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# If you had 10 bucks to bet...

Topic closed. 132 replies. Last post 6 years ago by JADELottery.

 Page 3 of 9

What do you think is the better bet?

 1 Play / 10 Draws [ 58 ] [64.44%] 10 Plays / 1 Draw [ 32 ] [35.56%] Total Valid Votes [ 90 ] Discarded Votes [ 3 ]
The Quantum Master
West Concord, MN
United States
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December 7, 2001
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 Posted: April 7, 2011, 12:15 pm - IP Logged

For review, http://en.wikipedia.org/wiki/Probability

We'll get to the answer after work today.

I'm on lunch break now.

Just look over the web page; you don't have to fully understand it, just review it.

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Order is a Subset of Chaos
Knowledge is Beyond Belief
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Jehocifer

Kentucky
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 Posted: April 7, 2011, 12:58 pm - IP Logged

tickets or chances per draw  10
possible combos of 5/39 numbers = 575757
MATCHES   ODDS              WINNING COMBOS    EXPECTED WINNERS
5/5    1 : 575757           1                    0.0000173684
4/5    1 : 3387             170                  0.0029526345
3/5    1 : 103              5610                 0.0974369395
2/5    1 : 10               59840                1.0393273551
_______________________________________________________________________
overall odds are 1 : 8.7           1.1397340000 total expected winners

True, adding a few more decimal places would change the results and show no differences in the two betting scenarios but usually 2 decimal places is close enough for my purposes.  As JADE said that's not the right way to prove it so I'll wait for The Mathematical Alpha Geek to post his calculations.  Thanks for the head-ups.

RJOh

The chart proves by betting the \$10 one time, the player can expect to match 2 numbers and because they should, it sightly increases the chances of matching 3, 4, or 5 numbers. However while the chances are slightly better, they still can only expect to get one 2 number match.

I'm assuming the intentions of the players are to win the jackpot so if the \$10 better craps out on the first drawing, the \$1 better still has 9 more chances to win. If the number of threads on "how I'll spend the winnings" are any indication, betting \$1 ten times will probably win this poll.

If mathematically one method is better than the other, I'll defer to the Alpha Geeks too.

NY
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 Posted: April 7, 2011, 7:48 pm - IP Logged

It is all about time. You can buy 1000 Pick 3 Tickets for one draw or you can buy 1000 and play for 1000 draws (quite a long time). One of them your going to win. The other you have the chance of winning 1000 times....or never. lol. To increase odds you always bet more on a single game. Even though each ticket has the same single chance of winning (1:1000) your chances have increased because you hold more than one ticket.

I guess that was a logical deduction. I don't know how to write out a formula, but I'm sure it's a simple one.

That's one way to  figure it out as a simple exercise in logic. If you play all 1000 possibilities you've got a 100% chance of winning. If you play 1 possibility 1000 times your chance of losing is .999 raised to the 1000th power, which is .3677, or 36.774%. meaning your chnce of winning is only 63.23%.

Here's another way to think about it without  doing the math. It's obvious that  playing a single ticket in multiple games gives you a chance of winning multiple times. There's no such thing as a free lunch, so what does the chance of winning multiple times cost? The cost is a reduced chance of winning even once.

Let's start small and play pick 1, a game that would have odds of 1 in 10 if anybody actually offered it, with two tickets.  If you buy one ticket for each of two games you've got a 90% chance of losing each time. That means the chance of losing both times is .9 * .9 = .81, so your chance of winning is .19 or 19%. Your chance of winning both times is .1 * .1 = .01.  When you play 2 tickets on a single game you've got 2 chances in 10 to win, which is 20%. You can't win twice, but you're 5% more likely to win. Of course that better chance of winning gets smaller as the odds of the game increase, but you'll always be trading the chance of winning multiple times for a slightly better chance of winning at all.

Just for fun, what happens after you find that your first of two pick 1 tickets is a loser? Doesn't that second ticket now gives you odds of 1 in 9? That's an 11% improvement over the 1 in 10 chance you'd have if that 2nd ticket was for a different drawing.

The Quantum Master
West Concord, MN
United States
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 Posted: April 7, 2011, 9:41 pm - IP Logged

Just have to get our Duckies in a row.

We'll post this in a few parts.

1 - The Calculation of Probabilities for matching a Pick to a Draw with Picking a Quantity of Plays.

2 - The Pick-Draw Symmetry.

3 - The Indeterminate Association Soloution.

Also, we'll relate this to RJOh's example.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
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 Posted: April 7, 2011, 10:03 pm - IP Logged

Just have to get our Duckies in a row.

