Mathematicians Wanted: Nevada Mule Deer Odds

[canyonhunter47]

I’m not familiar with Nevada’s system specifically but unless all those draw odds are independent (I.e. you could in theory draw more than 1 in a draw year) your draw odds in a given year are whatever the highest one you picked was. Say you picked 5 hunts with odds ranging from 5%-15%. Your odds of getting to go at all that year would be 15%. Now, over 5 years, your odds of never drawing with 15% odds is (1-draw_odds)^(years) or .85^5=44%. Conversely, your odds of drawing at least once during a 5 year period would be 56%

Edited: fixed phrasing and punctuation to be coherent

 

[Jaden Bales]

Im not familiar with Nevada’s system specifically but unless all those odds are independent (I.e. you could in theory draw more than 1 every year) your draw odds in a given year are whatever the highest one you picked was. Say you picked 5 hunts with odds ranging from 5%-15% your odds of getting to go at all would be 15%. Now, over 5 years, your odds of never drawing with 15% odds is (1-draw_odds)^(years) or .85^5=44%. Your odds of drawing would one or more times would be 56%

That makes sense. Goes back to that thing of, "more tag apps of 1% doesn't get you more than 1% odds."

Thanks for the insight!

 

[gonhunting247]

After reading all these posts, I think I'll go out turkey hunting and just wait for the results to come out

 

[Jaden Bales]

After reading all these posts, I think I'll go out turkey hunting and just wait for the results to come out

Hahaha. Yeah I think that's probably all of our best bet.

I ended up switching things around in a manner that makes me just expecting not to draw unless it's a real dandy of a tag.

 

[VANDAL]

I’m no statistician but I think I know the formula. If you put in for 5 choices, each of the hunts have an independent draw, so you can’t just add up each of the draw odds for the hunts. So if each hunts have a 20% draw probability, your chances of drawing 1 of the 5 choices would not be 100%.

Same logic as flipping a coin...if you flip a coin and it lands on heads 5 times in a row the probability of landing on tails the next flip is is still 50%. But the more times you flip the coin the better your chances are of getting tails.

This is the formula I use. Let’s assume each choice has a draw odd of 20%. You multiply your probabilities of not drawing against each other.

(1-.2)*(1-.2)*(1-.2)*(1-.2)*(1-.2) = chance you won’t draw.

Subtract the answer by 1 and that is the percent probability of drawing one of the five tags.

Hopefully that makes sense and helps.






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[VANDAL]

I didn’t give the answer in my post. So the probability of drawing one of the 5 tags if all hunts had draw odds of 20% would be 67%.




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[Jaden Bales]

I didn’t give the answer in my post. So the probability of drawing one of the 5 tags if all hunts had draw odds of 20% would be 67%.




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Man, that's kinda roughly what I ended up getting, ish, as well. Haha.

Waiting on the draws now!

 

[Dos Perros]

I’m no statistician but I think I know the formula. If you put in for 5 choices, each of the hunts have an independent draw, so you can’t just add up each of the draw odds for the hunts. So if each hunts have a 20% draw probability, your chances of drawing 1 of the 5 choices would not be 100%.

Same logic as flipping a coin...if you flip a coin and it lands on heads 5 times in a row the probability of landing on tails the next flip is is still 50%. But the more times you flip the coin the better your chances are of getting tails.

This is the formula I use. Let’s assume each choice has a draw odd of 20%. You multiply your probabilities of not drawing against each other.

(1-.2)*(1-.2)*(1-.2)*(1-.2)*(1-.2) = chance you won’t draw.

Subtract the answer by 1 and that is the percent probability of drawing one of the five tags.

Hopefully that makes sense and helps.






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This is not correct.

 

[Bubblehide]

Vandal, quoted above is as close to correct as one can get, and may be 100% correct.

Way back in the day, I was the department statistics coordinator at my university. I have had to much experience (in the very distant past) with advanced statistics, i.e., graduate courses. I have not been involved in statistics at that level in many years. I no longer have any statistics books at home. But I do intrepreate solved statistics at work on a very regular basis (computer programs do all the solving).

Years of applying increasing your odds of drawing:
A basic statistical model of draw odds is exactly the same as our lotteries. Your odds of drawing do not change because you play the lottery each week. Each weeks drawing has the same odds. So you never have an increased chance of winning. Same for our wildlife drawings (not taking into account preference points).

Does purchasing more tickets increase odds of winning:
Generally speaking, in a simple statistical model, such as our lotteries, purchasing more tickets DOES NOT make a SIGNIFICANT (this is a scientific term) difference in your odds of winning; the only time any significant difference exists from purchasing multiple tickets is when you have purchased a significant number of tickets. The problem with trying to figure out what number of tickets you need to purchase to make a significant difference, is that you need to know how many tickets are/will be sold, in advance of the drawing. In our lottery, this is never known ahead of time. In our wildlife drawings, we most often know the number of tickets that will be sold, but we are unable to get more than one drawing ticket. So in short, your odds do not change.

