Need a statistician here...

[BigPig]

OK, the rest of us regular rednecks, keep quiet here. I need an actual mathematical explanation to settle an argument.

Scenario:
You apply for 3 different hunts (say elk, deer, antelope for the sake of argument) that each have 5% draw odds. No points at work, just plain odds. What are the odds of drawing any one of those hunts? Is it 15% or still just 5%?

 

[BeanMan]

5%, statistically you evaluate them as separate events, not additive Just like flipping a coin. Flip a coin 4 times in a row and get heads, what is the probability of heads again? It’s still 50% each time.

 

[scottr]

5% is not the final answer. That is the percent chance on a single hunt. It does not account for the chance of drawing the other tags, which would increase the percentage.

The method I use when I dream of seeing green is calculating the chance of not drawing any tag.

Start off by calculating the chance of not drawing the first hunt (1-p). p being the chance of drawing hunt #1. Do the same for the other hunts. Then you multiply each of the chances of not drawing. Just as you could multiply 5%x5% to get the chance of drawing both tags. (1 out of 400 in this scenario.) And for fun if you add in the third hunt you have a 1 in 8000 chance of drawing all three tags.

For the OP's scenario you end up with (1-.05)(1-.05)(1-.05). The result is that ~85.7% of the time you will not draw any tag. Which translates to ~14.3% chance of drawing at least one tag.

 

[roadrunner]

Since this is going to get derailed, I will throw this in.

Say there are 6 tags and 30 people apply. It doesn't mean 6 groups and 1 out of 5 guys in each group are going to draw. It means 1 guy out of 30 will get tag no. 1, the next tag is going to 1 guy out of 29, the next will go to a 3rd guy out of 28 until all 6 tags are given out.

1 out of 5 is a 20% chance but 1 out of 30 is a 3.3%, the next opportunity is 3.4 until the last at 4%.

Then, if you consider the way NM does it with each choice being looked at before it goes to the next application, it gets even more complicated, or worse. Not sure which.

 

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This should do the trick

 

[BigPig]

Well Beanman and scoter's explanations both make mathematical sense, and I came to the exact 2 conclusions... but the two can't both be right. So which one is it? I see merit to both arguments.

 

[eelgrass]

BeanMan is correct. Three separate unrelated events. One does not influence the others.

 

[roadrunner]

BeanMan is correct. Three separate unrelated events. One does not influence the others.

That is meant to be understood...

 

[scottr]

You can add the probabilities of unrelated events, such as drawing a tag or flipping a coin. The OP's original question was odds of not drawing any tag. If he only put in for a single tag 5% would be the answer, however, you have to consider the odds of drawing a second or third tag.

For example, if you flip a coin you have 50/50 chance for heads or tails. Now consider if you had a 50% for a tag. You could make a tree chart that shows all outcomes and add the results. Roadrunner is correct in that this model is without replacement, meaning that once a tag is drawn it is not replaced in the pool. So in my calculations I underestimate slightly, but figure it's close enough for government work.

So you have 50% chance at tag #1, you have a 50% at tag #2, and 50% chance at tag #3. The chance to draw not one of those tags, would be the same chance as flipping 3 heads in a row. Which is 12.5% chance of not drawing anything or a 87.5% of flipping at least one tail or drawing one tag.

It is true that each event is independent, however, with proper analysis you can look at the cumulative chances.

BTW I have a masters of science in public policy analysis with focuses on mathematical modeling and econometrics, but have been running higher level statistics for most of my career.

 

[Sanjuanfly]

Beaman each coin flip is 50%.

What are the odds of flipping heads 3 times in a row?

 

[Bluehair]

This makes me want to hit the craps table.

 

[BuzzH]

A few people should have stayed awake during statistics...scottr has it right.

Only college math class I liked and made sense to me was probability/statistics. Trig, calculus, linear math, etc...about like watching paint dry and took them only because I had to.

