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Any Mathematicians Here? - 3/16/2007 4:45:23 PM

subfever

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What are the mathematical odds of any 4 selected and boxed (boxed = any random order of draw) numbers being drawn from a total field of 8 numbers?

What is the formula to arrive at the correct answer?

(I'm asking for the formula because I'm also interested in knowing the following mathematical odds of the above scenario: 4 of 9, 3 of 8, 3 of 9.)

Thanks in advance!

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RE: Any Mathematicians Here? - 3/16/2007 4:52:07 PM

farglebargle

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http://en.wikipedia.org/wiki/Mathematical_combination

http://en.wikipedia.org/wiki/Binomial_coefficient

8 choose 4 = 70 ( google )

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RE: Any Mathematicians Here? - 3/16/2007 4:53:17 PM

Bottomlftop

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Could you try clarifying the game? Odds aren't the same as calculating a combination (in this case they're close). That's just the number of different combinations of 4 items you can get out of 8 with order not mattering; the chance of drawing a correct 4 out of 8 is 1/2 * 3/8 * 2/8 * 1/8 = 6/1024 = 1/171 with replacement though I don't think that's exactly what he's looking for. 1/2 * 3/7 * 1/3 * 1/5 = 1/70 is without replacement (he doesn't specify) which is the reciprocal of the combination : ) Those are actually the probability of them happening. The odds are 1 to 170 and 1 to 69 respectively.

Edit ARGH kept screwing up it should be right now. Hahaha... I've had to edit this so many times my head isn't on the ball today...

The Reasoning: If your always putting the number drawn back in the box then the first draw has a 4 in 8 (1/2) chance you'll get one of them; For each one after there's one less correct choice (unless you can try to select the same number twice) so it's 3/8 then 2/8 then 1/8. Without replacement your looking to get a specific combination of 4 out of 8 so it's one out of 4C8(number of combinations) which is equal to 70 or (4/8 * 3/7 * 2/6 * 1/5) After each draw attempt there's one less correct draw and one less possible one to draw.

< Message edited by Bottomlftop -- 3/16/2007 5:43:35 PM >

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RE: Any Mathematicians Here? - 3/16/2007 5:47:30 PM

subfever

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quote:

Could you try clarifying the game?

Absolutely!

Let's say there are 8 marbles in a jar, numbered 1 through 8.

The jar is shaken, and you reach in the bowl blindfolded to select 4 marbles.

What are the mathematical odds that you are now holding, for example, numbers 1, 2, 3, and 4?

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RE: Any Mathematicians Here? - 3/16/2007 5:55:44 PM

Bottomlftop

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quote:

ORIGINAL: subfever

quote:

Could you try clarifying the game?

Absolutely!

Let's say there are 8 marbles in a jar, numbered 1 through 8.

The jar is shaken, and you reach in the bowl blindfolded to select 4 marbles.

What are the mathematical odds that you are now holding, for example, numbers 1, 2, 3, and 4?

Ok so if you read the part about without replacement you'll have it : ) Probability is 1 out 70. Odds are 1 to 69. My attempts at math today have been pretty rough... so you might want further verification.

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RE: Any Mathematicians Here? - 3/16/2007 5:56:48 PM

driversoft

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The odds of holding 1,2,3 & 4 (or any specific set) must be 1 in 2 (as there are eight numbers, and any of your choice of 4 will do to start with) x 3 in 7 x 2 in 6 x 1 in 5, or (1 x 3 x 2 x 1) / (2 x 7 x 6 x 5), or 6 in 420, or 1 in 70.

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RE: Any Mathematicians Here? - 3/16/2007 5:58:02 PM

Real0ne

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quote:

ORIGINAL: Bottomlftop

quote:

ORIGINAL: subfever

quote:

Could you try clarifying the game?

Absolutely!

Let's say there are 8 marbles in a jar, numbered 1 through 8.

The jar is shaken, and you reach in the bowl blindfolded to select 4 marbles.

What are the mathematical odds that you are now holding, for example, numbers 1, 2, 3, and 4?

