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This question is inspired by: A longer series is better for a better team: Can you see this at a glance?

Also obviously inspired by the NBA playoffs happening right now. :)

Suppose two teams are playing a series of $2k-1$ games, and the first team to win $k$ games wins the overall series. (No game can end in a tie.) Moreover, team A is "better" than team B.

  • Let $A_i$ denote the event that team A wins game $i$.

  • Let $A_{series}$ denote the event that team A wins the series, i.e., A wins $k$ or more games.

If the game results are i.i.d., and $P(A_i) = p > 1/2$ for any game $i$, then a longer series (larger $k$) increases A's chance of winning the series, i.e. $P(A_{series})$ is an increasing function in $k$. This is intuitively obvious, and a proof can be found in the link above (although that post asks an excellent question re: why such an "obvious" result requires an algebraically convoluted proof).

I wanna know under what conditions, i.e. under what probability model, would a longer match be BAD for the better team A. Perhaps some dependence that encodes "reversion to the mean" and/or (opposite of) "momentum"?

What I tried

My $0$-th attempt: If we allow "fatigue" in the form of decreasing $P(A_i),$ then a longer series can be bad for A, even when all $P(A_i) > 1/2.$ A simple 3-game example: if $P(A_1) = 1, P(A_2) = P(A_3) = 0.51$, then A always wins in a 1-game "series" ($k=1$) but B has a chance ($0.49^2$) in a 3-game series ($k=2$). @Henry in his answer gave an infinite length example. So I'm looking for something where the marginal probabilities $P(A_i)$ are constant (i.e. no fatigue).

My 1st attempt: let $P(A_1) = p > 1/2$, and thereafter every game $i$ result = game 1 result. This means: (1) $P(A_i) = p > 1/2$ (even though they are dependent), and yet (2) a longer series would give no advantage (nor disadvantage) to team A, because $P(A_{series}) = p$ regardless of $k$. However, I want a scenario where a longer series actually decreases $P(A_{series})$.

My 2nd attempt is something convoluted: the first 7 games ($k=4$) are played "normally" (i.i.d.) but then games 8 and 9 are always won by the loser of the best-of-first-7 series. I have not worked this out fully, but while this may give an example where $P(A_{series} | k = 4) > P(A_{series} | k = 5)$, I think this also implies the marginal probabilities $P(A_8), P(A_9) < 1/2$. So this isn't satisfying as a counterexample, since not only $P(A_i)$ are non-constant, team A actually becomes the worse team in a sense.

My 3rd attempt is a "surgical tweak" to the 2nd attempt: (1) if the first 7 games include exactly 4 wins by A, then games 8 & 9 are won by team B (therefore making B the overall winner), but (2) if the first 7 games have any other result, then games 8 & 9 are played i.i.d. but with an enhanced $P(A_8)=P(A_9) = p' > p$. By restricting the special dependence to an event of small enough probability, and balancing it out with $p' > p$ in the case (2), I think I can manage $P(A_8) = P(A_9) =p $. I have not worked out whether $P(A_{series} | k = 4) > P(A_{series} | k = 5)$ for some choice of $p, p'$. Also, this counterexample is too "artificial" for my taste.

What I seek is a probability model where:

  • For every game $i$, the marginal probability $P(A_i)$ is the same, i.e. $\forall i: P(A_i) = p > 1/2$.

  • There exists $k_0$ s.t. $P(A_{series} | k = k_0) > P(A_{series} | k = k_0 + 1)$.

    • Bonus if this is true for all $k_0$, or all sufficiently large $k_0$.
  • Aesthetic requirement :) - Any dependence is as "simple" as possible, i.e. I prefer not to have something like my 3rd attempt (or even more convoluted). I know this isn't a math requirement, and people can have different tastes... comments on this point are welcome.

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My other answer had the probability of Player A winning a point not depending on the game score, but depending on the number of points played so far. The example in this answer has the conditional probability of Player A winning a point depending on the game score, but the unconditional probability staying constant over time.

This example supposes that the probability that Player A wins a point is $0.6$. But Player A is sensitive to the score, especially after an even number of games:

  • If after an even number of games $2k$, Player A is losing overall then (through self-motivation) the probability that A wins the next game $2k-1$ is $1$
  • If after an even number of games $2k$, scores are even then (due to Player A choking or freezing, and this gets worse over time) the probability that A wins the next game is $0.6^{k+1}$
  • If after an even number of games $2k$, Player A is winning then (through relaxation) the probability that A wins the next game tends to increase and happens to be $1-\frac{0.4-\left(1- 0.6^{k+1}\right)\mathbb{P}(\text{A tying after }2k\text{ games})}{\mathbb{P}(\text{A winning after }2k\text{ games})}$
  • If after an odd number of games $2k-1$, the probability that A wins the next game is $0.6$, no matter what the score is at the time

