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This question is inspired by the ongoing baseball playoffs, but pertains to any tournament where 2 teams play a 7-game series, where the first to win 4 games is the overall (series) winner.

In times like these, the news coverage is full of useless statistics like "in the past 25 years, the team that wins game 3 (i.e. the 3rd game) has gone on to win the series 68% of the time". However, this does get me thinking...

  • In a symmetry sense, every game is equally important toward victory.

  • However, a series can end early, s.t. games 5, 6, 7 might not even be played. Given this asymmetry, it is correct to say that the winner of game 7 (if it is played) is always the series winner, but the winner of game 6 (if it is played) is not always the series winner.

Let $j \in \{1,2,...,7\}$ and $A_j$ be the event that game number $j$ is played AND team $A$ wins that game. Let $A_s$ be the event that team $A$ wins the series. My Question: What is $P(A_s | A_j)$?

Further thoughts:

For this question assume the chance of team $A$ winning any single game is $1/2$ and each game is independent.

Obviously each $P(A_s | A_j)$ can be calculated with a little bit of effort e.g. by (exhaustive) combinatorial counting. For something small like 7 games, this can be done by hand or using a small program. Moreover, the following are obvious:

  • $P(A_s | A_7) = 1$

  • $P(A_s | A_6) = 3/4$, because there is $1/2$ chance team $A$ had a 3-2 lead, in which case it wins after winning game 6, and a $1/2$ chance team $B$ had a 3-2 lead, in which case team $A$ has a $1/2$ to win the series (after winning game 6).

  • $P(A_s | A_1) = P(A_s | A_2) = P(A_s | A_3)$ by symmetry, since sudden ending cannot happen before game 4. [The symmetry exploited here is one can arbitrarily permute the results of these 3 games.]

However, is there a clever way to calculate $P(A_s|A_j)$ without resorting to (too much) explicit counting / chasing down the "tree-of-possibilities"?

Generalization to odd $N$ beyond $N=7$ would also be interesting.

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3 Answers 3

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The team that loses game $1$ now has to win at least $4$ out of $6$ games if we imagine all the games are played. We can read off from Pascal's triangle that happens $\frac {1+6+15}{2^6}=\frac {22}{64}$ of the time, so $P(A_s|A_1)=\frac {42}{64}$. This applies to game $4$ as well by your permutation argument for games $1,2,3$.

I don't see an easy way for game $5$.

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  • $\begingroup$ Good point that the symmetry extends to $P(A_s|A_4) = P(A_s|A_1)$, and indeed that value can be calculated by counting as you showed. For a 7-game series this leaves only $j=5$. But I'm still hoping someone comes up with a clever way to calculate these, esp. for arbitrary $N$. $\endgroup$
    – antkam
    Commented Oct 20, 2018 at 18:25
  • $\begingroup$ BTW typo: should be 22/64 & 42/64 instead of 21/64 & 43/64. :) $\endgroup$
    – antkam
    Commented Oct 20, 2018 at 18:28
  • $\begingroup$ @antkam: you are right about the typo. Fixed. Thanks $\endgroup$ Commented Oct 20, 2018 at 19:04
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I don't see a clever way here that doesn't require enumeration and is faster to come up with than enumeration takes to solve.

Visualize the state space as a five by five square with one missing corner, with states (m,n) signifying the score at that time, e.g. (2,1) would mean team A has won two games so far and team B has won one. You can have any combination of the five numbers from 0 to 4, except (4,4), hence 25 - 1 = 24 states. (0,0) is the initial state, and any state (4,n) or (m,4) is a terminal state. Terminal states have no outgoing edges, all non-terminal states (m,n) have one edge to (m+1,n) and one to (m,n+1).

To solve the problem, we need to assign two attributes to each state: the probability to reach and the probability to win. Both can be determined inductively:

$$P_R(m,n)=\cases{1 & if m,n = 0,0 \cr P_R(3,n)/2 & if m=4 \cr P_R(m,3)/2 & if n=4\cr (P_R(m-1,n)+P_R(m,n-1))/2 & otherwise}$$ $$P_W(m,n)=\cases{1 & if m=4 \cr 0 & if n=4 \cr (P_W(m+1,n)+P_W(m,n+1))/2 & otherwise}$$

Now we need to locate the subsets on which the conditional proabilities to win shall be calculated. Within layer $j$, identified by $m+n=j$, these are only those states which have an incoming edge in the direction of team A winning, i.e. from a parent (m-1,n), not (m,n-1). We can't use their probability to reach directly, however, since the condition requires us to reach these states via a team-A-winning edge only. We therefore also need a slightly modified probability to reach with a team A win: $$P_R'(m,n)=\cases{P_R(m-1,n)/2 & if n<4 and m>0 \cr 0 & otherwise}$$ We now have all the tools required to calculate the demanded probabilities: $$P(A_s|A_j)=\frac{\sum_{m+n=j}P_R'(m,n)*P_W(m,n)}{\sum_{m+n=j}P_R'(m,n)}$$ This yields the following results:

  1. P1 = 21/32 ~= 65.6%
  2. P2 = 21/32 ~= 65.6%
  3. P3 = 21/32 ~= 65.6%
  4. P4 = 21/32 ~= 65.6%
  5. P5 = 19/28 ~= 67.9%
  6. P6 = 3/4 ~= 75%
  7. P7 = 1/1 ~= 100%

This didn't feel very elegant, but at least it's a definitive answer. It also confirmed your intuitive observations.

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Here is an $\mathcal{O}(N^2)$ algorithm to determine $P(A_s|A_j)$ for all $1\leq j\leq N$.

Let ${n\choose\geq k}$ denote the sum $\sum_{i\geq k}{n\choose i}$. Note that you can easily calculate ${n\choose k}$ and ${n\choose\geq k}$ for all $0\leq k\leq n\leq N$ in $\mathcal{O}(N^2)$ time with Pascal's formula. Do note that this is probably going to overflow for large $N$ if you use finite sized integers.

We need to calculate $P(A_s|A_j)=P(A_s\cap A_j)/P(A_j)$. For this, note that the event $A_j$ can be split into two independent parts. Neither team has at least $(N+1)/2$ points after $j-1$ games, and team $A$ wins game $j$. The second part has probability $1/2$. The first one can be calculated as $\frac1{2^{j-1}}\sum{j-1\choose k}$ over all $0\leq k\leq j-1$ such that $2k<N+1$ and $2(j-1-k)<N+1$. We split into the cases $2(j-1)<N+1$ and $2(j-1)>N+1$.

$$P(A_j)=\left\{\begin{array}{ll}\frac12&\mbox{if}\ 2(j-1)<N+1\\\frac1{2^j}\left({j-1\choose\geq j-(N+1)/2}-{j-1\choose\geq(N+1)/2}\right)&\mbox{otherwise}\end{array}\right.$$

We can calculate $P(A_s\cap A_j)$ in a similar way. We get $\frac1{2^N}\sum{j-1\choose k}{N-j\choose\geq(N+1)/2-k-1}$ over all $0\leq k\leq j-1$ such that $2k<N+1$ and $2(j-1-k)<N+1$.

Example: Consider $N=7$ and $j=5$. We get $P(A_j)=\frac1{2^5}\left({4\choose\geq1}-{4\choose\geq4}\right)=\frac{14}{2^5}$ and $P(A_s\cap A_j)=\frac1{2^7}\left({4\choose1}{2\choose\geq2}+{4\choose2}{2\choose\geq1}+{4\choose3}{2\choose\geq0}\right)=\frac{38}{2^7}$. This gives $P(A_s|A_j)=\frac{38\cdot2^5}{14\cdot2^7}=\frac{19}{28}$.

Results for $N=7$: \begin{align*} P(A_s|A_{1})&=\frac{21}{32}&&=0.65625\\ P(A_s|A_{2})&=\frac{21}{32}&&=0.65625\\ P(A_s|A_{3})&=\frac{21}{32}&&=0.65625\\ P(A_s|A_{4})&=\frac{21}{32}&&=0.65625\\ P(A_s|A_{5})&=\frac{19}{28}&&\approx0.6785714285714\\ P(A_s|A_{6})&=\frac{3}{4}&&=0.75\\ P(A_s|A_{7})&=\frac{1}{1}&&=1\\ \end{align*}

Results for $N=21$: \begin{align*} P(A_s|A_{1})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{2})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{3})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{4})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{5})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{6})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{7})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{8})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{9})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{10})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{11})&=\frac{308333}{524288}&&\approx0.588098526001\\ P(A_s|A_{12})&=\frac{28007}{47616}&&\approx0.5881846438172\\ P(A_s|A_{13})&=\frac{27879}{47360}&&\approx0.5886613175676\\ P(A_s|A_{14})&=\frac{2115}{3584}&&\approx0.5901227678571\\ P(A_s|A_{15})&=\frac{2051}{3456}&&\approx0.5934606481481\\ P(A_s|A_{16})&=\frac{1939}{3232}&&\approx0.5999381188119\\ P(A_s|A_{17})&=\frac{1771}{2896}&&\approx0.6115331491713\\ P(A_s|A_{18})&=\frac{91}{144}&&\approx0.6319444444444\\ P(A_s|A_{19})&=\frac{75}{112}&&\approx0.6696428571429\\ P(A_s|A_{20})&=\frac{3}{4}&&=0.75\\ P(A_s|A_{21})&=\frac{1}{1}&&=1\\ \end{align*}

Results for $N=101$:

P(A_s|A_{1})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{2})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{3})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{4})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{5})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{6})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{7})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{8})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{9})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{10})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{11})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{12})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{13})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{14})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{15})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{16})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{17})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{18})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{19})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{20})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{21})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{22})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{23})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{24})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{25})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{26})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{27})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{28})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{29})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{30})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{31})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{32})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{33})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{34})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{35})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{36})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{37})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{38})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{39})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{40})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{41})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{42})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{43})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{44})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{45})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{46})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{47})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{48})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{49})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{50})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{51})&=\frac{171067743096724199353939462829}{316912650057057350374175801344}&&\approx0.5397946186936\\
P(A_s|A_{52})&=\frac{167221645255839744493109587}{309787536712665756499705856}&&\approx0.5397946186936\\
P(A_s|A_{53})&=\frac{19007527010746718867833116293}{35212516673005543476365885440}&&\approx0.5397946186936\\
P(A_s|A_{54})&=\frac{1075897755324992549264831795}{1993161321112934480078700544}&&\approx0.5397946186936\\
P(A_s|A_{55})&=\frac{3227693265967201901562863513}{5979483963323251947772837888}&&\approx0.5397946186937\\
P(A_s|A_{56})&=\frac{3227693265914715614499348377}{5979483963218279373645807616}&&\approx0.5397946186944\\
P(A_s|A_{57})&=\frac{3227693265626041035650015129}{5979483962640930215947141120}&&\approx0.5397946186982\\
P(A_s|A_{58})&=\frac{1661190563190372105517135}{3077449284583959931387904}&&\approx0.5397946187162\\
P(A_s|A_{59})&=\frac{3227693258794076002882461593}{5979483948977000150412034048}&&\approx0.5397946187892\\
P(A_s|A_{60})&=\frac{54706665066298548892112443}{101347184901899608710447104}&&\approx0.5397946190538\\
P(A_s|A_{61})&=\frac{54706663961717347878344251}{101347182692737206682910720}&&\approx0.5397946199213\\
P(A_s|A_{62})&=\frac{298247589765812802025}{552520490796066013184}&&\approx0.5397946225236\\
P(A_s|A_{63})&=\frac{896830351801089866538967}{1661428814574352043343872}&&\approx0.5397946297391\\
P(A_s|A_{64})&=\frac{81529996607932517384933}{151038912395153426612224}&&\approx0.5397946483793\\
P(A_s|A_{65})&=\frac{896829019834510497226711}{1661426150641193304719360}&&\approx0.5397946935459\\
P(A_s|A_{66})&=\frac{12285299517020779601695}{22759203293934330576896}&&\approx0.5397947967843\\
P(A_s|A_{67})&=\frac{896822195376530807267287}{1661412501725233924800512}&&\approx0.5397950204692\\
P(A_s|A_{68})&=\frac{13385262161331111469789}{24796914038664794734592}&&\approx0.5397954818273\\
P(A_s|A_{69})&=\frac{13384978909350336799453}{24796347534703245393920}&&\approx0.5397963910055\\
P(A_s|A_{70})&=\frac{13384443877831095755485}{24795277471664763305984}&&\approx0.5397981084554\\
P(A_s|A_{71})&=\frac{13383472373230368596701}{24793334462463308988416}&&\approx0.5398012273618\\
P(A_s|A_{72})&=\frac{188475665354635155899}{349154002765644562432}&&\approx0.5398066866246\\
P(A_s|A_{73})&=\frac{2385255518940145877}{4418646127201484800}&&\approx0.5398159187848\\
P(A_s|A_{74})&=\frac{2580396534879752675}{4780007755463262208}&&\approx0.5398310351967\\
P(A_s|A_{75})&=\frac{2578956557704761827}{4777127801113280512}&&\approx0.5398550478603\\
P(A_s|A_{76})&=\frac{2576736592893317603}{4772687871490392064}&&\approx0.5398921241604\\
P(A_s|A_{77})&=\frac{2573406645676151267}{4766027977056059392}&&\approx0.5399478681335\\
P(A_s|A_{78})&=\frac{197579984595643823}{365868791168499712}&&\approx0.5400296209049\\
P(A_s|A_{79})&=\frac{2561600037950329315}{4742414761604415488}&&\approx0.540146774738\\
P(A_s|A_{80})&=\frac{32302961280597037}{59785855291621376}&&\approx0.5403111007283\\
P(A_s|A_{81})&=\frac{32136304439508013}{59452541609443328}&&\approx0.5405371001734\\
P(A_s|A_{82})&=\frac{3546010590895109}{6556458151837696}&&\approx0.5408424043553\\
P(A_s|A_{83})&=\frac{3513754428103685}{6491945826254848}&&\approx0.5412482670286\\
P(A_s|A_{84})&=\frac{41836460476231}{77220354326528}&&\approx0.5417802189734\\
P(A_s|A_{85})&=\frac{41210275497799}{75967984369664}&&\approx0.5424689866361\\
P(A_s|A_{86})&=\frac{2378632522199}{4377702301696}&&\approx0.5433518220911\\
P(A_s|A_{87})&=\frac{2323380906455}{4267199070208}&&\approx0.5444744592949\\
P(A_s|A_{88})&=\frac{8648988203}{15843713024}&&\approx0.5458940205429\\
P(A_s|A_{89})&=\frac{25055140481}{45747490816}&&\approx0.5476833818443\\
P(A_s|A_{90})&=\frac{269915753}{490811392}&&\approx0.5499378323313\\
P(A_s|A_{91})&=\frac{770030395}{1393000448}&&\approx0.5527854611286\\
P(A_s|A_{92})&=\frac{55796071}{100279808}&&\approx0.5564038475223\\
P(A_s|A_{93})&=\frac{51981799}{92651264}&&\approx0.5610479205119\\
P(A_s|A_{94})&=\frac{1542073}{2719232}&&\approx0.5670987249341\\
P(A_s|A_{95})&=\frac{1396345}{2427776}&&\approx0.5751539680761\\
P(A_s|A_{96})&=\frac{65299}{111392}&&\approx0.5862090634875\\
P(A_s|A_{97})&=\frac{56651}{94096}&&\approx0.60205534773\\
P(A_s|A_{98})&=\frac{491}{784}&&\approx0.6262755102041\\
P(A_s|A_{99})&=\frac{395}{592}&&\approx0.6672297297297\\
P(A_s|A_{100})&=\frac{3}{4}&&=0.75\\
P(A_s|A_{101})&=\frac{1}{1}&&=1\\
$\endgroup$

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