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Campfire Kahuna
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Campfire Kahuna
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You're making this way too complicated.
It's impossible to know the number of applicants this year because the application period is still open. So, we must use last years number of 762 for an estimate. Whether they're single applicants or grouped in parties doesn't matter for a single applicant because the 762 is still the total people who apply. If a party of 4 draws, they'll get 4 tags out of the 762. It only affects the number of times they have to pull a number out of the hat. whether they have to draw once or 4 times is irrelevant. It's still 4 tags gone. It could only matter at the end if there are only 3 tags left and a party of 4 is drawn. Then the party is tossed out and they draw again. They won't go over the 125.
The odds of drawing 1 tag are 762/125, or 1 chance in 6.1. Since 5 are applying (assuming they're part of the 762), multiply that by 5, so the odds of getting at least 1 tag are 5 in 6.1, or 82%. A party of 2 won't affect the odds much. It only reduces the times they have to draw if the group comes up.
I'm assuming this is unit 51 bulls? Have you considered unit 48? The past drawing odds have been slightly higher and it's closer to Boise. It's the earlier season if that matters. Be prepared for some hard climbing, though. It's steep and the elk like to hang high.
“In a time of deceit telling the truth is a revolutionary act.”
― George Orwell
It's not over when you lose. It's over when you quit.
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This from my son with a master's degree in finance:
"It's very hard to directly figure out probability of success for 6 people out of 762 applying for 125 tags. It's far easiery to figure probability of FAILURE. You just take the probability of failure of each person (which is 83.6%) and multiply them together 6 times, for a combined probability of the group failing of 34.1%.
This means the probability of SUCCESS of at least one of the group getting a tag is the inverse, or 65.9%."
Overall, you have increased your chance of drawing a tag to over 50%.
Last edited by czech1022; 05/24/17.
All that is necessary for the triumph of evil is that good men do nothing -- Edmund Burke
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OP
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You're making this way too complicated.
It's impossible to know the number of applicants this year because the application period is still open. So, we must use last years number of 762 for an estimate. Whether they're single applicants or grouped in parties doesn't matter for a single applicant because the 762 is still the total people who apply. If a party of 4 draws, they'll get 4 tags out of the 762. It only affects the number of times they have to pull a number out of the hat. whether they have to draw once or 4 times is irrelevant. It's still 4 tags gone. It could only matter at the end if there are only 3 tags left and a party of 4 is drawn. Then the party is tossed out and they draw again. They won't go over the 125.
The odds of drawing 1 tag are 762/125, or 1 chance in 6.1. Since 5 are applying (assuming they're part of the 762), multiply that by 5, so the odds of getting at least 1 tag are 5 in 6.1, or 82%. A party of 2 won't affect the odds much. It only reduces the times they have to draw if the group comes up.
I'm assuming this is unit 51 bulls? Have you considered unit 48? The past drawing odds have been slightly higher and it's closer to Boise. It's the earlier season if that matters. Be prepared for some hard climbing, though. It's steep and the elk like to hang high. That's the road of logic I erroneously started down. You'll see the fallacy behind that thinking if you assume I theoretically have 8 relatives to put in. Using that logic, he would be guaranteed a tag...and he most certainly is not. You following me? I think I'm to a point where I understand you have to figure out the UNSUCESSFUL likelihood, before you can figure out the likelihood. Thanks for everyone that weighed in. I knew the Campfire would produce the correct answer, even if it was a bit painful getting there. Dave
If you're not burning through batteries in your headlamp,...you're doing it wrong.
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I feel like I need to say up front that I'm not a total-window licker...which probably means I am of course. I have a B.S. from UI, and am considered "reasonably" intelligent by my peers. That may just mean my peers are also dip-chits. I digress however.
I am trying to calculate the odds that my son will pull a particular elk tag in one of Idaho's controlled hunt areas. The situation/numbers are as follows.
- There are 762 applicants for a controlled hunt elk tag, with 125 tags available.
