# Two player football game -- probability.

A friend gave me the following problem to solve. I have been stuck at it for days. With little or no progress. Well, I did get an estimated answer by running a simulation but not the formal solution.

Two players are playing a game of football. The goals are located at positions $$0$$ (Player A) and $$100$$ (Player B), and the ball is initially at position $$50$$.

In one move, a player draws a number from the uniform random distribution $$(0, 100)$$ and kicks the ball by the same amount towards the other player's goal post.

If at any point the ball crosses a goal post the game ends and the player who made the last move wins. Find the probability of player A winning.

Edit:

I tried creating an infinite series of probabilities of winning in each move like in the first move there is a 50% chance that Player A wins. This quickly turned into a integral starting at the second move, i.e $$\int_x x * P(player-B's-win | ball-is-at x) * P(ball-at-x) \;\mathrm{d}x$$

• Who moves first? This is very important to know as if A moves first, he would have to get a number greater than 49 in order to win and after his turn, B would have to get a number greater than 49+A's number. So the winning probability of B decreases. Aug 13, 2021 at 11:25
• Yes player A goes first. One more thing to add is that the non-integral numbers can be drawn from the distribution (0, 100) Aug 13, 2021 at 11:32
• What is the estimate that came out of your simulation? Aug 13, 2021 at 11:36
• I ran the game around 1 million times and player A won 736789 times hence the probability is approx 0.736789. Aug 13, 2021 at 11:49
• @HennoBrandsma, I'm not sure what you mean. Your suggestion, $e^{-1}\approx0.367879$, was not at all close to what the OP got. In any event, see the answers that Mike Earnest and I posted nearly simultaneously. Aug 13, 2021 at 16:44

For any $$0\le x\le 1$$, let

• $$A(x)$$ be the probability A wins when it is A's turn, and the ball is $$100x$$ units from $$A$$'s goal.

• $$B(x)$$ be the probability A wins when it is B's turn, and the ball is $$100x$$ units from $$A$$'s goal.

You want $$A(\frac12)$$. We then have that $$A(x) = x + \int_x^{1} B(t)\,dt,\qquad A(1)=1\\ B(x) = \int_0^x A(t)\,dt\qquad\;\qquad B(0) = 0$$ If we take $$\frac{d}{dx}$$ of both equations, you get $$A'(x) = 1 - B(x),\\ B'(x) = A(x)\;\;\;\;\;\;\;$$ The general solution to that system of differential equations is $$A(x) = c_1\cos x-c_2\sin x\;\;\;\;\;\;\\ B(x) = c_1 \sin x + c_2 \cos x + 1$$ The initial conditions $$A(1)=1$$, $$B(0)=0$$ then imply $$c_1=\frac{1-\sin 1}{\cos 1}$$ and $$c_2=-1$$, which means that $$A(\tfrac12) = \frac{1-\sin 1}{\cos 1}\cdot \cos \tfrac12 + \sin \tfrac12\approx 0.736915$$

• Great minds think alike! Aug 13, 2021 at 16:41
• @BarryCipra Ya beat me to it! I even cheated and used Wolfram|Alpha to solve the system, but your answer shows it is totally doable by hand. Aug 13, 2021 at 16:42
• I would have been even quicker, but I kept screwing up the calculations doing them by hand. It helped a lot knowing what the target answer was. Aug 13, 2021 at 16:46

$${\cos(1/2)+\sin(1/2)\over1+\sin(1)}\approx0.7369152768$$

Let's normalize the football field to total "yardage" $$1$$, and suppose Player A has a distance $$x$$ to the endzone. If they kick the ball a random distance $$k$$, they will win with certainty if $$k\ge x$$ and with probability $$1-P(1-x+k)$$ if $$k\lt x$$, where $$P(1-x+k)$$ represents the probability that Player B wins if the ball lands short of the endzone, setting up B's attempt from a distance $$1-x+k$$ to the opposite endzone. Thus

$$P(x)=1-x+\int_0^{1-x}(1-P(1-x+k))dk=1-\int_0^{1-x}P(1-x+k)dk=1-\int_{1-x}^1P(u)du$$

It follows that $$P'(x)=-P(1-x)$$, hence

$$P''(x)=P'(1-x)=-P(x)$$

which means $$P(x)=A\cos x+B\sin x$$ for some coefficients $$A$$ and $$B$$. Now $$P(0)=1$$ (i.e., Player A is guaranteed a win when there's no distance to kick). Thus $$A=1$$, so it remains to find $$B$$. The equation

$$-\sin x+B\cos x=P'(x)=-P(1-x)=-(\cos(1-x)+B\sin(1-x))$$

which holds for all $$x$$, tells us, on setting $$x=0$$, that

$$B=-\cos1-B\sin1$$

so $$B=-\cos1/(1+\sin1)$$ and thus

$$P(x)=\cos x-{\cos1\over1+\sin1}\sin x={\cos x+\sin(1-x)\over1+\sin1}$$

Plugging in $$x=1/2$$ gives the aforementioned result.

Remark (added later): I really was surprised at the form of the answer; I can't recall ever seeing the sine and cosine of $$1$$ (and $$1/2$$) arise so naturally in a probability setting before. Incidentally, if you want to know where Player A should start from so that $$P(x)=1/2$$, the answer turns out to correspond to

$$\cos x={1+\sin1+\sqrt{(1-\sin1)(7-\sin1)}\over4}\approx0.707388183$$

or $$x=\arccos(0.707388183)\approx0.785000122$$. That is, to make the contest fair, Player A should make the first kick from ever so slightly inside their own twenty-two-and-a-half yard line.

• BTW, here is a similar problem where weird sine/cosine values show up in the solution: math.stackexchange.com/questions/3866128/… Aug 14, 2021 at 17:27
• @MikeEarnest, excellent! I see they arise in much the same way there as here, out of a differential equation. (And I am relieved, for the sake of my memory, to find that I had not provided an answer there.) Aug 14, 2021 at 21:09