Mr. Sweepy's Forums

 Interesting theory efd3467, but slightly contradictory. On one hand you state the odds remain the same, on the other, you say they've improved. The error enters with the equation (51/52)^20. You were right to state,"The odds of any single one of those draws resulting in the Ace of Spades being drawn is still 1 in 52.". The odds do not change, each draw is it's own, and the odds are 1:52.
On another note, I guess we would have to define what "considerably" means as the odds of some (many?) of the sweepstakes will be considerably worse than 1:1,000,000 depending on the prize, sponsor, and length of the sweepstakes. 0.0001% odds variation hardly has me quitting my job.
The odds are much better if you count winning any prize, rather than just the grand prizes only if those other prizes are part of the same sweep.
In response to chudder4444: Many experiments have been done with coin flipping, very well documented. Whether you flip 2 or 100 the odds are always 1:2,
regardless of which side you pick. If you flip a coin 1000 times and count the number of 'heads' it's going to be close to 50%. Try it, it's educational, or
go to this web site: http://science.kennesaw.edu/~jdemaio/1107/binomial_probability_distributio.htm.
In any case remember this; many have gone to Vegas and won on one spin of the roulette wheel. But the palatial casinos and hotels were built with the money of those that thought their odds would get better if they played more. Frankly, Vegas wouldn't even exist otherwise.

You're still trying to define the odds for any ONE event. Of course it's still 50% for each and every coin flip. If you flip the coin ONCE, odds are only 50% that you're going to hit a win (heads), whereas if you flip the coin 100 times the odds GREATLY IMPROVE that you're going to get a win. See, in your example again, you're talking about TOTAL WINS (heads) and of COURSE it's going to average out to 50% over a large sample. We only need ONE win (1 head) and sorry to say, the more times I flip a coin, the better odds I have of eventually getting a head (a win). It's simple common sense. We're not talking about rocket science and although probability and statistics was fun in college, this is a simple common sense math problem.

 

 You're correct, this is a simple common sense problem. The error in logic lies in the fact that there is no cummulative effect. This is where certain equations fail. To use the previous 'Coin Toss' example, what you're saying (correct me if I'm wrong) is that each time you flip the coin the odds get better of you getting a heads. Yes? So as you flip the coin and your odds get better and better and better, eventually, the odds will reach a point where they can no longer deny you and give up that elusive heads. And then you will continue to get heads over and over because you have flipped and flipped until the odds are overwhelmingly in your favor. And yet they are 1:2. They are 1:2 of getting heads if you flip it once, they are 1:2 of getting heads if you flip it 100 times (as you said, "...of COURSE it's going to average out to 50% over a large sample."). By the way, probability and statistics were never fun.

No, they don't get better and better for each INDIVIDUAL event. For each INDIVIDUAL event, the odds are still 50%. Taken as 100 flips the odds of winning are no longer 50% of a win, they're now approaching 100%. (That is getting ONE win in ONE HUNDERD tries). Now instead of only having a 50% chance of winning if I enter (flip one coin) once, because I entered 100 sweeps (flipped 100 times) my odds now approach 100% of winning at least once.

I think we're both right in what we're saying. You are only looking at an individual event whereas I'm looking at many events together, even though each invidual event is independent of each other.

Sorry for the confusion with my post -- I think I figured out how to explain it better.  There are really two separate and distinct concepts:

The odds applies to one distinct event: drawing one card, one spin of the roulette wheel, one sweepstakes.

The second concept applies to combining the odds of multiple distinct events: drawing one card over X number of trials, X number of spins of the roulette wheel, X number of sweepstakes.

 

The key point is that the second concept doesn't affect the first set of odds at all, but rather, it gives you the likelihood that an event will have occurred/not occurred after X number of trials. 

To use flipping a coin as an example: 

 1 flip has resulted in heads

• The odds says the odds of the next flip coming up heads is 1 in 2.

• The second concept looks only at the flip that have occurred and says:  "Yeah, it was 50/50 odds".

 5 flips has resulted in 5 heads

• The odds says the odds of the next flip coming up heads is 1 in 2.

• The second concept looks only at the flips that have occurred and says:  "It's somewhat unlikely that you would've gotten 5 heads in a row."

 1000 flips has resulted in 1000 heads

• The odds says the odds of the next flip coming up heads is 1 in 2.

• The second concept looks only at the flip that have occurred and says:  "You might want to check to make sure that's not a two headed coin, because this is getting outside the realm of possibilities."

 This is really cutting in to my sweep entry time!
I think the main difference is most people subscribe to probability theory instead of a more dynamic probabilistic logic.
The likelihood that an event will have occurred/not occurred after X number of trials is directly governed by the odds of it occurring/not occurring. So while the two concepts are distinct, they are not separate. As you so humorously pointed out, "You might want to check to make sure that's not a two headed coin, because this is getting outside the realm of possibilities.". Too true, because we understand the odds are 1:2 each time. It's more likely to be a 50/50 split because those are the governing odds of each flip.

I think this has gone far beyond wagging one's tail or shaking one's head - The most simplistic fact is, if you don't enter 1 time, or if you don't enter 100 times, you are not going to win even once. I enter one time, daily, weekly, or monthly as the rules permit. I do not enter multiple daily entry sweeps, If i win, that's great, if I don't then congrats to the person(s) that did.

Jim

You don't stop laughing when you get old, you get old when you stop laughing!

