# Probability

• May 1st, 2010, 01:28 PM
Probability
There will be two elections. Candidate Heads and candidate tails are running in each election.

Four people vote in each election.
Every voter has a fifty percent chance to vote Heads, and a fifty percent chance to vote tails.
What is the probability that the election results are identical in each election?
By "identical" I mean something like this...
One election shows 3 votes for Heads and 1 vote for tails,,,,, and the other election shows 3 votes for Heads and 1 vote for tails.

I thought of this problem yesterday and I think I have an answer. Here is my work.
Possible Cases for a given election

There are 6 50%H 50%T cases
THTH
HTHT
TTHH
HHTT
THHT
HTTH
There are 4 25% T 75% H cases
THHH
HTHH
HHTH
HHHT
There are 4 25% H 75% T cases
HTTT
THTT
TTHT
TTTH
There is 1 100% T case
TTTT
There is 1 100% H case
HHHH

So there are 16 cases total for each election. So I should find the probability that each case would occur in both elections, which could be found by squaring it.
50 50 case = (6/16)(6/16) = 14%
25T 75H case = (1/4)(1/4) = 6.25%
25H 75T case = (1/4)(1/4) = 6.25%
100% H case = (1/16)(1/16) = .39%
100% T case = (1/16)(1/16) = .39%
Then sum the probabilities for the final answer.

The probability that each election would have the same results is roughly 27.28%.
Did I make any mistakes in my process here?
• May 1st, 2010, 03:45 PM
Arcane_Mathematician
Re: Probability
Quote:

There will be two elections. Candidate Heads and candidate tails are running in each election.

Four people vote in each election.
Every voter has a fifty percent chance to vote Heads, and a fifty percent chance to vote tails.
What is the probability that the election results are identical in each election?
By "identical" I mean something like this...
One election shows 3 votes for Heads and 1 vote for tails,,,,, and the other election shows 3 votes for Heads and 1 vote for tails.

I thought of this problem yesterday and I think I have an answer. Here is my work.
Possible Cases for a given election

There are 6 50%H 50%T cases
THTH
HTHT
TTHH
HHTT
THHT
HTTH
There are 4 25% T 75% H cases
THHH
HTHH
HHTH
HHHT
There are 4 25% H 75% T cases
HTTT
THTT
TTHT
TTTH
There is 1 100% T case
TTTT
There is 1 100% H case
HHHH

So there are 16 cases total for each election. So I should find the probability that each case would occur in both elections, which could be found by squaring it.
50 50 case = (6/16)(6/16) = 14%
25T 75H case = (1/4)(1/4) = 6.25%
25H 75T case = (1/4)(1/4) = 6.25%
100% H case = (1/16)(1/16) = .39%
100% T case = (1/16)(1/16) = .39%
Then sum the probabilities for the final answer.

The probability that each election would have the same results is roughly 27.28%.
Did I make any mistakes in my process here?

Looks good. You may want to leave them in fraction form, as it gives an exact answer instead of an approximate one.
• May 1st, 2010, 10:41 PM
I realized that a much simpler way to state my question is this.

Throw 8 coins on a table and divide them into 2 groups of four. What is the probability that each group of four has the same number of heads and tails?

This is essentially what my original question asked.
• May 2nd, 2010, 04:04 PM
phyti
Re: Probability
Quote:

There will be two elections. Candidate Heads and candidate tails are running in each election.

Four people vote in each election.
Every voter has a fifty percent chance to vote Heads, and a fifty percent chance to vote tails.
What is the probability that the election results are identical in each election?
By "identical" I mean something like this...
One election shows 3 votes for Heads and 1 vote for tails,,,,, and the other election shows 3 votes for Heads and 1 vote for tails.

The probability that each election would have the same results is roughly 27.28%.
Did I make any mistakes in my process here?

There are only 2^4 = 16 possible sequences of H & T, so your fraction should reflect that. There would be 4 that contain 1 T.
For 1 election, wouldn't the answer be 1/4, and for 2 elections 1/16, since the elections are independent?
• May 2nd, 2010, 04:53 PM
Arcane_Mathematician
Re: Probability
Quote:

Originally Posted by phyti
Quote:

There will be two elections. Candidate Heads and candidate tails are running in each election.

Four people vote in each election.
Every voter has a fifty percent chance to vote Heads, and a fifty percent chance to vote tails.
What is the probability that the election results are identical in each election?
By "identical" I mean something like this...
One election shows 3 votes for Heads and 1 vote for tails,,,,, and the other election shows 3 votes for Heads and 1 vote for tails.

The probability that each election would have the same results is roughly 27.28%.
Did I make any mistakes in my process here?

There are only 2^4 = 16 possible sequences of H & T, so your fraction should reflect that. There would be 4 that contain 1 T.
For 1 election, wouldn't the answer be 1/4, and for 2 elections 1/16, since the elections are independent?

no, because he asked what are the chances that both elections will have the same results
• May 4th, 2010, 11:20 AM
phyti
ok, I read it as if he wanted the specific result of 1T and 3H twice.
• May 4th, 2010, 01:41 PM
Scifor Refugee
Perhaps I am misunderstanding the question, but since there are 16 possible outcomes for an election, it seems like the second election would have a 1/16 chance of being identical to the first, regardless of the specific results of the first election. So the answer would be 1/16.
• May 4th, 2010, 09:37 PM
MagiMaster
That would be true if the order of the coin tosses in each mattered. Since they don't, HTHH in the first would match HHHT in the second, and that makes the match more likely.
• May 4th, 2010, 10:02 PM
I wonder if it would be possible to develop a general formula that would compute the probability that the two sets of coins tossed have the same number of heads and tails.

For example, inputting 4 into the formula would output 70/256 or 27.28%

I would imagine it would drop off extremely fast.
For example, if you drop 20 coins and split them into 2 groups of 10, the odds will probably be less than 1% that each set has the same number of heads and tails.

I can't see a way of producing such a formula at the moment.
• May 4th, 2010, 11:04 PM
Arcane_Mathematician
1 coin is

2 coins is

3 coins is

4 coins is

There sure as hell is a pattern. I think I'll take a stab at it.

Could it beeee..... The chances of two independent and equally long strings of coin tosses yeilding the same number of heads and tails is: