How many permutations for two-heads-two-tails?
If I flip a fair coin two times, the probability of obtaining 1 head and 1 tail (in any order) is 50%.
If I flip the coin four times, is the probability of obtaining 2 heads and 2 tails (in any order) 50% as well?
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10 solutions
You got me fair a square err, I mean rounds! :)
Does it matter that you don't have four coins as the question suggests? You only have one coin flipped four times with each outcome a unique event. Just curious.
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Four coins flips once is basically the same as one coin flipped four times; both are four separate, independent coin flips.
your odds are still 50/50 with each coin flip
Intuitively I would guess it to be less than 50%. There is 1 “fair” coin in this example. The top statement states 50% chance when flipping twice. Based on the information given in that statement’s “claim” could I not get to 4 flips in increments of 2? Guess I’m looking at the information based on logic and not calculations. Do the calculations given in other discussions support the first statement? I didn’t look at the first statement as being actually factual, so ended up answering logic based on the initial “fact” presented. Help me get past this :)
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Yes, the first statement is factual on the basis of probability calculations (4 possibilities with 2 flips (whether 2 coins or 1 coin twice doesn't matter) (HH, HT, TH, TT) and 2 of the four contain a head-tail pair so 50%) Four flips contain more possibilities (HHHH (1 way 4 heads), HHHT, HHTH, HTHH, THHH (4 ways 3 heads), HHTT, HTHT, HTTH, THHT, THTH, TTHH (6 ways 2 heads), HTTT, THTT, TTHT, TTTH (4 ways 1 head), TTTT (1 way 0 heads), so there are 6 out of a total of 16 possible combinations of ways to get 2 heads and 2 tails, and 6 out of 16 is less than 50%)
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Thank you! "On the basis of probability calculations" helped me differentiate this from actual practice.
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@Chuck Carpenter – Good, I thought from what you said that piece might be the key.
It's easier if you focus on the last flip. If the first three flips were all the same (all heads or all tails), then there's no way to get the desired outcomes. If they were not all the same (two heads and one tail, or two tails and one head) then there's a 50% chance of getting the desired outcome. Hence, with no further calculation we know that the overall probability of getting the desired outcome is less than 50%.
I think this is wrong , the question states any order so order does not matter. There are only 5 possibilities (All head , All tail , 3 head 1 tail,1 head 3 tail, 2 head 2 tail ) so the probability shuld be 1/5 x 100 = 20 %
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No. By this logic, if you flip two coins you will only have 3 possibilities (all head, all tail, 1 head 1 tail), which means that there is only a 33% chance that you will flip 1 head and 1 tail, which is incorrect.
That is true only when each of the 5 cases are equally likely which clearly is not
Using binomial theorem, where p = P(Heads), q = P(Tails) for a given flip and p = q = 1/2,
( p + q ) 4 = p 4 + 4 p 3 q + 6 p 2 q 2 + 4 p q 3 + q 4 .
The middle term where the exponents of p and q are both 2 is the probability of two heads and two tails, hence
6 p 2 q 2 = 6 × ( 2 1 ) 2 × ( 2 1 ) 2 = 1 6 6 = 8 3 < 5 0 % .
Amazing is there a wiki for this?
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It is pretty useful. There's a wikipedia article if that's what you're after: https://en.wikipedia.org/wiki/Binomial_theorem or a wiki on Brilliant: https://brilliant.org/wiki/binomial-theorem-n-choose-k/
Very interesting approach, thanks.
You need to choose
2
of the
4
coin flips for obtaining heads. Therefore, the final probability is
(
2
4
)
×
2
4
1
=
8
3
<
5
0
%
.
Consider the problem where there are 10000 coins and you need to calculate the probability that there are EXACTLY 5000 heads. Intuitively, we know that it is far smaller than 2 1 . The more coins there are, the smaller the probability is. Therefore, we can pick the correct answer. If this solution is not convincing enough, we can calculate the answer, which is shown in many other solutions of this problem.
P(2H, 2T) = Number of ways of arranging 2 H's and 2 T's × ( P ( H ) ) 2 ( P ( T ) ) 2 .
= 2 ! 2 ! 4 ! × 2 1 2 × 2 1 2
= 4 2 4 × 2 4 1
= 8 3 < 2 1
Imagine a similar situation for 100 coins: the chance of a 50/50 split of heads and tails falls as the number of possible splits increases.
Assuming the first probability, the question is twice that so I imagined 0.5*0.5=0.25 and answered less than 50%. Plus, it just makes sense that the question is asking for something more difficult to obtain than the statement given, therefore it would have less of a chance of occurring.
While you have gotten the right answer, the exact probability is not 25%. Can you figure out why?
Enhalten. Probos sabinabil
To put it simply the there are more outcomes with four coins therefore having way more possible outcomes - instead of only two
Less than 50 percent. It is because it is very unlikely to flip 2 head-tails in a row(or not).
The question didn't say that the 2 heads (or 2 tails) must happen one after another.
No, Freddie is correct since there are more rolls and in the formula (2=y3#4=s5), it proves that there are less chances if there are more rolls(also %5ts)
Have a nice day!:)
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If you have four coins laid out in a row, there are 2 4 = 1 6 ways of choosing heads and tails, because each coin has 2 different outcomes: heads or tails. 6 of these ways will results in 2 heads and 2 tails (HHTT, HTHT, HTTH, THHT, THTH, TTHH), so therefore, 1 6 6 = 8 3 is the probability of flipping 2 heads and 2 tails, and 8 3 is less than 2 1 .