Our wildlife drawing are NOT a simple statistical model, in any state that I am aware of; they are complex models with the addition of preference points involved, splits between resident and nonresident allotments; allotments for preference points not being considered, and I am sure I have missed a few allotments, advantages and disadvantages, as I do not apply for drawing in all our states. My point here is that it is not practical for the average Joe to attempt to calculate draw odds based on their individual situation; but to rather recognize their situation based on preference points. What I mean here is you need to stay abreast of how many preference points one has, and the points required to draw a particular zone of choice along with any 2nd choice odds... Keep in mind that this is not static, the odds for any zone in any state can change at any time, as can the value/perceived value of a tag in any given zone., thus changing the number of people applying.

So to clarify that value, what I mean is that events happen that affect the population of the animal(s) you are drawing for. Such as, high winter kills, large die offs due to a variety of reasons, someone harvesting a ginormous anilam, the same is true for the opposite happening. But bottom line for trophy hunters is number or quality of mature animals in their given zone; also not static.

In short, if you are not using one of the draw services/draw odds services, you should be sticking with the simple known numbers, otherwise you are fooling yourself, as the only thing that makes a significant difference/increase in anyone drawing a tag is preference points, preference points, and preference points.

 

[MuleyFever]

I don’t know how you could figure accurate draw odds for NV or any state with the random aspect to it. Throw in NR quotas, people switching units and squared points and it gets impossible to figure in advance. With the squared points a handful of people jumping in for a unit could change the odds quite a bit from year to year. Someone with 15 points putting in adds 225 entries for that person.

 

[Jaden Bales]

Vandal, quoted above is as close to correct as one can get, and may be 100% correct.

Way back in the day, I was the department statistics coordinator at my university. I have had to much experience (in the very distant past) with advanced statistics, i.e., graduate courses. I have not been involved in statistics at that level in many years. I no longer have any statistics books at home. But I do intrepreate solved statistics at work on a very regular basis (computer programs do all the solving).

Years of applying increasing your odds of drawing:
A basic statistical model of draw odds is exactly the same as our lotteries. Your odds of drawing do not change because you play the lottery each week. Each weeks drawing has the same odds. So you never have an increased chance of winning. Same for our wildlife drawings (not taking into account preference points).

Does purchasing more tickets increase odds of winning:
Generally speaking, in a simple statistical model, such as our lotteries, purchasing more tickets DOES NOT make a SIGNIFICANT (this is a scientific term) difference in your odds of winning; the only time any significant difference exists from purchasing multiple tickets is when you have purchased a significant number of tickets. The problem with trying to figure out what number of tickets you need to purchase to make a significant difference, is that you need to know how many tickets are/will be sold, in advance of the drawing. In our lottery, this is never known ahead of time. In our wildlife drawings, we most often know the number of tickets that will be sold, but we are unable to get more than one drawing ticket. So in short, your odds do not change.

Our wildlife drawing are NOT a simple statistical model, in any state that I am aware of; they are complex models with the addition of preference points involved, splits between resident and nonresident allotments; allotments for preference points not being considered, and I am sure I have missed a few allotments, advantages and disadvantages, as I do not apply for drawing in all our states. My point here is that it is not practical for the average Joe to attempt to calculate draw odds based on their individual situation; but to rather recognize their situation based on preference points. What I mean here is you need to stay abreast of how many preference points one has, and the points required to draw a particular zone of choice along with any 2nd choice odds... Keep in mind that this is not static, the odds for any zone in any state can change at any time, as can the value/perceived value of a tag in any given zone., thus changing the number of people applying.

So to clarify that value, what I mean is that events happen that affect the population of the animal(s) you are drawing for. Such as, high winter kills, large die offs due to a variety of reasons, someone harvesting a ginormous anilam, the same is true for the opposite happening. But bottom line for trophy hunters is number or quality of mature animals in their given zone; also not static.

In short, if you are not using one of the draw services/draw odds services, you should be sticking with the simple known numbers, otherwise you are fooling yourself, as the only thing that makes a significant difference/increase in anyone drawing a tag is preference points, preference points, and preference points.

Holy cow. Thanks for the thought and effort you put into this response. Just the man we were looking for!

So if you were to boil this down to one sentence for the average joe... would you say a guy who wants to know drawing odds should just use whatever stat the draw services provide and go with that as their likelihood of drawing and call it macaroni?

It seems like all of this discussion has basically come down to the fact that it's damned complex and a guy should just budget/plan for drawing the tag or buy a preference/bonus point if the money and time isn't there.

 

[Bubblehide]

Jaden, At it's simplest, yes. But it could easily depend on individual circumstances. If you are applying as a resident, the draw odds are pretty easy to estimate accurately based on needed preferance points. If your not apply as a resident, go with published non resident odds or pay for a service. The actual formulas are rather complicated, and since actual applicants varies each year, odds are predicted using historical data.

 

[Jaden Bales]

Jaden, At it's simplest, yes. But it could easily depend on individual circumstances. If you are applying as a resident, the draw odds are pretty easy to estimate accurately based on needed preferance points. If your not apply as a resident, go with published non resident odds or pay for a service. The actual formulas are rather complicated, and since actual applicants varies each year, odds are predicted using historical data.

Awesome! Thanks for the info!