We'll post this in a few parts.

1 - The Calculation of Probabilities for matching a Pick to a Draw with Picking a Quantity of Plays.

2 - The Pick-Draw Symmetry.

3 - The Indeterminate Association Soloution.

Also, we'll relate this to RJOh's example.

Correction: Just a typo, Pick should really be Play.

Just have to get our Duckies in a row.

We'll post this in a few parts.

1 - The Calculation of Probabilities for matching a Play to a Draw with Picking a Quantity of Plays.

2 - The Play-Draw Symmetry.

3 - The Indeterminate Association Soloution.

Also, we'll relate this to RJOh's example.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
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 Posted: April 7, 2011, 10:26 pm - IP Logged

First, we have to calculate the Probability of M - matching numbers from P - played quantity in D – drawn quantity within an N - Numbers pool. Below is a diagram of how this looks graphically. M is how many individual combination numbers played match the drawn numbers. P is how many numbers you are playing per combination. D is the quantity of numbers being selected for a draw. N is the quantity of numbered balls in the bin for the drawing.

The probability of matching is, where C(?, ?) is the Combination function:

The total combinations of P – Play quantity that can be taken in the possible ways of M –Matching numbers.

C(P, M)

Times, the total combinations of the remaining N – Numbers pool minus the P – Played quantity taken by the remaining D – Drawn quantity minus the M – Matching quantity.

C(N - P, D - M)

Divided by the total possible combinations of N – Numbers in the pool taken in D – Drawn ways.

C(N, D)

The probability of matching M numbers between a Play and a Draw is (C(P, M) * C(N – P, D – M)) / C(N, D).

To find the Probability of matching numbers between a Q – Quantity of plays and a Draw is (Q * C(P, M) * C(N – P, D – M)) / C(N, D).

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
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 Posted: April 7, 2011, 11:03 pm - IP Logged

Let's apply the (C(P, M) * C(N – P, D – M)) / C(N, D) probability to the 5 / 39 lottery pointed out by RJOh.

We won't reduce the fraction to a decimal value, we'll just leave it as an integer fraction.

Running through the number by substitution, we can assign some of the variables: P = 5, N = 39, D = 5.

Next we can work the M variable for {0, 1, 2, 3, 4, 5}, then see if the sum of the fractions (probabilities) equals 1.

 M (C(P, M) * C(N – P, D – M)) / C(N, D) 0 278256 / 575757 1 231880 / 575757 2 59840 / 575757 3 5610 / 575757 4 170 / 575757 5 1 / 575757

Now, add up the fractions (1 / 575757) + (170 / 575757) + (5610 / 575757) + (59840 / 575757) + (231880 / 575757) + (278256 / 575757).

Becomes, (1 + 170 + 5610 + 59840 + 231880 + 278256) / 575757.

Finally, this is 575757 / 575757 or 1.

This is just to verify that the calculation of Probability is correct.

The Sum of all possible probabilities is 1.

Side note:

- You might be wondering why we are not using Odds, because Odds is a ratio of Successful Outcomes to Failure Outcomes.

- You can find Odds using the Probabilities by the relationship of S + F = 1, where F = 1 - S, then the Odds is O = S / F or O = S / (1 - S).

Wthin this though, there is an unseen Symmetry.

Looking at P and D, we can see that they are both Equal, P = 5 and D = 5.

They are equal and will always be equal because the Quantity of numbers you pick in a single play has to be the same as the Quantity of balls drawn.

So, with this fact we can say that P = D and can we say that the Play Quantity is Symmetrical with the Draw Quantity.

Using this we can now set both P and D equal to a general variable R, P = D = R.

The probability value, (C(P, M) * C(N – P, D – M)) / C(N, D), becomes, (C(R, M) * C(N – R, R – M)) / C(N, R).

To that point, the Quantity Played probability then becomes, (Q * C(R, M) * C(N – R, R – M)) / C(N, R).

Keep this final probability in mind for the next post.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
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 Posted: April 7, 2011, 11:53 pm - IP Logged

Ok, now this gets a bit more abstract.

When we posted the poll, we could have said, “What’s the better bet? – 1 Wingdebing / 10 Zibdo  or  10 Wingdebing / 1 Zibdo”

In that, could you say for absolute certainty that a Wingdebing is or really determine what a Zibdo is?

I could have reversed the order from what you may think it is.

Ultimately, you can’t say what they mean, and I won’t say either because I don’t speak Jibber-Jab.

