To make a touchdown (American football)

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Qaanol
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To make a touchdown (American football)

Postby Qaanol » Wed Nov 03, 2010 1:19 am UTC

Your football team takes the field on its own 20 yard line, with 80 yards to go. You have to call the plays in order to maximize your team’s chances of scoring a touchdown. Problem is, your team always has an expectation value of gaining 2 yards per play.

You start on first down and each play uses a down. If you advance the ball at least 10 yards from where you had your most recent first down, then you get a new first down. If it gets to be fourth down and you don’t make it to the first down marker (10 yard gain from your previous first down spot) then the other team gets the ball. Since you need a touchdown, that is as bad as losing the game.

If the ball breaks the plane of the end zone, that counts as being in the end zone. In other words, the 0 yard line is the end zone. The refs will always spot the ball on a yard line, meaning the result of each play must always put the ball on an integer yard line. Football may be a game of inches, but this puzzle isn’t.

You can design any play you want, provided the ball will end up on some yard line, and the expectation value is a gain of 2 yards. For example, you could guarantee a gain of 2 yards on a play. However if you do that four times in a row then you’ll only gain 8 yards, not enough for a first down. On the other hand you could go for a 10-yard play, with an 80% chance of going nowhere and a 20% chance of getting 10 yards and a first down.

Plays can be as complicated as you like, so long as they follow the rules. For example, you could have a play with 25% chance of 0 yards, 25% chance of 2 yards, and 50% chance of 3 yards gained.

What is the greatest probability you can have of scoring a touchdown, and can you find a play-calling strategy that achieves it?
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phlip
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Re: To make a touchdown (American football)

Postby phlip » Wed Nov 03, 2010 1:53 am UTC

Two clarificationis: First, does the play need to be the same every time? Or can the plays be varied on the fly depending on the outcomes of previous plays? Eg: on the third play, if we're 5yd away do this, if we're 10yd away do that...
Secondly, is a play allowed to lose ground? Does that count as a negative in the EV calculations?

Spoiler:
Given the most liberal answers to the two questions, my first guess is that you'd go with a p chance of advancing d/n yards, and a (1-p) chance of falling all the way back to your 0yd line. With d=the distance remaining to your marker, n=the number of plays left, and p calculated such that the EV is 2. Seems that makes the chance you'd reach the line be the highest, with the side-effect of making it so that if you win, you only just scrape in, but if you fail, it's catastrophic. Which I think is what you need to do, with a set EV... you could have 80% win 20% moderate loss, or 90% win 10% really bad loss... which have the same EV, but the latter has a higher chance-of-win.

I haven't crunched the numbers yet, so I'm not sure whether "p chance of advancing d/n, (1-p) chance of falling back to 0" or "p chance of advancing d, (1-p) chance of falling back to 0" is the better option... one has a higher p, but the other you only need to succeed 1/4 as many times...

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Re: To make a touchdown (American football)

Postby Qaanol » Wed Nov 03, 2010 4:27 am UTC

phlip wrote:Two clarificationis: First, does the play need to be the same every time? Or can the plays be varied on the fly depending on the outcomes of previous plays? Eg: on the third play, if we're 5yd away do this, if we're 10yd away do that...

Play-calling can be situational. So to answer the questions asked, no, the play need not be the same every time; and yes, the plays can be varied on the fly depending on the outcomes of previous plays.

phlip wrote:Secondly, is a play allowed to lose ground? Does that count as a negative in the EV calculations?

Let’s make two variations. In the first, the team refuses to lose ground, so the worst possible outcome is 0 yards gained. In the second, the team can lose ground, so long as they don’t lose more ground than there is on the football field. Note that if they end up losing *all* the ground, back to their own 0 yard line, the other team scores a safety (2 points) and gets the ball, so for our purposes that ends the game in a bad way.
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undecim
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Re: To make a touchdown (American football)

Postby undecim » Wed Nov 03, 2010 2:39 pm UTC

During each play, we can maximize out chances of getting a first down.

