# Homework Help: Coin Probability

1. Sep 4, 2011

### ArcanaNoir

1. The problem statement, all variables and given/known data

If n people flip a coin what is the probaility that one is at odds with the rest?
(answer is $n2^{1-n}$)
2. Relevant equations

3. The attempt at a solution

I tried (2/2)(1/2)(1/2)(1/2)..... and figure that equaled $\frac{1}{2^{n-1}}$ but that isn't the answer. :(

2. Sep 4, 2011

### process91

Try listing the possibilities for n=2 and n=3 explicitly, and see if you can see the pattern. What you have used for the formula doesn't correspond to what the question is asking.

For n=2, we have two possibilities: TH or HT, and obviously we don't care about the order.

For n=3, we have six possibilities: THH, HTH, HHT, HTT, THT and TTH and again we don't care about the order.

Try to convert these explicit cases into a formulaic representation.

3. Sep 4, 2011

### ArcanaNoir

Okay. for two people: HH, HT, TT = 1/3
three people: HHH, HHT, HTT, TTT = 2/4
Four people: HHHH, HHHT, HHTT, HTTT, TTTT = 2/5
five people= HHHHH, HHHHT, HHHTT, HHTTT, HTTTT, TTTTT = 2/6

So I'm thinking 2/(n+1), except that doesn't work for two people, but hey, I can specify 3 or more peoples easy. Problem is, this is not the answer either.

4. Sep 4, 2011

### I like Serena

You have the pattern XYYYYYY here.
How many ways to shuffle that?

5. Sep 4, 2011

### process91

Hmm... no that's not right. Sorry, I think you read further into my "we don't care about order" statement than I was intending. I probably shouldn't have written that in the first place. The probability for 2 people, for instance, would be 1/2, since it could happen by getting HT or TH, and the total possibilities are HH, TH, HT, TT.

I'll let you work with Serena on this, as I see she's helped you before.

6. Sep 4, 2011

### ArcanaNoir

Oh okay, order does matter.

For two people: (Shuffle XY)(P XY) = (2!/2!)(2/2)(1/2) = 1/2
Three people: (Shuffle XXY)(P XXY) = (3!/2!)(2/2)(1/2)(1/2) = 3/4

So for four peeps: (4!/3!)(2/2)(1/2^3)

So for n peeps: $$(\frac{n!}{(n-1)!})(\frac{1}{2^{n-1}}) = n2^{1-n}$$ correct?

7. Sep 4, 2011

### I like Serena

Yep!

Note that the formula breaks down for n=2.
It seems that the author of your problem has been sloppy.
(Or was there an extra condition that n>2?)

8. Sep 4, 2011

### ArcanaNoir

No extra condition. Author was sloppy. Book just came out this summer. Sloppy author is actually prof at MY university. Uck. I hate errors.

Um, on a shuffling side note (I don't see why I still don't get it) take a look at these:

a) 6,7,8,9
b) 6,7,8,8
c) 7,7,8,8
d) 7,8,8,8

Now, how many ways can I shuffle each?

For a) I tried 4!/(1!1!1!1!4!) but that equal 1. What gives?
For b) 4!/(1!1!2!2!)= 6 but isn't it 10? (by tabling)
For c) 4!/(2!2!2!)= 3 but isn't it 6?
For d) I tried 4!/(1!3!) = 4 which I think is correct.

9. Sep 4, 2011

### HallsofIvy

If you flip a coin n times the probability of m "heads" and "n- m" tails is
$$\left(\begin{array}{c}n \\ m\end{array}\right)\frac{1}{2^n}$$.

If I read your question correctly, you want the probability that either there are n-1 heads and 1 head or n-1 tails and 1 head.

10. Sep 4, 2011

### I like Serena

Well, there's shuffling and shuffling.
The correcting factor was because you were double-counting certain combinations.
That's not the case here.
I'll try to explain using your examples.

a) The permutions of 4 different specific numbers is simply 4! = 24

b) In this case you'd be double-counting the 8, so the number is 4! / 2! = 12
Actually, you'd have 4! / (1!1!2!).
Where did you get the last 2!? And how did you get 10?

c) Double-counting 7, and double-counting 8.
However, 7 and 8 are specific numbers, and not counted once for the one, and another time for the other number.
So: 4! / (2!2!) = 6.

d) Yes, this is correct.

11. Sep 4, 2011

### Ray Vickson

If b(n,k) = C(n,k)/2^n = probability of getting k heads in n tosses (a binomial probability), then you want b(1,n) + b(n-1,n) = 2*n/2^n.

RGV