mid-Ohio United States Member #9 March 24, 2001 19821 Posts Offline

Posted: April 8, 2011, 8:29 pm - IP Logged

Quote: Originally posted by JADELottery on April 8, 2011

Alright, using your example of hitting a target we’ll setup a simple expression of what we are proving by generalizing a similarity.

First, we have the target shown below. The target is the total possible outcomes for a lottery of N numbers taken R at a time.

Next we have an arrow that represents a selection that is to be made from that total possible set of outcomes. The selection could be a Play or it could be a Draw; either way, they are the same R selections from N numbers.

When it hits the target, it selects an outcome that can be expressed as a smaller area around the arrow shown below. That area represents one of the possible outcomes or the R selections from N pool of numbers.

When two arrows have overlapping areas there is said to be a match between the two selections. The following image shows what this looks like on our target.

We can setup an example of our two different types of strategies of playing 10 bucks. Below are the two different examples.

From those two examples, can you tell me which is which; Either 10 Plays / 1 Draw or 1 Play / 10 Draws?

You could not say exactly which is which based on those hits shown. That’s what the probability, (Q * C(R, M) * C(N – R, R – M)) / C(N, R), is generalizing and it is equally applicable to both examples of 1 Play / 10 Draws and 10 Plays / 1 Draw.

They both look the same because there is one target and 10 arrows in each circle (10 plays per one draw). If you are representing 10 arrows shot at 10 different targets (one play per draw for 10 draws) then you should have 10 large circles with one target and one arrow in each. That might be interpreted as hitting the target is more likely if all ten arrows were shot at the same target since it allows more misses of the same target.

* you don't need to buy more tickets, just buy a winning ticket *

They both look the same because there is one target and 10 arrows in each circle (10 plays per one draw). If you are representing 10 arrows shot at 10 different targets (one play per draw for 10 draws) then you should have 10 large circles with one target and one arrow in each. That might be interpreted as hitting the target is more likely if all ten arrows were shot at the same target since it allows more misses of the same target.

Your statements have many flaws.

They both look the same because there is one target and 10 arrows in each circle (10 plays per one draw).

Incorrect because the large circle is not a draw, it is the total possible outcomes and each arrow is a selection or sample from those outcomes. The draw is an arrow just like your play is an arrow.

If you are representing 10 arrows shot at 10 different targets (one play per draw for 10 draws) then you should have 10 large circles with one target and one arrow in each.

A misperception as it relates to the first incorrect assumption. There cannot be more than one target, the target is all the possible combinations that can be selected.

That might be interpreted as hitting the target is more likely if all ten arrows were shot at the same target since it allows more misses of the same target.

Your conclusion is therefore based on flawed logic and has not established the proper relationship to the 1 Play / 10 Draws and 10 Plays / 1 Draw Probability.

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I'll have to start over and read this thread again and then maybe I'll have a clearer understanding of your graphs.

RJOh

RJOh,

I think I know what might be keeping you from following Doug's logic. When you think of betting on 10 Draws, 1 Ticket each, you probably subconciously think of time, that the scenario will take 10 days to complete. On the other hand, when you think of purchasing 10 Tickets for 1 Draw, your mind sees it happening in 1 day. This can be confusing when trying to equate the two scenarios.

To make things simpler, because of parimutel complications, let's assume you buy ten unique Tickets for each of the 2 scenarios. For the 10 Tickets/1 Draw scenario, just imagine buying all 10 Tickets on the same day for one game in one state. Now you have 10 Tickets with odds for each component as outlined by Doug at the outset. Now comes the mind trick. For the 10 Draws/1 Ticket each scenario, just assume you purchase all $1 Tickets in 10 different (5,39) games in 10 different states ON THE SAME DAY! What are the odds for these 10 Tickets? The same as the other 10!

United States Member #75358 June 1, 2009 5345 Posts Offline

Posted: April 9, 2011, 1:04 am - IP Logged

After much thought I came to the following conclusion, and i'm sticking with it. Unless someone can convince me otherwise, this makes much sense.

The scenario shows one side having 10 bets in one draw, compared to one bet a draw for 10 consecutive draws. Here's my solution to this. I think the best way is to use simpler games like the P-3 with a 1000 to 1 odds for a straight bet to better illustrate this.

I believe that the scenario that involves 10 bets for one draw seems to be the best one, and I'll explain why.

If you exaggerate an enigma to a higher degree, sometimes it makes more sense.

To apply this to the p-3, one must know that it's a 10 to 1 ratio (10 bets or 1 bet)...10 to 1.....

