MT- M/S/G results 26

[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%.

 

[brownbear932008]

Nailed it!

But yet old guys feel they are entitled to a permit after 65. (Coming from an older guy just not nearly 65) Maine started that junk with residents. I can only imagine folks younger than 65 are having a hard time drawing there now with so few tags.

 

[Wildabeest]

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%.

Per my response above, multiplying the probability of winning a tag by the number of tags doesn’t work. You could easily get a result where your probability is greater than 100%.

For example, let’s say there are 50 prizes (ie. tags), 100 tickets sold (ie. total points in the draw) and you have 10 tickets (points). Your probability of winning the first drawn is 10/100=10%. Using your math, if there’s 50 tags to hand out then you’d say your probability of winning any one of those would be 0.1*50 = 5.0 (or 500%). That can’t be right because there’s definitely a scenario where they draw 50 tickets and none of them is yours.

Your probability of winning any one of those 50 tags is really (100% - (90% chance of losing ^ 10)) = 0.65, or 65%.

Note the example above is for independent draws meaning they put the winning ticket back in after each draw. But the logic also works for non-independent draws where the ticket is not put back in (ie. you can’t win multiple times) - you just need to adjust the probability of NOT winning for each subsequent draw and multiply those together instead of raising the original probability of NOT winning the first tag to the power of the number of tags. That would be (100% - (10/100 * 10/99 * 10/98 * 10/97 * … * 10/50)).

 

[MTTW]

So...it is really quite simple.

 

[SAJ-99]

Per my response above, multiplying the probability of winning a tag by the number of tags doesn’t work. You could easily get a result where your probability is greater than 100%.

For example, let’s say there are 50 prizes (ie. tags), 100 tickets sold (ie. total points in the draw) and you have 10 tickets (points). Your probability of winning the first drawn is 10/100=10%. Using your math, if there’s 50 tags to hand out then you’d say your probability of winning any one of those would be 0.1*50 = 5.0 (or 500%). That can’t be right because there’s definitely a scenario where they draw 50 tickets and none of them is yours.

Your probability of winning any one of those 50 tags is really 100% x (90% chance of losing ^ 10) = 0.65, or 65%.

Note the example above is for independent draws meaning they put the winning ticket back in after each draw. But the logic also works for non-independent draws where the ticket is not put back in (ie. you can’t win multiple times) - you just need to adjust the probability of NOT winning for each subsequent draw and multiply those together instead of raising the original probability of NOT winning the first tag to the power of the number of tags.

I think the point is it is a quick shortcut. The odds of not drawing are so large that it gets you close to the true number without having to do all the math, which isn't a strength of HT.

I wonder if they ever imagined have so many people jump in the bonus pool (pun intended) when it started? The more I look at the stats the more I think I will be withdrawing from applying for these tags going forward.

 

[NoWiser]

Per my response above, multiplying the probability of winning a tag by the number of tags doesn’t work. You could easily get a result where your probability is greater than 100%.

For example, let’s say there are 50 prizes (ie. tags), 100 tickets sold (ie. total points in the draw) and you have 10 tickets (points). Your probability of winning the first drawn is 10/100=10%. Using your math, if there’s 50 tags to hand out then you’d say your probability of winning any one of those would be 0.1*50 = 5.0 (or 500%). That can’t be right because there’s definitely a scenario where they draw 50 tickets and none of them is yours.

Your probability of winning any one of those 50 tags is really (100% - (90% chance of losing ^ 10)) = 0.65, or 65%.

Note the example above is for independent draws meaning they put the winning ticket back in after each draw. But the logic also works for non-independent draws where the ticket is not put back in (ie. you can’t win multiple times) - you just need to adjust the probability of NOT winning for each subsequent draw and multiply those together instead of raising the original probability of NOT winning the first tag to the power of the number of tags.

This is correct. But, when the odds are as low as sheep and moose tags, I think multiplying by the number of tags will still get you fairly close.

 

[Forkyfinder]

I think the point is it is a quick shortcut. The odds of not drawing are so large that it gets you close to the true number without having to do all the math, which isn't a strength of HT.

Yes. That is precisely the point.

 

[sclancy27]

Per my response above, multiplying the probability of winning a tag by the number of tags doesn’t work. You could easily get a result where your probability is greater than 100%.

For example, let’s say there are 50 prizes (ie. tags), 100 tickets sold (ie. total points in the draw) and you have 10 tickets (points). Your probability of winning the first drawn is 10/100=10%. Using your math, if there’s 50 tags to hand out then you’d say your probability of winning any one of those would be 0.1*50 = 5.0 (or 500%). That can’t be right because there’s definitely a scenario where they draw 50 tickets and none of them is yours.