Spoiler:
First play, we try for 4 yards (1/2 chance). If that fails, we try for 6 (1/3). If that fails, 8 (1/4), and if that fails, 10 (1/5)

If any of those plays succeeds, we can guarantee 2 yards for the rest of our plays to get to the next first down.

With this, we have P(First Down) = 1 - (4!/5!) = 4/5

We need 8 first downs, giving us (4/5)^8 or roughly 16.8% chance of getting a touchdown.
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Re: To make a touchdown (American football)

Postby Lopsidation » Wed Nov 03, 2010 8:58 pm UTC

Qaanol wrote:We need 8 first downs, giving us (4/5)^8 or roughly 16.8% chance of getting a touchdown.


We can do better than this (more than 20%!) with just the first play. 1/5+epsilon chance of going 80 yards ahead, 4/5-epsilon chance of going 20 yards back. (where epsilon=whatever it needs to be for the expected gain to be 2 yards)

My attempt:
Spoiler:
For the first three plays, advance 2 yards.
For the fourth, all-or-nothing first down (if we're distance D into the field, this has a chance of (2+D)/(4+D) of succeeding).
We attempt this 8 times in a row.
Chance:
(28/30)*(38/40)*(48/50)*...*(98/100) =~ 74.7%

Proof-ish of maximality:
Consider a quantum football team which, instead of making discrete plays, splits into various probability fragments.
Each play moves the center of gravity of the team ahead 2 yards. Therefore, we want as many plays as possible => as many first downs as possible. In fact, the probability of victory for any reasonable (read: all-or-nothing) strategy depends only on the expected number of first downs.

Each first down chance should also be all-or-nothing, bringing us either to the exact 10-yards-ahead line or back to the start. Other than that, it doesn't matter at all how we go about getting the first down, because the center of gravity of the team will always be 8 yards ahead no matter what happens.


EDIT: Oh, by the way,
Note that if they end up losing *all* the ground, back to their own 0 yard line, the other team scores a safety (2 points) and gets the ball, so for our purposes that ends the game in a bad way.

That doesn't really make a difference- I can just go back to the 0.000000000001 picoangstrom line instead.

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Re: To make a touchdown (American football)

Postby imatrendytotebag » Thu Nov 04, 2010 3:21 am UTC

One approach (In which the football team cannot move backward):

Spoiler:
First, we examine the optimal strategy for getting a first down. Suppose we only have one down, and D yards to go for the first. If [imath]D \leq 2[/imath], we can get the first with probability 1. If D>2, say we run a play that has Py probability of getting y yards. Then the probability of getting a first down is: [imath]\sum_{y \geq D}P_y \leq \frac{1}{D}\sum_{y \geq D}yP_y \leq \frac{1}{D} \sum_yyP_y = \frac{2}{D}[/imath]. So the best we can do is 2/D, which is attainable with an all-or-nothing play for first down.

Now suppose we have two downs, and D yards to go. If [imath]D \leq 4[/imath], we get the first with probability 1. If D > 4, say we run a play that has Py probability of getting y yards. Note that if we get more than D - 2 yards, we automatically get the first down. Otherwise, we attempt the remaining yards using an all-or-nothing strategy, as determined by the previous paragraph. So the probability of success is: [imath]\sum_{y \geq D - 2}P_y + \sum_{y < D - 2}P_y\frac{2}{D - y}[/imath]. We can remove the [imath]P_0[/imath] term by replacing it with [imath]1 - \sum_{y > 0}P_y[/imath], and we get [imath]\sum_{y \geq D - 2}P_y\frac{D - 2}{D} + \sum_{0 < y < D - 2}P_y\frac{2y}{D(D - y)} + \frac{2}{D}[/imath]. Now we note that, on the left sum, [imath]D - 2 \leq y[/imath] and on the right sum [imath]\frac{1}{D - y} \leq \frac{1}{2}[/imath], so we get:

