Calculation - Odds of drawing a tag?
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LopeHunter
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I try to estimate current year odds based on prior year odds taking into consideration how points, if any, might raise or lower my odds as well as tag quantity changes. I use x in 5000 odds as some sheep tags are worse than 1 in 1000 odds.
NV, NM and AZ offer multiple choices that are each considered if your application is drawn so that is a curve ball and I usually go with the best odds of the choices that would be considered.
So, I end up with around 70 applications of which 50 are not point savers. For the 50 "live" applications then I add up the numbers and divide by 5000. Since the odds are guesses as I go to apply since the Wild Cards are human behavior this year and Lady Luck the ability to drill down on the data or slice with precision is a leap in faith I do not take as to the precision and reliability of this method. My quick and dirty approach is not something that can be predicted in the way a coin toss or roll of the dice can so me trying to do the math to figure odds draw 0 tags, exactly 2 tags, etc, seems a futile exercise.
An example. Say the WY pronghorn 1st choice is 300 in 5000, the Second choice is 2500 in 5000, the CO bighorn is 25 in 5000, the ID bighorn is 100 in 5000, the AZ Coues is 3500 in 5000 and the AK mountain goat is 320 in 5000. I would add all those up including both bites of the apple at WY Pronghorn (since are combined less than 5000 in 5000) and get 6745. 6745/5000 is a bit over one tags. I might draw 5 tags with those 6 applications but the most likely outcome is 1 tag.
My predictive outcome for tags drawn using the above approach over the past 25 years for the 50 or so applications that are live each year is to draw around 2.5 tags. I have had as few as 1 tag drawn and as many as 5 tags. Last year I drew 5 tags though 3 tags were expected to be 5000 of 5000 odds (WY pronghorn 2nd choice and WY pronghorn leftover and WY deer 2nd choice) so drew 2 tags of the 47 or so where odds were under 5000 in 5000 which is what I would expect based on my system. The tags I drew that were less than 5000 in 5000 odds were MT pronghorn and WY elk.
So, flawed as this approach is in several ways, has been useful for me.
NV, NM and AZ offer multiple choices that are each considered if your application is drawn so that is a curve ball and I usually go with the best odds of the choices that would be considered.
So, I end up with around 70 applications of which 50 are not point savers. For the 50 "live" applications then I add up the numbers and divide by 5000. Since the odds are guesses as I go to apply since the Wild Cards are human behavior this year and Lady Luck the ability to drill down on the data or slice with precision is a leap in faith I do not take as to the precision and reliability of this method. My quick and dirty approach is not something that can be predicted in the way a coin toss or roll of the dice can so me trying to do the math to figure odds draw 0 tags, exactly 2 tags, etc, seems a futile exercise.
An example. Say the WY pronghorn 1st choice is 300 in 5000, the Second choice is 2500 in 5000, the CO bighorn is 25 in 5000, the ID bighorn is 100 in 5000, the AZ Coues is 3500 in 5000 and the AK mountain goat is 320 in 5000. I would add all those up including both bites of the apple at WY Pronghorn (since are combined less than 5000 in 5000) and get 6745. 6745/5000 is a bit over one tags. I might draw 5 tags with those 6 applications but the most likely outcome is 1 tag.
My predictive outcome for tags drawn using the above approach over the past 25 years for the 50 or so applications that are live each year is to draw around 2.5 tags. I have had as few as 1 tag drawn and as many as 5 tags. Last year I drew 5 tags though 3 tags were expected to be 5000 of 5000 odds (WY pronghorn 2nd choice and WY pronghorn leftover and WY deer 2nd choice) so drew 2 tags of the 47 or so where odds were under 5000 in 5000 which is what I would expect based on my system. The tags I drew that were less than 5000 in 5000 odds were MT pronghorn and WY elk.
So, flawed as this approach is in several ways, has been useful for me.
