Coin Probability: Odds of One vs Rest

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In summary: C(n,k)/2^n = probability of getting k heads in n tosses (a binomial probability), where G is a general generator.In summary, the probability of one person being at odds with the rest is (2*n/2^n) or (1-n)%
  • #1
ArcanaNoir
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Homework Statement



If n people flip a coin what is the probaility that one is at odds with the rest?
(answer is [itex] n2^{1-n} [/itex])

Homework Equations


The Attempt at a Solution



I tried (2/2)(1/2)(1/2)(1/2)... and figure that equaled [itex] \frac{1}{2^{n-1}} [/itex] but that isn't the answer. :(
 
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  • #2
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
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. :confused:
 
  • #4
You have the pattern XYYYYYY here.
How many ways to shuffle that?
 
  • #5
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
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: [tex] (\frac{n!}{(n-1)!})(\frac{1}{2^{n-1}}) = n2^{1-n} [/tex] correct?
 
  • #7
Yep! :smile:

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

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
ArcanaNoir said:
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.

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
ArcanaNoir said:

Homework Statement



If n people flip a coin what is the probaility that one is at odds with the rest?
(answer is [itex] n2^{1-n} [/itex])

Homework Equations





The Attempt at a Solution



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

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
 

FAQ: Coin Probability: Odds of One vs Rest

What is the meaning of coin probability?

Coin probability is the likelihood or chance of a specific outcome when flipping a coin. It is typically expressed as a percentage or a fraction.

How do you calculate the probability of getting heads or tails?

The probability of getting heads or tails when flipping a coin is 1/2 or 50%. This is because there are only two possible outcomes and each has an equal chance of occurring.

What is the difference between odds and probability in coin flipping?

Odds and probability are both ways of expressing the likelihood of an event occurring, but they are calculated differently. Probability is the number of favorable outcomes divided by the total number of possible outcomes. Odds, on the other hand, are the ratio of favorable outcomes to unfavorable outcomes.

Can you have a probability greater than 1 for coin flipping?

No, the probability of any event cannot be greater than 1 (or 100%). This would mean that the event is certain to occur, which is not possible in a random process like coin flipping.

How does the number of coin flips affect the overall probability?

The more times you flip a coin, the closer the actual probability will be to the theoretical probability. For example, if you flip a coin 10 times, you may get 6 heads and 4 tails, but if you flip it 100 times, you are more likely to get closer to a 50/50 split.

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