MT- M/S/G results 26

[BuzzH]

Yeah, that's totally true about the 10% meaning once every ten years. Each year is independent, so you have a 10% chance every year.

I don't quite agree with your assessment about each number picked is an individual event. In that example I gave, the 625 points/apps are placed in a list with the other 64,613 and given a random number. You need 1 of your 625 apps to be in the top 12. I know GoHunt and AI both says you have an 11% chance of that happening and you say you have a .96% chance of that happening. No idea lol

Thats why people are disappointed with the published draw odds by the various services like gohunt and the MTFWP "detailed" reports.

There is no way to predict draw odds with a squared points system, better off rolling chicken bones.

 

[SerialScience]

@BuzzH and @5yearcoueshunter

10% odds

What are my odds of drawing next year? 10%
Ok and then the year after? 10%

Ok but then what are my odds of drawing a tag once in 10 years? ~65%

P(not winning any given year) =1-.1=.9

P(not winning ever) = (.9)^10 = .3487

P(winning at least once) 1-.3487 =.6513

 

[sclancy27]

The specifics of how they draw the tags are the most important. Say there is a unit with 30 tags. The same unit has 300,000 total squared bonus points of applicants.

My guess of how they complete the draw is they assign each individual entry (total squared bonus pts) a randomly generated number. The lowest 30 of those 300,000 randomly generated numbers draws a tag. So, each entry has a 30/300,000 chance of a tag (.01%). So, a person has 10 pts. Since each random number is independent (ie RANDOM) you could calculate the chances of being one of the 30 lowest random numbers this way, 1-(.9999)^100 = .0099506% x 100 (10 pts) = .995066% chance. You could just say 100 x (30/300,000) = 1%, close enough. The NR 10% issue (and a single individual having multiple numbers within the top 30), complicates slightly, but from a practical perspective the above will get you close enough. This is pretty easy to figure out for each individual district once you calculate the total entries (squared bonus pts) in play. ****Obviously, you can only look at past draw odds and figure this out, but, the past is the best predictor of the future, so there's that.****

They could operate the draw differently (each tag is drawn independently of one another) but I think that is unlikely because the above is really simple for them to run (and probably more importantly pretty difficult to screw up).

Flame away.

 

[cgasner1]

The specifics of how they draw the tags are the most important. Say there is a unit with 30 tags. The same unit has 300,000 total squared bonus points of applicants.

My guess of how they complete the draw is they assign each individual entry (total squared bonus pts) a randomly generated number. The lowest 30 of those 300,000 randomly generated numbers draws a tag. So, each entry has a 30/300,000 chance of a tag (.01%). So, a person has 10 pts. Since each random number is independent (ie RANDOM) you could calculate the chances of being one of the 30 lowest random numbers this way, 1-(.9999)^100 - (.0001)^100 = .0099506% x 100 (10 pts) = .995066% chance. You could just say 100 x (30/300,000) = 1%, close enough. The NR 10% issue (and a single individual having multiple numbers within the top 30), complicates slightly, but from a practical perspective the above will get you close enough. This is pretty easy to figure out for each individual district once you calculate the total entries (squared bonus pts) in play.

They could operate the draw differently (each tag is drawn independently of one another) but I think that is unlikely because the above is really simple for them to run (and probably more importantly pretty difficult to screw up).

Flame away.

How many seagulls do I have to put in your house before you think someone is doing it intentionally?

 

[sclancy27]

How many seagulls do I have to put in your house before you think someone is doing it intentionally?

One

 

[BuzzH]

The specifics of how they draw the tags are the most important. Say there is a unit with 30 tags. The same unit has 300,000 total squared bonus points of applicants.

My guess of how they complete the draw is they assign each individual entry (total squared bonus pts) a randomly generated number. The lowest 30 of those 300,000 randomly generated numbers draws a tag. So, each entry has a 30/300,000 chance of a tag (.01%). So, a person has 10 pts. Since each random number is independent (ie RANDOM) you could calculate the chances of being one of the 30 lowest random numbers this way, 1-(.9999)^100 = .0099506% x 100 (10 pts) = .995066% chance. You could just say 100 x (30/300,000) = 1%, close enough. The NR 10% issue (and a single individual having multiple numbers within the top 30), complicates slightly, but from a practical perspective the above will get you close enough. This is pretty easy to figure out for each individual district once you calculate the total entries (squared bonus pts) in play. ****Obviously, you can only look at past draw odds and figure this out, but, the past is the best predictor of the future, so there's that.****

They could operate the draw differently (each tag is drawn independently of one another) but I think that is unlikely because the above is really simple for them to run (and probably more importantly pretty difficult to screw up).