WapitiBob
Well-known member
I have seen where the "correct way" mentioned was to calculate the odds of not drawing and take the inverse.
from my notes:
"For example you have a 25/26 (or 96.15%) chance of NOT drawing a deer tag, a 433/434 (or 99.77%) chance of NO elk tag, etc,etc,
Multiply all .9615 x .9977 x .9991 x .75 = .7188 or a 71.88% chance of not drawing any tag, with the inverse (100% minus 71.88%) being 28.12% chance of drawing at least one tag."
from my notes:
"For example you have a 25/26 (or 96.15%) chance of NOT drawing a deer tag, a 433/434 (or 99.77%) chance of NO elk tag, etc,etc,
Multiply all .9615 x .9977 x .9991 x .75 = .7188 or a 71.88% chance of not drawing any tag, with the inverse (100% minus 71.88%) being 28.12% chance of drawing at least one tag."
Schaaf
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Nerds
ImBillT
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WapitiBob is doing it correctly for the odds of drawing at least one tag.
D
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To be honest I'm not sure how accurate you can be determining the odds given all the unknowns, loopholes, asterisk in the draw systems or data, and human vagaries coupled with social media as well as hunting forums
Examples:
Alaska does a straight draw but you can put in 6 times for the same hunt code, the draw stats don't tell you how many times each person puts in for each unit and a person can obviously only draw one tag. You have have 24 apps for 4 tags with 100% odds because there are only 4 people are putting in for the 4 tags, or the odds could be ~17% because everyone is only putting in one app for the unit. I'm not sure how you accurately deal with that situation.
New Mexico your first three choice are looked at... so that's a curve ball
As to the vagaries, look at SE Montana after the world record bull was killed there, a big animal is killed or a unit is heavily shared on social media and pressure is pulled from one area to another. This is especially true of units where only a few tags are issued like sheep, goats, and moose. One year a ewe only unit in MT could have 75% draw odds, but if a dude in a bar starts talking about how much fun he had hunting it and all of a sudden 10 guys decided to burn their points that unit could become a 5% draw unit.
My approach is to look at the last 5 years of draw stats and then just bucket hunts into the following:
> 65% odds "Will probably draw this year",
<65%-40% "Will draw every other year"
<40%-20% "Will draw every couple of years"
< 20-5% "Might draw in my life time"
<5% "Once-in-a-lifetime/may will never draw"
I think to get more complicated than these buckets is just more brain damage then it's worth and in all reality probably isn't any more accurate.
Examples:
Alaska does a straight draw but you can put in 6 times for the same hunt code, the draw stats don't tell you how many times each person puts in for each unit and a person can obviously only draw one tag. You have have 24 apps for 4 tags with 100% odds because there are only 4 people are putting in for the 4 tags, or the odds could be ~17% because everyone is only putting in one app for the unit. I'm not sure how you accurately deal with that situation.
New Mexico your first three choice are looked at... so that's a curve ball
As to the vagaries, look at SE Montana after the world record bull was killed there, a big animal is killed or a unit is heavily shared on social media and pressure is pulled from one area to another. This is especially true of units where only a few tags are issued like sheep, goats, and moose. One year a ewe only unit in MT could have 75% draw odds, but if a dude in a bar starts talking about how much fun he had hunting it and all of a sudden 10 guys decided to burn their points that unit could become a 5% draw unit.
My approach is to look at the last 5 years of draw stats and then just bucket hunts into the following:
> 65% odds "Will probably draw this year",
<65%-40% "Will draw every other year"
<40%-20% "Will draw every couple of years"
< 20-5% "Might draw in my life time"
<5% "Once-in-a-lifetime/may will never draw"
I think to get more complicated than these buckets is just more brain damage then it's worth and in all reality probably isn't any more accurate.
ImBillT
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^^^Not to mention, the draw odds are LAST YEAR’s draw odds. There is no way of knowing what will actually happen THIS year. We can calculate how lucky, or unlucky we were, but it’s little more than a loose correlation to what might happen this year. Hunts with lots of tags, and lots of applicants are more stabile, while hubts with a handful of tags can swing from 80%+ to 5% or less simply because last year showed high draw odds and a handful of people decided put in for an “easy” draw.
