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KnackChap

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18 minutes ago, Mr. Comingle said:

Won't it double for every new application. 

For eg - if an ipo is 8 times oversubscribed, you have 1/8 chance of getting it. But if you have 2 applications, you have 2/8=1/4 chance of getting it. Similarly it goes up to 1/2 if you have 4 applications. 

I'm sure I'm doing this wrong but not sure where. 

That's correct but when you are dealing with such large numbers the probability doesn't always reflect the reality.

 

Expectation value is a better metric i.e. if you repeat the same N number of times what will your avg. value be and given that the odds remain the same, chances of getting an IPO will considerably drop (i shouldn't have said that it will be insignificant) as we don't know whether we will be dealt the avg. up or avg. down case.

 

Probability as a metric is good for simple instances and maybe I am over analysing this but there have been occasions where my friends applied for more lots than the subscription number and didn't get any.

Edited by KunjanPSD
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35 minutes ago, Mr. Comingle said:

Won't it double for every new application. 

For eg - if an ipo is 8 times oversubscribed, you have 1/8 chance of getting it. But if you have 2 applications, you have 2/8=1/4 chance of getting it. Similarly it goes up to 1/2 if you have 4 applications. 

I'm sure I'm doing this wrong but not sure where. 

Pretty sure that's not how it works .

Otherwise if you apply with 8 times for the above IPO then you should have 100% chance of getting an allotment which is not the case . Even if you apply with 8 pan there remains a small chance you may not get one .

Chances of getting success are calculated like this . 

1 -  chances of not getting subscriptions in all applications ( which is a function of and/x) 

For 4 applications - 7/8 * 7/8 * 7/8 * 7/8 = 0.586 

Hence chances of success 1 - .586 = .414 or 41.4% 

For 8 applications - .586 * .586 = .343

Chance of Success 1 -.343 = .637 or 63.7 %

Edited by Krazyniks
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19 minutes ago, Mr. Comingle said:

Won't it double for every new application. 

For eg - if an ipo is 8 times oversubscribed, you have 1/8 chance of getting it. But if you have 2 applications, you have 2/8=1/4 chance of getting it. Similarly it goes up to 1/2 if you have 4 applications. 

I'm sure I'm doing this wrong but not sure where. 

 

I am no statistician, but I think to improve your odds and half them, you would need to hold half of the pool. 

 

So, lets say there are 100 applications (not applicants) and 5 shares, the odds per application is 1 in 20. Now assume that you hold 50% of that application pool. The per application odds, remain at 1/20, but the probability of who holds those applications kicks in, and that is 50% in your favor. So, the only way to reduce your personal odds to 1/10 is to hold 50% of 100 applications...... I think. I could be wrong.:fear1:

 

 

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5 minutes ago, Krazyniks said:

Pretty sure that's not how it works .

Otherwise if you apply with 8 times for the above IPO then you should have 100% chance of getting an allotment which is not the case . Even if you apply with 8 pan there remains a small chance you may not get one .

Chances of getting success are calculated like this . 

1 -  chances of not getting subscriptions in all applications ( which is a function of and - * ) 

For 4 applications - 7/8 * 7/8 * 7/8 * 7/8 = 0.586 

Hence chances of success 1 - .586 = .414 or 41.4% 

For 8 applications - .586 * .586 = .343

Chance of Success 1 -.343 = .637 or 63.7 %

But the events are mutually independent.

Result of the first application doesn't affect the result of the 2nd, vice versa and so on.

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33 minutes ago, Krazyniks said:

Pretty sure that's not how it works .

Otherwise if you apply with 8 times for the above IPO then you should have 100% chance of getting an allotment which is not the case . Even if you apply with 8 pan there remains a small chance you may not get one .

Chances of getting success are calculated like this . 

1 -  chances of not getting subscriptions in all applications ( which is a function of and/x) 

For 4 applications - 7/8 * 7/8 * 7/8 * 7/8 = 0.586 

Hence chances of success 1 - .586 = .414 or 41.4% 

For 8 applications - .586 * .586 = .343

Chance of Success 1 -.343 = .637 or 63.7 %

Converse of not getting allotment in all applications is getting allotment in all IPOs.

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47 minutes ago, KunjanPSD said:

But the events are mutually independent.

Result of the first application doesn't affect the result of the 2nd, vice versa and so on.

True but these events occur at same time so when 1st application gets success - 2nd application is also successful or unsuccessful .

Similarly when 2nd application gets success - 1st application is either successful or unsuccessful .

Hence you can't just add success rates of both applications as it would lead to double counting .

 

21 minutes ago, KunjanPSD said:

Converse of not getting allotment in all applications is getting allotment in all IPOs.

No it means sum of total of getting allotment in either 1 or all of applications(let's say 4) .

You can get allotment in 1/2/3/4 applications or in none . 1 - Allotment in none would give you allotment in 1 or more or all 

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12 hours ago, Mr. Comingle said:

Which one has more chances statistically for getting IPO allotment?

Applying for 4 IPOs or applying for 1 IPO with 4 different accounts? 

I think the latter but my brain is not working to figure out why. 

Applying for 1 IPO with 4 different accounts would have much higher chances of success . Let's say all ipo are oversubscribed 8x . 

If you apply in 4 different ipo with 1 acc - chances of success remain 12.5% each for all IPOs . and results would be independent everytime , so you can't add the probabilities . 

If you apply with 4 application in 1 ipo your chances of success are 42% ( in which you get at least 1 or more allotment ) 

 

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51 minutes ago, Krazyniks said:

True but these events occur at same time so when 1st application gets success - 2nd application is also successful or unsuccessful .

Similarly when 2nd application gets success - 1st application is either successful or unsuccessful .

Hence you can't just add success rates of both applications as it would lead to double counting .

I actually meant adding and then subtracting the intersection. Sorry if I didn't make it clear.

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