Game Probability Formula

gameprobability

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Oct 19, 2014
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Hi,
I have a game where a player can score between 0 and 5 in a single game with the following probabilities
012345
P0P1P2P3P4P5
I am trying to come up with a formula that calculates
If a players plays x number of games, what is the probability that their combined score will be at least Y
Example
012345
4025155105
If a player plays 6 games what is the probability that their combined score will be greater than 5
Any help with this would be very much appreciated.
Thanks
 
"Score greater than 5" is the opposite of "score 5 or less" so the probability of "score greater than 5" is 1 minus the probability of "score 5 or less" and that is easier to calculate (although still tedious.)
One way of "scoring 5 or less in 6 games" is to score 0: to do that you have to score 0 in every game and the probability of that is \(\displaystyle P_0^6\)

Another way of "scoring 5 or less in 6 games" is to score 1: to do that you have to score 1 in one game and score 0 in the other 5 games. The probability of scoring 1 in a specific game is \(\displaystyle P_1\) and the probability of scoring 0 in the other 5 is \(\displaystyle P_1^5\). The probability or scoring 1 in a specific game and 0 in the other 5 is \(\displaystyle P_0^5P_1\). But that "specific game" could be any of the 6 games so the probability of scoring 1 in 6 games is \(\displaystyle 6P_0^5P_1\).

Another way of "scoring 5 or less in 6 games" is to score 2: Now it gets complicated! You could score twice in one game and 0 in the other 5. As above, the probability of that is \(\displaystyle 6P_0^5P_2\(\displaystyle . But you could also score once in two of the 6 games and 0 in the other 3. The probability of scoring once in each of two specific games and 0 in the other 4 is \(\displaystyle P_1^2P_0^4\). But there are \(\displaystyle \begin{pmatrix}6 \\ 2\end{pmatrix}= _6C_4= \frac{6!}{4! 2!}= 15\) ways ways to choose the two games so the probability of scoring once in two games and 0 in the other 4 games is \(\displaystyle 15P_1^2P_0^4\(\displaystyle . The probability of scoring two points in 6 games is the sum of those: \(\displaystyle 6P_0^5P_2+ 15P_0^4P_1^2\).

Another way of "scoring 5 or less in 6 games" is to score 3: Do the same- you could score 3 in a single game, 0 in the other 5. The probability of that is \(\displaystyle 6P_0^5P_3\). You could score 2 in one game, 1 in another, and 0 in the other 4. The probability of scoring 2 in a specific game, once in another specific game, and 0 in the other 4 is \(\displaystyle P_0^4P_1P_2\(\displaystyle . But there are \(\displaystyle \frac{6!}{1!1!4!}= 30\) ways of specifying those games so the probability of scoring 2 in one game and one in another is \(\displaystyle 30P_0^4P_1P_2\). Finally, you could score once in each of 3 games, 0 in the other three. The probability of that is \(\displaystyle \frac{6!}{1!1!1!3!}P_0^3P_1^3= 120P_0^3P_1^3\).

Continue in the same way with 4 or 5 scores in 6 games and add them all up.\)\)\)\)\)\)
 
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