Thanks for weighing in guys. I love math and I have no idea why I can't grasp an understanding of this. So if I put in for 20 different hunts and the chances of drawing each individual tag is 5% would the formula simply be .05^20? which is .358 or 36% odds of drawing one tag?
glass eye
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For many years CA would have Period 1 and Period 2 Tule Elk hunts. If last years statistics showed period 1 with the larger number of applicants, I knew it would flip so I would apply for whichever had last year's majority of applicants. True to form, every year I would be in the smaller app group. Still a very long shot but I did draw it eventually.
I slowed down and read WapitiBob's calculation and can now grasp that concept. most of the hunt's i'm applying for are less than 1% because many are sheep, goat, moose, etc. I do have a few in there that look to be between 5% and 12% which if calculating correctly based on WapitiBob's calc put my overall odds at around 45% of drawing a single tag.
This will help me consider what I'm putting in for in future years as I'd like to start pulling a single tag per year in an ideal scenario. Oh the excitement of application season! Good luck in the draws!
This will help me consider what I'm putting in for in future years as I'd like to start pulling a single tag per year in an ideal scenario. Oh the excitement of application season! Good luck in the draws!
WapitiBob
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I pulled that stuff from my substantial internet notes database, I flunked bonehead math in college.
EYJONAS!
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Wow........ just wow.
Aperventure
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since each are individual events, your odds are individual and can't be combined.
what are the odds when you flip a coin it lands on heads? 50%
what are the odds the next time you flip the same coin, it lands on heads? 50%
curious how wapatibob looks at it, I see the logic and agree that is reasonable. still, each are individual events.
what are the odds when you flip a coin it lands on heads? 50%
what are the odds the next time you flip the same coin, it lands on heads? 50%
curious how wapatibob looks at it, I see the logic and agree that is reasonable. still, each are individual events.
npaden
Well-known member
Okay, I will bet you $100 even up that you will flip at least one heads on the coin if you flip it 10 times in a row. 50/50 chance right?
How about if you flip 10 times in a row without flipping heads a single time I will pay you $1,000, but if you flip heads at least one time you pay me $100?
Aperventure
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Okay, I will bet you $100 even up that you will flip at least one heads on the coin if you flip it 10 times in a row. 50/50 chance right?
How about if you flip 10 times in a row without flipping heads a single time I will pay you $1,000, but if you flip heads at least one time you pay me $100?
ha! think i see the point you are trying to make, and I think you know am reasonable person would not take that bet.
said another way:
the history of each coin toss has no effect on the current toss
the result of each individual draw has no effect on the next draw
npaden
Well-known member
Agreed. And I wouldn't want to bet on each individual coin toss or each individual draw, but I would bet on the cumulative coin tosses or the cumulative draw results.
Using cumulative coin tosses or dice rolls is the easiest way to explain it, but the math is pretty much like wapitibob mentioned above.
Aperventure
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good perspectives here. thanks
LopeHunter
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So, seems your comfort level is, "Well, independent actions are involved so I will get no value in crafting an imperfect approach to provide an estimated outcome for the number of tags I am most likely to draw in the upcoming draws."
Interesting. And yet, Vegas often faces such uncertainty while looking at a series of independent events. Vegas is offering bets right now with odds for each NFL team to win the Superbowl to be played Feb 2020. How can they do that? Rosters are not set yet. The opponents are not known nor the order of the games at this point. Each game in the regular season is yet to be played. Each game is an independent event where no points or timeouts are carried over to the next game and there are only 3 outcomes: win, loss or tie. Players will be lost for the season. Might snow at a game or rain a lot. A lot is not known as we sit here in April trying to predict February. Those guys in Vegas setting odds must be losing a lot of money every year using such an imperfect approach.
LCH
Well-known member
Me simple.. Higher draw % = Better chance, More apps submitted = Better chance
Flip that coin as many times as possible.
Flip that coin as many times as possible.
ImBillT
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The confusion here is based on what question is being asked. The probability of flipping a coin heads up is 50%, and the probability of doing it twice in a row is 25%. If you flip a coin heads up on your first toss, your probability of flipping it heads up on your second toss is 50%. Wait what? I thought I just said that flipping heads up twice in a row only had 25% probability. Yes, but in the suggested scenario, I already flipped the coin for the first toss, and know the outcome, thus I eliminate all the possible scenarios in which my first toss was tails up, with half the possibilities eliminated I’m back at 50%.