Flame away.

Correct, you would have to know how they run the draw.

 

[cgasner1]

One

 

[SAJ-99]

Ok, I see what you're doing there. So if that's true, a lot of hunts would be better for even the top point holders if there was no point system. lol

Actual Moose hunt in MT: 12 permits
Applicants: 978
12/978 = 1.2% draw odds (no points)

25 point holder - 626 squared points (chances)
Total squared points from all applicants - 65,238
626/65,238 = .96%

Doesn’t you second example only have a single draw/ 1 permit vs the first example had 12?

 

[mtmuley]

View attachment 408073

He looks scared. mtmuley

 

[5yearcoueshunter]

Doesn’t you second example only have a single draw/ 1 permit vs the first example had 12?

Correct. Because Buzz says multiplying by 12 is inaccurate.

 

[mtmiller]

How many seagulls do I have to put in your house before you think someone is doing it intentionally?

The same number of Canadian geese

 

[DFS]

The tag numbers dropping precipitously means most people will never draw in a lifetime regardless of how you calculate things.

 

[NoWiser]

Multiplying by 12 is going to be pretty close to accurate, depending on how exactly they do the draw. You absolutely have to take the number of tags available into account.

 

[SAJ-99]

Correct. Because Buzz says multiplying by 12 is inaccurate.

Sorry, didn’t understand what you were trying to show. Technically it is wrong, but close enough.

 

[cgasner1]

He looks scared. mtmuley

You would be to if you got locked in @sclancy27 house

 

[sacountry]

When I drew my previous goat tag in 1978 I shot my billy near the top of the mountain that took me 3 hours to climb in -5* F temperature and knee deep snow. Early season scouting w/o snow I could climb that mountain in about an hour.

In the years immediately following my hunt, 4 of my friends all shot their goats in the same drainage that I shot mine, only they all shot their goats within a couple hundred yards of where we all had parked.

Dang....

 

[Wildabeest]

To solve for the # of permits available, you need to think of each permit as a separate contest, each with either independent or dependent odds. To get the probability of winning any of those contests, you need to invert that problem and calculate the odds of NOT winning any given contest and then multiply those probabilities together and subtract from 100%.

So assuming you had 10 permits available and a 5% chance of winning any given one (independent contests), then your probability of winning at least one of them would be 100% - (0.95^10) = 40%.

For tags draws, they are not independent from one tag to the next, so you’d need to recalculate your probably of NOT drawing for each subsequent available tag and then multiply those together and subtract from 100% vs. “base probability ‘to power of’ the number of tags” as in my simplified example above.

 

[Caseknife]

The tag numbers dropping precipitously means most people will never draw in a lifetime regardless of how you calculate things.

Nailed it!

 

[Beignet]

I’m guessing it’s a little what gambling feels like, but when I see unsuccessful my first reaction is to log back in and apply for the next thing immediately. Like “No good deer/elk tag, all in on M/S/G. Nada, let’s throw the deed down for deer B cause baby needs a new pair of shoes!”

 

[Forkyfinder]

Ok, I see what you're doing there. So if that's true, a lot of hunts would be better for even the top point holders if there was no point system. lol

Actual Moose hunt in MT: 12 permits
Applicants: 978
12/978 = 1.2% draw odds (no points)

25 point holder - 626 squared points (chances)
Total squared points from all applicants - 65,238
626/65,238 = .96%

Lol thats grade A buzzard math....


Notice how the bonus points math is entirely missing the quantity of permits? Just need to multiply by tags.... 0.96 × 12 = 11.5%.

Yes this is ignoring the loss of bonus points as the draw continues - but that only helps the point im making. Theoretically the odds of the "second" permit will be better, because some bonus points/applicants are gone after each time someone draws. But with small tag numbers and the high applicant count its not a crazy assumption, even if a lot of bonus points disappear. Ie - a 25 point holder draws the first moose tag - then the next tag drawn, for our 25 point holder will have 626/64612 = 0.0968, or .97%.