Need help calculating odds

[Smooth60]

let assume the juice is 10% , but you can get 7% on most sites.

you bet 10 games for $100 each.
6 win = +600
4 loss = -440
profit = +160

You missed my example earlier.

16 x 160 = 2560
16 x 1000 on Cleveland = ~ 10,000 And that works out to a 75% win %

The more games you bet per week worse off you are gonna be.

 

[RemyMartin]

You missed my example earlier.

16 x 160 = 2560
16 x 1000 on Cleveland = ~ 10,000 And that works out to a 75% win %

The more games you bet per week worse off you are gonna be.

I was replying to Teejay not you.

 

[GPIDEAL]

Assuming a fair coin is 50/50. Always. You don't bet that it will catch up.

Right.

Isn't each coin toss, an independent event?

 

[GPIDEAL]

every time you flip the coin it is an independent event so. one the 100th flip my odds of heads or tails for just that one coin is 50 50.

that is why if you play the lottery weekly or kust once in your life. your one ticket has the same chance of winning as someone who plays weekly.

That's true, BUT if you play weekly (vs. once in your life), your chances of winning anything increase.

 

[benstt]

That's true, BUT if you play weekly (vs. once in your life), you're chances of winning anything increase.

Yep.

If you play 6/49 once, your chance of winning the jackpot is 1 in 13million or so.

If you play once a week for 20 years, you're up to 1040 in 13 million or so.

 

[GPIDEAL]

Yep.

If you play 6/49 once, your chance of winning the jackpot is 1 in 13million or so.

If you play once a week for 20 years, your up to 1040 in 13 million or so.

Ha ha, that's right benstt.

So that's why you play several lines a week to get an edge, he he.


P.S. I had to correct my spelling of "you're" to "your".

 

[benstt]

So that's why you play several lines a week to get an edge, he he.

So you're up to say 5200 out of 13 million or so.

 

[Butler1000]

I understand that a coin flip done right now is a 50/50 regardless of what happened in the past. Yeah, I remember my math teacher saying the same thing.

But how does that explain batting avgs?

Most batters have a career batting avg of let's say about .270

At the beginning of the year, any batter can start hot or cold in their first 50 or 100 at bats (a batter will have about 600 ABs in a full season). One batter can be batting .150, while another is some reason red hot at .400.

Both batters have career .270 avgs. By year end, what will likely happen is the guy batting .150 will bat better and get back up to .270-ish, by batting .300 rest of the year. While the guy who got lucky hitting .400 in the first month will likely crash and burn and end up at .270, which means the rest of the season he bat maybe .240

Because the hot hitter will see Shittier pitches, probably resulting in frustration swings. The crappy one will see more fastballs. Also less chance of having to face the better reliever later.

You would need some serious chaos theory math to try to make sense of it.

 

[TeeJay]

That's true, BUT if you play weekly (vs. once in your life), your chances of winning anything increase.

False
People seem to assume this all the time (eg if I gamble every week or if I play lottery every week it increases my chance to win)
It is simply a failure to understand math that breeds this

Same idea as coin toss
If you throw 10 heads in a row you *expect* the 11th or 12th or 13th to be tails
It's simply not true, as each toss has a 50/50 shot

 

[TeeJay]

Yep.

If you play 6/49 once, your chance of winning the jackpot is 1 in 13million or so.

If you play once a week for 20 years, you're up to 1040 in 13 million or so.

No
You realize even lottery people state this is a fallacy
No idea how so many people can think this

 

[TeeJay]

let assume the juice is 10% , but you can get 7% on most sites.

you bet 10 games for $100 each.
6 win = +600
4 loss = -440
profit = +160

You left out all the odds (?) as well as the fact most people do not bet on a single game
If you bet on 10 games in a week I doubt all 10 would be individual bets

Even your math is wrong
If I bet $100, then "win" $100 I have broken even
If I bet $100 then lose $110 I am now -$110
So using your example I have spent $1000 ($100 x 10)
I win $600
I lose $440
+600 - 440 - 1000 = -$840
You are in quite the hole despite being correct on 60% of your bets

 

[Big Rig]

No
You realize even lottery people state this is a fallacy
No idea how so many people can think this

While each coin flip is independent, if you flip a coin twice you are more likely to get a head than if you flip it once, no? Same thing with lottery. The more often you play the more likely you win. Cannot be otherwise.

