Calculating draw odds.

[fez]

Quick question for you smart guys.

With 250 tags available and 1150 applicants it means I have a 21.7% chance at drawing. Now here comes my question. What are the odds that either me or my wife will draw? Is it as simple as adding another 21.7% or is it more complex than that? There are no bonus points to make it more difficult, just a total random system

Thanks for your expertise
fez

 

[Cornell2012]

If I remember stats correctly, it is slightly more complicated, but not much. I believe that the odds of either or both of you drawing are

P(you draw) + P(your wife draws) - P(both of you draw) = 21.7% + 21.7% - (21.7%)*(21.7%) = 38.7%

 

[TheDudeAbides]

Quick question for you smart guys.

With 250 tags available and 1150 applicants it means I have a 21.7% chance at drawing. Now here comes my question. What are the odds that either me or my wife will draw? Is it as simple as adding another 21.7% or is it more complex than that? There are no bonus points to make it more difficult, just a total random system

Thanks for your expertise
fez

Well,

There is a difference between draw odds and draw probabilities, in case you decide to go to Vegas.

Odds are normally expressed in a ratio and are expressed in successes : failures. In this case an individual will have 23 total chances. 5 chances of success and 18 chances of failure. The odds would be 5:18. If you calculate by 5/18 you will receive a higher percentage 5/18 ~ 27.8%, which isn't correct. It is also a reason why odds are used for gambling, to make your chances look better than they actually are.

Probability is normally expressed in a percentage. Probability is expressed in this case would be 5/23 or 21.7%. These are commonly mislabeled as draw odds, when in fact they are probabilities.

Your situation is interesting.


I would calculate it like this.

P(You Draw) + P(Wife Draw) - P (Tags that you both can't draw at the same time)

(P(250/1150) + P(250/1150)) - P (249/1150) = 251/1150 or 21.8%

Which to me makes sense, because your probability will be slightly better without being overly better.

 

[npaden]

I'm going with Cornell2012's math on this one.

 

[npaden]

Found an old thread where this was discussed in the terms of applying over several years for the same tag. Essentially the same thing you are talking about other than the fact that you could actually both draw.

Here's my math from that thread:

Each year does stand on it's own, however the probability of going years without drawing is different than each stand alone event.

An easy way to prove it is rolling dice. There is 1 out of 6 chance of rolling a specific number. Let's say we want to roll a 1. There is a 1 in 6 chance every time we roll that it will come up as a 1. 16.67%.

The chances each time you roll are the same, however if you roll the dice 6 times there is a 66.6% chance (1-(5/6)^6) that one of those rolls will come up a 1.

If you want we can put some money on it. There is a 1 in 6 chance that I will roll a 1 on the dice. I'll even give you better odds and only make you pay me $2 if I roll a 1 and I'll pay you $1 if I don't. The odds are dramatically in your favor. Now let's keep those odds the same way and let me roll the dice 10 times, or even better 50 times. The odds of me rolling a 1 out of 10 rolls increases to 83.8%, out of 50 times it goes to 99.9%.

Each roll is a stand alone event and the odds stand by themselves, but the your chances do increase with every roll of eventually rolling that 1.

mdunc8 didn't like my explanation and thought coin flips provided a more understandable scenario. Here is his explaination:

It's easy to follow using coin flips. Let's say Paden and I are the only two people in the raffle for the sheep. I'm "heads". He's "tails". Let's run it for three years. I think everyone will agree that we both a 50/50 change the first year. Pretty simple. There's 8 possible outcomes if you run the raffle three years: heads, heads, heads; heads, heads, tails; heads, tails, tails; tails, tails, tails; tails, tails, heads; tails, heads, tails; heads, tails, tails; and heads, tails, heads. There's only one possible run of events after three years where Paden (heads, heads, heads) or I (tails, tails, tails) will not draw a tag. Therefore, after three years, we each have increased our odds from 50% to 87.5%. However, our odds EACH year remain 50%.

Again this was calculating cumulative odds over a period of years, but you could apply the same math (pretty much) to both you and your wife's chances the same year.

Here's the link to that thread with some people telling me I'm wrong. (but I think I'm right)

http://onyourownadventures.com/hunttalk/showthread.php?246372-Raffle-hunts

 

[hank4elk]

Now that I have a pea brain sized headache,I'm vindicated on my draw probability figuring each year.
Or a guess as it is known.This is because you never really know how many folks will apply for that tag ,this year.

 

[rmyoung1]

I'm going with Cornell2012's math on this one.

