2
$\begingroup$

I have a logic question regarding the approach of a combinatorics problem:

"Two chess players, 𝐴 and 𝐡, are going to play 7 games. Each game has three possible outcomes: a win for 𝐴 (which is a loss for 𝐡), a draw (tie), and a loss for 𝐴 (which is a win for 𝐡). A win is worth 1 point, a draw is worth 0.5 points, and a loss is worth 0 points.

Assume that they are playing a best-of-7 match, where the match will end when either player has 4 points or when 7 games have been played, whichever is first. For example, if after 6 games the score is 4 to 2 in favor of 𝐴, then 𝐴 wins the match and they don’t play a 7th game. How many possible outcomes for the individual games are there, such that the match lasts for 7 games and 𝐴 wins by a score of 4 to 3?"

This question has being asked and answered here original question

Since the original answer is rather long, hence, I am using only part of the answer:
Player $A$ wins 3 games, draws 1 game and loses 2 games gives $\binom{6}{3} \binom{3}{1} \binom{2}{2}$; for my question.

The logic from the posted answer is that player $A$ plans out his game 3 wins, 1 draw, 2 loses and winning the last game, hence, $n$ is adjusted after a grouped outcome. However, in real life situation shouldn't it be $\binom{7}{1} \binom{6}{1} \cdots \binom{1}{1}$ where the outcome is adjusted after every game and player $A$ is only able to do $n-1$ "choose" 1 till the 7th game?

For clarity $n$ is from $\binom{n}{k}$, and my question is not about questioning the correctness of the answer from the original post but the approach and logic in live settings.

Kindly advise

Added info for clarity;
Sorry for the lack of vocabulary, there are probably better words/phases than my choice of "sequential choices". I'll try to express it in my best effort.

$\require{cancel}$ There are 7 games of which the players are to play in sequence, one game after another till the game ends at the 7th game. Therefore, shouldn't it be$\cancel {\binom{7}{1} \binom{6}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1}}$ when the choices are not determined until the current game ends. The logic from Joffan and user2661923 answers makes sense to me where the 7 games are played simultaneously.

Added info after seeing @user2661923's Addendum:
Thanks to user2661923's Addendum, I realized that my alternate view of updating after each game should be using win,lose, and tie(WLT) instead(my original intuition = "strike-thru"). And is very tedious and risk over and under count.

$\endgroup$
3
  • $\begingroup$ In the linked problem, Joffan selected the positions of the wins, ties, and losses among the first six games. That determines the sequence of events. $\endgroup$ Apr 22, 2022 at 12:10
  • $\begingroup$ @N.F.Taussig My answer is supposed to also critique the OP's (i.e. original poster's) thinking. I am having trouble doing that because I don't understand the OP's thought processes. Are you able to understand his thinking? If so, could you leave me a comment, explaining your take on his thinking, so that I can react to it? Alternatively, could you post a separate answer where you dissect the OP's thinking? $\endgroup$ Apr 22, 2022 at 12:16
  • $\begingroup$ @N.F.Taussig and to the OP: Okay, I think I see where the OP is headed now. It will take me a few minutes. I will add an Addendum, where I react to your approach, and then leave a comment, when done. $\endgroup$ Apr 22, 2022 at 12:29

1 Answer 1

3
$\begingroup$

Just added an Addendum to my answer, to explore the analytical approach suggested by the original poster.


How many possible outcomes for the individual games are there, such that the match lasts for 7 games and 𝐴 wins by a score of 4 to 3?"

The logic from the posted answer is that player $A$ plans out his game 3 wins, 1 draw, 2 loses and winning the last game, hence, $n$ is adjusted after a grouped outcome. However, in real life situation shouldn't it be $\binom{7}{1} \binom{6}{1} \cdots \binom{1}{1}$ where the outcome is adjusted after every game and player $A$ is only able to do $n-1$ "choose" 1 till the 7th game?

With respect to the paragraph immediately above, I do not understand what you are trying to say. If you could edit your posting to re-word your analysis, then I can take a crack at examining your approach.

I had trouble following both the OP's analysis and the analysis of the linked question. So, I will answer the question from scratch.

I am assuming that the outcome of (for example) A winning Game-1 and drawing Game-2 is considered distinct from A drawing Game-1 and winning Game-2.


Since the match went $7$ games, there are $2$ mutually exclusive cases, that will be explored separately:

  • A Tied Game-7.
    The variable $R$ will be used for enumeration here.
  • A Won Game-7.
    The variable $S$ will be used for enumeration here.