 

[RemyMartin]

You left out all the odds (?)

Even your math is wrong
If I bet $100, then "win" $100 I have broken even
If I bet $100 then lose $110 I am now -$110
So using your example I have spent $1000 ($100 x 10)
I win $600
I lose $440
+600 - 440 - 1000 = -$840
You are in quite the hole despite being correct on 60% of your bets

What odds, we are talking point spread here not money line.

Your calculation is wrong, I think you've never place a bet on sports betting, go check it out before you talk again.
maybe you are one of those suckers betting on pro-line.

 

[TeeJay]

Your calculation is wrong, I think you've never place a bet on sports betting, go check it out before you talk again.
maybe you are one of those suckers betting on pro-line.

So you are going to claim spending $1000 to "win" $600 is a profit? Really?

Out of curiosity what is wrong with Pro-Line? Simple ease of access and do not require references to bet

 

[RemyMartin]

So you are going to claim spending $1000 to "win" $600 is a profit? Really?

Out of curiosity what is wrong with Pro-Line? Simple ease of access and do not require references to bet

I didn't say $600 profit, I said $160 is the profit, read my post again. The math is simple and easy.

Pro-line charge too much juice.

 

[GPIDEAL]

False
People seem to assume this all the time (eg if I gamble every week or if I play lottery every week it increases my chance to win)
It is simply a failure to understand math that breeds this

Same idea as coin toss
If you throw 10 heads in a row you *expect* the 11th or 12th or 13th to be tails
It's simply not true, as each toss has a 50/50 shot

Tee Jay

Let's say you have a 50/50 chance of winning 100 bucks (disregard the entry cost of the raffle). You have two people that play. One plays only once, the other plays every week. Are you telling me that the person who plays more doesn't have a better chance of winning something than the person who only played once?

 

[benstt]

No
You realize even lottery people state this is a fallacy
No idea how so many people can think this

This is pretty much exactly how it works for the jackpot. I'm very familiar with lottery math.

There is a very slight correction for the remote chance you'll win multiple jackpots, but that can be ignored for practical purposes.

 

[Submariner]

Tee Jay
Let's say you have a 50/50 chance of winning 100 bucks (disregard the entry cost of the raffle). You have two people that play. One plays only once, the other plays every week. Are you telling me that the person who plays more doesn't have a better chance of winning something than the person who only played once?

Each draw is an independent event. Buying a single ticket in 6/49 this week gives you a 1 in 14 million chance to win. Next week, buying a single ticket in 6/49 gives you a 1 in 14 million chance to win. Since each event is independent, the odds are not additive. In other words, the odds are not (1 in 14 million + 1 in 14 M = 2 in 14 Million). The odds are not 2 in 14M, they remain at 1 in 14M.

So now, let's say two friends each buy a ticket this week for 6/49. Each has a 1 in 14M chance to win. Next week, only one friend buys a ticket. His chance of winning is 1 in 14M, while his buddy who did not play has 0% chance. So the ticket buying friends chances are infinitely better of winning a prize in week two vs his abstaining friend. But over the course of the two draws, the ticket buying friends chance of winning is still just 1 in 14M, same as his friend who only bought a ticket in week one.

Also, recognize that buying 10 tickets this week is different then buying one ticket for 10 weeks. If you buy 10 tix this week, the odds are additive since you are entering several times in the same event. Thus, your odds improve to 10 in 14 million (or 1 in 1.4M). If you play one ticket each week for ten weeks, then you are participating just once in 10 independent events and your odds stay at 1 in 14 million, not just for each draw, but over the course of the entire 10 draws.

