Math guys - How often will I lose both sides of this bet?

[Emily_Haines]

Will there be a score in 1st 7min 30sec of 1st qtr
NO +145

Will there be a score in 1st 6min 00sec of 1st qtr
YES +138

[FourLengthsClear]

Cannot say without knowing the projected total.
For the 'average' NFL matchup though you would lose both around 13% of the time i.e the first score would be between 6.00 and 7.30.

[Emily_Haines]

It's a NCAA game

I think it would be profitable betting both sides. I look at the first 30 games from last weeks card and 3 out of the 30 had a score in the 90 second window. I imagine that is going to be fairly accurate if I did a larger sample. For this to be unprofitable scores would have to come more than 25% of the time in that window.

[mathdotcom]

You should think about both these bets individually

[Justin7]

A typical ncaaf game has a total of about 52, and each average score is about 5.7 points. I would expect about 9 scores in that game, over 60 minutes. If you are looking at a 1 minute window (and that window is long enough into the game that field position is relatively random), you might guess there is about a 9/60 chance -- a little more than 1 in 7 of a score during that minute.

[RickySteve]

A typical ncaaf game has a total of about 52, and each average score is about 5.7 points. I would expect about 9 scores in that game, over 60 minutes. If you are looking at a 1 minute window (and that window is long enough into the game that field position is relatively random), you might guess there is about a 9/60 chance -- a little more than 1 in 7 of a score during that minute.

This is a decent first approximation but would you really use a number this rough to put your own money at risk?

[Justin7]

It is a starting point. A better way would be with a database... Scoring distributions are weird at the start and end of each half. This time point is almost outside the "weird distribution" time... right on the edge.

[wrongturn]

The issue is what the chance of the first score falls between 6:00 and 7:30 is. It should be less than 9/60?

[samserif]

Small refinement: instead of basing scores per minute off the NCAA average game total score, perhaps use the over/under for this particular game, assuming it's a good expectation of this particular total. But yeah, better to look at the actual distributions (from a database) than to use broad averages.

[laxbrah420]

...

[laxbrah420]

...

[Justin7]

The issue is what the chance of the first score falls between 6:00 and 7:30 is. It should be less than 9/60?

Oops, I misread original problem. I did a 1-minute interval, not 90 second.

[mathdotcom]

For two bets to be +EV together, at least one of them must be +EV. odds are it is just one of them and not both. So evaluate them individually, not this bullshit 90 second window.

[RickySteve]

For two bets to be +EV together, at least one of them must be +EV.

False.

[Frats]

False.

For a situation like this, I don't see why his statement wouldn't be true. You might have a tough time figuring out which is +EV, but I don't see how it would be possible for the two of them put together to be +EV without one of them being +EV.

[hutennis]

For two bets to be +EV together, at least one of them must be +EV.

It is not true.

It is NOT impossible to generate positive expected value out of a series of negative expected value bets. No need for even one +EV bet to be there.

[mathdotcom]

http://forum.sbrforum.com/players-talk/499911-very-common-misconception-among-posters-middles.html

It is the same idea here.

[samserif]

For two bets to be +EV together, at least one of them must be +EV. odds are it is just one of them and not both. So evaluate them individually, not this bullshit 90 second window.

But EV has nothing to do with the question being asked, right? The question is what's the value of P(~A & ~B), where ~A="no (scoring) events in the first six minutes" and ~B="at least one event between 6 and 7.5 minutes". If you can break the first 7.5 minutes down into discrete intervals and approximate the probability of a scoring event during each interval, you can figure out the probability of this conjunction.

[Frats]

But EV has nothing to do with the question being asked, right? The question is what's the value of P(~A & ~B), where ~A="no (scoring) events in the first six minutes" and ~B="at least one event between 6 and 7.5 minutes". If you can break the first 7.5 minutes down into discrete intervals and approximate the probability of a scoring event during each interval, you can figure out the probability of this conjunction.

And what he's saying is correct (in my mind). If a bet on the 1:30 interval is going to be +EV, then it means that either one or both of the under/over would also have to be +EV (and likely even more +EV than the method that the player was suggesting). I don't see a statistical way for that to be untrue in this situation.

[That Foreign Guy]

It is not true. It is NOT impossible to generate positive expected value out of a series of negative expected value bets. No need for even one +EV bet to be there.

Unless the -EV bets are positively correlated I don't see how this is possible.

Anyway, this original proposal can be viewed as one bet "first score in the 6:00-7:30 window" Yes / No (if the player dutches).

If this bet is profitable it is likely because the distribution of first score times is weighted to one side or the other. If that is true then betting that side of the two possible half-bets in the OP is +EV.