- My son and I put in for the elk tag on a "party hunt" together, which means that if one license gets drawn, then both get drawn. It also means we only get once chance I believe.
- Additionally, Idaho law allows for a parent/grandparent to "gift" a controlled hunt tag to a son/grandson under the age of 16.
- My wife, both my in-laws, and both my parents will also independently put in for the tag. (5 people, with 5 more independent "chances"). It should be assumed any/all would give the tag to my son if their license was drawn.
The IDFG web-site lists the "drawing odds" at 16% for this particular hunt, but I THINK that's a misnomer. I have long believed that meant I personally had a 16% chance at drawing the tag, but I'm now second-guessing that assumption. My question is this...what is the percentage of likelihood that my son will have this tag in his pocket this fall, given the aforementioned parameters?
Bonus points if you can explain it in laymens terms that my dumb-azz can understand, though at this point I'm beginning to believe that might be asking too much.
Dumb in Idaho
Thanks for asking this question, I'm loading up for my kid too (one of the drawbacks of this new system where parents and grandparents can draw for a minor child is that the drawing odds have gotten noticeably worse) and was struggling with the same question but the best I could remember from my statistics course was that The answer seemed complicated. Thanks math guys for the help!
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I'm with pharmseller on this. 16% each time approx with no cumulative effect. You can flip a coin 100 times
and get heads. But the odds for the 101st flip don't improve for tails. Each flip is 50% and each draw is 16%.
Last edited by 257wby; 05/24/17.
"I call that bold talk for a one-eyed fat man." --Robert Duvall.
"Fill your hand, you son-of-a-bitch!" --John Wayne.
~Molɔ̀ːn Labé Skýla~
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I feel like I need to say up front that I'm not a total-window licker...which probably means I am of course. I have a B.S. from UI, and am considered "reasonably" intelligent by my peers. That may just mean my peers are also dip-chits. I digress however.
I am trying to calculate the odds that my son will pull a particular elk tag in one of Idaho's controlled hunt areas. The situation/numbers are as follows.
- There are 762 applicants for a controlled hunt elk tag, with 125 tags available.
- My son and I put in for the elk tag on a "party hunt" together, which means that if one license gets drawn, then both get drawn. It also means we only get once chance I believe.
- Additionally, Idaho law allows for a parent/grandparent to "gift" a controlled hunt tag to a son/grandson under the age of 16.
- My wife, both my in-laws, and both my parents will also independently put in for the tag. (5 people, with 5 more independent "chances"). It should be assumed any/all would give the tag to my son if their license was drawn.
The IDFG web-site lists the "drawing odds" at 16% for this particular hunt, but I THINK that's a misnomer. I have long believed that meant I personally had a 16% chance at drawing the tag, but I'm now second-guessing that assumption. My question is this...what is the percentage of likelihood that my son will have this tag in his pocket this fall, given the aforementioned parameters?
Bonus points if you can explain it in laymens terms that my dumb-azz can understand, though at this point I'm beginning to believe that might be asking too much.
Dumb in Idaho
Thanks for asking this question, I'm loading up for my kid too (one of the drawbacks of this new system where parents and grandparents can draw for a minor child is that the drawing odds have gotten noticeably worse) and was struggling with the same question but the best I could remember from my statistics course was that The answer seemed complicated. Thanks math guys for the help! Be prepared for the morality police to make an arrest. I also think you'll find the "odds" have been on the decline long before the "gifted" kid-tag issue in Idaho which only went into affect two years ago. In fact, I know you will because I took the time to do some research on the matter. I THINK it's largely an issue of more and more people being savvy about the drawing odds game played across most of the west. The internet and associated availability of information has changed hunting forever more, and this is just one more sign of that.
If you're not burning through batteries in your headlamp,...you're doing it wrong.
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You're right about the odds but I thought I noticed another dip in the odds 2 years ago, but I didn't do the work, maybe it just felt that way. Yes, the morality police are probably bitter about me loading up for my kid but, like you, I'm just trying to work the system within the rules they make up.
