Strategy for a Lottery-Style Draw Application

[Zog]

Here is a problem that should be simple for most anyone on this forum but I thought it might be fun to throw it out there . . .

Background: There is a certain 4 day (or so) river float in Montana that is so popular that about 20 years ago the state decided to limit floaters by issuing float permits on a random lottery-style application/draw basis. Everyone applies for a permit by Feb 1 and sees the results by Mar 1 each year. There is no limit to the numbers of applicants, but only one application per person.

When you apply, you apply for three launch dates in order of your preference, and if you win you get only one of those dates. Peak dates run May 15 through August 15. I plan to apply for peak dates only, so that is not a variable for my question.

If you draw, your launch permit can be used for a party of up to 15 floaters. I plan to get 10 family/friends to apply this year, so if one of us draws we all get to float together.

Now for my simple question: Would our collective draw odds be better if we all apply for the same three date preferences, or better if we all apply for different dates?

Assume for my question that all seven days of the week are equally desirable for all applicants, even though that may not be the case.

Thanks if anybody wants to field this! For you it is probably a no-brainer, as they say.

 

[StoneTemplePython]

You're talking about applying for peak dates only. Can we assume that a large number of people will be in the drawing on any given peak day relative to the number of slots awarded for that day?

 

[Zog]

Thanks for responding!

The reason for peak dates is weather and water level. April has too much chance of winter weather, and by July water levels have usually dropped. I went to the state's website and found a few numbers . . .

Total 2017 season; April 15th to October 15th - 650 permits awarded to 10,077 applicants. 5599 floaters made the trip. Average party size was 7.8. (These don't exactly divide out but that's what they published. They are close.) Watercraft by month - Apr 333, May 912, Jun 1164, July 169. By July they asked for and received voluntary cancellations due to low water, which continued for the rest of the season. What they don't report, however, is applicants by month - the one number you are asking about. I assume, based on the numbers above, they award more draws in peak months. So let's assume the applicants are proportional to the draws for each month, so 10,077/650 = 16.6 applicants per draw.

Thanks again

 

[StoneTemplePython]

Another hanging item: what is the value of winning twice or three times or .. more? My inference right now is that you really just want one big outing and that's the goal here. I don't know whether you can sell extra tickets on craigslist or ebay or stubhub or whatever. This actually could matter quite a bit -- e.g. if you can't sell on secondary market but would be interested in going twice... then spreading out your bets over multiple days would make sense.

- - - -

The idea is you have up to 10 random variables ##X_1, X_2, ..., X_9, X_{10}## -- these are the 10 dates you can apply on. They are independent identically distributed (iid). They represent the number of applicants for a given day.

If you allow the assumption that a large enough number of people apply on each date, then we can basically view this as sampling with replacement which makes things nice and easy.

If your main interest is just in probability of winning at least once, overall it's the same in either case.

Things could get interesting if for some reason very few people applied on some day (hence my sampling with replacement approximation wouldn't be good) or for reasons related to winning multiple times (a good problem to have, I know) and what you can / want to do about it.

- - - -
edit: I'm never totally happy with answering these kinds of questions because they feel underspecified. In some sense the variance in your probability of winning goes down in the spread out case, but I'm not sure that this is meaningful to the user -- on average you have the same chances of winning in each case, the variance (unlike say fluctuations in bitcoin prices) is hidden from the end user, etc.

That said, the general argument is that if the mean is the same, people prefer a lower variance outcome, so splitting things up would smooth out the distribution of chances of winning each year.

 

[Zog]

Another hanging item: what is the value of winning twice or three times or .. more? My inference right now is that you really just want one big outing and that's the goal here. I don't know whether you can sell extra tickets on craigslist or ebay or stubhub or whatever. This actually could matter quite a bit -- e.g. if you can't sell on secondary market but would be interested in going twice... then spreading out your bets over multiple days would make sense.

- - - -

The idea is you have up to 10 random variables ##X_1, X_2, ..., X_9, X_{10}## -- these are the 10 dates you can apply on. They are independent identically distributed (iid). They represent the number of applicants for a given day.

If you allow the assumption that a large enough number of people apply on each date, then we can basically view this as sampling with replacement which makes things nice and easy.

If your main interest is just in probability of winning at least once, overall it's the same in either case.

Things could get interesting if for some reason very few people applied on some day (hence my sampling with replacement approximation wouldn't be good) or for reasons related to winning multiple times (a good problem to have, I know) and what you can / want to do about it.

- - - -
edit: I'm never totally happy with answering these kinds of questions because they feel underspecified. In some sense the variance in your probability of winning goes down in the spread out case, but I'm not sure that this is meaningful to the user -- on average you have the same chances of winning in each case, the variance (unlike say fluctuations in bitcoin prices) is hidden from the end user, etc.

