# Predicting the outcomes of a subset of chess games correctly

Suppose $n$ games of chess are played. In how many ways can I predict the outcomes of $m$ of the games ($A$ wins, $B$ wins, there is a draw) correctly?

Here's my solution.

I can choose the $m$ games in one of $n \choose m$ ways. For each of these choices, there are $n - m$ games remaining whose outcomes I predict incorrectly, i.e. in one of the two outcomes that the game did not result in. There are two choices for each of the $n -m$ games, giving $2^{n-m}$.

Hence the answer is $$2^{n-m}{n\choose m}$$

Please tell me if there any mistakes here.

[Source: Challenge and Thrill of Pre-college Mathematics, chapter Permutations and Combinations, exercise 9.2, problem no. 7]

• The redaction of the problem is terrible... what means "predict correctly"??? Oct 13 '14 at 15:45
• @Masacroso to predict a game's outcome correctly means to predict how it will end, i.e. whether $A$ wins, $B$ wins or there is a draw. Oct 13 '14 at 15:46
• Then just exist one correct prediction but if we live in a multiverse. LOL. Oct 13 '14 at 15:48
• @Masacroso what if I predict some of the games correctly, but not all? Oct 13 '14 at 15:50
• Anyway @Soham the redaction is far to be enough clear. Doesnt make sense too me as it is written by now. Not your fault. Oct 13 '14 at 15:55

Oh, ok, the other problem @Soham that you link in a comment is more clear. This is a binomial distribution.

Think like this: you have a sequence of independent events. Any of them can take only two values: good or bad (if they are more than two possible values then is a multinomial distribution).

Then based in it "atomic" probability for a isolated event the probability for some sequence where a concrete number $m$ will be good and other $n-m$ bad will be:

• the probability for the event to be good (that is $p=\frac{1}{3}$, in this case to predict correctly the result of a game, powered to the number with correct predictions,i.e.,$(\frac13)^m$
• multiplied by the cases where the predict ISNT correct, what is $(1-p)=\frac23$, powered to it number of cases, i.e., $(\frac{2}{3})^{n-m}$
• this multiplied by the possible number of permutations of these two groups on the sequence, i.e., $\binom{n}{m}$

$$P(X=m)=\binom{n}{m}\left(\frac13\right)^m\left(\frac{2}{3}\right)^{n-m}$$

But the problem isnt asking for any probability for a concrete number of positive predictions... it is asking for the number of ways that you can predict it. The number of ways for $m$ correct predictions and $n-m$ incorrect is just the number of permutations,i.e.,$\binom{n}{m}$.

But, still, the question isnt enough clear... because I dont know exactly what is asking for. It can be interpreted, too, as the different values that can have the complete sequences (values: win, lose, draw). From this interpretation the number of possible predictions will be

$$\binom{n}{m}1^m2^{n-m}$$

But it can be interpreted too as the number of sequences that can lead to a prediction of $m$ correct predictions and $n-m$ incorrect. In this case is just the total number of possible different sequences, i.e. $3^n$.

• Thank you. I think I can explain it to you like this: suppose there are four games, and the outcome is $AADB$ ($A$ wins the first and second games, the third is a draw and the fourth is won by $B$). I can get three correct in these ways: $$\color{red}{B}ADB, \color{red}{D}ADB\tag{first one wrong}$$ $$A\color{red}{B}DB, A\color{red}{D}DB\tag{second one wrong}$$ $$AA\color{red}{A}B, AA\color{red}{B}B\tag{third one wrong}$$ $$AAD\color{red}{A}, AAD\color{red}{D} \tag{fourth one wrong}$$ Oct 13 '14 at 18:24
• @Soham Chowdhury: yes, I see what you are asking. My problem of interpretation comes from my realistic interpretation of the thing. Because future isnt prefixed in reality any prediction that you do (anything, if you play, if you make one) can lead finally to $m$ correct predictions and $n-m$ incorrect predictions. Oct 13 '14 at 18:37
• I get what you mean. Thanks! :) Oct 13 '14 at 18:38