Probability of Going 3/3???

I know this has been discussed before, but does anyone have any source for the percentage of people that have passed the CFA while going 3/3? Other stats, such as average time to complete the CFA or percentage who enroll and eventually finish would be useful. Thanks for the help.

Good question, if Level I: 35% L II: 42% L III: 53% so i’d just say around 8% chances to do 3/3.

you cannot simply multiple the independent probabilities of each event because there are a substantial number of candidates who fail at each level that never end up obtaining the charter. there is no way that only 8% of CFA chartholders went 3/3. i would say it is in the 25-35% range.

Yeah, but less than 8% or so of people who sit for LI actually go 3/3.

not true there will be a bias as the # of people who go 2/2 are also more likely to go 3/3 than those who have needed 4 attempts for L1 and 3 attempts for L2.

In 2006, ~50k sat for LI, 40% (20k) passed. In 2007, ~25k sat for LII, 40% passed again, that means that from those LI’s only 8k passed in the unlikely event of all of them sitting for exam. 2008, 14k sitting and 53% rate, so ~4.5k passed. That’s 9% of those starting in 2006. Not lets see what bias brings. Let’s assume just for kicks that out of those 10k passing in LII, everyone was 2/2, and then everyone who passed LIII was 3/3. 7.5k out of 50k starting. 15%. Seeing that a) the candidate pool of just-passed is probably a bit weaker than skipped-a-year and retakers; b) there is a number of candidates who skips, I would guess the 3/3 population to be around 5%.

JDV, ever tried to come with an answer to this question ? Ponpon

no shows dont count in the pass rate. i remember 1/4 of the seats empty at level 1.

First of all, I think the question that the OP wants to answer is the following: “What percentage of Level I examinees successfully pass all three levels on their first try?” I don’t think there is sufficient public data to answer this. But I would guess the figure is in the 15 - 20% range.

joemontana makes a good point. The fact that the 34% who pass L1 include retakers means that the 3/3 percent could actually be lower than 8%. If the percentage who pass L1 on the first try is substantially lower than 34%, that could bring the percentage down substantially. We definitely don’t have enough information to work out the answer here. At the very least, we would need some kind of correlation information on how previous pass-on-first-attempt correlates to pass-on-next attempt, and we would need some estimate of how many of those 34% of L1 passers are in fact first time takers.

Its pretty simple: Whenever one takes the L1 exam, there is a 35% chance that he clears it. There is around a 45% chance to clear L2 on the next attempt (because only people who clear L1 can take L2) . And then 55% chance of L3. So the probability is .35 * .45 * . 55 ~ 8.6%

^ It’s not that simple. As bchadwick stated, we need to know more about what % of those passing is attributable to retakers. For example, if the pass rate for FIRST TIMERS was 30% for each level then the probability would suddenly be 0.30^3 = 2.7%

the question depends on whether the retakers have a better passing rate over the new entrants. The retakers are already familiar with the material, so it is reasonable that they have a better passing rate. Let’s assume the number of new entrants for L1 is 1. my estimate of the number of retakers is 0.7. assume the passing rate for new entrants is Y and the passing rate for retakers is r*Y, where r is larger than 1. (1*Y+0.7*r*Y)/(1+0.7) should equal 0.35. let’s solve Y for r = 1.2, 1.4, and 1.6 r= 1.2, Y=32.3% r=1.4, Y=30% r=1.6, Y=28% similarly for L2 and L3, the two deciding factors are the proportion of retakers to new entrants and the advantage of retakers over new entrants. if we assume the retakers have only minor advantage and r = 1.2. using the same scaling factor for all levels, we would get 32.3%, 41.5%, and 50.8% respectively for the first time passing rate for the three levels. the 3X3 passing rate would be 6.8%

btw i totally agree with our mentor JDV that “3-3 is an achievemnet to be privately proud of, not boast about” my hats off to those who finish the exam under heavy work load and family responsibilities, 3-3 or no.

mcpass Wrote: ------------------------------------------------------- > not true there will be a bias as the # of people > who go 2/2 are also more likely to go 3/3 than > those who have needed 4 attempts for L1 and 3 > attempts for L2. Of course! P(3/3 given 2/2) is greater than or equal to 0. In reality will be >0. P(3/3 given they needed 4 attempts for L1 and 3 for L2) = 0. Sorry, I agree with your logic just ina pedantic mood.

So I guess we’ll never know…interesting to discuss though. Thanks.

I’m sure CFAI has this statistic… they just choose not to release it. Let’s suppose that 35% of test takers pass level 1, but only 10% of first time L1 takers pass level 1. That would mean that, at most, 10% of test takers go 3/3, because only 10% get through the first hurdle with a chance to do it. If you then assume that passing L1 on the first try correlates perfectly with passing L2 and L3 on the first try (extreme and unlikely case), you find that 10% of test takers go 3/3, because they would then go on to pass L2 and L3. If the correlation is zero, you get 10%*42%*53% = 2.2%. The correlation between a first pass L1 and a first pass L2 is probably more than zero, less than one, although probably not too high. So the three variables you’d need (in addition to raw percentages) are: 1) Percent passing on first try 2) Correlation between a first try pass on L1 and first try pass on L2 3) Correlation between a first try pass on L1 and L2 and a first try pass on L3 You might be able to reduce the uncertainty to a relationship between two of these three variables by assuming that it all multiplies out to an average pass time of 4.5 attempts, or whatever that figure is. My guess is that there are a lot of failures at L1 that aren’t so correlated with ability, because people underestimate the intensity of the exam and come back as Rocky for a second go at it.

There is no way to answer the question without making assumptions. Correlations of passing Level 1, 2 and 3 lead to one answer. Another way to solve the problem would be to draw Euler diagram which helps come up with a range of probabilities of going 3/3. However, I don’t see value in solving the problem.