# Coming back from behind

July 11, 2020 |

## Alex Castaldo writes:

Heres the skinny. from math puzzles volume 1, by  presh talwalkar. doc here. from nature walk. originally to stretch aubrey's mind . odds of a comebak victory

Consider 2 teams a and b that are completely evenly matched. given that a team is behind in score at half time, what is the prob that a team will overcome the deficit and win the game. assume the first halve and the second half are taken to be independent events. Presh solves it as follows logically:

Since the two teams are evenly matched, it is equally likely that the team will score enuf points to overcome the deficit or that it will not score enuf points. fo example the event of falling behind 6 pts in a half game happens with the same prob as gaining 6 pts in a half game. He concludes prob is 0.25

Now we posted the empirical resutls from basektaball games and many others have given the empiriclal results for football games … and i gave some results for the markets.. this seems to be of interest to everyone , had the most views of any posts, and it was good for 7 or 8 points today.. lets have your discussion and solution of this problem. presh says the answer is 0.25 both empirically (NFL in 1995) and logically.

## Jared Albert writes:

In a game with two teams where in the first round, the team 1 advantage varies from flat to all the points available in the second round, the probability of  team 0 coming from behind to win are in array with 20 available points in the second round:

[0.49, 0.306, 0.22, 0.129, 0.09, 0.03, 0.018, 0.011, 0.004, 0.002, 0.002, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]

For example, if the teams are even going into the second round with 20 available points, .490 chance that team0 wins; with a one point advantage to team1 at the start of round2, team0 wins .306 of the time;

2 points to team1, team0 wins .220 of the time etc

Here's the montecarlo:

import numpy as np

np.random.seed(10)

out_list = []

out_list = []

count = 1000

win = 1

lose = 0

team0_start = 0

team1_start = 0

size=20

def runs():

z = np.sum(np.random.choice([win, lose], size=size, replace=True, p=None))

return z

def outcome(team1_start, count = count, team0_start=team0_start):

l= []

for _ in range(count):

team0_end = runs() + team0_start

team1_end = runs() + team1_start

came_from_behind = team0_end > team1_end

l.append(came_from_behind)

#print(f'l: {l}')

outcome = sum(i > 0 for i in l)

return(outcome)

for i in range(size):

out_list.append(outcome(team1_start=i)/count)

print(f'outlist: {out_list}')

## Victor Niederhoffer writes:

up your alley i  think. we have done something similar for market with real empirical results. the  unconditional prob is much less than20%

## Stephen Stigler writes:

I am sure you know but I repeat anyway:

1) the simple calculations ignore correlation between teams.

2) they also ignore information on the distribution of changes

3) Calculations using the distribution of changes are not hard.

4) But the information about the probability of extreme events is not well determined so they can be inaccurate

5) In any case  markets unlike sports are not zero sum games.

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