Ok so if you read the part about without replacement you'll have it : ) Probability is 1 out 70. Odds are 1 to 69. My attempts at math today have been pretty rough... so you might want further verification.

thats right according to my calculator too....

here is another way to set it up:

Permutations i get:

                        8!                  8!   8*7*6*5*4*3*2*1                       40320
8P4       =       -----       =       ----       =       --------------------------          =    ----------    =        1680
                     (8-4)                 4!                               4*3*2*1                          24

then combinations: 8C4 = 1680 / 24 = 70

Permutations:

                 n!
n_P_k = --------
              (n - k)!

Combinations:

                  n!
n_C_k = ----------
              k!(n - k)!

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RE: Any Mathematicians Here? - 3/16/2007 5:59:43 PM

Bottomlftop

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quote:

ORIGINAL: driversoft

The odds of holding 1,2,3 & 4 (or any specific set) must be 1 in 2 (as there are eight numbers, and any of your choice of 4 will do to start with) x 3 in 7 x 2 in 6 x 1 in 5, or (1 x 3 x 2 x 1) / (2 x 7 x 6 x 5), or 6 in 420, or 1 in 70.

Odds are the number of successful possibilities vs the number of unsucessful ones. So it's not 1 in 70 (the probability) but 1 to 69.

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RE: Any Mathematicians Here? - 3/16/2007 6:04:26 PM

subfever

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quote:

ORIGINAL: farglebargle

http://en.wikipedia.org/wiki/Mathematical_combination

http://en.wikipedia.org/wiki/Binomial_coefficient

8 choose 4 = 70 ( google )

Thanks, I'll have to study those links.

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RE: Any Mathematicians Here? - 3/16/2007 6:17:24 PM

farglebargle

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That's what I SAID!

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RE: Any Mathematicians Here? - 3/16/2007 10:14:38 PM

subfever

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Wow... lots of replies came within a very short time span. I didn't even see half of them when I posted my prior message.

Thanks to all!

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RE: Any Mathematicians Here? - 3/16/2007 11:13:00 PM

Termyn8or

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If you thought a picture was worth a thousand words, how many are a formula worth ?

OK, the first thing to realize when you box a four digit event like that, IIRC you are making 24 bets at once.

The thing here is that what you wrote indicates that the win is contingent on ALL of the numbers coming in. This leaves you at a certain disadvantage, because if three of your numbers come in, you get nothing.

BTW, what you mentioned is a nine digit game, there are ten digits, and you wrote 1 through 9, I am assuming you mentioned it for a reason. Whatever game this is, OK.

This makes it more complex to figure out the actual odds, when it is like a four digit lottery, it is easy to figure out. With four digits there are 10,000 possible outcomes, and if you bet on one of the your odds are 10,000 to 1. If you box your bet, your odds are 10,000 to 24. That is about 416.66 to 1.

If the game is limited to nine digits, the odds change. First of all your odds are no longer 10^4 for any given four digit straight, it is 9^4, which is 6,561 to 1. If you box it, you similarly make 24 bets at once and the odds drop to 273.375 to 1.

Now you wrote what you wrote, you did not say whether individuakl results were exclusive, that is like in a lotto type drawing. The digit 1 is like a "digit" 47 in such a game, the same number cannot come up twice in the same result. That changes the structure again.

Now if this is true, your base odds of any digit coming in straight is hard to figure. Purely mathematically it would seem to be 4 times 3 times 2, 24 to 1. However if your bet is boxed you do not get the total advantage of making 24 bets at once. This is because some of the 24 outcomes cannot happen. Also the bettor may be barred from betting the same number twice in the same draw in such a game, it couldn't win anyway.

So, if the game has the exclusivity of results of a lotto game, the odds really are that low. You actually have around a 50/50 chance of picking four numbers from a field of nine. But it is not 4 in 9, it is a bit more complicated than that.

So, that is the crux of it, can, for example "2222" come up or not ?. That changes things quite a bit in a game like this. Need to know.

T

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