That said, the general argument is that if the mean is the same, people prefer a lower variance outcome, so splitting things up would smooth out the distribution of chances of winning each year.

Yes I think you are correct. Granted I was underspecified. Your inference is correct - there is no value in any over-drawn float permits. The applicant must be on the float and there is actually a monitor at the put-in ramp to assure that the applicant is present with identification.

I did find more data - they actually published applicants/draws per day for the entire five months but it is really too much data for a forum like this and for you. If you graph the number of applicants for each given launch date it pretty much resembles a noisy normal distribution peaking on Jun 15, being rather flat in early May and flat again in late July.

For this discussion I chose the middle 10 days; five days either side of the peak, and here are the sets of (applicants, draws) - Jun 11(177,7); 12(306,8); 13(244,8); 14(282,7); 15(345,8); 16(217,8); 17(157,8); 18(197,7); 19(287,8); Jun 20(215,8). For 2018, we are not told which days will be 8 draws and which will be 7. Also I cannot predict which day in 2018 will be a peak application again; you can see it is quite variable..

With this new data I believe I reasoned out an answer with my simple math . . .

On peak day the odds were 345 applicants / 8 draws; 1 draw in 43. A team of ten applying for that day would have odds of (345/10)/8 or 1 in 4. Correct? (Close?)

Now, for the ten day period above, there were 2427 applications for 77 draws. 2427/77 would be 1 in 31. A team of ten's odds would be (2427/10)/77; 1 in 3. So I now believe we want to be sure to avoid applying for the same days, unless we somehow knew the lowest-applicant days. On the other hand, I believe that would also even itself out, because we may choose either a high day or a low day.

Sorry if my calc's are over-simple. You were a bit over my head (take as a compliment).

THANKS for entertaining my question!

 

[StoneTemplePython]

It's interesting in how much data there is and that they actually check on who is using the passes with IDs, etc. -- this must be quite a trip.

On peak day the odds were 345 applicants / 8 draws; 1 draw in 43. A team of ten applying for that day would have odds of (345/10)/8 or 1 in 4. Correct? (Close?)

If you play poker, this is basically poker maths. (Like say you have 4 hearts and you want to know the probability that neither of the turn or the river card are a heart, i.e. that you don't improve to a flush -- that is the sort of calculation here.)

What I think you've written is ok as a union bound (read optimistic estimate). With a large enough spread of applicants, I in effect modeled it as sampling with replacement, giving you

##1 - \big(\frac{345-10}{345}\big)^8 \approx 0.210##

the actual calculation, recognizing they are sampling / 'dealing a card without replacing it' is

##1 - \frac{\binom{345-10}{10}}{\binom{345}{10}}\approx 0.211##

but close enough
- - -
I did not realize how few draws there would be on some of these peak days. The fact is that there are only 7 or 8 draws on some of the days of interest, but you have the ability to submit 10 applications? This makes the conclusion simple: definitely spread your bets out over multiple days.

 

[Zog]

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It's interesting in how much data there is and that they actually check on who is using the passes with IDs, etc. -- this must be quite a trip.

If you play poker, this is basically poker maths. (Like say you have 4 hearts and you want to know the probability that neither of the turn or the river card are a heart, i.e. that you don't improve to a flush -- that is the sort of calculation here.)

What I think you've written is ok as a union bound (read optimistic estimate). With a large enough spread of applicants, I in effect modeled it as sampling with replacement, giving you

##1 - \big(\frac{345-10}{345}\big)^8 \approx 0.210##

the actual calculation, recognizing they are sampling / 'dealing a card without replacing it' is

##1 - \frac{\binom{345-10}{10}}{\binom{345}{10}}\approx 0.211##

but close enough
- - -
I did not realize how few draws there would be on some of these peak days. The fact is that there are only 7 or 8 draws on some of the days of interest, but you have the ability to submit 10 applications? This makes the conclusion simple: definitely spread your bets out over multiple days.

Only one application per person, which can be used for up to 15 floaters. In my case I have a team of 10 who will combine forces and float together if only one draws.

I hate to advertise the best places, but you helped me out, so here is a link for you. You should definitely check it out. 60 miles of float with only one put-in and one take-out. Beautiful country.

I don't like how closely it is managed; it makes it like a Disney park, but I have to understand that without management, it would be total chaos trying to float the Smith during summer. They also do need to guard against people who would apply just to sell their draw to somebody. If only a few people did so it would be ok, but if it got out of hand the trip would only be for the highest bidders.

They do have a much-too-elaborate annual report. .

Edit - Oops sorry if I violated forum rules by talking about river trips instead of physics.
THANKS again!

http://stateparks.mt.gov/smith-river/