Close Game Parlay

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"Straight Cash Homie"
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Ok, the basic principle is to take two separate games and make 2 parlays from one team in each game as one parlay. I will parlay the favorite of one game to the underdog on the RL in the second game and do this for two parlays. I’m trying to shoot for the middles and get the added pay of the parlay. I will use games that start at least 3 hours apart so I can hedge and attempt to middle the second game if needed. I will have at least half of one parlay locked up after the first game if not half of both parlays if the favorite wins by 1 run. If the favorite wins by more than 1 run in the first game or the underdog wins I will then middle the second game to cover the fact that one parlay will already be lost. This limits loses to 1 unit max while still providing a second chance for the middle and a profit. Approximately 4 units maximum possible win with only 1 unit maximum lose and only one of the games need the favorite to win by 1 run to have a profit every series. If the first game has the favorite win by 1 run both parlays are live and at least one will cash for a guaranteed profit regardless of the second games outcome. There are 5 ways to make a profit and 4 ways to lose a max of 1 unit with every possible outcome considered for both games. I think home favorites will be the best plays since they bat last. If they are losing or tied in the bottom of the 9<SUP>th</SUP> the first run to take the lead locks in the 1 run win for the favorite, which is the desired outcome (unless the hit is a home run).

Example:
Game 1: Team A -140 plays Team B +1 1/2 -150
Game 2: Team C -130 plays Team D +1 1/2 -160

Parlay #1: (Team A -140 to Team D +1 1/2 -160)
Parlay #2: (Team C -130 to Team B +1 1/2 -150)

No matter what happens in game #1 I will have at least one part of a parlay won (Team A or Team B +1 ½). If Team A wins by 1 run then at least one parlay is guaranteed to win and I can let the second game ride and hope for another 1 run game to win both parlays, but have a profit locked in regardless of the outcome in the second game. If Team A wins by more than 1 run, Parlay #2 is lost and I only have parlay #1 to go. I would then bet Team C to win the original amount lost on Parlay #2 and still have a chance for the middle while only risking a lose of approximately one bet the whole time. If Team B +1 ½ wins the game outright I will bet Team D +1 ½ in the second game to attempt the middle again and limit loses.

I calculated these parlays using opening RX lines and a parlay calculator, but actually changed my picks. I did this at work and hope anyone interested understands or asks questions. This is still a good example and my actual bets are at the bottom. Lines have moved since opening, but I just wanted to uses the lines to show the calculations of how many UNITS could be won or lost on each particular bet. I’m only going to place $400 to the side for trying this and will start each bet as $20 risk on parlays and to win $20 on side bets for the second game if needed. The risk is pretty low since the most that could be lost is $20 bucks, but I’m not tying up a lot of funds with this. I will post actual bet amounts and odds when bets are placed.

Opening Lines:
Toronto -145
Oakland +1 ½ -170
San Diego -125
San Francisco +1 ½ -160

Parlay #1 (San Diego –125 to Oakland +1 ½ -170) = 1.85
Parlay #2 (San Francisco +1 ½ -160 to Toronto –145) = 1.75


Watch for the following to decide late game plays and see what can be won or lost depending on outcome.

Game #1 - Toronto wins by more than 1
(Parlay #1 is lost/Middle Game #2)
Play San Diego –125 for 1 Unit

Game #2
If San Diego wins by more than 1:
Parlay #1 & #2 is Lost and San Diego –125 Wins (-1)
If San Francisco wins:
Parlay #1 is Lost, San Diego –125 is Lost, and Parlay #2 Wins (1.75 – 2.25) = (-0.5)
If San Diego wins by 1:
Parlay #1 is Lost, but San Diego –125 and Parlay #2 Wins (2.75 –1) = (1.75)


Game #1 - Oakland wins
(Parlay #2 is lost/Middle Game #2)
Play San Francisco +1 ½ -160 for 1 Unit

Game #2
If San Diego wins by more than 1:
Parlay #1 Wins, but Parlay #2 and San Francisco +1 ½ Loses (1.85 – 2.6) = (-0.75)
If San Francisco wins:
Parlay #1 & 2 Loses, but San Francisco +1 ½ -160 Wins (-1 Units)
If San Diego wins by 1:
Parlay #2 is lost, but Parlay #1 & San Francisco +1 ½ -160 (2.85 –1 = 1.85)


Game #1 - Toronto wins by 1
(Guaranteed Profit)
Both parlays can win (Let’em Ride)

Game #2
San Diego wins by more than 1: (1.85 – 1) = (0.85)
San Francisco wins: (1.75 – 1) = (0.75)
San Diego wins by 1: (1.85 + 1.75) = (3.6)
 

"Straight Cash Homie"
Joined
Nov 12, 2006
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First Play

Parlay #1 (Oakland +1 1/2 -140 to LA Angels -115) $20 pays $44.10
Parlay #2 (Indians +1 1/2 -210 to Toronto -173) $20 pays $26.59

Will look to hedge or let it ride after Toronto/Oakland game.

Parlay Bankroll $400.00
 

"Straight Cash Homie"
Joined
Nov 12, 2006
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Parlay #2 Loses -$20

Added Side Play:
Cleveland Indians +1 1/2 -200 $40 pays $20

I had to guess that Oakland +1 1/2 was safer than Toronto to win the game since the game was tied after 8.

Still a chance for a middle in the second game and a profit for the series.

If LA wins by 2 or more runs total lose will be (-$15.90)
If LA wins by 1 run profit will be ($44.10)
If Cleveland Wins total lose will be (-$20)
 

"Straight Cash Homie"
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Parlay #1 (LA Angels -108 to Detroit +1 1/2 -175) $20 pays $40.53
Parlay #2 (Boston -120 to Cleveland +1 1/2 -200) $20 pays $35.00
 

"Straight Cash Homie"
Joined
Nov 12, 2006
Messages
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Side Play

Parlay #2 Loses -$20

Added Side Play:
Boston Redsox -110 $22 pays 20

Middle chance in the second game would actually pay for the first series lost and still show a 1 unit profit if Boston wins by 1 run. That's 1 win in 4 games and a 1 unit profit.
 

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