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Statistics geeks! Answer my call!

crimson117

Platinum Member
How accurate is this claim? If it's wrong, what's wrong with it? If it's right, is there better math to back it up? (assuming the coin has a perfectly even chance to be heads or tails)

Actually, if you did 1000 flips of a coin, there are 1001 possible outcomes, and the chance of getting exactly 500 head and 500 tails are the same as the chance of getting exactly 0 heads and 1000 tails.

10 flips have an equal chance (1 in 11) to come up with any one of these 11 combinations:
h/t
0/10
1/9
2/8
3/7
4/6
5/5
6/4
7/3
8/2
9/1
10/0
 
its wrong.

take a much simpler example. Flip the coins twice, you are saying that there is the same chance of hitting 2tails, and 1H/1T which is incorrect.

The four possible ourcomes are
HH
HT
TH
TT
So hitting 2 tails = 25%, which hitting 1H and 1T is 50%.
 
wow dude if you really can't just SEE that is wrong, i feel sorry for you.

thats like saying everytime you drive your car, you have a 50% chance of getting into an accident, because you either WILL get into one or you WONT get into one.

 
the probability of getting exactly 500 heads and 500 tails is going to be close to the probabilty of all heads or all tails. both very close to nil, I can't remember the exact equation to put that into. but I could have my calculator flip a coin 1000 times for you
 
oh and your reasoning is completely wrong. there are MANY MANY different ways you can get 500 heads and 500 tails ...

you could get 500 tails in a row then 500 heads in a row.
you could get 499 tails in a row, then 500 heads in a row, then 1 tail.
you could get 2 tails in a row then 2 heads in a row, and repeat that until you have 1000 flips.

there is only one way to get 0 heads and 1000 tails, and thats by getting 1000 tails in a row.
 
Originally posted by: purbeast0
oh and your reasoning is completely wrong. there are MANY MANY different ways you can get 500 heads and 500 tails ...

you could get 500 tails in a row then 500 heads in a row.
you could get 499 tails in a row, then 500 heads in a row, then 1 tail.
you could get 2 tails in a row then 2 heads in a row, and repeat that until you have 1000 flips.

there is only one way to get 0 heads and 1000 tails, and thats by getting 1000 tails in a row.

i found it, 1000 C 2, 499,500 ways
 
Originally posted by: scorpmatt
the probability of getting exactly 500 heads and 500 tails is going to be close to the probabilty of all heads or all tails. both very close to nil, I can't remember the exact equation to put that into. but I could have my calculator flip a coin 1000 times for you

No that's incorrect, there are much better chances of getting 500 tails and 500 heads than 1000 of one or the other. The reason is quite simple, there are many more "ways" to make a 50/50 split than there are to make a 0/1000 split.

Jaybert gave a good example of this. If the order mattered and you wanted a specific sequence of heads and tails then you would be correct.

To use something similar to Jaybert's example with 4 flips:
The only way to get 4 heads is: HHHH -> (1/2)^4 = 1/16
The only way to get 4 tails is: TTTT -> (1/2)^4 = 1/16
There are several ways to get 2 of each: HHTT, HTHT, HTTH, THHT, THTH, TTHH -> 6*(1/2)^4 = 3/8

There is only one way to get the sequence HTHT -> (1/2)^4 = 1/16

Hope that clears things up for you. BTW, this is a probability problem, not a statistics problem.

edit: fixed a typo, the only way to get HTHT should be (1/2)^4 not (1/3)^4
 
Originally posted by: RaynorWolfcastle
Originally posted by: scorpmatt
the probability of getting exactly 500 heads and 500 tails is going to be close to the probabilty of all heads or all tails. both very close to nil, I can't remember the exact equation to put that into. but I could have my calculator flip a coin 1000 times for you

No that's incorrect, there are much better chances of getting 500 tails and 500 heads than 1000 of one or the other. The reason is quite simple, there are many more "ways" to make a 50/50 split than there are to make a 0/1000 split.

Jaybert gave a good example of this. If the order mattered and you wanted a specific sequence of heads and tails then you would be correct.

To use something similar to Jaybert's example with 4 flips:
The only way to get 4 heads is: HHHH -> (1/2)^4 = 1/16
The only way to get 4 tails is: TTTT -> (1/2)^4 = 1/16
There are several ways to get 2 of each: HHTT, HTHT, HTTH, THHT, THTH, TTHH -> 6*(1/2)^4 = 3/8

There is only one way to get the sequence HTHT -> (1/3)^4 = 1/16

Hope that clears things up for you. BTW, this is a probability problem, not a statistics problem.

you learn probabilty in statistics, probability is one of the basis of statistics
 
Originally posted by: RaynorWolfcastle
Originally posted by: scorpmatt
the probability of getting exactly 500 heads and 500 tails is going to be close to the probabilty of all heads or all tails. both very close to nil, I can't remember the exact equation to put that into. but I could have my calculator flip a coin 1000 times for you

No that's incorrect, there are much better chances of getting 500 tails and 500 heads than 1000 of one or the other. The reason is quite simple, there are many more "ways" to make a 50/50 split than there are to make a 0/1000 split.

