What are the odds?

Tikay10

You have to name the number of runners in 3 MTT's, & in the right order.

The number of runners are;

426

755

147

What are the odds of getting that exactly correct?

 

weecheez1

Astronomical 

unknown_username

In Response to What are the odds?:
You have to name the number of runners in 3 MTT's, & in the right order. The number of runners are; 426 755 147 What are the odds of getting that exactly correct?  
Posted by Tikay10


Somewhere between

160,000,000/1 and 4,800,000/1

Depending on what ranges we estimate the field sizes to be. And Im probably wrong.

Phantom66

Could do with knowing the min/mean/max entries for each comp over a decent sample size.

Tikay10

^^^

Well we don't have that, but just assume each one has a Guarantee, & meets it exactly.

To meet Guarantee, it would need;

334

500

150

Phantom66

I would say if guesses were random in the possible range of entries for each tourney then something like 1 in 25 million for a single set of guesses.

A bit more info on past events would enable people to get a lot closer than that though and guess within narrower ranges, probably under 1 in a million.

aussie09

1,000,000 to 1

aussie09

boxster?

Tikay10

^^^^

Yes, I was trying to figure what the chances were of Boxster predicting all three exactly correctly.

markycash

In Response to Re: What are the odds?:

^^^ Well we don't have that, but just assume each one has a Guarantee, & meets it exactly. To meet Guarantee, it would need; 334 500 150
Posted by Tikay10

In that case it would depend on how large the standard deviation was in relation to the mean number of entrants.

If the average number of entrants is taken to be exactly the number to meet the guarantee and the standard deviation is 100 then the odds would be roughly 100/1 x 100/1 x 100/1. So 1,000,000/1 seems a reasonable number however I have just plucked the standard deviation figure of 100 out of the air as an example, although the actual number would probably be in that ball park range.

Chris_Mc

did boxter guess the exaxt number?

Darkangel7

This time Yes

Chris_Mc

high five that man!

suzy666

way less than 1/1000000.

the odds of predicting the number of entrants for 3Ts are ~this

but the BH heads is less.

more like 1/10  x   1/10  x  1/4 max  =1/400,

the number of head take in each T only varires by a little

aussie09

the question tikay asked was about odds of runners (not heads) in three MTTs

guessing runners is about 1M to 1

(guessing most heads taken is about 500 to 1)

unknown_username

why am I the only one who thinks its 25,050,000 /1

334*500*150=25,050,000

that's assuming all guarantees can be up or down 50%

unknown_username

334 -500-150 these are the guarantees.

assume they can to 50% short, or 50 above.

take the 1st guarantee.

50% lower is 167, 50% higher is , call it  500., between 167  and 500 , there is 334 choices.

apply the same to the next 2 MTTs

(My logic is flawed, I guess, but im not sure where)

the choices of numbers are 334*500*150=25,050,000.