0

A one on one basketball tournament is happening in your community. Each game has a point system. 3 points for a win, 1 point for a tie, and 0 points for a loss. You have a 50% chance on winning, 40% chance of losing and a 10% of tying. We will say all these events are independent.

What's the probability by the end of 6 games you have at least 12 points.

This is my work for this question. Let X be the number of points after 6 games, then

$ P(X \ge 12)=1-(3(\frac{1}{2})^6+1(\frac{1}{10})^6+0(\frac{4}{10})^6)=\frac{238281}{250000}$

I am not confident in it though. Can somebody explain how I am approaching this right or wrong..

RobPratt
  • 50,938
  • Can you explain your reason to obtain that probability? – toric_actions Feb 11 '20 at 21:20
  • I am not sure if correct but basically at least usually you subtract from one and since 6 games are played .5 of the time your going to win so *$\frac{1}{2}^6$ and this argument holds for the other probabilites. I multiplied each probability by total points being earned –  Feb 11 '20 at 21:23
  • I think you are confused between expectation and probability. – toric_actions Feb 11 '20 at 21:34
  • Should I not be including the point values in my calculation then? –  Feb 11 '20 at 21:36

2 Answers2

1

The correct probability is $277/800$. One way to calculate it is as follows. First, list the possible $(W,T,L)$ triples with $W + T + L = 6$, $3W + 1T + 0L \ge 12$, and $W, T, L \ge 0$: $$\{(3,3,0),(4,0,2),(4,1,1),(4,2,0),(5,0,1),(5,1,0),(6,0,0)\}$$

Now the desired probability is $$\sum_{(W,T,L)} \binom{6}{W,T,L} 0.5^W 0.1^T 0.4^L = \frac{277}{800}.$$ Here, $$\binom{6}{W,T,L}=\frac{6!}{W!T!L!}$$ is a multinomial coefficient, which can also be written as $$\binom{6}{W}\binom{6-W}{T}\binom{6-W-T}{L}=\binom{6}{W}\binom{6-W}{T}\binom{L}{L}=\binom{6}{W}\binom{6-W}{T}.$$

An alternative approach uses recursion. Let $w(p,g)$ denote the probability of at least $p$ points in the next $g$ games. We want to compute $w(12,6)$. By conditioning on the outcome of the next game, we obtain: $$ w(p,n) = \begin{cases} 1 &\text{if $p\le 0$ and $g=0$}\\ 0 &\text{if $p>0$ and $g=0$}\\ 0.5 w(p-3,g-1) + 0.1 w(p-1,g-1) + 0.4 w(p,g-1) &\text{otherwise} \end{cases} $$ The resulting values of $w$ are: $$\begin{matrix} p\backslash g &0 &1 &2 &3 &4 &5 &6 \\ \hline 0 &1.00000 &1.00000 &1.00000 &1.00000 &1.00000 &1.00000 &1.00000 \\ 1 &0.00000 &0.60000 &0.84000 &0.93600 &0.97440 &0.98976 &0.99590 \\ 2 &0.00000 &0.50000 &0.76000 &0.88800 &0.94880 &0.97696 &0.98976 \\ 3 &0.00000 &0.50000 &0.75000 &0.87600 &0.93920 &0.97056 &0.98592 \\ 4 &0.00000 &0.00000 &0.35000 &0.63500 &0.80960 &0.90496 &0.95392 \\ 5 &0.00000 &0.00000 &0.25000 &0.51500 &0.71350 &0.84076 &0.91528 \\ 6 &0.00000 &0.00000 &0.25000 &0.50000 &0.68950 &0.81675 &0.89606 \\ 7 &0.00000 &0.00000 &0.00000 &0.20000 &0.44750 &0.65275 &0.79526 \\ 8 &0.00000 &0.00000 &0.00000 &0.12500 &0.32750 &0.53250 &0.69866 \\ 9 &0.00000 &0.00000 &0.00000 &0.12500 &0.31250 &0.50250 &0.66263 \\ 10 &0.00000 &0.00000 &0.00000 &0.00000 &0.11250 &0.30000 &0.49663 \\ 11 &0.00000 &0.00000 &0.00000 &0.00000 &0.06250 &0.20000 &0.37625 \\ 12 &0.00000 &0.00000 &0.00000 &0.00000 &0.06250 &0.18750 &0.34625 \\ \end{matrix} $$ In particular, $w(12,6)=277/800$.

RobPratt
  • 50,938
  • I get what your saying until the "Now the desired..." are we plugging in the values above for W, T, L? Cause I tried that for the first one and got nothing close to what you got –  Feb 11 '20 at 21:47
  • 1
    I updated my answer just now with more detail. – RobPratt Feb 11 '20 at 21:58
  • I like your solutions, but I think you have the probabilities of losing and tying switched in your calculations. It shoud be $P(loss) = 0.4$ and $P(tie) = 0.1$. – awkward Feb 12 '20 at 14:04
  • Yes, thanks for pointing it out. I was fooled by the inconsistent ordering in the problem statement. I corrected my answer just now, and it matches yours. – RobPratt Feb 12 '20 at 18:03
1

Here is an approach using a probability generating function. Readers unfamiliar with generating functions may find many applicable resources in the answers to this question: How can I learn about generating functions?

A probability generating function for the total number of points in six rolls is $$f(x) = (0.4 + 0.1 x + 0.5 x^3)^6$$ On expansion of the polynomial, $$f(x) = 0.004096\, +0.006144 x+0.00384 x^2+0.032 x^3+ \\0.03864 x^4+0.019224 x^5+0.100801 x^6+0.0966 x^7+ \\0.03603 x^8+0.166 x^9+0.120375 x^{10}+0.03 x^{11}+ \\0.1525 x^{12}+0.075 x^{13}+0.009375 x^{14}+0.075 x^{15}+ \\0.01875 x^{16}+0.015625 x^{18}$$ (This is easy if you use a computer algebra system or Wolfram Alpha, otherwise a bit tedious.) The coefficient of $x^n$ in this polynomial is the probability of winning $n$ points. So if we want the probability of winning $12$ or more points, we can just sum up the coefficients of $x^{12}$ through $x^{18}$:

$$0.1525+0.075+0.009375+0.075 +0.01875+0.015625 = \boxed{0.34625}$$

RobPratt
  • 50,938
awkward
  • 15,626