I misunderstood how it worked - apparently, it’s a straightforward survivor pool, with the possible exception that you can pick the same time multiple times (in different games) with the same entry; each entry is a separate thing.
It’s hard to say what is optimal, as (a) there is no optimal number of winners a person might choose in a given week, and (b) one bad week early in the season makes it that much harder on the remaining picks.
My thinking is one pick is definitely NOT optimal–you’ll almost certainly get eliminated in a few weeks at best. And a hundred picks aren’t optimal–you’ll lay out a shipload of money and you’re still not guaranteed a win. So some number in between those two extremes is optimal. Is it six? Is it fourteen? What is it?
If we assume the chance of picking one game per week correctly (using the obvious “pick the heaviest favorite” strategy) is 75% (which I think is a reasonable expectation for the very heaviest favorite in a particular week, but obviously you’d want to check the actual number before investing money), and that you need to survive to week 15 to win, your chance of doing so with one pick is 1.78%.
If you have multiple picks, I don’t think you ever want to double up, because the risk of being wiped out on one bad game is too high, so your second pick will always be on a different game. That seems like it should roughly double your odds to about 3.5%. But each additional pick has diminishing returns, since now you’re having to pick teams which aren’t so heavily favored. Even if you have only six picks, now that there are only twelve games some weeks, you’re going to be making picks that are basically coin flips. Doing the actual math would be ungodly complicated, but my guesstimate is the optimum number is 3 or 4.
Of course, that’s if you want to maximize your chance of winning. Obviously your potential return on investment decreases slightly with each additional $50 pick you buy, so from that point of view the optimum number is one. But if $50 buys you a 2% chance at $10,000 and $200 buys you a 7% chance, the $200 seems like a better deal, even if the odds are technically worse. Actually, those odds all seem implausibly favorable; if accurate, it suggests that a great many participants are irrationally buying too many tickets and basically just throwing extra money into the pool. If that is the case, and this question isn’t hypothetical, please PM me next year! 
The week that this type of pool ends varies wildly not “very late”. Two or three huge upsets early on and it can be all over in week 2. I’ve seen it.
I doubt your friend spent much time studying football or making hunches. Probably stuck with the strong favorites, as most people know that Vegas does the work for you.
Incidentally, as someone mentioned, location of the pool can be a way to game it. Suppose the person running the pool happens to be from Green Bay Wisconsin. Say 80% of the entries are Packer fans - very likely. But your friend isn’t. When the time and odds are right, he bets against the Packers but most of the others don’t because of team loyalty. Boom.
Finally, as you can surmise from the lack of a definitive answer on this thread there is no optimal number of picks. Think of it this way…if you spend $50 × 1000 or $50 × a million you could be out a TON of money and some schmoe with one pick gets lucky via ignorance all season and wins the whole pot.
To summarize…one pick, 10 picks, 57 picks, 213 picks. All the same. The more you risk, the better your odds. Someone please prove me wrong! SlicedAlone?
Excellent point about using the locale to your advantage. Only really works, though, when the Packers are strong.
How’s this for a strategy? Buy 8 tickets, for example, but scrupulously avoid the two most heavily favored favorites each week, spreading your picks among the 4 or 5 next most heavily favored teams, thus letting most others get eliminated when their favorites lose while you, through your large number of picks, cling to a few picks. The question of how many picks you need for this strategy is still open
That works only if everyone else is backing the most heavily backed 2 teams and the odds of the team(s) you back are not too far behind.
In any given week you want to maximise the probablity your team wins divided by the expected number of tickets left if they do, and to note these are not independent. If you choose a heavily backed team if you win there will be more tickets left than if you choose a less well backed team. Some sites of survivor pools tell you how many people have voted for each team in which case you can calculate this (if you using betting odds or an analytics site to estimate win probability).
Unless you are buying a ridiculous number of tickets I think the optimum number of tickers to buy to maximise expected revenue is 1, with one ticket you make the optimum decision with more than one for some tickets you are making the 2nd or third best choice. You think part of the value of the ticket is the fun of the game and you want to be reasonably sure of playing at least a few weeks then buying multiple tickets makes sense.
Buying millions of tickets can guarantee a “win” but at that point you are betting on the number of winners you might have 90% of the tickets, if you are the only winner you get your stake back plus 11%, if you have to share with smeone else you lose 44% of yor stake if you have to share with more than one other person you lose more.
There are practical considerations. The person(s) running the survival pool might well object to someone buying up a ridiculous number of tickets. Maybe that number is 100 and maybe it’s two dozen. Unless someone has tried, it’s unknowable what that point would be, but presumably it exists.
I think it’s fairly likely that there are many, probably most, backing the two most heavily favored teams, and teams #3 and 4 are not too far behind them in terms of points, so are safer bets as far as eliminating competitors go.
In our suicide pool, you win with a tie. You only lose if your pick loses, and a tie isn’t a loss.
I suspect the optimal betting strategy is heavily dependent on how many other players there are, and how many tickets they typically buy. People with multiple tickets can play more complex strategies to maximize the number of weeks they stay in the pool, so you’d need to compete with that. But it’s like the old joke, “I don’t need to outrun the bear…” The goal is to be the last man standing, so you don’t need the optimal strategy to stay in for as long as possible, just long enough to beat everyone else.
My plan is a single ticket, and betting the best option in a given week based on the online betting sites. One wrinkle in our pool, though, is that you’re only allowed to pick a given team once per season. So after a few weeks, you’re likely going to be picking lesser favorites to win. But everyone else has the same problem.
After reading all these comments it’s obvious that to answer the question the OP asked…there is no optimal number of picks. Period. One pick is fine if you’re a casual player. A thousand picks is great if you’re really wealthy and enjoy playing this stuff. 53 picks is great if you were born in 1953. Anyone have any actual facts to refute this?
After reading this I feel that there is no objectively correct answer, because optimum strategy depends on observing and reacting to what strategies others are following. Picking the heaviest favorite every week maximizes your chances of surviving that week, but if everyone else is doing the same thing, it’s not the optimal choice.
Additional data: Here is historical win percentage versus the line. Roughly 50/50 around 0, 55% up through -3, 65% up to -5, 75% up to -9, and 85% above that until you get to wildly high numbers.
Seems to be settled law, but excellent question SlicedAlone.