Let's exaggerate the 10 bets, but nine times it's weight. Now the figure is 90 bets. Let's go one step further and make the nine's weight ten times heavier making it 900.

The "1" in the 10 to "1" ratio is now 90 bets.

So now we have a better picture, a more exaggerated picture. Sort of looking at something through a magnifying glass.

Now ask the same question, but instead of saying 10 bets in a draw, or one bet in each of 10 draws..... You should say.....900 bets in one draw, or 90 bets in ten draws.

Who in their right mind would choose 90 bets in each of 10 draws?.....RIGHT??????

Knowing that there would be a very good chance that I'd win that 500 bucks with 900 bets in one draw, Id be a total fool to pick the 90 bets for 10 separate draws....don't you think?

Please show me the flaw in that logic.

Everything is still the same except that it's proportionally higher in count. 10 is now 900, and 1 is now 90.

900 to 90 is the same as 10 to 1.....They're both 10 times higher than the other but the only difference is the story they tell.

United States Member #75358 June 1, 2009 5345 Posts Offline

Posted: April 9, 2011, 1:27 am - IP Logged

Ran out of time to edit, but I wanted to say that's it not exactly 100% proportional, but close enough. Pretty much a 10 to one ratio compared to a 9 to 1....Just wanted to clear that up....lol

Texas United States Member #92330 June 5, 2010 887 Posts Offline

Posted: April 9, 2011, 3:40 am - IP Logged

Quote: Originally posted by joker17 on April 9, 2011

Ran out of time to edit, but I wanted to say that's it not exactly 100% proportional, but close enough. Pretty much a 10 to one ratio compared to a 9 to 1....Just wanted to clear that up....lol

Having not won with ten lines in a draw and not having won with one line in ten draws, I think I would go with ten draws with one draw each. At least that way I still have something to look forward to, rather than having played my ten lines on one draw without winning and continuing with the next nine drawings thinking "what if".

mid-Ohio United States Member #9 March 24, 2001 19821 Posts Offline

Posted: April 9, 2011, 11:50 am - IP Logged

Quote: Originally posted by jimmy4164 on April 9, 2011

RJOh,

I think I know what might be keeping you from following Doug's logic. When you think of betting on 10 Draws, 1 Ticket each, you probably subconciously think of time, that the scenario will take 10 days to complete. On the other hand, when you think of purchasing 10 Tickets for 1 Draw, your mind sees it happening in 1 day. This can be confusing when trying to equate the two scenarios.

To make things simpler, because of parimutel complications, let's assume you buy ten unique Tickets for each of the 2 scenarios. For the 10 Tickets/1 Draw scenario, just imagine buying all 10 Tickets on the same day for one game in one state. Now you have 10 Tickets with odds for each component as outlined by Doug at the outset. Now comes the mind trick. For the 10 Draws/1 Ticket each scenario, just assume you purchase all $1 Tickets in 10 different (5,39) games in 10 different states ON THE SAME DAY! What are the odds for these 10 Tickets? The same as the other 10!

I hope this helps.

--Jimmy4164

For the 10 Draws/1 Ticket each scenario, just assume you purchase all $1 Tickets in 10 different (5,39) games in 10 different states ON THE SAME DAY! What are the odds for these 10 Tickets? The same as the other 10!

Thanks for your efforts to help me understand Doug's logic but the above scenario wasn't in the question. it was a 10/1 or 1/10 in the same game. I think I understand his logic now.

It turns out both scenarios are 10 trys to match a combination but from opposite directions. In the first the player actively seeks the target with ten trys and in the second the player is the target and he hopes one of the 10 winning numbers finds him.

This could cause me to rethink my logic of not playing a set of numbers more than once.

* you don't need to buy more tickets, just buy a winning ticket *

For the 10 Draws/1 Ticket each scenario, just assume you purchase all $1 Tickets in 10 different (5,39) games in 10 different states ON THE SAME DAY! What are the odds for these 10 Tickets? The same as the other 10!

Thanks for your efforts to help me understand Doug's logic but the above scenario wasn't in the question. it was a 10/1 or 1/10 in the same game. I think I understand his logic now.

It turns out both scenarios are 10 trys to match a combination but from opposite directions. In the first the player actively seeks the target with ten trys and in the second the player is the target and he hopes one of the 10 winning numbers finds him.

This could cause me to rethink my logic of not playing a set of numbers more than once.

"This could cause me to rethink my logic of not playing a set of numbers more than once."

It could cause me to drink.

And for that I'm thankful.

.

"The only thing necessary for evil to triumph is for good men to do nothing"