$\underline{\text{Case 1: A tied Game-7}}$
This implies that after $6$ games, A had scored (3 + 1/2) points. This implies that during the first $6$ games, A won $k$ games, where $k$ is some element in $\{1,2,3\}.$

So, there will be $3$ subcases, based on how many of the 1st $6$ games that A won. I will use the variables $R_1, R_2,$ and $R_3$ to represent the respective (mutually exclusive) enumerations, re $1,2,$ or $3$ wins.

$R_1 = \binom{6}{1} \times \binom{5}{5} = 6.$
The 1st factor identifies the game won, and the 2nd factor identifies the games tied.

$R_2 = \binom{6}{2} \times \binom{4}{3} = 60.$

$R_3 = \binom{6}{3} \times \binom{3}{1} = 60.$

$$R = R_1 + R_2 + R_3 = 126.$$


$\underline{\text{Case 2: A won Game-7}}$
This implies that the score was tied after Game-6.
This implies that A tied an even number of the 1st $6$ games.

This implies that during the first $6$ games, A tied $k$ games, where $k$ is some element in $\{0,2,4,6\}.$

So, there will be $4$ subcases, based on how many of the 1st $6$ games that A tied. I will use the variables $S_0, S_2, S_4,$ and $S_6$ to represent the respective (mutually exclusive) enumerations, re $0,2,4,$ or $6$ ties.

$S_0 = \binom{6}{0} \times \binom{6}{3} = 20.$
The 2nd factor refers to the number of games that A won.

$S_2 = \binom{6}{2} \times \binom{4}{2} = 90.$

$S_4 = \binom{6}{4} \times \binom{2}{1} = 30.$

$S_6 = \binom{6}{6} \times \binom{0}{0} = 1.$

$$S = S_0 + S_2 + S_4 + S_6 = 141.$$


Final computation:

$$R + S = 126 + 141 = 267.$$


Addendum
If I understand the OP's (i.e. original poster's) alternative approach, he wants to examine the outcomes of each game sequentially. Then, after each game, he wants to create pertinent branches that reflect the plausible results of the games just played. Then, he wants to compute an overall enumeration, based on this approach.

Although this approach is extremely clumsy (i.e. cumbersome), it is possible, as long as you are extremely careful.

The overall constraint is that player A won 4-3, with the match lasting 7 games.

I will map out satisfying sequences, so that a sequence like WTL... refers to player A winning Game-1, tieing Game-2, and losing Game-3. Then, the overall enumeration will be the number of satisfying sequences.

There are $3$ choices for how the sequence begins:

  • W
  • T
  • L

Moving into Game-2, you again have the same $3$ choices, so now you have $9$ branches.

Moving into Game-3, you again have the same $3$ choices, so now you have $27$ branches.

These branches may be listed as:

WWW, WWT, WWL
WTW, WTT, WTL
WLW, WLT, WLL

TWW, TWT, TWL
TTW, TTT, TTL
TLW, TLT, TLL

LWW, LWT, LWL
LTW, LTT, LTL
LLW, LLT, LLL


Now, each of the $27$ possible outcomes of the first $3$ games, as shown above, will be individually explored. For each such outcome, the possible sequences through game 6 will be listed and counted. There is no need to take the sequence beyond game 6, since the sequence through game 6 uniquely determines the game 7 outcome.

Note that for the sequence to go $7$ games, with player A the winner, the number of points accumulated by player A after $6$ games must be either $3$ or $3.5$.

I will use the variable $O_k$ to represent the number of sequences based on the first $3$ games having outcome $k$ above, where $k \in \{1,2,3,\cdots, 27\}.$

$O_1:~$ WWW
WWWTLL, WWWLTL, WWWLLT, WWWLLL
$O_1 = 4.$

$O_2:~$ WWT
WWTwLL, WWTLWL, WWTLLW
WWTTTL, WWTTLT, WWTLTT
WWTTLL, WWTLTL, WWTLLT
$O_2 = 9.$

$O_3:~$ WWL
WWLWTL, WWLWLT, WWLTWL, WWWTLW, WWWLWT, WWWLTW
WWLTTT
WWWWLL, WWWLWL, WWWLLW,
WWWTTL, WWWTLT, WWWLTT
$O_3 = 13.$

$O_4:~$ WTW
$O_4$ represents having exactly (2.5) points after $3$ games. Therefore, since this is the same as $O_2$, the $9$ Game-4 through Game-6 continuations that were listed for $O_2$ must exactly match the satisfying Game-4 through Game-6 continuations for $O_4$.
I will refer to this type of shortcut as
$O_4 = O_2 = 9.$

$O_5:~$ WTT
$O_5$ represents having $2$ points after Game-3, which matches the number of points, after Game-3, for $O_3$.
Therefore,
$O_5 = O_3 = 13.$