Ouch, my head hurts from reading the odds and trying to determine the statistical probabilities.  I prefer to keep it short and simple:

Enter, enter often, and sit back and wait for your name to be drawn or not drawn.

The only thing that is guaranteed is "don't entry" and you will not win.

marli, I'm with you.

It's all Luck.  Mr. Sweepy is very Lucky as are some of the other people on here and they win a lot. I keep trying but it's getting to be a lost cause.

The bottom line is....Persistance. I just won the Grand Prize on Dr Pepper/Arby's Academy of Country Music Awards sweepstakes, and I play diligently everyday and every chance I get. I have won alot of sweeps and instants and have only been doing this for about a year. I spend 3 hours or better everyday, entering every one I can. Persistance! It will eventually payoff.

Again you got lucky.  I enter persistently EVERY DAY.  Spend hours on here and nothing in 17 months now.  I don't even, at this point believe half the crap about people winning over and over and over!!!!!!!1

Your odds for each sweepstakes is independent thus it is true that your odds of winning any given sweeps is low but if you enter "enough of them" then your odds could be a lot better for winning ANY given sweepstakes.  

 Let's take an example, if the odds of winning a sweeps is 1 in 100 then your odds of winning that sweeps is very low.  If you enter 100 sweeps then your odds of winning 1 of them is quite good but there is no way to state which.  In fact, you are favored to have at least 1 win.

 To find the value of your sweeps entry you take the total prize pool, let's say it is $100 and there are 100 entries and determine how many times you'd have to enter to get an expected win which would be 100.  In this case, it would be a par to "insure" you against a win at $1 (plus whatever premium is used).  So you should be "willing" to spend not more then $1 to enter this contest.  This is just an average, it might take you 200 or 300 entries before you win.  In a random distribution there would probably be "somebody" who entered several hundred times that did not win.

Think about this, every entrants odds of winning a sweeps that has 100 million entrants is virtually zero but it is guranteed that 1 of them will win.  If the prize pool is 100 million dollars then each entry is worth $1. Here is another way to look at it, if we do not count cost of time to enter, and you enter a sweeps with a 1 million payout with avg 200 million entrants then if you played this game "forever" then your expected winnings would be near infinite.

 So, what everyone want is a HIGH ARV and a low # entries.   If you want to figure out the value of your entries at par, let's say you've entered sweeps with 25,000 entrants on average with a $25,000 ARV and you've entered 1,000 sweeps then expected value of your entries is $1,000. So let's say you did this every year then you would be expected to make 1k per year sweeping.  Your odds of winning any given sweeps is 1 in 25,000.  Your odds of winning ANY sweeps over the year is 1 in 25. So you can see it all makes sense and everything adds up.  

If we take the Sweepy's example of 1 in 25 for the year of winning a big prize and we had say 100 people who used this strategy.  Most of these people should at the 25 year mark have winnings around 25k.  Some will have winnings more then 25k in the first couple years. The bad news is that in a random distribution some people might have to play for 75 or 100 years before they reached the 25k.  Someone might say well what if I won 25k the first year then should I quit playing while ahead? Not really, the odds are not changed.  So, when people say I am lucky or unlucky then if you know the expected value "theoretical value" of all of your entries compared to your winnings then that is a way to see. This is similar to concept to "Sklansky bucks" concept in poker. The important thing is to see if there is a positive exp value or a negative exp value in any case to see if it is wise to play that game. Sweeps should always have positive ev.

Examples

I have a report that the overstock $10,000 bailout had 250,000 entries. Thus we know that each entry has a probability 1 out of 250,000 of winning (as the total probabiltiy must be 1). Thus each entry was worth 4 cents.This tells us what we should be willing to pay for each entry (the value of them would go down) if hypothetically we were offered that option.   This can also be used to show us how much total added value you get when you enter a sweeps multiple times.  Say if you entered that sweeps 100 times then your total equity was around $4. 

Let's take the HGTV Green Home Giveaway, I think it had 16 million entries and it was worth around 750k -- I'll round the prize to a million.  Thus each entry was worth around 6 cents.

What if they had given you the option to buy $1,000 worth of entries at par value which would be approximately 17,000 entries?  Should you do that?  Well, your probability of winning the contest is still very miniscule at .001% probability. Mathematically the equity works out but you'd need at least a million dollars and probably more to ever have a decent chance of coming out -- and the ability to play the game at least a thousand times.

Let's look at the case were the prize pool is split equally among 10 places. Your odds are 10x greater of winning a prize but I guess the equity drops to 1/10th. So, oh well, but with more prizes you should have a smoother equity curve.

Since you don't have to pay for entries to sweeps, instead of having a "bankroll" you have a time equity.  Imagine if you were to enter sweeps eday to hit the 1 in 25 odds per year of hitting a big prize (10k or better).  You'd need at least 25 to 50 years to have a "good" chance of redeeming the exp value of your sweeps.  So, just like it would not make sense to buy $1,000 worth of tickets at "fair value" unless you were allowed and had the bankroll to play thegame -- then it may not make sense to play a lot of sweeps if your not going to be hitting a winning probability that you can be happy with.

It's also possible to figure up your exp. hourly wage/rate by multiplying expected value of a sweepstakes entry by the number of sweeps entered per hour.  Example, if average sweeps entry has a value of 4 cents and you enter 100 per hour then hourly rate is $4 per hour, expected.

What does all this mean? Be lucky? Too much time on my hands.  Find sweeps with a higher exp rate then I have.