[dynamite140]

Don't bash me for this please

Assuming you are betting on a game whether its NFL or NBA and the juice is -110. It could also be on baseball where the sides are pretty much close to even or -110 on both sides. Not sure if it matters to put the juice but what are the odds of you losing both games assuming you bet them at the same time?

My logic is if i bet 2 games and its not those moneylines where one team is favored or underdog, is my logic correct that the majority of the time you will go 1-1 assuming you bet 2 games at once.

Is there a percentage of the time where you would go 1-1, 0-2 and other times 2-0?

I'm assuming everything is even and there is so sort of edge on one side.

[MarketMaker]

2-0 = .25 (.5 *.5)
1-1 = .5 (.5 *.5) + (.5 * .5)
0-2 = .25 (.5 *.5)

[wrongturn]

and it has nothing to do whether you bet them at once or not.

[arwar]

take note .25 + .5 + .25 = 1 even if markets notation is a confusing.

say you want bet 3 games the odds are 1/2*1/2*1/2 = 1/8 of winning all three

stated another way there are 8 possible out comes for betting 3 games, one is winning all three and one is losing all three.

the other 6 possible outcomes are the various combinations of winning 1 or 2 and losing 1 or 2.

[thebayareabeast]

Take a course in statistical probability

[Dark Horse]

Chance of losing both games is (1-p1) x (1-p2), where p is your expected winning percentage for each bet

So if you expect to win both games 55% of the time, the chance of losing both is (1-0.55) x (1-0.55) = 0.2025 or 20.25%
The chance of winning one of these games and losing one is 0.55 x (1-0.55) = 24.75% for both W-L and L-W, so 49.50%
This means that the chance to win both should be what is left: 100%-69.75% = 30.25%
Check: 0.55 x 0.55 = 0.3025 or 30.25%

So betting with a 55% winning expectation means that nearly half the time the result over two bets will be 1-1.
Yet it shows a 10% better chance of winning both bets than of losing both.

For expected profit you should factor in the juice.

For longer streaks: http://www.sportsbookreview.com//Betting+Too...alculator.aspx

[HedgeHog]

Assuming you have no edge, your win % are as follows:

2-0 25%
1-1 50%
0-2 25%

[Optional]

and it has nothing to do whether you bet them at once or not.

Not unless the outcomes are correlated in some way.

What's your thinking with that? Are you hoping that if always betting in pairs of games you will have a sort of natural hedge against losses?

[csm506]

Good question but your chances are the same

[Tomahawk]

If the chance to win a bet is 50% then to lose 2 bets out of 2 bets is 25%.

You're right that the most chance is that you will go 1-1, since:
-you have a 25% chance to win the 1st and lose the 2nd
-you also have a 25% chance to lose the 1st and win the 2nd

25%+25% gives out 50%