322 consecutive "heads": formula for calculating?
- Thread starter bobW
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arthur ohlsten
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let h be the probability of a head
let t be the probability of a tail
[for a true coin both probability's are 1/2]
[h+t]=1
for n tosses we have
[h+t]^n=1^n
by the binomial expansion
h^n+nh^(n-1)t /1! +.... t^n =1
probibility of all heads for n=322:
h^322 answer
if h=1/2
322 heads in 322 tosses = 1/2^322
probability of 321 heads and 1 tail
322h^321 t^1 /1!
etcetera
Arthur
let t be the probability of a tail
[for a true coin both probability's are 1/2]
[h+t]=1
for n tosses we have
[h+t]^n=1^n
by the binomial expansion
h^n+nh^(n-1)t /1! +.... t^n =1
probibility of all heads for n=322:
h^322 answer
if h=1/2
322 heads in 322 tosses = 1/2^322
probability of 321 heads and 1 tail
322h^321 t^1 /1!
etcetera
Arthur
pka
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Re: 322 consecutive "heads"
If we flip any coin n times the probability of getting n heads is \(\displaystyle \left( {\frac{1}{2}} \right)^n\).
If we flip it 10 times the probability of getting the string HHHHHHHHHT is \(\displaystyle \left( {\frac{1}{2}} \right)^{10}\).
In fact that is the probability of getting any particular string of ten flips.
Is “a coin toss game” a particular game?bobW said:
If we flip any coin n times the probability of getting n heads is \(\displaystyle \left( {\frac{1}{2}} \right)^n\).
If we flip it 10 times the probability of getting the string HHHHHHHHHT is \(\displaystyle \left( {\frac{1}{2}} \right)^{10}\).
In fact that is the probability of getting any particular string of ten flips.
arthur ohlsten
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the probability of any specific sequence of 10 tosses is [1/2]^10 for a "good" coin
for instance hhhhhhhhht
this is different from ," what is the probability of only 1 tail and 9 heads in 10 tosses?"
then you have
hhhhhhhhht or hhhhhhhhth or hhhhhhht hh etc 10 [1/2]^10
[h+t]^10=
h^10/0! ten heads
10 h^9t/1! nine heads 1 tail
10[9]h^8t^2/2! eight heads 2 tails
etcetera down to
t^10 ten tails
Arthur
for instance hhhhhhhhht
this is different from ," what is the probability of only 1 tail and 9 heads in 10 tosses?"
then you have
hhhhhhhhht or hhhhhhhhth or hhhhhhht hh etc 10 [1/2]^10
[h+t]^10=
h^10/0! ten heads
10 h^9t/1! nine heads 1 tail
10[9]h^8t^2/2! eight heads 2 tails
etcetera down to
t^10 ten tails
Arthur
pka
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Actually the myth of the “a good coin” was exploded by two physicists at Cal Tech (I think) about 10 years ago. They showed if one “flips” a coin, send spinning and catch it on the fly, then there is no way to bias a coin unless it has two identical sides. It is a easy though experiment: as it spins in the air weight has nothing to do with its spin.arthur ohlsten said:
Moreover, I said nothing about there being one tail and n consecutive heads. And neither does the problem. From the statement it sounds as if this is a waiting time problem until the first success, in this case a tail. What is the probability of 323 flips until the first tail.
arthur ohlsten
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the cal tech conclusion is true if you catch it in the air. If you let it land, then I believe you can have a h=7/8 and t=1/8 as a example. I am not sure of this as I never tested the theory, but it is true if you use a random number generator rather than tossing a coin, for the coin test.
Arthur
Arthur
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and if you let in land
It can roll-of under the couch and never know what happened (quantum cat)
or it can land on mud on the edge and stay there
possibilities ...possibilities.....
It can roll-of under the couch and never know what happened (quantum cat)
or it can land on mud on the edge and stay there
possibilities ...possibilities.....