View Full Version : Football Probability Problem

David Sklansky
09-11-2002, 03:27 PM
You score a touchdown being down by fourteen with four minutes to go. Now you are down by eight. One point conversions will be assumed to be 100%. If you go into overtime you are even money. You should go for two here as long as your chances for success are above what? Either estimate it intuitively or calculate it exactly.

09-11-2002, 03:51 PM
Don't we need to know the probability of making another TD? Or did you want us to estimate that as well?

~Magic Man

09-11-2002, 04:06 PM
on how many time outs you have left, whether or not you are going to attempt an on-side kick and if this is an NFL or college game. Seems to me like the 2 point conversion rate is 50% in the NFL and for a likely on-side kick scenario the recovery rate for the kicking team is about 20%. I am also going to assume you will say all this is irrelevant so I will estimate above 25%.


09-11-2002, 04:18 PM
Assuming that in either case you will get the td, and forgetting about fg possibilities.

If you go for kicks, you have a .50 chance of winning in ot.

If you go for two-point conversions you have x% chance of winning on on the 1st TD(since if you got the 2 points here you would kick the second time), + x% * .50 chance of winning(missing the first one, but getting a tie with the second and winning in OT).

The break even point would be

.50 = X% + .50*X%
Giving you .50 = 1.5X or X= .3333333%
So if your chances at a 2 point conversion are higher than 1/3 you should go for it.

09-11-2002, 04:41 PM
Just realized my notation was confusing...
X=1/3 not 1/3%(would actually be 33 1/3 %)

Ed Miller
09-11-2002, 05:02 PM
Let's assume that you are guaranteed to score another touchdown, as if you don't, you lose 100% of the time.

That being the case, let's call the probability that you will convert a 2-point conversion x.

If you go for 2 the first time, the probability you will win is:

P(win) = x + 0.5(1 - x)x = -0.5x^2 + 1.5x
(either you hit your 2-pointer the first time (x) or you miss your 2-pointer, hit the next 2-pointer, and win in OT (0.5(1 - x)x))

Thus, you are looking for x such that,
-0.5x^2 + 1.5x > 0.5

solving as an equality:
x^2 - 3x + 1 = 0
x = (3 +- sqrt(9 - 4)) / 2

only the minus solution is between 0 and 1, so
x = 1.5 - 0.5 * sqrt(5) = 1.5 - 1.12 = 0.38 or 38%

You must succeed at least 38% of the time to want to go for 2.

09-11-2002, 05:19 PM
I agree with your solution and found the flaw in mine. I was double dipping...Forgot to take into account that the second scenario would only occur (1-x) times.

09-11-2002, 06:38 PM
In order to win by kicking the extra point, you must do ALL of the following:

1. score 7 more points
2. prevent another unanswered score by the other team
3. win in overtime.

In order to win by making 2-point conversion, you must do ALL of the following:

1. score 7 more points
2. prevent another unanswered score by the other team
3. make 2-point conversion

Note that we have to do 1 and 2 in both cases, and the only difference is winning in overtime vs. making the 2-point conversion. Since winning in overtime is 50%, we should go for the 2-point conversion as long as we can make it 50% of the time.

Note that it doesn't depend on the probability of making a td, onside kicks, fgs, or anything else.

09-11-2002, 06:55 PM
(Lorinda comes to mind) who may not be familiar with American football (as well as for many in this country), a touchdown is worth 6 points plus whatever you make on the conversion. You can elect to kick the ball through the uprights for 1 extra point. This is customary and almost always successful, making the touchdown worth 7 points. You can instead decide to go for a 2-point conversion by passing or running the ball into the endzone. This would make the touchdown worth 8 points if successful, but otherwise only 6.

09-11-2002, 07:38 PM
I think we need to make a few more assumptions:

1-The opponents will not score again at all in regulation.
2-We will score exactly 1 more TD in regulation (no FG, safties, etc)

Now, if we just kick the XP both times we have a 50% chance of winning the game. So we need to find a scenario where we win in regulation more than we lose.

Let X equal the probability we succeed. Then 1-X equals the probability we fail. We win in regulation if we succeed on the first try, but lose in regulation if we fail on both. (If we fail on the first try, but succeed on the second try, we will be no worse off than if we just kicked both times, so we can ignore this factor).

We need the probability of success on the first try to be greater then the probability of failure on both tries combined. So X>(1-X)^2.

Solving gives us X&gt;.382 (or <font color="red"> 38.2% </font color> )

Plug it back in to verify:

38.2% (X) of the time, we win by succeeding on the first try.



38.2% of the time, we lose by failing twice.

09-11-2002, 08:11 PM
I ignored the fact that you can go for the 2-point conversion and miss, but still make one later and win in OT. There is also the possibility of making 2 scores in regulation.

09-11-2002, 09:36 PM
If we make 2 more scores in regulation, we will win no matter what we do as long as one score is a touchdown, so we don't have to consider this case. Similarly we can ignore the case where we don't make another score, since then we always lose. I think majorkong has set up and solved the proper equation except there is still one small thing missing: we can win by making the 2-point conversion, getting 2 field goals, and then winning in OT (something the Bears might do). Then the necessary 2-point success rate can be slightly smaller than 38%.

09-11-2002, 09:38 PM
Let's assume that you are guaranteed to score another touchdown, as if you don't, you lose 100% of the time.

Unless we make the 2 point conversion, score 2 field goals, and then win in OT.

09-12-2002, 09:40 AM
David -

Didn't you ask this question last year?

Anyway, if you go for two, you have two ways to win
- make it on the first try, kick on the second, and win
- miss the first try, make a 2-pt conversion on the second, and score first in overtime.

The sum of these two probabilities has to exceed 50% to make going for two better than kicking. Put another way:

p + [p(1-p)]/2 &gt; 1/2 -&gt;
2p + p(1-p) &gt; 1

This breaks even around p = 0.4.


09-12-2002, 07:45 PM
I worked out and got the same answer as majorkong - the only thing I have to add is that by going for a 2 pt. conversion it also opens up the possibility of kicking 2 field goals to tie in reg. and then winning in OT. So if p is the probability of making the 2 pt. conv. and q is the probability of making 2 more field goals (without other team scoring or our team making TD) then we have the added probability of p*q/2 of winning, which means that maybe an even slightly lower probability than 38% would warrant going for 2 (probably just a point or two less).

10-09-2002, 01:33 PM
since the chance to make 2 should be the same if you score again, id take the 100% PAT, then worry about it if you score again. otherwise, if you miss, youll be forced to go for 2 if you score again. id rather wear down the defense on another drive if i could. then it's an option, not madatory on the 2nd score.