 

[sticksender]

Depends on the conditions of the draw algorithm. For example do the 3 draws happen to be Utah draws occurring in the same year? In that case, you can't draw more than one limited hunt per year, so the probability of drawing "at least one" must exclude the probability of drawing two and drawing three, since that is not allowed. If no such conditions exist, meaning that you could legally draw all three, then ScottR's calculation above is correct.

 

[schoolhousegrizz]

Scottr nice work, tell us the mathematical equation you used to get the 87.5% chance of getting 1 heads in 3 flips, thanks! I know the equation you use to get the 12.5% just 1/2 x 1/2 x 1/2. The chance of getting heads / possible outcomes X each other three times.

 

[grizzly]

Scottr is correct.

Another way to think about it. We're not affecting the chance of any particular draw, just that any one of them will hit.

If you have to flip heads on your next coin toss, would you rather do it with one coin in your hand or a dozen coins. The odds of each individual coin being heads is still 50/50, but the odds of getting AT LEAST one heads with a dozen coins in a single handful is much better at 99.97%.

 

[grizzly]

Scottr nice work, tell us the mathematical equation you used to get the 87.5% chance of getting 1 heads in 3 flips, thanks!

1-(.5 x .5 x .5)= .875 which is 87.5%

 

[schoolhousegrizz]

Oh, so you just subtract the chance you have of getting it all three times from 100%?
Thanks Grizz!

 

[grizzly]

Oh, so you just subtract the chance you have of getting it all three times from 100%?
Thanks Grizz!

You're welcome. Remember the multiple is based on the odds of NOT drawing. For example if you had a 3% chance of drawing, you'd multiply it by .97... those just happen to be the same .50 when discussing coin flips.

 

[schoolhousegrizz]

Got it, thanks!

 

[RSC68]

I'm an engineer and have taken many math classes in my past, but damn does my head hurt after following this thread of discussions!!

 

[RSC68]

I've had an 80% to supposedly 100% chance (according to GoHunt) to draw Premium LE deer tags in the past and I'm still waiting for that to happen.....

 

[roadrunner]

I'm an engineer and have taken many math classes in my past, but damn does my head hurt after following this thread of discussions!!

Yeah, I purged all that stuff over 20 years ago now, just do the simple algebraic type calcs.

The thing about draw hunts, is it only matters on how the state does it and what the actual numbers truly are relative to applicants against available tags. It usually isn't a simple coin toss.

How one state does it may vary how another does it. Exp: the only thing you can do in NM is choose hunts that don't "fill up" as quick by how many 2nd and 3rd choices get them to make yourself feel better about a better odd of drawing.

 

[eelgrass]

I put in for 20 different hunts. Each hunt has one tag available and 20 applicants. I have a 100% odds of getting a tag?

 

[BuzzH]

I put in for 20 different hunts. Each hunt has one tag available and 20 applicants. I have a 100% odds of getting a tag?

No, another person that slept through statistics...

 

[BigChiefJ]

For the OP, you have a 5% chance to draw a hunt. Now to get fancy for odds and probability, you can plug the numbers to see what you chances of getting 2 or 3 draws. I'm not a calculator, but it's way less than 5%.

Numbers are so easy, but drawing a tag can be hard.

 

[notdonhunting]

Scottr and beanman
Is a realist

Roadrunner and bluehair
Is a optimist

 

[Longun]

My Head Hurts..I'm going to town to by my general Elk and Deer tag..that'll ease the pain...

 

[roadrunner]

Scottr and beanman
Is a realist

Roadrunner and bluehair
Is a optimist

One approach describes the chance of falling into one of three categories. It does not describe the actual chance of pulling the tag in that category. If there are 5 tags and 100 applicants, you don't have a 50% chance of drawing the tag. To pull the tag, you only have 5 shots at having your name pulled from the hat against the other 99 names. Your chances become better each time, but it's a negligible increase.

That's the reality...