Yeah, Cornell2012 got it right. When in doubt about math, believe the Ivy League graduate

 

[LopeHunter]

Your scenario was one of the easier situations. Altering the formula for states that have caps for non-residents for a hunt unit or for a species or provide a preference or a bonus or look at multiple choices if an application is chosen or have some tags in a random draw while the rest are in a bonus or preference system can result in a reach for a big bottle of aspirin.

I keep a spreadsheet with my first choice odds as best as I can estimate them looking at the most recent year's results. States such as AZ or NV can make this tough to nail down but can get in the ballpark so know if closer to 1:10 or 1:500.

I also track how many desirable tags are offered to non-residents for my first choice. Sometimes is 1 tag at most (sheep in Idaho for a non-resident) for the hunt choice and other times will be a dozen or more.

The thing is, you can have hunt codes where the odds yo-yo as applicants chase the best odds on a sheep tag from last year but so did a lot of other people so is a crap shoot looking in the rearview to guess what is around the bend. One cover article in a hunting magazine about the big critter shot in Unit X or someone saying Unit X is a sleeper unit or is now the Number 3 unit in a state will also blow the odds up compared to last year.

I apply for about 80 hunts in over 10 western states. Most of those hunts are 1 in 500 type odds so when the dust settles I usually have a very desirable rut tag for some critter and maybe one or two more tags that are fun hunts with lots of public land and few hunters. I have up to 23 preference/bonus points for some species in some states so in a few cases intentionally am not being drawn for a tag in the current year due to concerns over the current hunt quality or prefer to sit on the virtually guaranteed tag (a risky strategy since states often change things) for the future.

Good luck on your draws and enjoy your hunts.

 

[BuzzH]

The only draw odds I worry about are:

If you don't apply your odds are 100% that you will never draw.

 

[TheDudeAbides]

Yeah, Cornell2012 got it right. When in doubt about math, believe the Ivy League graduate

He might be, I'm just a little skeptical because if you had 6 people in your group your probability would be 1.32 - 0.0001, which means you would expect to draw and I do not see that as being true.

Most of the stats that I do for work are known numbers and I use excel to calculated standard deviations, so I am by no means an expert and if Cornell2012 can correct my ways; I'm open to it.



My interruption of the draw would be correlated to your probability of drawing a spade or a face card in a deck of cards.

(P(Spade) 13/52 + P(Face Card) 12/52) - P(Space Face Card) 3/52 = 11/26

In the case it would be taking out all of the tags that can be mutually drawn by both.

You could make the case for only taking out one tag that can be mutually drawn between the two, but the issue I have with that is that if your group has 6 people then your probability would be over 100%. I don't find that to be true in a drawing for 250 tags with 1150 applicants.

 

[TheDudeAbides]

I used a similar method in Mesquite, Nevada at the $2 blackjack table.

Play by the book and double your bet every time that you lose.

Get free drinks and do not leave broke.

$2 -> $4 -> $8 -> $16 -> $32 -> $64

I calculated that there was a 3% chance of you losing 6 times in a row playing by the book.

I found that 3% once, because there was a non so smart person who took the dealers bust card 4 times when the dealer was showing a 4,5, and 6; add a dealer blackjack, and a 20 and that was all that she wrote on that night.

 

[BuzzH]

In party applications, the "group" is entered into the draw with a single number. The group/party has 250/1150 odds of drawing.

In the case of 6 people entering as individuals, the math holds up that at least one of the six will draw.

The reason most people don't draw is because they don't put in. Followed by, they don't understand the draw process. Followed by they don't understand math/statistics. Followed by they try to compare draw odds to dice, decks of cards, coin flips, and other non-sense.

How about just put in and be done with it.

 

[jtm307]

In the case of 6 people entering as individuals, the math holds up that at least one of the six will draw.

The probability of at least one of six drawing a tag when the probability for one is 21.7% is 80%.

 

[Cornell2012]

He might be, I'm just a little skeptical because if you had 6 people in your group your probability would be 1.32 - 0.0001, which means you would expect to draw and I do not see that as being true.

My way gets substantially messier when you start getting above 2 or 3.

Basically, when you have two people drawing, you can break it down like this: The probability of not drawing is 1 - P(drawing). The probability of drawing a tag consists of two parts, the probability of ONLY you drawing, and the probability of both of you drawing. So if you want to find the odds of either or both of you drawing, you want to add up P(only you draw) + P(only other person draws) + P (both draw). Based on what we know, P(drawing) = P(only you draw) + P(both draw) = P(only other person draws) + P(both draw).