 

[RemyMartin]

If you buy 2 weeks the odds is 2 in 14 million.

 

[benstt]

Each draw is an independent event. Buying a single ticket in 6/49 this week gives you a 1 in 14 million chance to win. Next week, buying a single ticket in 6/49 gives you a 1 in 14 million chance to win. Since each event is independent, the odds are not additive. In other words, the odds are not (1 in 14 million + 1 in 14 M = 2 in 14 Million). The odds are not 2 in 14M, they remain at 1 in 14M.

Your chance of winning on the second ticket indeed remains 1/14M. However, your odds of winning on either draw, before the first draw is made, is additive.

I'll write here about lottery strategies, which looks at the odds of different series of ticket buys. This is done prior to any draw, to determine what the expected odds or return is on a strategy as a whole.

The math for the two draw strategy goes like this. The key thing about probability is to remember:

Pr[Event A or B] = Pr[Event A] + Pr[Event B] - Pr[Event A and B].

The last part is like a correction for winning multiple jackpots, which either is not necessary for tickets in the same draw, or ends up being miniscule in over multple draws.

Often you can tackle a problem the other way:
Pr[Event A or B] = 1 - Pr[ not A and not B]

Probability of winning either week = 
       = 1 - Probabililty of losing both weeks
       = 1 - (1 - 1/14M)(1 - 1/14M)
       = 1 - 1 + 2(1/14M) - (1/14M)*(1/14M)
       = 2 x 1/14M  approximately    
      (the correction for winning twice is very small)

So, the odds of winning over the two weeks is 2/14M approximately. If you lose the first week, your odds collapse in a sense when going into the second week, down to 1/14M.

So now, let's say two friends each buy a ticket this week for 6/49. Each has a 1 in 14M chance to win. Next week, only one friend buys a ticket. His chance of winning is 1 in 14M, while his buddy who did not play has 0% chance. So the ticket buying friends chances are infinitely better of winning a prize in week two vs his abstaining friend. But over the course of the two draws, the ticket buying friends chance of winning is still just 1 in 14M, same as his friend who only bought a ticket in week one.

Think about it this way. One of them has a strategy to buy one ticket in week one, but abstain in week two. The other has a strategy to buy a ticket each week.

Over the course of the two draws, the odds of their two strategies winning are different. One is 1/14MM, the other is 2/14MM roughly, from the calculation above.

This doesn't mean the odds of the second week's ticket are not 1/14M when viewed in isolation.

Also, recognize that buying 10 tickets this week is different then buying one ticket for 10 weeks. If you buy 10 tix this week, the odds are additive since you are entering several times in the same event. Thus, your odds improve to 10 in 14 million (or 1 in 1.4M). If you play one ticket each week for ten weeks, then you are participating just once in 10 independent events and your odds stay at 1 in 14 million, not just for each draw, but over the course of the entire 10 draws.

The odds of 10 distinct tickets winning the jackpot in a single draw and the odds of 10 tickets over 10 draws are very close to each other.

The math goes like this for the two strategies:

Probability of a jackpot in 10 tickets in a single draw 
                = Prob Ticket 1 or 2 or 3 up to 10 win
                = Prob(Ticket 1 wins) + Prob(Ticket 2 wins) + ... + Prob(Ticket 10 wins)
         (this works because multiple tickets cannot win, so no need to correct for multiple jackpots)
                = 10 x 1/14M

Probability of a jackpot in 10 tickets over 10 draws 
               = 1 - Probability Zero wins in the 10 tickets over 10 draws
               = 1 - (1 - 1/14M)^10
               = 10 x 1/14M approximately

Again, this doesn't mean the odds of a single ticket in a single draw is not 1/14M. It just means that the strategy of buying 10 distinct tickets in a single draw is very much the same as 10 tickets over 10 draws.