Cross
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I did a quick spreadsheet... by taking the chance of failure to draw a tag the first time and running it out for 125 events, then multiplying the odds all the way through.... 66.1% chance to end up with a tag. There have been a few ways to get there, but this must be the answer. Here is how I got there.
"What is my chance to win once in three draws of a one-in-five chance to win?"
The chance of winning on a single draw is 20%. Thus, the chance of losing is 80%. The chance of losing all three is .8 x .8 x .8 = .512. Thus, the chance of not losing all three is 1 - .512 = .488. So the probability of winning at least once is 48.8%.
Use the method above, but with 6 chances to win out of 762, then 6 chances to win our of 761, and so on...
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If 125 tags are available for 762 people, each person has a chance of 125/762 or .1640 on the first draw, which can be expressed as 16.4%. Remember that if 125 tags were split between 125 people then each would have a chance of 125/125 = 1 which is 100% chance of winning. Or look at it the other way, 762 people with 125 tags, 762/125 = 6.096, or about 6 people vying for each tag. 100% success divided by 6.096 = 16.40 or again 16.4%. I suppose there is some algorithm that shows your odds decreasing with each unsuccessful (for you) draw. When the first ticket is drawn, then 124 people are vying for 761 tags, or 124/761 = .1629, then 123/760 = .1618, etc. until there is one tag left for 638 people or 1/638 = .0015. Not sure how 6 applicants vying for one of 762 tags would be calculated, I left college to join the Army before we got to that part of the class.  This is the closest to being correct. Odds are different than percentages. Anyway you cut it if there are 125 tags available for 762 people so you have a 16.4% chance of drawing a tag. In math to figure % you start with the number you are starting with and divide it by the number you are arriving at. Easy example: You have 10 pieces out of 100 of something. 10 divided by 100 is .10 thus you have 10% of the pieces. In the above example you have 125 tags (pieces) out of the total number of 762. 125 divided by 762 is .164 or 16.4% As far as the bold part, since each additional person will give their tag to the posters son, you increase the number of tags available, but not the number of applicants. So counting the OP there are now 6 additional tags available. 131 tags divided by 762 equals a now .171 or 17.1% chance of getting a tag. I never spent a day in college but enjoy math, and this is how my simple logical mind looks at this. I'm certain I'm correct, but I have been wrong once or twice before in my life. 
One is alone in a land so vast, there is only the mountains, the wind, and the eyes of God.
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You are unfortunately incorrect this time.
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You are unfortunately incorrect this time. Nope I don't think I am. I do know this however, it cannot be 66%. If you're the last 2 people standing there and there are only 2 names left in the hat, you have exactly a 50% chance of getting drawn. So how the heck do you have a 66% chance starting out as 1 of 762 or 6 of 757 vying for 125 tags? That's some fancy math. I'm just a dumb farm boy but the day 6 joined together efforts, out of 762 going for 125 available tags equals a 66% of winning, I'll go fly with monkeys. Either that or I'm leaving today yet for Vegas. Even if there were just 2 applicants per tag (250 applicants and 125 tags) you can't get above a 50% chance on any single draw as a single applicant, and that's on the first draw. As the number of tags available is lowered the % keeps going down.
One is alone in a land so vast, there is only the mountains, the wind, and the eyes of God.
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You're making this way too complicated.
It's impossible to know the number of applicants this year because the application period is still open. So, we must use last years number of 762 for an estimate. Whether they're single applicants or grouped in parties doesn't matter for a single applicant because the 762 is still the total people who apply. If a party of 4 draws, they'll get 4 tags out of the 762. It only affects the number of times they have to pull a number out of the hat. whether they have to draw once or 4 times is irrelevant. It's still 4 tags gone. It could only matter at the end if there are only 3 tags left and a party of 4 is drawn. Then the party is tossed out and they draw again. They won't go over the 125.