By definition of what we want to know, we are looking for the odds that one or both of you draws

P(you only) + P(other only) + P(both) = P(either)​

by a bit of algebra, we can look at this like

P(either) = P(drawing [in this case the 27.1%]) + P(other only)​

We also know that P(other only) + P(both) = P(drawing)
Rearrange to get:

P(other only) = P(21.7%) - P(both)​

Finally add everything together:

P(either) = P(21.7%) + P(21.7%) - P(both draw)​

This gets kind of disgusting to expand with 3 or more, as each P(21.7%) gets broken into a several more pieces:

P(21.7%) = P(only you) + P(you + only other person 1) + P(you + only other person 2) + P(all three)​

It is possibly to math this into the form you want, but it isn't pretty. I'm a bit rusty, but there is probably an iterative form of this that makes it look cleaner.

The easiest way to figure this sort of thing out though, is to just figure out the probability that nobody draws, and then subtract that from 100%.

ex: if you have 6 people applying, the probability that nobody draws is (1-P(drawing)^6, so the odds that somebody draws is:

1 - (1-P(drawing))^6 -> if we plug in 21.7% for P(drawing), we get 1 - (78.3%)^6 = 1 - (23.0%) = 77% chance of drawing

(note that this is not 100% precise because this math is not taking into account the fact that one person can't "win" more than one tag, but for the numbers we're talking about, it isn't too far off)

edit: for those who are curious, the more precise method of taking into account no duplicate winners gives a probability of drawing at 77.1%

​

 

[TheDudeAbides]

My way gets substantially messier when you start getting above 2 or 3.

Basically, when you have two people drawing, you can break it down like this: The probability of not drawing is 1 - P(drawing). The probability of drawing a tag consists of two parts, the probability of ONLY you drawing, and the probability of both of you drawing. So if you want to find the odds of either or both of you drawing, you want to add up P(only you draw) + P(only other person draws) + P (both draw). Based on what we know, P(drawing) = P(only you draw) + P(both draw) = P(only other person draws) + P(both draw).

By definition of what we want to know, we are looking for the odds that one or both of you draws

P(you only) + P(other only) + P(both) = P(either)​

by a bit of algebra, we can look at this like

P(either) = P(drawing [in this case the 27.1%]) + P(other only)​

We also know that P(other only) + P(both) = P(drawing)
Rearrange to get:

P(other only) = P(21.7%) - P(both)​

Finally add everything together:

P(either) = P(21.7%) + P(21.7%) - P(both draw)​

This gets kind of disgusting to expand with 3 or more, as each P(21.7%) gets broken into a several more pieces:

P(21.7%) = P(only you) + P(you + only other person 1) + P(you + only other person 2) + P(all three)​

It is possibly to math this into the form you want, but it isn't pretty. I'm a bit rusty, but there is probably an iterative form of this that makes it look cleaner.

The easiest way to figure this sort of thing out though, is to just figure out the probability that nobody draws, and then subtract that from 100%.

ex: if you have 6 people applying, the probability that nobody draws is (1-P(drawing)^6, so the odds that somebody draws is:

1 - (1-P(drawing))^6 -> if we plug in 21.7% for P(drawing), we get 1 - (78.3%)^6 = 1 - (23.0%) = 77% chance of drawing

(note that this is not 100% precise because this math is not taking into account the fact that one person can't "win" more than one tag, but for the numbers we're talking about, it isn't too far off)

edit: for those who are curious, the more precise method of taking into account no duplicate winners gives a probability of drawing at 77.1%

​

I'm sold. I like it.

 

[TheDudeAbides]

In party applications, the "group" is entered into the draw with a single number. The group/party has 250/1150 odds of drawing.

In the case of 6 people entering as individuals, the math holds up that at least one of the six will draw.

The reason most people don't draw is because they don't put in. Followed by, they don't understand the draw process. Followed by they don't understand math/statistics. Followed by they try to compare draw odds to dice, decks of cards, coin flips, and other non-sense.

How about just put in and be done with it.

But.....Where is the fun in that?

#AllNumbersLivesMatter

 

[jtm307]

What a great thread.

 

[BuzzH]

The probability of at least one of six drawing a tag when the probability for one is 21.7% is 80%.

Yes, and theres a chance that all six will draw, 5 of them, 2 of them, none of them...its random.

I've been on the good and bad side of chasing odds. I applied for a turkey permit in MT the first year a unit opened and offered 25 tags. Myself and 2 friends applied individually. Total of 28 people applied. Two out of 3 of us didn't draw (me being 1 of the 3 total applicants that didn't draw).

Not really good "odds" of that happening.

I also had a 3 year stretch applying in a pronghorn unit in Wyoming with about 1050 people applying for 1000 tags...didn't draw 3 years in a row.

But, I've also drawn tags where I was the only one that drew out of 10, 20, 50, 100, 200+ applicants.

Best way to draw is to apply.

When total applicants exceeds available permits, someone is going to be sitting the pine...