1 - rY) (1 - rP), from that, X would derive misleadingly high odds of X’s being admitted to at least one. This calculation would work only if the admissions process were totally random, so that students were admitted by drawing names out of hats. </p>

<p>Clearly this calculation could not be expected to work for a student with a 2.6 GPA in an easy curriculum at an undistinguished high school, with 1480 SAT I (CR + M + W), 1360 SAT II (total of 3 combined), no EC’s, and no hardships to overcome whatsoever.</p>

<p>On the other hand, I’ll go out on a limb a little, to say that I don’t think that this calculation will work for the “typical” Ivy applicant, either. The admissions rates in the 7-10% ballpark include a certain number of students whose odds of admissions are exceptionally high, either due to stellar academic qualifications (really, really stellar), world-class EC’s, or status as a 7th generation legacy from family of major donors or as the best quarterback who meets Ivy-agreed admissions policy . . .</p>

<p>However, I do think that there is an element of randomness in the process (some disagree). Therefore, I think that <em>if</em> a student could know his/her own personal pH, pY, and pP, the student could figure out how much the odds of at least one acceptance will go up, when another school is added to the list, and the multiplicative calculation will work.</p>

<p>With regard to item c) on siserune’s list: I admit to one assumption, namely that the overall probabilities and the conditional probabilities for a pair of schools can all be calculated from the 2 by 2 tables of observed outcomes. In fact, this is the sort of situation where I think that an infinite number of trials (i.e., an infinite number of students submitting applications to both schools) would be required to determine the true probabilities. However, if the common expectations about samples apply here, then I believe that if you aggregate 5 or 10 years of the total numbers of yes/yes, yes/no, no/yes, no/no decisions at Harvard and Yale, you will probably have a fair estimate of the probabilities and conditional probabilities.</p>

<p>siserune, the closest I can figure about our disagreement on point c), about the 2 by 2 tables, is that you are thinking about an individual applicant, while I am considering the entire population of cross-applicants. Is that it? If not, how do you derive the conditional probabilities for an entire population of applicants, from the observed numbers of zero, single- and cross-admits? I doubt that you can construct a realistic 2 by 2 table that gives conditional probabilities that are identical to the raw probabilities–I think this could only be done with an unrealistic underestimation of the number of cross-admits.</p>

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<p>I haven’t read this entire post, so if this has been mentioned, my apologies.</p>

<p>Our high school has has a very high acceptance rate for those students applying to Yale, Brown, Columbia and Cornell, (25%), but is much lower (10-16%) for Harvard, Princeton, Dartmouth and Penn. This is per our Naviance websites. So obviously there is something about the record and relationships between those ivies, and our high school that makes our school’s applicants more attractive.</p>

<p>So, I would say, it isn’t a math problem exclusively. But, like everything else in the universe, there are relationships that make some things work better than others.</p>

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<p>What I think bears repeating is that an 8% admission rate doesn’t translate into an 8% chance of admission for any particular applicant. The actual odds for a particular applicant are very difficult to identify. There may be two colleges with an 8% admission rate, and the same applicant may have very different chances of admissions at the two because of his personal characteristics.
As for the math problem, it seems to me that it is trivially true that whatever his odds are of admission to each of a group of schools, his odds of being admitted to at least one of the group is greater than his odds of being admitted to any particular one. How much more is the question–if he’s a D student, and the group is Harvard, Yale, and Princeton, the increase in odds is infinitesimal–maybe nonexistent.
This means that if you have a larger group of schools to apply to, your odds of being admitted to at least one of them is increased, as long as there is not some kind of diminishing return that hurts your chances (like lower quality applications).</p>

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<p>I quote this because I also think it bears repeating. I’d like to hear from anyone here just how it is that an applicant knows, before submitting applications, what the applicant’s numerical probability of admission is at each college.</p>

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That’s simple! You start a “What Are My Chances” thread here on CC, and then average all the predictions. The more predictions there are, the more likely the resulting average is to be accurate. Right?</p>