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# frequencies of numbers in Pick 3

Topic closed. 9 replies. Last post 13 years ago by your wrong.

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S.Windsor, CT
United States
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May 4, 2004
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 Posted: May 6, 2004, 5:32 pm - IP Logged

Can somebody tell me what the std.dev. will be for the mean of 10 per number when drawing 10 000 times?

Australia
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December 22, 2003
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 Posted: May 9, 2004, 12:48 am - IP Logged

Bertil

I think you may have to rephrase what your asking. Standard Deviation (square root of variance) and 10,000 Draws if that's what you mean - I'm OK with.

However, "... will be for the mean of 10 per number ..." you've lost me.

Which Lotto game are you referring to?

Regards

Colin

S.Windsor, CT
United States
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 Posted: May 9, 2004, 3:52 pm - IP Logged

Colin,

I'm refering to a Pick 3 type game drawn 10 000 times in which #330 came up only 6 times instead of 10. The mean frequency of 10 must have std.dev. which I now think will be 3.16 if the binomial formula applies. If not, please tell me how to find the std.dev.

Bertil

Australia
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December 22, 2003
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 Posted: May 9, 2004, 11:16 pm - IP Logged

Bertil

Sorry, I mustn't of put my reading glasses on. I've quite deliberately have not got involved in discussing Pick3 or Pick4 games because I haven't done any work there. I have tended to look at them as three or four Power Ball's put together and the strategy to deal with that I leave to others at this stage.

In Access there are Functions STDEV, STDEVP, VAR and VARP which I have found totally useless in providing anything meaningfull to work with. It's just like the Sum of the occurrences over a large number of draws - should be all close to each other but useless to work anthing else out from. (apart from the performance of your RNG).

But it's Pick3/Pick4 - shoot me down. I won't bite back.

Regards

Colin

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Wheaton
United States
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May 21, 2004
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 Posted: May 21, 2004, 8:04 pm - IP Logged

Standard Deviation

Chance variation is conveniently measured by standard deviation, denoted by SD. Roughly, SD measures how far a typical occurrence of a random process will be from the average. For example, in many situations, about 68 percent of observations will be within one SD of the average. Often, the SD is difficult to compute, but there is simple formula for the SD of the number of occurrences of an dvent in repeated, independent trials, whether it be heads in coin tosses or "7" in dice rolls. If p is the chance of occurrence in one trial, then 1 - p is the chance of nonoccurrence, and the formula for the SD of, say, the number of occurrences in N trials is as follows:

SD=/N x P x (1-p)

Since the chance that a coin lands heads is 1/2, to find the SD for the number of heads in, say, 100 coin tosses, we get:

SD =/100 x 1/2 x 1/2=5

Since the average number of heads in 100 tosses equals 100 x 1/2 = 50, the observed number of heads in 100 tosses will typically be within one SD of the average, between 45 and 55.

When rolling a pair of dice, we have seen that the chance of rolling "7" is 1/6 and the chance of not rolling "7" is 5/6. Thus, for my computer simulation of 72,000 rolls,

SD = /72,000 x 1/6 x  5/6 = 100

Since the average number of occurrences of "7" in 72,000 dice rolls is 72,000 x 1/6 = 12,000, a typical number of "7"s in 72,000 rolls would be within one SD of the average, or between 71,900 and 72,100 occurrences. In my 10 computer simulations of 72,000 dice rolls, 8 were in the one SD range

San Diego
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May 1, 2004
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 Posted: May 21, 2004, 10:25 pm - IP Logged

I think "your wrong" must be working for Microsoft, writing help files, because the above post is perfectly accurate and utterly useless to Bertil's question.

United States
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April 24, 2004
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 Posted: May 22, 2004, 5:23 am - IP Logged

Yup dragon. I think i might of read that information from my probability book and everything matches word for word. lmao

S.Windsor, CT
United States
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May 4, 2004
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 Posted: May 22, 2004, 7:42 am - IP Logged
Quote: Originally posted by your wrong on May 21, 2004

Standard Deviation

Chance variation is conveniently measured by standard deviation, denoted by SD. Roughly, SD measures how far a typical occurrence of a random process will be from the average. For example, in many situations, about 68 percent of observations will be within one SD of the average. Often, the SD is difficult to compute, but there is simple formula for the SD of the number of occurrences of an dvent in repeated, independent trials, whether it be heads in coin tosses or "7" in dice rolls. If p is the chance of occurrence in one trial, then 1 - p is the chance of nonoccurrence, and the formula for the SD of, say, the number of occurrences in N trials is as follows:

SD=/N x P x (1-p)

Since the chance that a coin lands heads is 1/2, to find the SD for the number of heads in, say, 100 coin tosses, we get:

SD =/100 x 1/2 x 1/2=5

Since the average number of heads in 100 tosses equals 100 x 1/2 = 50, the observed number of heads in 100 tosses will typically be within one SD of the average, between 45 and 55.

When rolling a pair of dice, we have seen that the chance of rolling "7" is 1/6 and the chance of not rolling "7" is 5/6. Thus, for my computer simulation of 72,000 rolls,

SD = /72,000 x 1/6 x  5/6 = 100

Since the average number of occurrences of "7" in 72,000 dice rolls is 72,000 x 1/6 = 12,000, a typical number of "7"s in 72,000 rolls would be within one SD of the average, or between 71,900 and 72,100 occurrences. In my 10 computer simulations of 72,000 dice rolls, 8 were in the one SD range

S.Windsor, CT
United States
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May 4, 2004
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 Posted: May 22, 2004, 7:50 am - IP Logged

I used the same formula to calculate the std.dev. for 10 00 draws of a pick 3 game when I got 3.16 as the square root of 9.99. But the lottery expert at my state lottery claimed a more complex formula must be used, but he would not tell me its form.I'm confident the binomial formula applies here.

Bertil

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Wheaton
United States
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May 21, 2004
18 Posts
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 Posted: May 22, 2004, 8:23 am - IP Logged

The binomial probability distribution is a discrete distribution. An experiment qualifies as a binomial if there are exactly two outcomes, the probability of success remains the same from trial to trial, the trials are independent, and a fixed number of trials. The number of successes, probability of success, probability of failure, and number of trials must be known to calculate binomial probabilities. One common formula used for binomial probabilities is:

P(x successes) = Cn,x px qn-x where n represents the number of trials, x is the number of successes, p is the probability of success, and q represents the probability of failure.

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