Jaybert gave a good example of this. If the order mattered and you wanted a specific sequence of heads and tails then you would be correct.

To use something similar to Jaybert's example with 4 flips:
The only way to get 4 heads is: HHHH -> (1/2)^4 = 1/16
The only way to get 4 tails is: TTTT -> (1/2)^4 = 1/16
There are several ways to get 2 of each: HHTT, HTHT, HTTH, THHT, THTH, TTHH -> 6*(1/2)^4 = 3/8

There is only one way to get the sequence HTHT -> (1/3)^4 = 1/16

Hope that clears things up for you. BTW, this is a probability problem, not a statistics problem.

Thanks, purbeast0 for your second comment and mcbain for the above, that did clear it up for me...

My understanding is now that there is only one sequence of 1000 throws that leads to 0 heads and 1000 tails (you have to throw 1000 tails), while there are many sequences of throws that could lead to 500/500. (you could throw 500 heads, then 500 tails, or 498 heads, then 500 tails, then 2 heads, etc.).
 
Originally posted by: scorpmatt
i found it, 1000 C 2, 499,500 ways
Ummm... no, actually the probability would be
(1000 C 500)*.5^500*(1-.5)^(1000-500)
= (1000 C 500)*.5^1000

You'd probably have to use some approximation to solve this because you're multiplying a huge number with a tiny one.

FWIW, this is a simple Bernoulli trials experiment, you can read up on Bernoulli distributions here if you like
 
It's correct IF he really meant to say...

Actually, if you did 1000 flips of a coin, there are 1001 possible outcomes, and the chance of getting 500 heads in a row and then 500 tails in a row are the same as the chance of getting all 1000 tails.

Otherwise... he's VERY wrong.
 
Originally posted by: crimson117
Originally posted by: RaynorWolfcastle
Originally posted by: scorpmatt
the probability of getting exactly 500 heads and 500 tails is going to be close to the probabilty of all heads or all tails. both very close to nil, I can't remember the exact equation to put that into. but I could have my calculator flip a coin 1000 times for you

No that's incorrect, there are much better chances of getting 500 tails and 500 heads than 1000 of one or the other. The reason is quite simple, there are many more "ways" to make a 50/50 split than there are to make a 0/1000 split.

Jaybert gave a good example of this. If the order mattered and you wanted a specific sequence of heads and tails then you would be correct.

To use something similar to Jaybert's example with 4 flips:
The only way to get 4 heads is: HHHH -> (1/2)^4 = 1/16
The only way to get 4 tails is: TTTT -> (1/2)^4 = 1/16
There are several ways to get 2 of each: HHTT, HTHT, HTTH, THHT, THTH, TTHH -> 6*(1/2)^4 = 3/8

There is only one way to get the sequence HTHT -> (1/3)^4 = 1/16

Hope that clears things up for you. BTW, this is a probability problem, not a statistics problem.

Thanks, purbeast0 for your second comment and mcbain for the above, that did clear it up for me...

My understanding is now that there is only one sequence of 1000 throws that leads to 0 heads and 1000 tails (you have to throw 1000 tails), while there are many sequences of throws that could lead to 500/500. (you could throw 500 heads, then 500 tails, or 498 heads, then 500 tails, then 2 heads, etc.).

yea no problem. lol i find it funny how you called RaynorWolfcastle McBain haha, nice.
 
Originally posted by: jaybert
its wrong.

take a much simpler example. Flip the coins twice, you are saying that there is the same chance of hitting 2tails, and 1H/1T which is incorrect.

The four possible ourcomes are
HH
HT
TH
TT
So hitting 2 tails = 25%, which hitting 1H and 1T is 50%.

Get back to work, kid! 😉
 
Originally posted by: crimson117
Actually, if you did 1000 flips of a coin, there are 1001 possible outcomes, and the chance of getting exactly 500 head and 500 tails are the same as the chance of getting exactly 0 heads and 1000 tails.

I stopped reading right here because this is wrong. The chances of getting 500 heads and 500 tails (assuming a fair coin of course) is about 10^29 times as much as getting exactly 0 heads and 1000 tails.
 
Originally posted by: scorpmatt
Originally posted by: purbeast0
oh and your reasoning is completely wrong. there are MANY MANY different ways you can get 500 heads and 500 tails ...

you could get 500 tails in a row then 500 heads in a row.
you could get 499 tails in a row, then 500 heads in a row, then 1 tail.
you could get 2 tails in a row then 2 heads in a row, and repeat that until you have 1000 flips.

there is only one way to get 0 heads and 1000 tails, and thats by getting 1000 tails in a row.

i found it, 1000 C 2, 499,500 ways


nope, it's 1000 C 500.
 
500/500 is far more likely than 0/1000. With 0/1000 you can basically fail with 1 flip. With 500/500 you still have a chance at getting it up until you get 501 of something....
 
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