# Re: some help if you would

dtsteele@xxxxxxxxx wrote:
what are the true odds against hitting the following parlays assuming
you are betting the point spread? (a link to a site with this info
3 teams
4 teams
5 teams
10 teams

tia

cheers,
dan

Relatively simple math.
You multiply the "expectation" by itself for a 2-teamer,
and by itself again for a 3-teamer, and by itself again
for a 4-teamer, etc.
So, ASSUMING an expectation of 50% ATS - which it would
be if say you were flipping a coin - then for a 2-teamer
the "true odds" of hitting it are 50% X 50% = 25%
This makes sense when you merely look at the chances by
scoring ALL the possible results:
Team 1 wins and Team 2 wins = WINNER
Team 1 wins and Team 2 LOSSES = loser
Team 1 LOSSES and Team 2 wins = loser
Team 1 LOSSES and Team 2 LOSSES = loser
Thus you have a 1 in 4 chance of winning BOTH, or 25%
of winning the 2-team parlay....or expressed another
way, the odds are 3 to 1 AGAINST (ie. 3 bad results, 1 good)
So, at 3 to 1 odds against, as I understand your question,
the "true odds" or BREAK-EVEN point would be *IF* the
payoff for each winner is 3 TIMES the cost of any loss.
Thus, over time, for every 3 losses you will win ONCE
on average and that ONE WINNER needs to pay \$300 if each
loser parlay costs you -\$100...
However, the books will NOT give you "true odds"
They need to take their cut (vig)
They do this by taking it from the WINNING payout.
Instead of paying you a fair \$300 on a winning 2-team
\$100 parlay, most books only pay \$260 or so...

For a 3-teamer parlay: 50% X 50% X 50% = 12.5% (or 1 in 8)
Most books only pay \$600 or so for a winning \$100 3-teamer.

For a 4-teamer parlay: 50% X 50% X 50% X 50% = 6.25%
which is a 1 in 16 chance.
Most books only pay \$1000 or so for a winning \$100 4-teamer.
.

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