 

[Specialk]

The bottom line is odds are just odds, and don’t really mean anything. The best part about Nevada is you can draw any tag with 0 points....happens all the time. Put your name in the bucket and make sure to put your ‘dream hunt’ choices first

 

[Jaden Bales]

The bottom line is odds are just odds, and don’t really mean anything. The best part about Nevada is you can draw any tag with 0 points....happens all the time. Put your name in the bucket and make sure to put your ‘dream hunt’ choices first

I hear ya there!

 

[RCB]

I'm assuming these tags are drawn sequentially, i.e. first they take the first choice draw. If you fail there, you go to second choice, then third, and so on. (This is how it works where I hunt). If so, then the formula is

C1 + (1-C1)C2 + (1-C1)(1-C2)C3 + (1-C1)(1-C2)(1-C3)C4 + (1-C1)(1-C2)(1-C3)(1-C4)C5

i.e. the probability that you draw the first, plus the probability that you don't draw the first but do draw the second, plus the probability that you don't draw first or second but do draw the third, and so on.

Important point: C2 is the probability of drawing the 2nd choice tag as a 2nd choice - NOT the probability of drawing it as a first choice. And so on. These stats may not be published, but the latter terms are probably very low (e.g. what's the chance that there are any C5 tags left after 4 rounds?). The last few terms might be effectively 0.

If instead the draws are independent - i.e. all are drawn at the same time, so that you could end up with anywhere between 0 to 5 mule deer tags (which seems unlikely but I don't know NV rules) - then the formula is
1- (1-C1)(1-C2)(1-C3)(1-C4)(1-C5)
Don't take the average.

 

[Jaden Bales]

I'm assuming these tags are drawn sequentially, i.e. first they take the first choice draw. If you fail there, you go to second choice, then third, and so on. (This is how it works where I hunt). If so, then the formula is

C1 + (1-C1)C2 + (1-C1)(1-C2)C3 + (1-C1)(1-C2)(1-C3)C4 + (1-C1)(1-C2)(1-C3)(1-C4)C5

i.e. the probability that you draw the first, plus the probability that you don't draw the first but do draw the second, plus the probability that you don't draw first or second but do draw the third, and so on.

Important point: C2 is the probability of drawing the 2nd choice tag as a 2nd choice - NOT the probability of drawing it as a first choice. And so on. These stats may not be published, but the latter terms are probably very low (e.g. what's the chance that there are any C5 tags left after 4 rounds?). The last few terms might be effectively 0.

If instead the draws are independent - i.e. all are drawn at the same time, so that you could end up with anywhere between 0 to 5 mule deer tags (which seems unlikely but I don't know NV rules) - then the formula is
1- (1-C1)(1-C2)(1-C3)(1-C4)(1-C5)
Don't take the average.

That's good to know man! Thanks for that... That first equation may work for NV.

So they basically have a raffle. When you have 0 points, you are given a number in the bin 1 time. They randomly draw numbers, and if your number is drawn, they look at your 1st choice. If there are no tags available, they move to your 2nd choice, and so on through your 5 choices. In Nevada, unlike an Oregon or Colorado, you really have a super high chance of drawing that 3-5 choice, and that's usually what guys draw because they swing for the fences on the first couple choices since there's always a chance to draw randomly.

After each year a guy receives a bonus point, their bonus points are squared and turned into names in the hat. So if you have 4 points, you have 16 chances to get drawn in the raffle and so on. Someone just starting has 1 chance.

Make sense?

 

[RCB]

Interesting. More complicated than what I wrote, then. I don't think you could get an exact answer without knowing more information. The formula I wrote might make for a decent approximation though...

 

[Jaden Bales]

Awesome! I know "gut feeling" isn't a great statistical indicator of accuracy, but I felt good when I got a 20% odd of drawing when my chance draws last year were .065%, 1%, 6.8%, 9% and 14%. Haha.

We'll see if I get lucky on May 24th.

 

[S.Clancy]

How would one calculate the odds of drawing 1 of your 5 choices in Nevada?

Better question, what are the odds anyone wants to try and tackle this math?

Here's the situation.

I got all trigger happy and applied for Nevada before mapping out my hunting budget for the year. Went to get a refund (since app date is April 29) and that is not possible. Totally get it. No problems there. So, I'll stick with my plan to try and pull a tag and if I do, I'll just sell my body or something to make sure I can go. No problems there either.

Now, I am trying to figure out what the chances of drawing just ANY of my units using GoHunt's per unit draw odds and costing me another...costing me a good chunk of change to do the hunt.

I come up with the following formula using this article: https://math.stackexchange.com/questions/91998/probability-of-winning-a-prize-in-a-raffle

C = Individual Unit Draw Odds
5 = Total Draw Choices

(C1 + C2 + C3 + C4 + C5)/5 = Average Draw Odds (ADO)

1 - ADO = Average Fail Odds (AFO)

(AFO)^5 = Odds of Failing to Draw (OFD)

1 - OFD = Odds of Drawing Tag

Bottom Line: If your 5 choices average out to 10% draw odds, you actually have 40% draw odds of pulling ONE tag of the 5 that year.

Does that make sense to anybody?

This is the way to do it!!!