Help me with a good comeback for this statement - Page 4 - PeachParts Mercedes-Benz Forum

Chas H · #46 07-28-2007, 12:44 AM

Quote:

Originally Posted by Botnst

You're probably thinking about the Gambler's Fallacy -- assuming that one random event can predict another. And you are correct that any event is separate and does not effect any subsequent event (if it is a fair system, but let's just assume it is).

Yess, they're independent events and there are rules for combining independent events to produce a combined probability.

Think of how the probability of the single event is computed -- as a product of fractions. repeating any single event will return the same probability for each single event. Just as a coin toss is always 50-50.

But what is the probability of getting only 1 head in 3 tosses? = 1/2 * 1/2 * 1/2= 1/8. And so forth.

What is the probability of getting at least 1 head in 3 coin tosses? = 1- (1/2 * 1/2 * 1/2) = 7/8ths chance.

And so on.

Now use the same logic with the lottery. It has a 1/300M chance of paying off for you in one event.

B

The coin toss is not relevent to the lottery because it is a single event. The lottery is 6 mutually exclusive events. That is, once a number has been picked it can not be picked again. The odds in tossing a coin are not mutually exclusive in that the same result could happen every time.

Matt L · #47 07-28-2007, 01:22 AM

Quote:

Originally Posted by Chas H

The coin toss is not relevent to the lottery because it is a single event. The lottery is 6 mutually exclusive events. That is, once a number has been picked it can not be picked again. The odds in tossing a coin are not mutually exclusive in that the same result could happen every time.

This is not correct.

You can make an ordering of every possible lottery outcome (i.e., a list). Then you can toss a coin enough times to create a binary number long enough to pick something from the list. That shows that tossing a coin is sufficient to the task of picking a lottery number.

The converse is easy. You look at the first ball drawn. If it's even, you call heads. If it's odd, you call tails. That shows that drawing a lottery is sufficient to the task of flipping a coin.

They're the same thing.

OMEGAMAN · #48 07-28-2007, 01:25 AM

Just tell her the only reason she buys lotto tickets is because her Pinto wont make it to central city.

Chris Bell · #49 07-28-2007, 01:34 AM

Quote:

Originally Posted by Botnst

I assume the lottery is based on those numbered balls.

Here's a test: http://www.webmath.com/lottery.html

I entered a simple lottery with the following constraints:

You must choose a sequence of 5 numbers correctly to win.
The lowest number you can choose is 2
The highest number you can choose is 55
A given number can only be chosen once per try (per lottery ticket, etc.)

The resultant chance of winning is 1/54 × 1/53 × 1/52 × 1/51 × 1/50 = 1/379,501,200 or one chance in about 380 million.

Now let's say the lottery ticket costs $5 and I buy 1 ticket per week, 52 tickets/ year for say, 20 years. 52 * 20 * $5 = $5,200 spent on tickets.

How has purchasing 1020 tickets improved my odds?

Basically, 380,000,000/1,000 = 380,000. In other words, after 20 years of buying weekly tickets my chances of winning are about the same as dying from a fireworks discharge in the USA (http://www.nsc.org/lrs/statinfo/odds.htm)

Somebody who knows how to compute compound interest from regular deposits can probably provide you with the 20 year investment return from $5/week. I'll bet it's $15k-$20k at say, the S&P 20 yr average change.

5 dollars a week at 12.9% (if my memory is correct here I believe that is the long term S&P 500 average) for 20 years works out to $24,498.53. Not to bad considering the actual cash deposits work out to only $5,200.

tankdriver · #50 07-28-2007, 09:17 AM

Quote:

Originally Posted by Matt L

The lottery's probability isn't similar at all. Fully half of the bets are siphoned off, leaving the rest to pay out to the winner.

That's like having half of the wheel showing numbers on which you can't even bet.

I wrote probably, not probability. As in the lottery probably also has a negative expectation like roulette.

Botnst · #51 07-28-2007, 09:59 AM

Quote:

Originally Posted by Chas H

The coin toss is not relevent to the lottery because it is a single event. The lottery is 6 mutually exclusive events. That is, once a number has been picked it can not be picked again. The odds in tossing a coin are not mutually exclusive in that the same result could happen every time.