 

[sticksender]

For the OP, you have a 5% chance to draw a hunt. Now to get fancy for odds and probability, you can plug the numbers to see what you chances of getting 2 or 3 draws. I'm not a calculator, but it's way less than 5%.

Here's that answer. Entering three independent draws, each offering a 1 in 20 chance to draw (5%) and assuming a pure random draw, there are a total of 8000 possible outcomes (20x20x20).

6859 times out of 8000.....draw none = 85.7375%
1 time out of 8000.....draw all three = 0.0125%
57 times out of 8000.....draw two = 0.7125%
1083 times out of 8000.....draw one = 13.5375%

Soooo....1+57+1083 = 1141 occurences out of 8000, or a 14.2625% probability, of drawing "something".

 

[grizzly]

One approach describes the chance of falling into one of three categories. It does not describe the actual chance of pulling the tag in that category. If there are 5 tags and 100 applicants, you don't have a 50% chance of drawing the tag. To pull the tag, you only have 5 shots at having your name pulled from the hat against the other 99 names. Your chances become better each time, but it's a negligible increase.

That's the reality...

After the first name is drawn (5/100)... It's 4/99 (4.04%)... And then 3/98 (3.06%). Your odds of each tag decrease over time, not become better, but the overall odds are unchanged.

_____________

Imagine a room where are there are 100 applicants for 5 tags and the most any person can draw is one tag (no duplicates).

You can put in as many bonus point or squared bonus point schemes as you want. Everybody can have one ticket, or 100 tickets, or a thousand tickets. Or some guys get 1 ticket and others get 100 tickets. The draw system you choose really doesn't matter...

Because the final total overall draw odds will be 5%. Exactly 5 people will leave with a tag in their pocket and 95 will not. Everything else is just fluff. Point systems don't affect overall draw odds, they just try to balance it out in an attempt at fairness.

If you truly want to improve overall odds, you have to either increase tags or decrease applicants.

 

[roadrunner]

Here's that answer. Entering three independent draws, each offering a 1 in 20 chance to draw (5%) and assuming a pure random draw, there are a total of 8000 possible outcomes (20x20x20).

6859 times out of 8000.....draw none = 85.7375%
1 time out of 8000.....draw all three = 0.0125%
57 times out of 8000.....draw two = 0.7125%
1083 times out of 8000.....draw one = 13.5375%

Soooo....1+57+1083 = 1141 occurences out of 8000, or a 14.2625% probability, of drawing "something".

Common sense tells me there is nothing for second place. If your name ain't the first one out of the hat with a ratio if 1:20, you lose. You increase the distribution of the ratio to 100:2000 and you get additional tries, but it's still pretty small. There's no multiple combinations. You either get it or you don't. Those are the same combinations for each species. Having multiple outcomes only matters if each outcome equates to some kind of win or benefit.

But then again, I did stats over twenty years ago, passed the class, and haven't looked back...

 

[BigPig]

Here's that answer. Entering three independent draws, each offering a 1 in 20 chance to draw (5%) and assuming a pure random draw, there are a total of 8000 possible outcomes (20x20x20).

6859 times out of 8000.....draw none = 85.7375%
1 time out of 8000.....draw all three = 0.0125%
57 times out of 8000.....draw two = 0.7125%
1083 times out of 8000.....draw one = 13.5375%

Soooo....1+57+1083 = 1141 occurences out of 8000, or a 14.2625% probability, of drawing "something".

This is the most logical explanation yet. But mathematically, how are you getting the 57 and the 1083 time?

 

[sticksender]

This is the most logical explanation yet. But mathematically, how are you getting the 57 and the 1083 time?

There's mathematical shorthand to get those, but when you only do it that way, you may not gain a grasp on the logic. Better way is to picture it first, then calculate it. Sort of like the way poker-players, if they're good, can sum all possible hands for a given situation.