The overall effect is to keep the probability that Player A wins a particular point at $0.6$, but surprisingly the probability that Player A is ahead after an odd number of games keeps falling (i.e. would win a series of that length), below $0.5$. I would expect it to be possible to extend the example, and perhaps the limit of the probability that A wins a long series is just over $0.4729$

Games   Prob A is   Prob        Prob A is   Prob A wins Prob A wins Prob A wins
played  losing      series      winning     next game   next game   next game 
so far  series      tied        series      if behind   if tied     if ahead
0       0           1           0                       0.6        
1       0.4                     0.6         0.6                     0.6
2       0.16        0.48        0.36        1           0.36        0.742222222
3       0.4672                  0.5328      0.6                     0.6
4       0.18688     0.38656     0.42656     1           0.216       0.772747187
5       0.48994304              0.51005696  0.6                     0.6
6       0.195977216 0.351566124 0.452456660 1           0.1296      0.792252266
7       0.501980370             0.498019630 0.6                     0.6
8       0.200792148 0.334642831 0.464565021 1           0.07776     0.803302033
9       0.509413153             0.490586847 0.6                     0.6
10      0.203765261 0.325848417 0.470386322 1           0.046656    0.810040466
11      0.514410894             0.485589106 0.6                     0.6
12      0.205764358 0.321140653 0.473094989 1           0.0279936   0.814309532
13      0.517915128             0.482084872 0.6                     0.6
14      0.207166051 0.318607172 0.474226777 1           0.01679616  0.817082862
15      0.520421846             0.479578154 0.6                     0.6
16      0.208168739 0.317259969 0.474571292 1           0.010077696 0.818915974
17      0.522231458             0.477768542 0.6                     0.6
18      0.208892583 0.316564635 0.474542781 1           0.006046618 0.820143698
19      0.523543073             0.476456927 0.6                     0.6
20      0.209417229 0.316225505 0.474357266 1           0.003627971 0.820975122
21      0.524495478             0.475504522 0.6                     0.6
| cite | improve this answer | |
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  • $\begingroup$ I have produced a note on this at se16.info/hgb/Playing_more_can_be_worse_for_better_players.html , showing that if the constant marginal probability of the better player winning each point is $p$ (e.g. $0.6$ here) then the probability of the better player winning the match can approach $2p-1$ (e.g. $0.2$) $\endgroup$ – Henry Dec 31 '18 at 22:16
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To save typing, let's define

  • $p_k$ as the probability that Player A wins game $k$
  • $q(j,k)$ as the probability that after $k$ games, Player A has won exactly $j$ of them
  • $r_{2k-1}=\sum_{j=k}^{2k-1} q(j,k)$ is the probability that Player A has won the majority of an odd number $2k-1$ games

so we have the recurrence $$q(j,k)= p_k \,q(j-1,k-1) +(1-p_k)\,q(j,k-1)$$ starting at $q(0,0)=1$ and $q(j,0)=0$ for $j \not =0$, and given $p_1,p_2,\ldots,p_k$ we can calculate all $q(j,k)$

You seem to want

  • $p_k \gt \frac12$ and $r_{2k-1} \gt \frac12$ for all positive integer $k$ since Player A is the better player
  • $r_{2k-1} \gt r_{2k+1}$ for all positive integer $k$ to suggest Player A's chances get worse with a longer series

Since playing two extra games only makes a difference to the series if the score was already almost even, I think $$r_{2k-1} - r_{2k+1} = (1-p_{2k})(1-p_{2k+1})q(k,2k-1) - p_{2k} p_{2k+1} q(k-1,2k-1)$$ and so for the longer series to be worse for the better player you want this to be positive and so $$\dfrac{(1-p_{2k})(1-p_{2k+1})}{p_{2k} p_{2k+1} } > \dfrac{q(k-1,2k-1)}{q(k,2k-1)}$$

One possible approach would be to set $p_{2k+1}=p_{2k}$ and then to choose $p_{2k}$ such that $$\frac12 < p_{2k} < \dfrac{1}{1+\sqrt{\frac{q(k-1,2k-1)}{q(k,2k-1)}}}$$ where the right hand side is greater than $\frac12$ since $q(k-1,2k-1) < q(k,2k-1)$ because A is always better than B and so is always more likely to just win than to just lose

The following rather arbitrary example starts with $p_1=0.6$ and sets, using a weighted average from the inequality, $p_{2k+1}=p_{2k}=0.9\times\frac{1}{1+\sqrt{\frac{q(k-1,2k-1)}{q(k,2k-1)}}}+0.1\times\frac12$. It illustrates the point that it is not difficult to use this rule to create a long sequence, which I think meets all of your final points, and in particular the result of each game does not depend on the results of the previous games

