Probability of getting better than a certain score

Question

I apologize in advance if this is a duplicate, I suspect it must be but I don't know how to search for it.

Imagine you play a game 7 times; each time you add your score. The number of ways to get a particular score (the numerator of the probability fraction) is displayed below:

$$ \begin{array}{cccc} \begin{array}{c|c} \text{Score} & \text{Combos} \\ \hline 1 & 720 \\ 2 & 1080 \\ 3 & 1792 \\ 4 & 3648 \\ 5 & 7176 \\ \hline \\ \end{array} & \begin{array}{c|c} \text{Score} & \text{Combos} \\ \hline 6 & 11136 \\ 7 & 17040 \\ 8 & 26460 \\ 9 & 40104 \\ 10 & 50776 \\ \hline \\ \end{array} & \begin{array}{c|c} \text{Score} & \text{Combos} \\ \hline 11 & 59016 \\ 12 & 60660 \\ 13 & 55800 \\ 14 & 40592 \\ 15 & 24284 \\ \hline \\ \end{array} & \begin{array}{c|c} \text{Score} & \text{Combos} \\ \hline 16 & 10968 \\ 17 & 3328 \\ 18 & 776 \\ 19 & 80 \\ 20 & 4 \\ \hline \text{Total} & 415440 \end{array} \end{array} $$

I know how to compute the $EV$; $EV = 11.091555$.

How would you calculate the probability of getting less than some particular total score $n$? Is the EV enough? I don't need the answer, just the procedure is enough.

Update:

• The runs are completely independent.
• I need to calculate the probability for various $50 \leq n \leq 80$

Answer 1

We may also use probability generating functions:

$P(x)=\left(4 \, x^{20} + 80 \, x^{19} + 776 \, x^{18} + 3328 \, x^{17} + 10968 \, x^{16} + 24284 \, x^{15} + 40592 \, x^{14} + 55800 \, x^{13} + 60660 \, x^{12} + 59016 \, x^{11} + 50776 \, x^{10} + 40104 \, x^{9} + 26460 \, x^{8} + 17040 \, x^{7} + 11136 \, x^{6} + 7176 \, x^{5} + 3648 \, x^{4} + 1792 \, x^{3} + 1080 \, x^{2} + 720 \, x\right)/415440$

and the probability can be calculated from $P(x)^7$

E.g. for a score less than $n=61$, summing the coefficients of the terms $x^7\ldots x^{60}$ in the above polynomial gives $\displaystyle \frac{844291674672075639073225007734969}{65179080904297424559820836480000000}\approx 0.0129534148527156$

As an aside, the expected value can also be calculated from the p.g.f,

$\displaystyle P'(1)=\frac{1151969}{103860}\approx 11.0915559406894 $

Answer 2

The EV is not enough for less than some particular score, but it can give you a bound (via the Markov inequality, which says that $P[X \geq c] \leq E[X]/c$ (this upper bounds the probability the total score is greater than $n$).

The answer is that the probability when k games are played (in this case, $k=7$) is $P(\text{total score} \leq n) = \sum_{i=0}^n \sum_{x_1+x_2+\ldots+x_k=i, x_i \geq 1, x_i \in \mathbb{Z}} P(\text{game 1 score = } x_1 \cap \text{game 2 score = } x_2 \cap \ldots \cap \text{game k score = } x_k)$. If the runs are independent, then the probability becomes a product, which is still not nice. If $n$ is small though, this approach may be feasible.

The inner sum has $\binom{k-1}{i-1}$ terms, so there are $\sum_{i=0}^n \binom{k-1}{i-i}$ terms in this sum, which is not nice. Simulating it would likely be easier.

Answer 3

Per the advice in this thread, I ran a simulation with 10,000,000 trials. It took only a few seconds, not enough for coffee. Here are the results:

$$ \begin{array}{cccc} \begin{array}{c|c} \text{Score} & \text{Times} \\ \hline 33 & 1 \\ 35 & 1 \\ 36 & 2 \\ 37 & 2 \\ 38 & 2 \\ 39 & 4 \\ 40 & 13 \\ 41 & 9 \\ 42 & 23 \\ 43 & 53 \\ 44 & 79 \\ 45 & 129 \\ 46 & 199 \\ 47 & 314 \\ 48 & 462 \\ 49 & 791 \\ 50 & 1089 \\ 51 & 1625 \\ 52 & 2459 \\ & \\ \end{array} & \begin{array}{c|c} \text{Score} & \text{Times} \\ \hline 53 & 3687 \\ 54 & 5185 \\ 55 & 7505 \\ 56 & 10323 \\ 57 & 14374 \\ 58 & 19584 \\ 59 & 26890 \\ 60 & 35304 \\ 61 & 46301 \\ 62 & 60692 \\ 63 & 77926 \\ 64 & 98238 \\ 65 & 122342 \\ 66 & 150361 \\ 67 & 182853 \\ 68 & 218069 \\ 69 & 256349 \\ 70 & 298477 \\ 71 & 340846 \\ & \\ \end{array} & \begin{array}{c|c} \text{Score} & \text{Times} \\ \hline 72 & 384231 \\ 73 & 424468 \\ 74 & 462215 \\ 75 & 494474 \\ 76 & 519681 \\ 77 & 537268 \\ 78 & 544470 \\ 79 & 543348 \\ 80 & 529893 \\ 81 & 509777 \\ 82 & 477961 \\ 83 & 440529 \\ 84 & 397553 \\ 85 & 349523 \\ 86 & 301595 \\ 87 & 254746 \\ 88 & 208387 \\ 89 & 168148 \\ 90 & 131310 \\ & \\ \end{array} & \begin{array}{c|c} \text{Score} & \text{Times} \\ \hline 91 & 100616 \\ 92 & 74686 \\ 93 & 53593 \\ 94 & 37876 \\ 95 & 26259 \\ 96 & 17534 \\ 97 & 10996 \\ 98 & 6855 \\ 99 & 4158 \\ 100 & 2487 \\ 101 & 1321 \\ 102 & 724 \\ 103 & 403 \\ 104 & 195 \\ 105 & 93 \\ 106 & 34 \\ 107 & 17 \\ 108 & 9 \\ 109 & 3 \\ 110 & 1 \\ \end{array} \end{array} $$

Here is a link to the code (in Java), if you want to run something similar, although ideone doesn't like how long 10,000,000 takes to run, so I reduced it to 1,000,000 trials.