I do not understand very well, but I speak of this formula, have you tried? The formula clearly says that the exact result is [TEX]\frac{1}{3}[/TEX] and should move p1 and p2 until it approaches [TEX]\frac{1}{3}[/TEX].


[QUOTE=Godzilla;448569][B]Now the new formula is :[/B]

[TEX]p1 * p2 = product[/TEX]

[B]formula :[/B]


[TEX]\frac{(p1)\pi}{product} * \frac{p2}{3\pi} = 0.33333333 = \frac{1}{3}[/TEX] [B][U]always[/U][/B] [TEX]\frac{1}{3}[/TEX]

or this (it is equal)

[TEX]\frac{(p2)\pi}{product} * \frac{p1}{3\pi} = 0.33333333 = \frac{1}{3}[/TEX] [B][U]always[/U][/B] [TEX]\frac{1}{3}[/TEX]


And i think that in few test is possible to know the factors [TEX]p1[/TEX] and [TEX]p2[/TEX] , because the result is always [TEX]\frac{1}{3}[/TEX] .

What do you think about it ?

thanks[/QUOTE]

[QUOTE=Godzilla;448586]I do not understand very well, but I speak of this formula, have you tried? The formula clearly says that the exact result is [TEX]\frac{1}{3}[/TEX] and should move p1 and p2 until it approaches [TEX]\frac{1}{3}[/TEX].[/QUOTE]

the pi's cancel out the p1*p2 and product cancel out you get left with 1/3 regardless of p1 and p2 assuming their product is not 0.

I realized, but you did you realize that we can get 0,33 or 0.0033 anyway?

Now

[TEX]\frac{n}{product*3} = 0,33 or0,0033 or 0,00033 etc...[/TEX]

Now [TEX]n[/TEX] is near at [TEX]p1 [/TEX] or [TEX]p2[/TEX]

[QUOTE=Godzilla;448590]I realized, but you did you realize that we can get 0,33 or 0.0033 anyway?

Now

[TEX]\frac{n}{product*3} = 0,33 or0,0033 or 0,00033 etc...[/TEX]

Now [TEX]n[/TEX] is near at [TEX]p1 [/TEX] or [TEX]p2[/TEX][/QUOTE]

if we knew either p1 or p2 we could know both so it's irrelevant. and knowing one of these is needed to set n to that value. if it were true than n/product =1,100,1000 etc. ... if the ratio is more than 1 we have n>=100*product = 10*p1*10*p2.

[QUOTE=science_man_88;448591]if we knew either p1 or p2 we could know both so it's irrelevant. and knowing one of these is needed to set n to that value. if it were true than n/product =1,100,1000 etc. ... if the ratio is more than 1 we have n>=100*product = 10*p1*10*p2.[/QUOTE]

but if i take

[TEX]211 * 17 = 3587[/TEX]

[TEX]\frac{37}{3587 * 3} = 0,003438...[/TEX]

[TEX]37[/TEX] is near [TEX]17[/TEX]


it isn't irrelevant for me.

[QUOTE=Godzilla;448596]but if i take

[TEX]211 * 17 = 3587[/TEX]

[TEX]\frac{37}{3587 * 3} = 0,003438...[/TEX]

[TEX]37[/TEX] is near [TEX]17[/TEX]


it isn't irrelevant for me.[/QUOTE]

oops made a mistake myself 1,1/100 1/1000 etc. but the point is I don't see how getting close is a formula for it 36 will be closer still ... as the fraction will approach 12/3587.

[QUOTE=science_man_88;448597]oops made a mistake myself 1,1/100 1/1000 etc. but the point is I don't see how getting close is a formula for it3 6 will be closer still ... as the fraction will approach 12/3587.[/QUOTE]


I think One is more big when [TEX]\frac{n}{product * 3} = 0,33 [/TEX] and the other One is more small when [TEX]\frac{n}{product} = 0,33 [/TEX]

[QUOTE=Godzilla;448599]I think One is more big when [TEX]\frac{n}{product * 3} = 0,33 [/TEX] and the other One is more small when [TEX]\frac{n}{product} = 0,33 [/TEX][/QUOTE]

n=product if n/(product *3) = 1/3; n= 1/100 * product if n/(product*3) = 1/300 etc.

[QUOTE=science_man_88;448600]n=product if n/(product *3) = 1/3 n= 1/100 * product if n/(product*3) = 1/300 etc.[/QUOTE]

ok, but for me, it is a higher probability because there are only n+10*10*10*10...etc and only one of this or pearls two are correct.

[QUOTE=Godzilla;448602]ok, but for me, it is a higher probability[/QUOTE]

probability has nothing to do with this for these ratios to work I've listed the exact ratio of n and product.

[QUOTE=science_man_88;448603]probability has nothing to do with this for these ratios to work I've listed the exact ratio of n and product.[/QUOTE]

There are multiple ratio min e max for n-digits-product . Two for One ratio. Well.