Wednesday, April 16, 2008

The odds of getting an H-1B visa

I just read about this. US Citizenship and Immigration Services (USCIS) received 163,000 applications for H1-B visas by April 7. Since that number exceeds the visa cap, the government will not consider any more applications this year (Source: USCIS, via the H1B data blog). 31,200 of the applications were from individuals with an advanced degree. The government quotas are: 20,000 for applicants with an advanced degree and 65,000 for the rest. (I wrote about the H1-B visa system a couple of weeks ago.)

The USCIS will hold two lotteries this week. The first one is for applicants with an advanced degree from a US institution (MA or higher). Applicants who are not selected in that first lottery will be pooled with the rest of the applications in the second lottery.

Unless my math is failing me, the probability of getting a work visa is then 80.4% for advanced-degree holders, and 45.5% for the rest of the applicants.

UPDATE: I'm really behind on this. USCIS already conducted the lottery (they did it on April 14, two days ago). Lucky applicants should get a notification by early June. I really recommend reading H1B data if you want timely information.

6 comments:

Anonymous said...

the odds for an individual having a masters quota getting selected should be p(a)+ p(b) where
p(a) = 20,000/31,200 : event of getting selected in the masters lottery
p(b) = number of ways that a given element occurs in the selection of 65000/ number of ways of selecting 65000 elements out of a lot of 143,000 elements

ie(163,000-20,000)

The probability of a non-masters candidate getting selected is p(b)

Francisco said...

Only people who are not selected in the masters lottery participate in the general lottery. The probability for an MA holder of getting a spot through the second lottery is then

(1-p(a))*p(b)

(in your notation)

Therefore the probability of getting selected through some lottery for an MA holder is

p(a) + (1-p(a))*p(b) =

20,000/31,200 +

(1-20,000/31,000)*(65,000/143,000)=

0.64 + 0.36*0.45 = 0.804, i.e.

80.4%, as I say in the text.

Voilà.

The key is that the second lottery is conditional on the result of the first lottery, for an MA holder.

Review my answer and let me know if I made any mistake.

Thanks for visiting EconWeekly!

Anonymous said...

Your answer seems to be fine enough. In math, there can be more than one way to reach the same answer, as you are aware of. My best wishes to all the anxious students on OPT waiting for their H1B!!! Thanks and good luck with the blog!!!

Rahul said...

hi,

i really dont know about this visa plese write somethig more about it as i know about the H 1b visa by this visa man can cary his wife and childern both .




***********************************


Rahul Kumar


Massachusetts Treatment Centers

sudheer said...

now a days getting H1-B is very difficult.Bcoz of huge competition.
=================================
sudheer
http://www.worldinfo.com

Mandy said...

Hi Francisco, thank you for the article. It's really helpful.
I heard a comment that applicants from larger company will have higher chance of winning the H1B visa than applicants from smaller company. Do you know if it is true or not?
But this H1B lottery is a computer-generated random selection. Wouldn’t that mean every applicant will have the same chance of winning the H1B?
I am not in the drawing this year but I will be applying for H1B next year. I am trying to decide if I should go with a larger company that sponsor more H1B or a small company that I like better.