        Prob A          Prob A 
Game    wins this       winning series 
        game            if stop here 

1       0.6             0.6
2       0.5454592       
3       0.5454592       0.5950459
4       0.5288103       
5       0.5288103       0.5926991
6       0.5224735       
7       0.5224735       0.5911743
8       0.5189403       
9       0.5189403       0.5900499
10      0.5166238       
11      0.5166238       0.5891616
12      0.5149601       
13      0.5149601       0.5884287
14      0.5136932       
15      0.5136932       0.5878058
16      0.5126882       
17      0.5126882       0.5872645
18      0.5118665       
19      0.5118665       0.5867864
20      0.5111790       
21      0.5111790       0.5863584
22      0.5105929       
23      0.5105929       0.5859713
24      0.5100859       
25      0.5100859       0.5856180
26      0.5096418       
27      0.5096418       0.5852932
28      0.5092487       
29      0.5092487       0.5849928
30      0.5088976       
31      0.5088976       0.5847133
32      0.5085817       
33      0.5085817       0.5844522
34      0.5082954       
35      0.5082954       0.5842072
36      0.5080346       
37      0.5080346       0.5839765
38      0.5077955       
39      0.5077955       0.5837585
40      0.5075755       
41      0.5075755       0.5835519
42      0.5073722       
43      0.5073722       0.5833557
44      0.5071834       
45      0.5071834       0.5831688
46      0.5070077       
47      0.5070077       0.5829904
48      0.5068436       
49      0.5068436       0.5828199
50      0.5066899       
51      0.5066899       0.5826564
52      0.5065455       
53      0.5065455       0.5824996
54      0.5064096       
55      0.5064096       0.5823489
56      0.5062814       
57      0.5062814       0.5822038
58      0.5061601       
59      0.5061601       0.5820640
60      0.5060452       
61      0.5060452       0.5819290
62      0.5059361       
63      0.5059361       0.5817986
64      0.5058324       
65      0.5058324       0.5816725
66      0.5057337       
67      0.5057337       0.5815504
68      0.5056395       
69      0.5056395       0.5814320
70      0.5055495       
71      0.5055495       0.5813172
72      0.5054635       
73      0.5054635       0.5812058
74      0.5053811       
75      0.5053811       0.5810975
76      0.5053022       
77      0.5053022       0.5809922
78      0.5052264       
79      0.5052264       0.5808897
80      0.5051536       
81      0.5051536       0.5807899
82      0.5050836       
83      0.5050836       0.5806927
84      0.5050162       
85      0.5050162       0.5805979
86      0.5049512       
87      0.5049512       0.5805054
88      0.5048886       
89      0.5048886       0.5804151
90      0.5048282       
91      0.5048282       0.5803269
92      0.5047698       
93      0.5047698       0.5802407
94      0.5047133       
95      0.5047133       0.5801565
96      0.5046587       
97      0.5046587       0.5800740
98      0.5046059       
99      0.5046059       0.5799934
100     0.5045547       
101     0.5045547       0.5799144
| cite | improve this answer | |
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  • $\begingroup$ This is super neat, but sadly not exactly what I want. My $0$-th attempt is a 3-game example where decreasing $p_k$ leads to the longer series favoring B. You have (very impressively!) generalized this to infinite length! I love your example. However, what I was seeking was an example where all $p_k$ are constant, i.e $\forall k : p_k = p>1/2.$ Obviously this will require the games to be dependent. Thanks for your solution though -- after my 3-game example I was wondering if an infinite-length example exists, and you have found it. $\endgroup$ – antkam May 24 '18 at 1:28
  • $\begingroup$ @antkam - I picked up on your word "fatigue" to suggest player A weakens over time while remaining better $\endgroup$ – Henry May 24 '18 at 7:19
  • $\begingroup$ yeah, sorry about that. i should have put more emphasis on the constant $p_k$ requirement. very nice example of fatigue though. $\endgroup$ – antkam May 24 '18 at 11:06
  • $\begingroup$ I have added a different answer which aims at the constant $p$ $\endgroup$ – Henry May 24 '18 at 15:54
  • $\begingroup$ I have produce a note on this at se16.info/hgb/Playing_more_can_be_worse_for_better_players.html , showing that having the probability the better player wins the $n$th point $p_n = \frac12 +\frac{p_1-\frac12}{n}$ will mean that the probability the better player wins the match will reduce towards $0.5$ as $n$ increases $\endgroup$ – Henry Dec 31 '18 at 22:20

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