[imath]\sum_{y \geq D - 2}P_y\frac{D - 2}{D} + \sum_{0 < y < D - 2}P_y\frac{2y}{D(D - y)} + \frac{2}{D} \leq \sum_{y \geq D - 2}P_y\frac{y}{D} + \sum_{0 < y < D - 2}P_y\frac{2y}{2D} + \frac{2}{D} \leq \sum_{y > 0}yP_y\frac{1}{D} + \frac{2}{D} = \frac{4}{D}[/imath]

So the best we can do is [imath]\frac{4}{D}[/imath], with the best play an all-or-nothing (for D-2) on the second down.

This train of logic shows that the optimal strategy is what undecim suggested.
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Re: To make a touchdown (American football)

Postby undecim » Thu Nov 04, 2010 1:56 pm UTC

Another strategy is to just go for a touchdown without worrying about first down. In the case that the team cannot move backwards, we can try 4 times for an 80 yard play with 1/40 chance of success. With 4 downs, this gives us only 9.6% chance of winning.

However, if we allow ourselves to lose ground, we can attempt the above play 3 times, and on the fourth play, have a try for -20 (39/50 chance) or 80 (11/50), And unless I'm mistaken, this gives us a 27.7% chance of victory, which is a better strategy if we allow ourselves to lose ground.

EDIT: Even better, If we are allowed to lose ground, we can maximize our chances of getting 2.5 yards each play, which is enough to consistently get our first down. On each play, we make either 2.5 yards or lose the g‌ame, with P(Success)=(4+y)/(5+y), where y is the number of yards to lose the g‌ame, and using this python code:

Code: Select all

c=1
for i in range(32):
   y = 2.5 * i + 20
   c *= (4 + y)/(5 + y)
print (c)

That gives us 55.1% chance of victory, which is an edge over our opponents.
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Re: To make a touchdown (American football)

Postby Veritas » Fri Nov 05, 2010 6:37 pm UTC

undecim wrote:EDIT: Even better, If we are allowed to lose ground, we can maximize our chances of getting 2.5 yards each play, which is enough to consistently get our first down. On each play, we make either 2.5 yards or lose the g‌ame, with P(Success)=(4+y)/(5+y), where y is the number of yards to lose the g‌ame, and using this python code:

Code: Select all

c=1
for i in range(32):
   y = 2.5 * i + 20
   c *= (4 + y)/(5 + y)
print (c)

That gives us 55.1% chance of victory, which is an edge over our opponents.


I don't see where you get P(Success) = (4+y)/(5+y)

I get (2+y)/(2.5+y)

e.g. (2+20)/(2.5+20) = 44/45
44/45 * 2.5 + 1/45 * -20 = 2

This equation provides a 72.7% chance of scoring a touchdown. *edit* this does not meet the requirement of landing on an integer yard line for ever play.

*edit* I think I found a better solution, and one that does meet all requirements (assuming negative yardage is allowed)

For the first two downs, you gain 2 yard, this leaves you at the 24 yard line on the first iterations, or on the y yard line on subsequent iterations.
For the next two downs, attempt to gain 3 yards with P(success) = (2+y)/(3+y), with failure resulting in a safety for the other team.

Successful completion of the algorithm results in a first down, and you just iterate after that.

I get a final probability of a touchdown as 74.2%

*edit 2*
Better yet, gain 2 yards on the first 3 plays of every set of downs and go for the "1st down or lose" play on 4th down each time.

The results is change of success of 74.7%

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Re: To make a touchdown (American football)

Postby jbwraith » Thu Dec 23, 2010 11:46 pm UTC

Anyone want to send these to football teams and make the games more interesting?

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Re: To make a touchdown (American football)

Postby mike-l » Fri Dec 24, 2010 2:14 pm UTC

jbwraith wrote:Anyone want to send these to football teams and make the games more interesting?


Heh. Except the expected gain on a given play is not constant, though it would be interesting seeing the QB run back to the goal line then hail mary every single play.
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