Three heads up

[vze28h54 at verizon.net (Chris)]

At 06:21 PM 10/16/2003 -0400, Kent McLean wrote:

>Look at it this way:
>
>Toss 1:    H-H-H
>Toss 2:    H-H-T
>Toss 3:    H-T-H
>Toss 4:    H-T-T
>Toss 5:    T-T-T
>Toss 6:    T-T-H
>Toss 7:    T-H-T
>Toss 8:    T-H-H

But we are not concerned with the ORDER in which the coins fall as heads or 
tails.

Thus, for our purposes,  2 Hs and a T are the same whether they fall as in 
Toss 2: , Toss 3: or Toss 8:
By the same token (ha) 2 Ts and an H are the same whether they fall as in 
Toss 4: , Toss 6:  or Toss 7:

Therefore, Toss' 2, 3, and 8 count as ONE  (since we do not care about the 
order in which the coins fall).
               Toss' 4, 6, and 7 count as ONE.
All Hs count as ONE
All Ts count as ONE

We are not concerned with the total number of combinations based on order 
in which the coins fall.

So we end up with the possibilities being       0       Heads
                                                 1       Head
                                                 2       Heads
                                                 3       Heads

Thus, 1 in 4 seems to fit.

(Nothing will get a non statistician as confused as listening to 6 
statisticians giving six different opinions as to the correct statistical 
analysis.)

Chris


>All Heads turns up only in Toss 1, out of 8 possible results.
>So, it's 1 chance in 8. Or 1/2*1/2*1/2, or 1/(2^3).
>
>Comprenez-vous?
>
>Kent