The odds of drawing 1 tag are 762/125, or 1 chance in 6.1. Since 5 are applying (assuming they're part of the 762), multiply that by 5, so the odds of getting at least 1 tag are 5 in 6.1, or 82%. A party of 2 won't affect the odds much. It only reduces the times they have to draw if the group comes up.
. This from my son with a master's degree in finance:
"It's very hard to directly figure out probability of success for 6 people out of 762 applying for 125 tags. It's far easiery to figure probability of FAILURE. You just take the probability of failure of each person (which is 83.6%) and multiply them together 6 times, for a combined probability of the group failing of 34.1%.
This means the probability of SUCCESS of at least one of the group getting a tag is the inverse, or 65.9%."
Overall, you have increased your chance of drawing a tag to over 50%. You guys are too educated for your own good. What is 1 chance in 6? 16%. What is 83.6% failure off of 100? What are you left with? 16.4%. Just because the number changes from 1 to 6 doesn't mean you multiply it 5 or 6 times, they are individuals.........you ADD them to the total. You are not taking 6 individuals that are all applying 5 or 6 times each. Each applicant has a single chance on each draw. Where the heck is the power of multiplication coming from?????? So on Rock Chucks post, 762/125 equals 6 or 1 chance in 6. Now you add 5 or 6 more chances to the total depending on whether you are counting the OP or not . Same as making the available tags now 130 or 131. Same with Czech's post. These individuals don't have the power of multiplication, they are individuals ADDED to the total. Not multiplied. If I have 10 of 100 I have 10%. If I add 10 more to that I have 20 or 20%%. Using the logic of these 2 posts above and many more on here, I would instead take that additional 10 and multiply it by what I started with giving me 100% chance of winning even though I'm actually now only 20 out of 100. If my number I'm working with is less than half of the total, I know right off the bat I'm less than 50%. I don't need to apply any math to figure that out.
One is alone in a land so vast, there is only the mountains, the wind, and the eyes of God.
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Campfire Kahuna
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You're making this way too complicated.
It's impossible to know the number of applicants this year because the application period is still open. So, we must use last years number of 762 for an estimate. Whether they're single applicants or grouped in parties doesn't matter for a single applicant because the 762 is still the total people who apply. If a party of 4 draws, they'll get 4 tags out of the 762. It only affects the number of times they have to pull a number out of the hat. whether they have to draw once or 4 times is irrelevant. It's still 4 tags gone. It could only matter at the end if there are only 3 tags left and a party of 4 is drawn. Then the party is tossed out and they draw again. They won't go over the 125.
The odds of drawing 1 tag are 762/125, or 1 chance in 6.1. Since 5 are applying (assuming they're part of the 762), multiply that by 5, so the odds of getting at least 1 tag are 5 in 6.1, or 82%. A party of 2 won't affect the odds much. It only reduces the times they have to draw if the group comes up.
. This from my son with a master's degree in finance:
"It's very hard to directly figure out probability of success for 6 people out of 762 applying for 125 tags. It's far easiery to figure probability of FAILURE. You just take the probability of failure of each person (which is 83.6%) and multiply them together 6 times, for a combined probability of the group failing of 34.1%.
This means the probability of SUCCESS of at least one of the group getting a tag is the inverse, or 65.9%."
Overall, you have increased your chance of drawing a tag to over 50%. You guys are too educated for your own good. What is 1 chance in 6? 16%. What is 83.6% failure off of 100? What are you left with? 16.4%. Just because the number changes from 1 to 6 doesn't mean you multiply it 5 or 6 times, they are individuals.........you ADD them to the total. You are not taking 6 individuals that are all applying 5 or 6 times each. Each applicant has a single chance on each draw. Where the heck is the power of multiplication coming from?????? So on Rock Chucks post, 762/125 equals 6 or 1 chance in 6. Now you add 5 or 6 more chances to the total depending on whether you are counting the OP or not . Same as making the available tags now 130 or 131. Same with Czech's post. These individuals don't have the power of multiplication, they are individuals ADDED to the total. Not multiplied. If I have 10 of 100 I have 10%. If I add 10 more to that I have 20 or 20%%. Using the logic of these 2 posts above and many more on here, I would instead take that additional 10 and multiply it by what I started with giving me 100% chance of winning even though I'm actually now only 20 out of 100. If my number I'm working with is less than half of the total, I know right off the bat I'm less than 50%. I don't need to apply any math to figure that out. Many should ask advice, not give it...