To that end, when we look at the probability, (Q * C(R, M) * C(N – R, R – M)) / C(N, R), the Q variable is not Associated with anything.

It can not, by the very essence of Quantum Mechanics, be Associated with any value of P and D, because, behold, we removed them and replaced it with R.

Now, can you say for absolute certainty what value R is only Associated with, P or D; I’d say not.

We can then say that in this state as an unsolved probability the Q value is in an Indeterminate Association.

It is only when we apply the probability that Q’s association is determined, and if P = D = R, then we could with the pervious Play-Draw Symmetry assign Q to the D value.

This means the Q - Quantity of D - Draws as it relates to finding the Probability of matching a single P – Play is the same as finding the Probability relating to the Q – Quantity of Plays matching a single D – Draw.

The best we could say is finding the probability of multiple Q – Quantity of some selection A from N items taken R at a time matching a single selection B from a similar set of N items taken R at a time is (Q * C(R, M) * C(N – R, R – M)) / C(N, R).

In that, we now have a generalized probability where A selection could be Plays or Draws and B selection could be a Draw or a Play, respectively.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

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 Posted: April 8, 2011, 12:58 am - IP Logged

Don't you want to expand on that by detailing all the 2nd tier prize probabilities in a lotto game?

Just kidding!

Nice work Doug!

The Quantum Master
West Concord, MN
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 Posted: April 8, 2011, 1:03 am - IP Logged

Don't you want to expand on that by detailing all the 2nd tier prize probabilities in a lotto game?

Just kidding!

Nice work Doug!

Thanks.

It's been a long night.

While I was doing this, I was also doing my laundry.

I have to check the final load now.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

mid-Ohio
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 Posted: April 8, 2011, 8:49 am - IP Logged

Now I'm more confused.  How would you use that formula to evaluate taking 10 random shots at an invisible target all at once verses taking 10 random shots one at a time at an invisible target that changes position after each shot?

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

The Quantum Master
West Concord, MN
United States
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 Posted: April 8, 2011, 11:20 am - IP Logged

Now I'm more confused.  How would you use that formula to evaluate taking 10 random shots at an invisible target all at once verses taking 10 random shots one at a time at an invisible target that changes position after each shot?

That's an intersting anecdotal summary of confusion.

In order to clear up the confusion, we need a common frame of reference.

So, please relate the following key words to the posts we made: shots, invisible, position.

How do you quantify or equate those to our post?

shots = ?, invisible = ?, position = ?

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

mid-Ohio
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 Posted: April 8, 2011, 12:35 pm - IP Logged

That's an intersting anecdotal summary of confusion.

In order to clear up the confusion, we need a common frame of reference.

So, please relate the following key words to the posts we made: shots, invisible, position.

How do you quantify or equate those to our post?

shots = ?, invisible = ?, position = ?

shots = picking lottery combinations to play

invisible = combination will be from any of 575,757 possible combinations but unknown at the time

position = combination different each drawing thus it's new/has positional change each time

Playing lottery is similar to trying to shoot a moving target in the dark.  Using a gun that can cover the whole area with a burst of unlimited fire power will hit the target for sure but if you only get ten shots, is it a better strategy to have a burst of ten shots or take one shot at a time?  If you don't know what you're trying to hit, does it make any differences?

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

The Quantum Master
West Concord, MN
United States
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 Posted: April 8, 2011, 5:25 pm - IP Logged

shots = picking lottery combinations to play

invisible = combination will be from any of 575,757 possible combinations but unknown at the time

position = combination different each drawing thus it's new/has positional change each time

Playing lottery is similar to trying to shoot a moving target in the dark.  Using a gun that can cover the whole area with a burst of unlimited fire power will hit the target for sure but if you only get ten shots, is it a better strategy to have a burst of ten shots or take one shot at a time?  If you don't know what you're trying to hit, does it make any differences?

Well, we are almost there.

We have something to go on.

We're getting something ready, so it may be a bit.

Hopefully, we can make a better understand of what we've shown as proof.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.

Order is a Subset of Chaos
Knowledge is Beyond Belief
Wisdom is Not Censored
Douglas Paul Smallish
Jehocifer

The Quantum Master
West Concord, MN
United States
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December 7, 2001
3675 Posts
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 Posted: April 8, 2011, 8:00 pm - IP Logged

Now I'm more confused.  How would you use that formula to evaluate taking 10 random shots at an invisible target all at once verses taking 10 random shots one at a time at an invisible target that changes position after each shot?

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.

Presented 'AS IS' and for Entertainment Purposes Only.
Any gain or loss is your responsibility.