Unlikely World Series Title Pattern: 1917, 1918, 2004, 2005

In summary, the conversation discussed a problem posted from another board about the odds of two teams winning titles in 1917 and 1918, then not winning another title until going back-to-back in 2004 and 2005. The problem was posed in light of the upcoming World Series with Chicago, who haven't won since 1918, playing against Boston, who hadn't won since 1917 until last year. The original poster gave a calculation for the odds, but was unsure of its accuracy and was seeking feedback. They also discussed the gradual increase in team numbers from 1918 to 2004 and expressed interest in finding a general formula for any number of teams. The conversation ended with someone sharing the answer they
  • #1
eldrick
52
0
someone posted an interesting problem from another board.
now, the upcoming world series have chicago playing & they haven't won it since 1918.
last year's winner, boston, hadn't won since 1917 !
the problem posed was:
"what are the odds of two teams (not specifically having to be boston/chicago ) winning titles in 1917 & 1918, then not winning another title between them until they go back-to-back again in 2004 & 2005."
apparently 16 teams present in 1917 - 1918 & with franchise expansion there are ~ 30.
he gave the answer ( which i am having trouble working out ), which i'll post as soon as we get a coupla ( or even 1 ! ) reply.
many thanks :smile:
 
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  • #2
When did franchising start?
 
  • #3
EnumaElish said:
When did franchising start?

i'm not overly sure, but a gradual increase in team nos. from 1918 to 2004 can be assumed.

try the calculation with the assumption of following no. of teams thruout
1) 16 teams
2) 24 teams
3) 30 teams.

a general formula is of interest, for any N teams you wish to use.

my rough methodology was:

- assume that 2 teams winning in 1917 /1918 is a given occurence with probability of 1

- probability of the 2 teams not winning from 1918 - 2004 ( which is 86y ) is probabilty of only N - 2 teams winning for 86y = [(N - 2) / N]^86

- probability of 2 teams winning back-to-back = (1/N) * (1/N) which you multiply by 2, as it can happen in 2 ways = 2* (1/N)^2

then multiply all the aspects mentioned in bold to get a formula for the overall probability.

i'm just wondering if there is a flaw in my workings ?
 
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  • #4
one thing that perplexed me was if you work out the odds for N = 3 to 10 ( each of 3, 4, 5,... 10 ) & say just 5y instead of 86y, you get an odd pattern.
someone try it & please explain it to me.:bugeye:
 
  • #5
the answer i was given to the original question was:

18,000,000 to 1

i hope this may help
 

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