WapitiBob’s calculations are correct. He is examining the probability that something will occur at least once over a series of n trials, which is equal to the percentage of times the event would occur if he examined n number of events an infinite number of times.
Aperventure is changing the question and examining the probability that something will occur one time, in a single trial, which is equal to the percentage of times the desired event would occur over an infinite number of events.
So more simply, Bob’s calculations answer the question, “if I pick ten marbles from a jar, how often is at least one of them blue?” And Aperventure’s answers the question “if I choose one marble from a jar, how often is it blue?” They are two different questions.
Obviously the percentage of time that the desired event occurs at least once when you repeatedly select a bundle of events(WapitiBob’s examination that something will occur at least once over a period of trials) is much higher than the percentage of times that the event occurs.
A scenario that exemplifies the importance of asking the right question is the old game show where the contestant is asked to choose one of three doors, then the host reveals a “loosing” door and the contestant is asked if he wants to change his original choice. One must be careful with what question they answer. What is the contestant’s odds of choosing the correct door on the first choice? 33%. What is the contestant’s odds of choosing the incorrect door on the first choice? 66%. If the contestant chooses to toss a coin to decide whether to leave his original choice changed or unchanged what are the odds of winning? 50%. Why? The contestant who chooses to stick with the original choice MUST choose CORRECTLY on the first choice to win. The odds of choosing correctly are 33%. The contestant that chooses to change his door on the second choice MUST choose INCORRECTLY on the first choice.(the host will reveal a non-winning door, so in order to move to a winning door, the contestant must have chosen incorrectly initially) Thus his odds are 66%. What is actually happening is that the contestant is choosing whether to win by picking the winning door(stay), or to win by picking the loosing door(change). By using a coin instead of logic to decide how to “make the second choice” half the second choices will be stay(33%) and half the second choices will be change(66%)
(.66*.5)+(.33*.5)=.5 and thus the odds would be 50%, which is also the odds choosing the correct door in the second round.
WapitiBob’s calculations are correct. He is examining the probability that something will occur at least once over a series of n trials, which is equal to the percentage of times the event would occur if he examined n number of events an infinite number of times.
Aperventure is changing the question and examining the probability that something will occur one time, in a single trial, which is equal to the percentage of times the desired event would occur over an infinite number of events.
So more simply, Bob’s calculations answer the question, “if I pick ten marbles from a jar, how often is at least one of them blue?” And Aperventure’s answers the question “if I choose one marble from a jar, how often is it blue?” They are two different questions.
Obviously the percentage of time that the desired event occurs at least once when you repeatedly select a bundle of events(WapitiBob’s examination that something will occur at least once over a period of trials) is much higher than the percentage of times that the event occurs.
A scenario that exemplifies the importance of asking the right question is the old game show where the contestant is asked to choose one of three doors, then the host reveals a “loosing” door and the contestant is asked if he wants to change his original choice. One must be careful with what question they answer. What is the contestant’s odds of choosing the correct door on the first choice? 33%. What is the contestant’s odds of choosing the incorrect door on the first choice? 66%. If the contestant chooses to toss a coin to decide whether to leave his original choice changed or unchanged what are the odds of winning? 50%. Why? The contestant who chooses to stick with the original choice MUST choose CORRECTLY on the first choice to win. The odds of choosing correctly are 33%. The contestant that chooses to change his door on the second choice MUST choose INCORRECTLY on the first choice.(the host will reveal a non-winning door, so in order to move to a winning door, the contestant must have chosen incorrectly initially) Thus his odds are 66%. What is actually happening is that the contestant is choosing whether to win by picking the winning door(stay), or to win by picking the loosing door(change). By using a coin instead of logic to decide how to “make the second choice” half the second choices will be stay(33%) and half the second choices will be change(66%)
(.66*.5)+(.33*.5)=.5 and thus the odds would be 50%, which is also the odds choosing the correct door in the second round.
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