GoFuckYourself.com - Adult Webmaster Forum - View Single Post - Here's a braintease I made up (Math Question)

ucv.karl · 01-13-2007, 07:50 AM

The probablity of filliping a coin and getting 2 heads (HH) is:

(1/2)*(1/2)=1/4

Getting HT = (1/2)*(1/2)

Getting TH = (1/2)*(1/2)

Getting TT = HH = 1/4

So 50% of the time (1/4+1/4=1/2) you'll get the TH, HT combination and 25% of the time you'll get HH and 25% of the time you'll get TT.

Now just extending this to the problem from the initial post.

The probablity of getting 2 heads (HH) by rolling 2 times on this 5 sided dice is
(1/5)*(1/5)=(1/5)^2

The probablity of getting anything but heads (XXXXXX) and rolling 6 times is

(4/5)*(4/5)*(4/5)*(4/5)*(4/5)*(4/5)=(4/5)^6

Now the probablity of getting 2 out of 8 times of this 5 side dice is just
mutlitplying the above probabilities. (HHXXXXXX, where 'X' means anything but Heads)

=(1/5)^2 * (4/5)^6

I get the same answer as Myst.

(hopefully there aren't too many typos)

01-13-2007, 07:50 AM
ucv.karl Confirmed User Join Date: Jul 2006 Location: Phx,Az Posts: 498	The probablity of filliping a coin and getting 2 heads (HH) is: (1/2)(1/2)=1/4 Getting HT = (1/2)(1/2) Getting TH = (1/2)(1/2) Getting TT = HH = 1/4 So 50% of the time (1/4+1/4=1/2) you'll get the TH, HT combination and 25% of the time you'll get HH and 25% of the time you'll get TT. Now just extending this to the problem from the initial post. The probablity of getting 2 heads (HH) by rolling 2 times on this 5 sided dice is (1/5)(1/5)=(1/5)^2 The probablity of getting anything but heads (XXXXXX) and rolling 6 times is (4/5)(4/5)(4/5)(4/5)(4/5)(4/5)=(4/5)^6 Now the probablity of getting 2 out of 8 times of this 5 side dice is just mutlitplying the above probabilities. (HHXXXXXX, where 'X' means anything but Heads) =(1/5)^2 (4/5)^6 I get the same answer as Myst. (hopefully there aren't too many typos) __________________ It's better when you can Switch. ICQ: 263079754 Last edited by ucv.karl; 01-13-2007 at 07:53 AM..