Winning

Winning.  Have you ever wondered about just how many games must your team win to know, guaranteed, it has won a certain number of consecutive games?  This is, not likely to have won a certain number of consecutive games.  But actual consecutive wins.  Thinking a bit combinatorially we can determine this with a simple formula.  Notation:

n = number of games in a season
r = number of consecutive wins desired

Let n/r = m R k.  That is m is the integer divisor of n by r, which is 0, 1, 2,…, and k is the remainder, 0 ,1,…,  r-1. 

For example 53/5 = 10 R 3, or 21/6 = 3 R 3. Then we have the minimum number of games that must be won to guarantee r consecutive wins W sometime during a season is given by

W = m(r-1) + k + 1

In the table below, we give some examples for various sports.Of course, when r = 2, that value is the next highest number greater than half the number of games when n is even.  

Number of games played	Run of consecutive wins	Minimum wins needed	Winning percentage	Sport
160	2	81	50.6%	Baseball
80	2	41	51.3%	Basketball
38	2	20	52.6%	Soccer*
16	2	9	56.3%	Football (American)
160	3	108	67.5%	Baseball
80	3	55	68.8%	Basketball
38	3	27	71.1%	Soccer*
16	3	12	75.0%	Football (American)
160	4	121	75.6%	Baseball
80	4	61	76.3%	Basketball
38	4	30	78.9%	Soccer*
16	4	13	81.3%	Football (American)
160	5	129	80.6%	Baseball
80	5	65	81.3%	Basketball
38	5	32	84.2%	Soccer*
16	5	14	87.5%	Football (American)
* A win is exactly that; tie counts as a non-win

Now to solve the same problem over two seasons or more you merely select twice the season length.  Thus, while to win five consecutive wins over two seasons of 80 games each, take n to be 160, which gives the requirement to 129 wins. There is a slight possibility that a team could win the last three games of one season and the first two of the next season to get to five.  This forces the win requirement to be slightly smaller than double the previous value. 

The required number of wins is perhaps higher than you thought, but it is the guarantee that drives it upwards.  Treating this problem probabilistically is rather different. 

P.S. Thanks to John A. for reviewing the problem for clarity