Betting NO on the window is only profitable if you believe the 6-7:30 min window is a dip in the first scoring time distribution which is an odd position to take without significant data.

[hutennis]

If this comment

For two bets to be +EV together, at least one of them must be +EV.

addresses specific OP question, I don't want to get involved.

If it is a general statement, then it is not true.

[mathdotcom]

I am talking straight bets. Obviously a highly correlated parlay of two -EV lines (ie. spread of -20 and O20.5 when market lines are -19 and 20).

Amazing how many guys yell false but don't have a counter-example.

[samserif]

I am talking straight bets. Obviously a highly correlated parlay of two -EV lines (ie. spread of -20 and O20.5 when market lines are -19 and 20).

Amazing how many guys yell false but don't have a counter-example.

I agree that you can't get EV>0 from multiple EV<0 bets. The simplistic counterexample of a parlay doesn't work: a parlay isn't multiple bets but instead a single bet on the outcome of multiple events. (Subtle but important concept.) I was just pointing out that the OP asked for a frequency, not an EV. In fact, the payoffs could have been omitted from the example and the question would still be valid.

[hutennis]

I agree that you can't get EV>0 from multiple EV<0 bets.

Yes you can.

http://en.wikipedia.org/wiki/Parrondo's_paradox

FAIR WARNING.

This is NOT a blueprint for magically turning shi*t into gold!!!!
Not a gambling holy grail either.

This is very sophisticated stuff!!!!

But, in principal, string of -EV bets can and will produce +EV results.

Old mantra stating the opposite is wrong.

[wrongturn]

I think the two games in that paradox are slightly "correlated" when played in that sequence, although the correlation is not obvious. It is an interesting read, although not understanding everything there.

[mathdotcom]

I see that you get horny reading about paradoxes, but this is a gambling forum. Please show us how to apply this in a gambling market so we can beat bookmakers by betting any side of any game at any price. If those are the simplest examples of that paradox, then that tells you something about their relevance.

[samserif]

Interesting, thanks! But it appears that this only works for certain examples of highly correlated games, where either information from one game gives you a EV>0 for the second game or where by alternating games you end up making a bet when the short-term EV is postive although the long-term isn't. So, you end only betting only when EV is (temporarily) positive. From the Wikipedia page: " The individual games are losing only under a distribution that differs from that which is actually encountered when playing the compound game". In other words, the compound game has the property of "pulling out" the EV>0 outcomes of the overall EV<0 games.

I can't think of any practical examples where this can be applied. Anyone? It sounds to me like announcing that I've found a way to make money betting ML on Kansas (or any other terrible team): wait till they play a Div-II school. Tah dah, I've created EV>0 out of an EV<0 bet.

[hutennis]

@mathdotcom

If this comment

Originally Posted by mathdotcom
For two bets to be +EV together, at least one of them must be +EV.

addresses specific OP question, I don't want to get involved.

If it is a general statement, then it is not true.

This.

You made a general statement being absolutely sure that it is true without any contingencies.
I pointed out that general statement is not true.
I'm right and you are wrong.
If you don't agree, take it up with scientific community.

Also, scientific community agrees, that word "paradox" does not really apply here in its traditional sense.
Basically, it is proven concept.
Word "paradox" is used to emphasize a counter intuitive nature of it.

As far as applications of it in a speculative markets such as sports betting, a lot of people are hard at work
trying to develop such an applications.
It's not easy, you know. Its sophisticated stuff.

It's so unfortunate, that relevance of genius discovery is a suspect as far as you are concerned.
Well, world will just have to deal with.

@samserif

It is already being used to manage financial risks
That's a lot.

[mathdotcom]

hutennis

This is a gambling forum. When you find me a bookmaker that offers two straight -EV bets that you can turn into +EV by betting both of them, then I will give you a round of applause and retire from posting. Until then maybe you should post on a theoretical gambling forum. I prefer the actual gambling forums.

[RickySteve]

hutennis

This is a gambling forum. When you find me a bookmaker that offers two straight -EV bets that you can turn into +EV by betting both of them, then I will give you a round of applause and retire from posting. Until then maybe you should post on a theoretical gambling forum. I prefer the actual gambling forums.

I will post such wagers if you promise to kill yourself.

[mathdotcom]

Thanks Ricky you contribute a lot, as usual

[Patrick McIrish]

Actually I'd like to see it as well. If in what we do here we're dealing with straight bets yes than at least one would have to be +EV. If not educate me on what I'm missing. Thanks.

[Jontheman]

False.

However, in the context set by the original poster it's true. Maybe you need to stop wanking over your own knowledge of the existence of irrelevant theoretical maths and start concentrating on the practical question being discussed.