This week the #puzzle is: How Much Does Game 1 Matter? #probabilities #combinatorics #PascalsTriangle #animation

You and your opponent are beginning a best-of-seven series, meaning the first team to win four games wins the series. Both teams are evenly matched, meaning each team has a 50 percent chance of winning each game, independent of the outcomes of previous games.

As the team manager, you are trying to motivate your team as to the criticality of the first game in the series (i.e., “Game 1”). You’d specifically like to educate them regarding the “probability swing” coming out of Game 1—that is, the probability of winning the series if they win Game 1 minus the probability of winning the series if they lose Game 1. (For example, the probability swing for a winner-take-all Game 7 is 100 percent.)

What is the probability swing for Game 1?

And for extra credit:

Instead of a best-of-seven series, now suppose the series is much, much longer. In particular, the first team to win N games wins the series, so technically this is a best-of-(2N−1) series, where N is some very, very large number.

In the limit of large N, what is the probability swing for Game 1 in terms of N? (For full credit, I’m expecting an answer that is rather concise!)

How Much Does Game 1 Matter?

Intermission

---

This is blog post no. 1002! I forgot to celebrate at no. 1000!
🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰🇩🇰

---

Highlight to reveal (possibly incorrect) solution:

Program PDF

Just for funzies, I created a video!

Thoughts:

• Notation: What is the probability of winning a best of 7, if the score is already a-b? It’s p_w(a,b).
• The probability swing of game 1 is p_s(1) = p_w(1,0) – p_w(0,1).
• As noted, p_s(7) = p_w(4,3) – p_w(3,4) = 1 – 0 = 1 = 100%. (7 = 4 + 3.)
• Let’s first go through all the probabilities. I begin with the probabilities related to 7 played games, then 6 etc. In each case I calculate the probability based on a) the chance of winning a game is 0.5, b) the probabilities related to winning if this game is won/lost.

(a,b)	pw(a,b)
(4,3)	1 (1)
(3,4)	0 (2)
(4,2)	1 (1)
(3,3)	0.5 * pw(4,5) + 0.5 * pw(3,4) = 0.5 * 1 + 0.5 * 0 = 0.5 (3)
(2,4)	0 (2)
(4,1)	1 (1)
(3,2)	0.5 * pw(4,2) + 0.5 * pw(3,3) = 0.5 * 1 + 0.5 * 0.5 = 0.75
(2,3)	0.5 * pw(3,3) + 0.5 * pw(2,4) = 0.5 * 0.5 + 0.5 * 0 = 0.25
(1,4)	0 (2)
(4,0)	1
(3,1)	0.5 * pw(4,1) + 0.5 * pw(3,2) = 0.5 * 1 + 0.5 * 0.75 = 0.875
(2,2)	0.5 * pw(3,2) + 0.5 * pw(2,3) = 0.5 * 0.75 + 0.5 * 0.25 = 0.5
(1,3)	0.5 * pw(2,3) + 0.5 * pw(1,4) = 0.5 * 0.25 + 0.5 * 0 = 0.125
(0,4)	0
(3,0)	0.5 * pw(4,0) + 0.5 * pw(3,1) = 0.5 * 1 + 0.5 * 0.875 = 0.9375
(2,1)	0.5 * pw(3,1) + 0.5 * pw(2,2) = 0.5 * 0.875 + 0.5 * 0.5 = 0.6875
(1,2)	0.5 * pw(2,2) + 0.5 * pw(1,3) = 0.5 * 0.5 + 0.5 * 0.125 = 0.3125
(0,3)	0.5 * pw(1,3) + 0.5 * pw(0,4) = 0.5 * 0.125 + 0.5 * 0 = 0.0625
(2,0)	0.5 * pw(3,0) + 0.5 * pw(2,1) = 0.5 * 0.9375 + 0.5 * 0.6875 = 0.8125
(1,1)	0.5 * pw(2,1) + 0.5 * pw(1,2) = 0.5 * 0.6875 + 0.5 * 0.3125 = 0.5
(0,2)	0.5 * pw(1,2) + 0.5 * pw(0,3) = 0.5 * 0.3125 + 0.5 * 0.0625 = 0.1875
(1,0)	0.5 * pw(2,0) + 0.5 * pw(1,1) = 0.5 * 0.8125 + 0.5 * 0.5 = 0.65625
(0,1)	0.5 * pw(1,1) + 0.5 * pw(0,2) = 0.5 * 0.5 + 0.5 * 0.1875 = 0.34375
(0,0)	0.5 * pw(1,0) + 0.5 * pw(0,1) = 0.5 * 0.65625 + 0.5 * 0.34375 = 0.5

(1) The series is decided with a win.
(2) The series is decided with a loss.
(3) The series is decided in the last game, 50/50.

• It is reassuring, that p_w(a,a) = 0.5.
• p_s(1) = p_w(1,0) – p_w(0,1)
• p_s(1) = 0.65625 – 0.34375 = 21/32 – 11/32
• p_s(1) = 0.3125 = 5/16
• Win the 1st game, and the probability of winning the series rises by more than 30%.

Another set of thoughts:

• My pdf shows a few ways to look at this problem. A full series walks through this graph (page 1), starting at (0,0) and ending at some (4,b) or (a,4).
• Page 2 tilts the graph.
• If I extend the graph to include all the series, that have 7 games (e.g. (5,2)), what I’m looking at here is actually all the paths from (0,0) to any state with 7 played games. (See video above.) E.g. the path from (0,0) to (5,2) could go through (0,1), (1,1), (2,1), (2,2), (3,2) and (4,2). In reality, the 7th game wouldn’t be played, but for analysis purposes I need both paths going beyond (4,2) — the one going to (5,2) and the one going to (4,3).
• A win is a path with at least 4 wins.
• How many paths are there with 7 wins? 1.
• How many paths are there with 6 wins? 7.
• … 5 wins? 7!/5!2! = 21.
• … 4 wins? 7!/4!3! = 35.
• But the question was actually, how many paths are there, beginning in (1,0), with 4-7 wins? So, 3-6 wins out of 6.
• 1 + 6 + 6!/4!2! + 6!/3!3! = 1 + 6 + 15 + 20 = 42
• Also, counting losses:
• 1 + 6 + 6!/4!2! = 1 + 6 + 15 = 22
• There are 2⁶ = 64 paths in all
• The probability of winning from (1,0), p_w(1,0), is 42/64 = 21/32 = 0.65625. This checks out.
• For symmetry reasons, the probability of losing from (0,1) is also 0.65625 (it’s the probability of the other side winning from a similar position as already analyzed), so the probability of winning, p_w(0,1), is 0.34375.
• What we also see here is p_w(1,0) – p_w(0,1) = ((1 + 6 + 15 + 20) – (1 + 6 + 15)) / 2⁶ = 20/2⁶.
• In a first to 4 / best of 7 game, p_s(1) = 6!/3!3! / 2⁶.
• In a first to N / best of 2N-1 game, p_s(1) = (2N-2)! / [((N-1)!)² * 2^2N-2].
• BTW, we have reproduced a couple of lines of Pascal’s Triangle.

And for extra credit:

Desmos

Thoughts:

• In a first to N / best of 2N-1 game, p_s(1) = (2N-2)! / [((N-1)!)² * 2^2N-2].
• What are some values for larger N?
• N = 100, p_s(1) = 0.0566316.
• N = 1000, p_s(1) = 0.0178479.
• N = 10000, p_s(1) = 0.00564211.
• N = 100000, p_s(1) = 0.00178413.
• Desmos suggests tending towards 1/√π√N
• It seems we’re tending towards 0.