$O_6:~$ WTL
WTLWWL, WTLWLW, WTLLWW, WTLWTT, WTLTWT, WTLTTW
WTLWTL, WTLWLT, WTLTWL, WTLTLW, WTLLWT, WTLLTW
WTLTTT
$O_6 = 13.$

$O_7:~$ WLW
$O_7 = O_3 = 13.$

$O_8:~$ WLT
$O_8 = O_6 = 13.$

$O_9:~$ WLL
WLLWWT, WLLWTW, WLLTWW
WLLWWL, WLLWLW, WLLLWW
WLLWTT, WLLTWT, WLLTTW
$O_9 = 9.$

Running total of:
$O_1 + O_2 + \cdots + O_9$
$= 4 + 9 + 13 + 9 + 13 + 13 + 13 + 13 + 9 = 96.$


$O_{10}:~$ TWW
$O_{10} = O_2 = 9.$

$O_{11}:~$ TWT
$O_{11} = O_3 = 13.$

$O_{12}:~$ TWL
$O_{12} = O_6 = 13.$

$O_{13}:~$ TTW
$O_{13} = O_3 = 13.$

$O_{14}:~$ TTT
$O_{14} = O_6 = 13.$

$O_{15}:~$ TTL
$O_{15} = O_9 = 9.$

$O_{16}:~$ TLW
$O_{16} = O_6 = 13.$

$O_{17}:~$ TLT
$O_{17} = O_9 = 9.$

$O_{18}:~$ TLL
TLLWWW, TLLWWT, TLLWTW, TLLTWW
$O_{18} = 4.$

Running total of:
$O_{10} + O_{11} + \cdots + O_{18}$
$= 9 + 13 + 13 + 13 + 13 + 9 + 13 + 9 + 4 = 96.$

Running total of:
$O_{1} + \cdots + O_{18}$
$= 96 + 96 = 192.$


$O_{19}:~$ LWW
$O_{19} = O_3 = 13.$

$O_{20}:~$ LWT
$O_{20} = O_6 = 13.$

$O_{21}:~$ LWL
$O_{21} = O_9 = 9.$

$O_{22}:~$ LTW
$O_{22} = O_6 = 13.$

$O_{23}:~$ LTT
$O_{23} = O_9 = 9.$

$O_{24}:~$ LTL
$O_{24} = O_{18} = 4.$

$O_{25}:~$ LLW
$O_{25} = O_9 = 9.$

$O_{26}:~$ LLT
$O_{26} = O_{18} = 4.$

$O_{27}:~$ LLL
LLLWWW
$O_{27} = 1.$

Running total of:
$O_{19} + O_{20} + \cdots + O_{27}$
$= 13 + 13 + 9 + 13 + 9 + 4 + 9 + 4 + 1 = 75.$

Final total of:
$O_{1} + \cdots + O_{27}$
$= 192 + 75 = 267.$

$\endgroup$
14
  • $\begingroup$ thank you for the reply. As mentioned, I am asking in regards to the logic in "live" setting. From your example $ S_2 = \binom{6}{2} \binom{4}{2}$, those are player $A$'s predetermined choices rather than the sequential $n-1$ choices. fyi, your corrected answer of adding $S_0$ corresponds to the answer from the original post by Joffan. And as mentioned, I'm not questioning whether the answers are correct but rather; whether predetermined choices or choices in sequence is appropriate in this situation. $\endgroup$ Apr 22, 2022 at 12:00
  • $\begingroup$ @ManOnTheMoon I don't understand what you intend by the phrase sequential n-1 choices. This is why I could not comment on it. I will be happy to examine any alternative approach that you have. However, first, you have to edit your posting to make your alternative approach, much, much, clearer. I can not react to your alternative analysis until I understand your thought processes. I have never seen any analysis on any problem of this nature that involved sequential choices. You will need to make your thinking crystal clear. $\endgroup$ Apr 22, 2022 at 12:05
  • $\begingroup$ @ManOnTheMoon While this solution is phrased differently from Joffan's solution, both Joffan and user2661923 selected the positions of the wins, ties, and losses in the first six games. That determines the sequence of events. $\endgroup$ Apr 22, 2022 at 12:13
  • $\begingroup$ @user2661923, thank you for helping me understand the approach, I've added more information in my question. Hope it gives more clarity on what i am trying to express. $\endgroup$ Apr 22, 2022 at 12:26
  • $\begingroup$ What the OP is saying is that the outcome of the first match limits the number of possible sequences for the remaining matches, that the outcome of the first two matches limits the number of possible sequences for the remaining matches, and so forth. Of course, you counted all those possibilities in your calculations, without those case dependencies. $\endgroup$ Apr 22, 2022 at 12:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .