Couple questions/thoughts: Let’s say I know 50% of answers stone cold. The other 50% I am familiar enough with to eliminate one of the choices, hence I have 50% probability of guessing correctly. Does this make me good for 75% on the exam? I thought I saw a thread somewhere where some said this logic doesn’t hold up. Btw, the 75% comes from 50%(100%)+50%(50%)= 75% I think the logic is fine. Also, if you can eliminate one choice, do you try to guess correctly or always guess “A” for example (or the first available one if you’ve eliminated A already). By doing this you try to systematically(no username pun intended) guarantee yourself 50% when you don’t know the answer vs falling prey to picking an answer that “sounds correct” but is intended to fool an unprepared candidate. Or do you try to guess correctly? Also, if a hypothetical question answer table says: ROI P/E 4% 10 4% 12 5% 12 Do you assume the middle column is correct because it contains the 2 “grouped” choices. For example, if someone only knows how to calc ROI @4% they want to force them to guess between the PE being 12 or 10. Conversely, if someone only knows how to calc the PE of 12, they want to force them to guess between the ROI of 4 or 5 percent. By having this type of question asking method the answer is defaulted to the 4% and 12 answer. Anyways, just wanted to throw this stuff out there… I feel like these things might be able to make a slight difference (or I just believe in ridiculous shit). Btw, I’ve been studying hard… just want ppl’s thoughts on these things.
First, guessing on 50% even if you eliminated 1 choice will not garantee you getting half them right as you should know, it is a distribution, an easy one I belive, it will be symetrical and you have a 50% right would be in the center or the distribution, 0 on one end and 100 on the other end P(0) = 0 P(1) = 0 P(2) = 0 P(3) = 0 P(4) = 0 P(5) = 0 P(6) = 0 P(7) = 0 P(8) = 0 P(9) = 0 P(10) = 0 P(11) = 0 P(12) = 0 P(13) = 0 P(14) = 0 P(15) = 0 P(16) = 1.0617e-12 P(17) = 5.246e-12 P(18) = 2.419e-11 P(19) = 1.044e-10 P(20) = 4.2282e-10 P(21) = 1.6107e-9 P(22) = 5.784e-9 P(23) = 1.9615e-8 P(24) = 6.2932e-8 P(25) = 1.9131e-7 P(26) = 5.5187e-7 P(27) = 0.0000015125 P(28) = 0.0000039434 P(29) = 0.0000097904 P(30) = 0.000023171 P(31) = 0.000052321 P(32) = 0.00011282 P(33) = 0.00023247 P(34) = 0.00045811 P(35) = 0.00086386 P(36) = 0.0015597 P(37) = 0.0026979 P(38) = 0.0044729 P(39) = 0.0071107 P(40) = 0.010844 P(41) = 0.015869 P(42) = 0.022292 P(43) = 0.030069 P(44) = 0.038953 P(45) = 0.048474 P(46) = 0.057958 P(47) = 0.06659 P(48) = 0.073527 P(49) = 0.078029 P(50) = 0.079589 P(51) = 0.078029 P(52) = 0.073527 P(53) = 0.06659 P(54) = 0.057958 P(55) = 0.048474 P(56) = 0.038953 P(57) = 0.030069 P(58) = 0.022292 P(59) = 0.015869 P(60) = 0.010844 P(61) = 0.0071107 P(62) = 0.0044729 P(63) = 0.0026979 P(64) = 0.0015597 P(65) = 0.00086386 P(66) = 0.00045811 P(67) = 0.00023247 P(68) = 0.00011282 P(69) = 0.000052321 P(70) = 0.000023171 P(71) = 0.0000097904 P(72) = 0.0000039434 P(73) = 0.0000015125 P(74) = 5.5187e-7 P(75) = 1.9131e-7 P(76) = 6.2932e-8 P(77) = 1.9615e-8 P(78) = 5.784e-9 P(79) = 1.6107e-9 P(80) = 4.2282e-10 P(81) = 1.044e-10 P(82) = 2.419e-11 P(83) = 5.246e-12 P(84) = 1.0617e-12 P(85) = 0 P(86) = 0 P(87) = 0 P(88) = 0 P(89) = 0 P(90) = 0 P(91) = 0 P(92) = 0 P(93) = 0 P(94) = 0 P(95) = 0 P(96) = 0 P(97) = 0 P(98) = 0 P(99) = 0 P(100) = 0 using the numbers above you can calculate the probabilty of you getting x many correct out of a 100 you guessed on randomly after eliminating 1 choice ----------------- regarding the chosing the same answer for each time you random guess if the answers are randomly set by the CFAI, and they probably use some random program to switch the options around, so that C on your paper is not the C on my paper in that case, guessing randomly or selecting the same answer would give you the same result if it is on a test of 1000000 questions and assuming you are a true random number generator (which you cant be, cause nothing is, and esp you cause you will be biased) if you start getting biased towards one of them, and they are truly random, you would be hurting your chances, so best stick to one of them again all this is theoretical statistics stuff that would apply under large numbers, in a test of a 120 questions, even if the CFAI is using a random order of questions, they may not end up being so random lool off to bed, i hope something in here made sense
Yo Gulf, get your @ss back to studying.
^please move in with me, i will pay you to beat me with a club every time I do stupid shit insted of studying you would get rich.
Gulf, I’m guessing you just threw that in excel. What Function are you using? 1-.5^nth? Somehting like that? To me, this is a binomial distribution where its 1 or 0, right or wrong. Hence at every question my odds are 1:1 therefore in aggregate my expected return is 50% Your distribution actually proves that. 50% is the mean/median/mode> so you should expect to get 50% right with some variance. 68% of a outcomes are between 46-55, so we have a standard deviation of about 5% Btw, I didn’t truly calc the std deviation, I just grabbed the 10 around the mean which happened to be 68% which is about 1 std deviation At first glance, it appears that the odds of getting 50% in correct are low. But getting 46-55 are infact very high
binomial. Use the formula from level 1 book with permutation and stuf. I just used an online calc
That was complete gibberish. I feel dumber from just reading those long few paragraphs.