两名玩家A、B系列对局：两类获胜规则下A的获胜概率求解

Let's break down both match formats step by step, using probability state transitions (a straightforward way to model these sequential game scenarios):

1. 率先连胜3局者赢得系列赛 #

First, let's define states based on the current winning streak of each player (since the game ends as soon as either player gets 3 consecutive wins):

• P(a, b) = Probability A wins the series when A has a current streak of a wins, B has a current streak of b wins.
• Boundary conditions:
  • P(3, b) = 1 (A already has 3 straight wins, so they've won the series)
  • P(a, 3) = 0 (B has 3 straight wins, so A loses)

For non-terminal states, the probability transitions as follows:
If A wins the next game, their streak increases by 1 and B's streak resets to 0; if B wins, their streak increases by 1 and A's streak resets to 0. So:
P(a, b) = p * P(a+1, 0) + q * P(0, b+1)

We need to solve for P(0, 0) (the starting state). Let's simplify by defining shorthand states:

• x = P(0,0) (starting state)
• y = P(1,0) (A just won 1 game, B has 0 streak)
• z = P(2,0) (A just won 2 games, B has 0 streak)
• u = P(0,1) (B just won 1 game, A has 0 streak)
• v = P(0,2) (B just won 2 games, A has 0 streak)

Now write the transition equations:

1. x = 0.6y + 0.4u
2. y = 0.6z + 0.4u
3. z = 0.6*1 + 0.4u (A wins the next game to finish the series, or loses to reset to B's 1-streak state)
4. u = 0.6y + 0.4v
5. v = 0.6y + 0.4*0 (B wins the next game to finish the series, so A's chance drops to 0)

Let's solve step by step:

• From equation 5: v = 0.6y
• Substitute v into equation 4: u = 0.6y + 0.4*0.6y = 0.84y
• Substitute u into equation 3: z = 0.6 + 0.4*0.84y = 0.6 + 0.336y
• Substitute z and u into equation 2: y = 0.6*(0.6 + 0.336y) + 0.4*0.84y
  • Expand: y = 0.36 + 0.2016y + 0.336y
  • Simplify: y - 0.5376y = 0.36 → 0.4624y = 0.36 → y ≈ 0.7786
• Calculate u: u = 0.84 * 0.7786 ≈ 0.6540
• Finally, substitute into equation 1: x = 0.6*0.7786 + 0.4*0.6540 ≈ 0.7288

Result: Player A has approximately a 72.9% chance to win the series under this format.

2. 率先比对手多赢2局者赢得系列赛 #

This is a classic "win by 2" format, similar to tennis tiebreaks or volleyball deciding sets. Again, we'll use state transitions:

• Define x = Probability A wins from a tied state (starting state, 0 games difference)
• y = Probability A wins when leading by 1 game
• z = Probability A wins when trailing by 1 game

Boundary conditions:

• If A leads by 1 and wins the next game, they win the series; if they lose, it's tied again: y = 0.6*1 + 0.4x
• If A trails by 1 and wins the next game, it's tied; if they lose, they lose the series: z = 0.6x + 0.4*0
• From the starting state, A either leads by 1 or trails by 1: x = 0.6y + 0.4z

Substitute y and z into the equation for x:
x = 0.6*(0.6 + 0.4x) + 0.4*(0.6x)

• Expand: x = 0.36 + 0.24x + 0.24x
• Simplify: x - 0.48x = 0.36 → 0.52x = 0.36 → x = 36/52 = 9/13 ≈ 0.6923

Alternatively, this matches the classic formula for "win by 2" scenarios: A's win probability is p²/(p² + q²). Plugging in the numbers: 0.6²/(0.6² + 0.4²) = 0.36/(0.36 + 0.16) = 9/13 ≈ 0.6923.

Result: Player A has approximately a 69.2% chance to win the series under this format.

内容的提问来源于stack exchange，提问作者Siyang Li