Mark Begich, Joaquin Jackson, and Heller resistance... Three huge reasons to worry about the NRA.
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You're making this way too complicated.
It's impossible to know the number of applicants this year because the application period is still open. So, we must use last years number of 762 for an estimate. Whether they're single applicants or grouped in parties doesn't matter for a single applicant because the 762 is still the total people who apply. If a party of 4 draws, they'll get 4 tags out of the 762. It only affects the number of times they have to pull a number out of the hat. whether they have to draw once or 4 times is irrelevant. It's still 4 tags gone. It could only matter at the end if there are only 3 tags left and a party of 4 is drawn. Then the party is tossed out and they draw again. They won't go over the 125.
The odds of drawing 1 tag are 762/125, or 1 chance in 6.1. Since 5 are applying (assuming they're part of the 762), multiply that by 5, so the odds of getting at least 1 tag are 5 in 6.1, or 82%. A party of 2 won't affect the odds much. It only reduces the times they have to draw if the group comes up.
. This from my son with a master's degree in finance:
"It's very hard to directly figure out probability of success for 6 people out of 762 applying for 125 tags. It's far easiery to figure probability of FAILURE. You just take the probability of failure of each person (which is 83.6%) and multiply them together 6 times, for a combined probability of the group failing of 34.1%.
This means the probability of SUCCESS of at least one of the group getting a tag is the inverse, or 65.9%."
Overall, you have increased your chance of drawing a tag to over 50%. You guys are too educated for your own good. What is 1 chance in 6? 16%. What is 83.6% failure off of 100? What are you left with? 16.4%. Just because the number changes from 1 to 6 doesn't mean you multiply it 5 or 6 times, they are individuals.........you ADD them to the total. You are not taking 6 individuals that are all applying 5 or 6 times each. Each applicant has a single chance on each draw. Where the heck is the power of multiplication coming from?????? So on Rock Chucks post, 762/125 equals 6 or 1 chance in 6. Now you add 5 or 6 more chances to the total depending on whether you are counting the OP or not . Same as making the available tags now 130 or 131. Same with Czech's post. These individuals don't have the power of multiplication, they are individuals ADDED to the total. Not multiplied. If I have 10 of 100 I have 10%. If I add 10 more to that I have 20 or 20%%. Using the logic of these 2 posts above and many more on here, I would instead take that additional 10 and multiply it by what I started with giving me 100% chance of winning even though I'm actually now only 20 out of 100. If my number I'm working with is less than half of the total, I know right off the bat I'm less than 50%. I don't need to apply any math to figure that out. Many should ask advice, not give it... ALL he is doing is eliminating 5 or 6 other people he's competing against out of the total group of 762 because those other 5 or 6 are now all one in the same. It's that simple. Odds might be one thing, but when you're talking percentages, it is what it is. What started out at 16.4% is now just slightly higher than that. His total number is still WAY, WAY less than half of the total number competing for 125 tags. . Now my dumb farm boy logic tells me that if a bucket is way less than half full, then it's way less than 50%. At that level it will never be 66% or 82% full, ever. It wasn't when men were walking around in animal skin sandals, and it isn't now.
One is alone in a land so vast, there is only the mountains, the wind, and the eyes of God.
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TTT just because it's that time of year again...
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TTT just because it's that time of year again... So.... tell us... did the boy or one of the others get drawn last year ???
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He/we did NOT in fact draw...though he/we were able to take some critters during the general season hunts.
Dave
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