Your probability of winning any one of those 50 tags is really (100% - (90% chance of losing ^ 10)) = 0.65, or 65%.

Note the example above is for independent draws meaning they put the winning ticket back in after each draw. But the logic also works for non-independent draws where the ticket is not put back in (ie. you can’t win multiple times) - you just need to adjust the probability of NOT winning for each subsequent draw and multiply those together instead of raising the original probability of NOT winning the first tag to the power of the number of tags. That would be (100% - (10/100 * 10/99 * 10/98 * 10/97 * … * 10/50)).

Wouldn't that be (100% - (90% chance of losing ^ 50)) =.995 or 99.5%.......?

 

[Big Fin]

I wonder if they ever imagined have so many people jump in the bonus pool (pun intended) when it started? The more I look at the stats the more I think I will be withdrawing from applying for these tags going forward.

The answer to that is NO.

The committee was shown the historical application trends for each species. The numbers we were shown were based on the old system of having to front the money, via a check, and mail in an application. The legislature changed all of that, as they did with squaring bonus points, something the Committee rejected when seeing the math behind such idea. I testified against the legislative changes, as did a few others who sat on the Committee. No surprise, but legislators were not interested in data that did not support their narrative.

We were shown the miniscule improvement by squaring bonus points, especially for species with so few tags, such as M/G/S. It does almost nothing to improve the odds of those who get in on the ground floor, yet serious decreases the odds of those who got in at a later date.

No matter what math one subscribes to for the exercise of the most recent posts, if we keep losing herds and having fewer tags, it is a moot issue. In third grade, when I was taught fractions, it became very obvious that increasing the numerator is a great way to increase the ending result. Better draw odds won't come from finding obscure ways to slow down the growth in the denominator, rather finding ways to increase the numerator (tags issued).

 

[BuzzH]

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%.

math...11% moose odds are found in fantasyland, certainly not in Montana.

 

[duckhunt]

Think they have a Dairy Queen for the boomers

DQ double cheeseburgers are the bomb.

 

[It’s Bucky]

I should have near max bonus points for all 3 species. In college, however, being poor for a few years and not knowing the rules, I didn’t apply 2 years in a row and at the time, you lost all bonus points if you did that. What a bummer that was. I reached out about getting them back now that the rules have changed and they said I was SOL. The price of admission was small but I was young and dumb.

 

[sclancy27]

I should have near max bonus points for all 3 species. In college, however, being poor for a few years and not knowing the rules, I didn’t apply 2 years in a row and at the time, you lost all bonus points if you did that. What a bummer that was. I reached out about getting them back now that the rules have changed and they said I was SOL. The price of admission was small but I was young and dumb.

I did the same thing.

 

[5yearcoueshunter]

Per my response above, multiplying the probability of winning a tag by the number of tags doesn’t work. You could easily get a result where your probability is greater than 100%.

For example, let’s say there are 50 prizes (ie. tags), 100 tickets sold (ie. total points in the draw) and you have 10 tickets (points). Your probability of winning the first drawn is 10/100=10%. Using your math, if there’s 50 tags to hand out then you’d say your probability of winning any one of those would be 0.1*50 = 5.0 (or 500%). That can’t be right because there’s definitely a scenario where they draw 50 tickets and none of them is yours.

Your probability of winning any one of those 50 tags is really (100% - (90% chance of losing ^ 10)) = 0.65, or 65%.

Note the example above is for independent draws meaning they put the winning ticket back in after each draw. But the logic also works for non-independent draws where the ticket is not put back in (ie. you can’t win multiple times) - you just need to adjust the probability of NOT winning for each subsequent draw and multiply those together instead of raising the original probability of NOT winning the first tag to the power of the number of tags. That would be (100% - (10/100 * 10/99 * 10/98 * 10/97 * … * 10/50)).

Ok, thanks for that explanation.

So using this real world example I had earlier, looks like odds of drawing is at least 10.92%? (this would assume no chances/tickets are taken out after each successful pull)



View attachment 408121

 

[SAJ-99]

math...11% moose odds are found in fantasyland, certainly not in Montana.

Looking at 2026 results for 100-50 with 12 tags, the 1 applicant with 26 points looks like he was close to that, no? You have remember that still is close to 90% not drawing any of those tags.

Someone with 1 point drew. The odds on that were incredibly small. Probably need an audit on that.

 

[antlerradar]

I should have near max bonus points for all 3 species. In college, however, being poor for a few years and not knowing the rules, I didn’t apply 2 years in a row and at the time, you lost all bonus points if you did that. What a bummer that was. I reached out about getting them back now that the rules have changed and they said I was SOL. The price of admission was small but I was young and dumb.

you were never dumb