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.