Cute Applicable Math Question

[ML4L]

Search:

"Way Tougher Two Round Game Theory Problem"
Last 6 months
General Theory

ML4L

[DrSavage]

Let's say probability of winning a freezeout is x
you win immidiately if you flip tails twice and you are in the exact same situation if you flip one tails and one heads
therefore:
x = 0.6 ^ 2 + (0.6 * 0.4 * 2) x
or
x = 0.36 + 0.48 x
therefore x = 0.36/0.52 = 0.6923

[Terry]

This is a simple game to simulate. (No comments needed on the quick & dirty code.)

================================================== ===========================
RANDOMIZE TIMER: DIM NumFlips, Awon, Bwon, i AS LONG
PRINT "Working..."

BRofA = 2: BRofB = 2: Awon = 0: Bwon = 0

FOR i = 1 TO 20000000

Flip = INT(RND(1) *10)
IF Flip < 6 THEN BRofA = BRofA + 1 ELSE BRofB = BRofB + 1
IF Flip < 6 THEN BRofB = BRofB - 1 ELSE BRofA = BRofA - 1

IF BRofA = 0 THEN Bwon = Bwon + 1: BRofA = 2: BRofB = 2
IF BRofB = 0 THEN Awon = Awon + 1: BRofA = 2: BRofB = 2

NEXT i

PRINT "Player A (60% advantage) won "; Awon; " out of "; Awon; Awon + Bwon; "completed games."
PRINT 100 * (Awon / (Awon + Bwon)); "%"
================================================== ==========================

Repeated runs give 69.2% with the next digit varying from 0 to 5.

[slider77]

I would ratio the winning streaks required for each player.

P(Player 1 Streak) = .6^2 = .36
P(Player 2 Streak) = .4^2 = .16

The game must start at the same "node point" - i.e. the start of the game and a win/loss are at the same point.

So, P(Player 1 Wins) = .36/(.36+.16) = .692