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What are the odds of hittingfive, six, and seven -200 favorites in a row. Thank you and an explanation would help, if not the answer will suffice. Thanks guys.
xyz
SBR Wise Guy
02-14-08
521
#2
Assume -200 is the true odds for the favorite (i.e. the no vig price), then the probability for the favorite to win is 2/3. So winning 5 in a row would be (2/3)^5=13.1%, 6 in a row is (2/3)^6=8.7%, and 7 in a row is (2/3)^7=5.8%. Unless you are playing at Matchbook, the -200 you get at a book is likely worst than the no vig price, which will bring down those percentages.
Comment
twister
SBR Sharp
09-09-08
405
#3
Originally posted by xyz
Assume -200 is the true odds for the favorite (i.e. the no vig price), then the probability for the favorite to win is 2/3. So winning 5 in a row would be (2/3)^5=13.1%, 6 in a row is (2/3)^6=8.7%, and 7 in a row is (2/3)^7=5.8%. Unless you are playing at Matchbook, the -200 you get at a book is likely worst than the no vig price, which will bring down those percentages.
Just to expand on this, incase it wasn't obvious, is that anytime you want to know the probability of 'event x' occuring 'y times in a row', you just take the probability of 'event x' occuring once, and apply 'y' as the power.
i.e a coin landing on heads, 14 times in a row, is:-