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My exam had the following prompt:

A best-of-seven playoff is a competition between two teams head-to-head which must win four games to win the series. Four is chosen as it would constitute a majority of games played; whoever has won four games before all seven games have been played, all other games are omitted. Note that NBA finals are played based on best-of seven games series.

We have two competing teams in a best-of-seven games series: Team A and Team B. The probability of Team A winning a game is $p$, and Team B winning a game is $1-p$ (no draw games) where $0 < p < 1$.

Hint: The winner of the series has to win the last game.

One of the questions is based on this prompt and reads:

We know that Team A won the series in five games (i.e, won 4-1). What is the probability of Team A losing the first game?

And here's how I solved it, but I only got 5/10 points for it, and I'm not sure what I did wrong exactly. The professor hasn't released the solutions yet, so I figured I'd ask here.

My solution:

There is only one such outcome: LWWWW.

Probability = $(1-p)p^4$

Why is this wrong? Since the prompt says the winner of the series has to win the last game, and we're told team A wins in 5 games, we're asked to find the probability of it losing the first game under these terms. Thanks!

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  • $\begingroup$ It's worth noting that you answered the question "what is the probability that team A will win the series in 5 games...". However the question you were asked was "given that team A won series in 5 games...". The question, "What is the probability of Team A losing the first game?" is confusing because it is a question about the past asked in a way that suggests a future outcome. I've found that a number of math problems on MSE and elsewhere revolve around fairly subtle English constructions. $\endgroup$
    – Χpẘ
    Jul 28, 2017 at 19:46

3 Answers 3

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Since we know the series lasted only five games, there are four possible outcomes:

$$ 1. LWWWW\\ 2. WLWWW\\ 3. WWLWW\\ 4. WWWLW $$

Thus, the probability that they Team A lost game one is $1/4$.

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  • $\begingroup$ But if team A lost game one, how can you have arrangements 2, 3, and 4? $\endgroup$
    – Aleksandr Hovhannisyan
    Jul 28, 2017 at 18:38
  • $\begingroup$ @AleksandrH No, we don;t know that A lost game one ... we just know they lost one game $\endgroup$
    – Bram28
    Jul 28, 2017 at 18:38
  • $\begingroup$ But it says "What is the probability of Team A losing the first game?" $\endgroup$
    – Aleksandr Hovhannisyan
    Jul 28, 2017 at 18:39
  • $\begingroup$ @AleksandrH you have to divide the number of good options (here 1) by the total number of options (4), because these options all have the same chance of occurring given that $A$ won in $5$ games. $\endgroup$
    – Henno Brandsma
    Jul 28, 2017 at 18:40
  • $\begingroup$ @AleksandrH Right, the probability of that ... we are not certain of that. But given the information that A won in 5, we know that it has to be one of these 4 arrangements $\endgroup$
    – Bram28
    Jul 28, 2017 at 18:41
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You are calculating $P(LWWWW)$, but you need to calculate the conditional probability $P(LWWWW|\text{Team A won in 5})$

That is, you are provided with information (Team A won in 5 games), and that means that the probability of event $LWWWW$ to happen (well, have happened) wil go up: it will be higher as compared to not knowing anything at all (other than the basic value of $p$ and that this is a best of 7)

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  • $\begingroup$ Thank you! This makes sense. $\endgroup$
    – Aleksandr Hovhannisyan
    Jul 28, 2017 at 18:38
  • $\begingroup$ @AleksandrH You're welcome! This is exactly why probability is so tricky .... $\endgroup$
    – Bram28
    Jul 28, 2017 at 18:39
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GIven that we are in the situation that $A$ won in $5$ games, only 4 sequences of matches can have occurred:

$LWWWW, WLWWW, WWLWW, WWWLW$, all with the same probability.

So the chance we had the first is $\frac{1}{4}$.

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