What you are describing is the "Gambler's Fallacy" and you are absolutely correct -- in a fair system random events cannot affect each other. But if you think about it, the probability of a combined event is itself composed of the constituent random events.

On the previous page (beginning at #35, IIRC) we discussed this in agonizing detail using coin tosses as a logical simplification of the more complex lottery balls. But since the lottery balls system is itself composed of independent events, it is no different in principle from a coin toss.

(We'd just need more coins to replicate it -- like representing a decimal number or operation using binary numbers -- the operations and values are equal.)

Chas H · #52 07-28-2007, 10:46 AM

Quote:

Originally Posted by Matt L

This is not correct.

You can make an ordering of every possible lottery outcome (i.e., a list). Then you can toss a coin enough times to create a binary number long enough to pick something from the list. That shows that tossing a coin is sufficient to the task of picking a lottery number.

The converse is easy. You look at the first ball drawn. If it's even, you call heads. If it's odd, you call tails. That shows that drawing a lottery is sufficient to the task of flipping a coin.

They're the same thing.

No they are not. Once a numbered ball is picked out it can not be picked again. The choices of numbers is reduced by 1 every time one is picked. The odds are computed as -N being the number of balls- 1/N X 1/N-1 x 1/N-2 and so on.
The odds in a coin toss are 1/2 X 1/2 x1/2.
They completely different actions and have completely different calculations for the odds.

Chas H · #53 07-28-2007, 10:50 AM

[quote=Botnst;1575868]What you are describing is the "Gambler's Fallacy" and you are absolutely correct -- in a fair system random events cannot affect each other. But if you think about it, the probability of a combined event is itself composed of the constituent random events.quote]

No need to think about it, the method of computing the odds in a pick six game has been established for a very long time. As has the method for computint the odds in a coin toss. The 2 methods are very different because the 2 events are very different.

Botnst · #54 07-28-2007, 10:53 AM

Quote:

Originally Posted by Chas H

No they are not. Once a numbered ball is picked out it can not be picked again. The choices of numbers is reduced by 1 every time one is picked. The odds are computed as -N being the number of balls- 1/N X 1/N-1 x 1/N-2 and so on.
The odds in a coin toss are 1/2 X 1/2 x1/2.
They completely different actions and have completely different calculations for the odds.

I'm not sure about this. What is the difference between a coin toss and a numbered ball selection? You cannot choose the coin a second time, either.

B

Chas H · #55 07-28-2007, 11:21 AM

Quote:

Originally Posted by Botnst

I'm not sure about this. What is the difference between a coin toss and a numbered ball selection? You cannot choose the coin a second time, either.

B

Sure you can, that's what is being done. The odds of tossing three heads in a row are the odds of each event multiplied together IOW 1/8. The action uses the same coin.
In the pick 6 the choices are reduced by one every time a ball is picked.
These two actions are common and basic to statistics.

Botnst · #56 07-28-2007, 12:23 PM

Quote:

Originally Posted by Chas H

Sure you can, that's what is being done. The odds of tossing three heads in a row are the odds of each event multiplied together IOW 1/8. The action uses the same coin.
In the pick 6 the choices are reduced by one every time a ball is picked.
These two actions are common and basic to statistics.

Okay, we agree on that.

Would you say that the computation follows in the same manner for a die (a six-sided object)?

Matt L · #57 07-28-2007, 01:12 PM

Chas, where is the fallacy in my proof above? Unless you can refute it, I think the equivalence stands.

Chas H · #58 07-28-2007, 02:13 PM

Quote:

Originally Posted by Matt L

Chas, where is the fallacy in my proof above? Unless you can refute it, I think the equivalence stands.

I think I explained it rather well to B. and that explanation will have to be to you also.

Botnst · #59 07-28-2007, 02:16 PM

Quote:

Originally Posted by Chas H

I think I explained it rather well to B. and that explanation will have to be to you also.

I think that Matt & I believe that your argument is flawed or that we don't understand it. Further, I believe that Matt & I are in agreement and so we are puzzled. So we are taking two approaches to trying to understand it.

B