What is the probability that if you pick 22 correct college winning teams (all outcomes must be correct)?
If you were able to place a bet for $5 on a 22-team parlay. What are the odds of getting all 22 correct? Also, how many different bets (combintations) would you have to place to guarentee a win on that 22-team parlay?
By: egthareal
About the Author:
By: egthareal
About the Author:
No related posts.
Related posts brought to you by Yet Another Related Posts Plugin.
One Response to “What is the probability that if you pick 22 correct college winning teams (all outcomes must be correct)?”
Leave a Reply
Live Parlay Odds
Parlay Odds
Pages
Tags
Alot Of People
Bet
Bet On Soccer
Betting
Betting Football
Betting Games
Betting Man
Betting Nfl
Betting Online
Betting On Sports
Betting Sports
Betting Uk
Bowling
Colts
Football Betting Tips
Game
Gas Bills
Horse Betting
Limit Poker
Mlb
Nfl Games
parlay bets
Parlay Odds
Pats
People Talking
Pluses
Poker Betting
Realistic Answer
Sarah Palin
Soccer Betting
Sports Betting
Sports Check
Sports Online
Spread Betting
Superbowl
Super Bowl
Superbowl Bet
Superbowl Betting
Superbowl Party
super bowl xliv
Texas Holdem
Uk Betting
Upcoming Sports
Virus Protection
Water Bills
Online Poker
If there are 22 teams playing then there are 11 games. Assuming that the palray is such that there WILL be a winner (ie no draws) then there are two possible results for each game .. a win or a loss (Team A Wins – Team A loses). So there are a total of 2^11 possible combinations each round.
If a draw is possible then there are 3^11 possible combinations W – D – L on each game (Team A Wins – Team A Draws -Team A Loses).
P(all correct) = 1/2^11
= 1/2048
? 0.05%
If a draw is possible then there are 3^11 possible combinations W – D – L on each game (Team A Wins – Team A Draws -Team A Loses).
Then P(all correct) = 1/3^11
= 1/177147
? 0.0006%