Nevada Odds Question for Math People

[rmyoung1]

I’m not a mathematician but I’ve asked myself a lot of these same questions. Here’s the deal: Your odds of drawing will never be better than the odds of drawing your easiest choice. That’s what @npaden correctly stated above. The folks at toprut go into some detail on this but I’ll throw out a simple example.

Let’s say you only rank two choices.Your first choice has a 1% chance at your point level and the odds on your second choice are 3%. Your odds of pulling a tag are 3%, the same as the odds of your easiest choice. The odds of drawing your first choice, specifically, are 1%. The odds of getting your second choice, specifically, now are the difference. In this case 2% since 3%-1% is 2%. So... in total, you’ll have a 3% chance of drawing - 1% chance of choice #1 and 2% chance of drawing choice #2. Hope that helps!

 

[ImBillT]

For a single species in NV specifically, your total odds are equal to the highest odds hunt code that you applied for. Getting those correct is EXTREMELY difficult, and depends upon the order and point totals of ALL APPLICANTS, as well as which other applicants ACTUALLY DRAW ahead of you if there is more than one tag for any of those hunt codes. The logic is as follows.

For the first point, we will assume that the odds are actually correct. If you have .9, 1.0, 1.1, 1.2, and 1.3 percent draw odds on your hunt choices 1-5, your total probability of drawing is 1.3%. This is because the results are not independent. They have a single drawing, and go from top to bottom. They don’t assign you, or anyone else, a new random number for your second choice. If your random number wasn’t good enough for your fifth choice(it wasn’t in the top 1.3% of random numbers) then it definitely isn’t good enough for your first(top .9%). Again, this assumes that the odds you’re working with are correct. It does not take into account any flaws in the number you are presented.

On to the second point(extremely difficult to accurately calculate your true odds). With only one choice, and no points, or with points, but with only one tag, the probability of drawing is simple. Apps/tags in the first case. Your points/total points in the second case. For hunt codes with more than one tag and point system, things get slightly tricky. The probability of you drawing the first tag is your points/total points, but for the second tag, we have to know how many points the first tag holder had. If it was a first time applicant, then only one point has been removed from the pool -> your points/(total points -1) On the other hand, if the first tag holder had 15 points, then in NV, 256(15^2=255 +1 for this year) points were removed from the pool. That makes the formula for the second tag your points/(total points-256). To calculate this without actually knowing who drew, you must assign the probability that the first tag will go to someone with 0,1,2,etc points. If they simply assign 256 random numbers to the person with 15pts, then use his lowest number, it actually ends up being the same probability as if he’d literally had his name in 256 times(but could only draw a single tag before being removed from consideration) so no worries about the semantics. That only covers the point complication. Next is the complication of multiple choices, choice order, and if there are more hunt codes than choices, who chose what. Now you have to assign a probability to the likelihood that someone who applied for the same hunt that you did, will actually draw some hunt that you did not and thus be removed from the pool, AND the probability that someone with a better random number than you, and the same hunt choices, drew a very easy to draw hunt as one of their higher choices, essentially wasting their luck, and leaving the more desired hunt available a little longer. All of this can be calculated, but why not just write some simple code to simulate the draw a few million times, and simply report the percentage of times that a each outcome occurred? That’s what GoHunt claims to do. It’s also why the don’t list a probability at point levels that weren’t on any applications for specific hunts last year.

So....assuming that the probabilities you were working with were correct, and 1.3% was your highest probability of any of your five choices for a single species in NV, then the probability that you’ll draw any of the five choices is 1.3%

Now let’s examine multiple species. Let’s say that you apply for five species in NV, and that your highest draw odds is exactly 1% on all five. These are independent events. You’re assigned a new random number for each species, and the results of each drawing have no bearing on the results of any of the others. In this case the calculation is the probability that you won’t draw anything is equal to the probability that won’t draw each code times the probability that you want draw the rest of the codes. At 1% for all it would be 1-.01=.99, so all codes have a 99% chance that you won’t draw, so .99x.99x.99x.99x.99=.951, or a 95.1% chance you won’t draw, or a 4.9% chance that you will draw at least one tag. The best odds of each species could be different ie. 1%, 3%, 4%, 2%, 1%, .99x.97x.96x.98x.99=.894, or a 10.6% chance of drawing at least one tag.

If you take it to multiple states and multiple years, they are all independent as well, so let’s pretend that you apply for ten sheep tags every year that each have exactly 1% draw odds(that’s pretty good for sheep) and that never changes over the span, and that you apply for thirty years. That’s 300 attempts. .99^300=.049, or a 95.1% chance that you’ll draw at least one sheep tag. Keep in mind that many sheep tags are well under 1%(more like .01%)

 

[MTHeifer]

Right, your odds are better than 1.3% because of that scenario, but since there are so few tags for bighorn sheep and so many applicants, they are just barely above that value... still probably less than 2%. I don't think you could come up with the exact odds unless you had the statistics for every 1st through 5th choice on every application. If there were hundreds of tags or fewer applicants, then the odds would be more affected by the scenario you described.