WIN TWICE: So we look and see that each draw consists of 19 different "losing tickets" plus one winner. And we know there are only 3 different ways to win two tags, as follows: win the first draw and the second draw, win the second draw and the third draw, or win the first draw and the third draw. And whichever draw you lose has 19 different "losing tickets", meaning 19 different ways to lose one and win two. So the total number of possible "hands" or combinations in which you could win 2 tags is: 3x19=57

WIN ONCE: Same logic as above. There are 3 ways to win one tag, as follows: win the first draw and lose the second and third, win the second and lose the first and third, and win the third and lose the first and second. Each one of those 3 draws has 19 different losing tickets, or ways to lose. But the "hands" we want to add up here are the hands in which you lose twice and win once. So considering these 3 combinations in which you lose 2 draws and win one, we simply multiply that out. For the 3 draws to produce 1 winner and 2 losers, there are 19x19 different possible combinations of losing hands. So the total number of "hands" in which you'd win only once is 3x19x19=1083.

 

[roadrunner]

There's mathematical shorthand to get those, but when you only do it that way, you may not gain a grasp on the logic. Better way is to picture it first, then calculate it. Sort of like the way poker-players, if they're good, can sum all possible hands for a given situation.

WIN TWICE: So we look and see that each draw consists of 19 different "losing tickets" plus one winner. And we know there are only 3 different ways to win two tags, as follows: win the first draw and the second draw, win the second draw and the third draw, or win the first draw and the third draw. And whichever draw you lose has 19 different "losing tickets", meaning 19 different ways to lose one and win two. So the total number of possible "hands" or combinations in which you could win 2 tags is: 3x19=57

WIN ONCE: Same logic as above. There are 3 ways to win one tag, as follows: win the first draw and lose the second and third, win the second and lose the first and third, and win the third and lose the first and second. Each one of those 3 draws has 19 different losing tickets, or ways to lose. But the "hands" we want to add up here are the hands in which you lose twice and win once. So considering these 3 combinations in which you lose 2 draws and win one, we simply multiply that out. For the 3 draws to produce 1 winner and 2 losers, there are 19x19 different possible combinations of losing hands. So the total number of "hands" in which you'd win only once is 3x19x19=1083.

Please show the calculation that depicts the winning outcome with a cage tumbler filled with 20 ping pong balls, 19 marked "L" and one marked "W". Cage is tumbled for 10 seconds and 20 guys get one chance each to pull a ball out. Once the ball is pulled it does not go back in. The cage is not tumbled again. Three tumble cages, one marked "D", one marked "A", and one marked "E". Each 20 guys gets to pull one ball from each cage.

 

[Buckjunkie]

4 out of 3 people struggle with math.??

 

[sticksender]

roadrunner, you've just described a scenario exactly matching the OP's, so that has already been answered above.

 

[roadrunner]

roadrunner, you've just described a scenario exactly matching the OP's, so that has already been answered above.

I suppose, guess that's what I get for listening to the radio the same time I read and posted... ?

 

[jls456]

Wow.
I just remember standing in line at McKnights sporting goods in St. George to get my license with my Dad ang Grandpa...
$50.00 for a NR license if I remember..
They use to let schools out for Deer season....

 

[lostinOregon]

I've had an 80% to supposedly 100% chance (according to GoHunt) to draw Premium LE deer tags in the past and I'm still waiting for that to happen.....

That’s because the odds were from last years drawing and have no relevance on this years drawing.

Rich

 

[Mbogo]

I've had an 80% to supposedly 100% chance (according to GoHunt) to draw Premium LE deer tags in the past and I'm still waiting for that to happen.....

They've done studies you know. 60% of the time it works every time . . .

 

[Zeke]

Some of you brilliant folks are forgetting the “x” factor.

If MY name is on a random application , I’ll have exactly ZERO chance to draw.

I’m sure I cannot be alone when feeling like this!

Zeke

 

[Joe2Kool]

This should help...

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