Combinatorics Confusion

[MIT_Aspirant]

Question - A fair coin is flipped three times. What is the probability that the coin lands on heads exactly twice?

Now, based on the Anagram model, the number of desired outcomes would be - 3!/2! = 3

However I would think that the total number of possible outcomes would be 3! = 6, however that is not true. The total number of possible outcomes in this case for the 3 coins flipped would be = 8.

Can someone help me understand how can we calculate 8 other than manually work out all the possible scenarios? Also why does this case does not fall into regular Anagram model which states there are 3! ways of arranging a 3 letter word?

[gmatkid]

Question - A fair coin is flipped three times. What is the probability that the coin lands on heads exactly twice?

Now, based on the Anagram model, the number of desired outcomes would be - 3!/2! = 3

However I would think that the total number of possible outcomes would be 3! = 6, however that is not true. The total number of possible outcomes in this case for the 3 coins flipped would be = 8.

Can someone help me understand how can we calculate 8 other than manually work out all the possible scenarios? Also why does this case does not fall into regular Anagram model which states there are 3! ways of arranging a 3 letter word?

I can see why you're confused: combinatorics and probability are not the same thing! You don't use the Anagram model to find probability.

[RA]

Total outcomes = (Possible outcomes for the first flip) X (Possible outcomes for the second flip) X (Possible outcomes for the third slip)

Now we know each flip has 2 possible outcomes, therefore total outcomes = 2 X 2 X 2 = 8

Hope that helps.

[RonPurewal]

Total outcomes = (Possible outcomes for the first flip) X (Possible outcomes for the second flip) X (Possible outcomes for the third slip)

Now we know each flip has 2 possible outcomes, therefore total outcomes = 2 X 2 X 2 = 8

Hope that helps.

heh heh, 'slip'. cute.

the above is totally correct.

as for why you can't use the anagram model - you're not finding rearrangements of anything. if you had 3 different coins (say, nickel, dime, quarter) and you were deciding in which order to insert them into a vending machine, then that would be an "anagram model" situation, because you're actually finding the number of different orderings of the three coins.
in this problem, on the other hand, you're not finding the number of orderings of a set of 3 distinct items, so the anagram method doesn't apply. instead, because you're combining the outcomes of different steps of a process (1st flip, 2nd flip, 3rd flip) to find aggregate outcomes, you use consecutive multiplication.

--

to the other poster: it's quite possible for the anagram method to be used in a probability problem, actually. if your set of "total outcomes" or your set of "successes" can be counted by rearranging items, then you can use the anagram method for that;
remember that the numerator and denominator in probability problems, individually, are the result of counting. since combinatorics is the mathematics of counting, it's no surprise that combinatorial methods can apply.

[Guest]

I get really confused too, but anytime it says "probability", I just find the probability of each thing happening...

1/2 heads the first time
1/2 heads the second time
1/2 heads the third time

= 2*2*2 = 8 chances

Then I think about it a bit: you can flip 2 heads if you flip it on A and B, on A and C, or on B and C. That = 3, so the answer is 3/8.

If you want a bit more of an explanation of the permutation, you can think of it in terms of the second part. There, we have 3 shots of hitting heads twice... so it is 3c2 or 3! / (3-2)!2! which = 3!/(1)!2! = 3. I hope that helps!

[bekkilyn]

Since the initial question that started the discussion concerning total possible outcomes was about the probability of the events, let me see if I can answer that one. (I need the practice!)

The probability of the coin landing on heads "exactly twice" is the same as the probability of the coin landing on heads only once, or even three times, since each coin toss is independent and has an equal chance of landing on heads or tails.

The probability would be P(HHT) + P(HTH) + P(THH) = (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) + (1/2)(1/2)(1/2) = (1/8) + (1/8) + (1/8) = 3/8 = 0.375 = 37.5%

[san]

Possible outcomes from fliping 3 coins.

HHH
HHT
HTT
TTT

We are not concerned with which coin (1,2 or 3) produced H or T.

so we have atleast 2 Head combination out of total 4 combinations. Hence 2/4 = 1/2

[san]

sorry i misread the question.

I thought 3 coins are used. Slip is right.

[RonPurewal]

Possible outcomes from fliping 3 coins.

HHH
HHT
HTT
TTT

We are not concerned with which coin (1,2 or 3) produced H or T.

so we have atleast 2 Head combination out of total 4 combinations. Hence 2/4 = 1/2

nope, incorrect. you can't consider "0 heads", "1 head", etc. on equal footing with each other, because they aren't equally likely. if you're going to use the basic formula, which is "probability = number of successes ÷ number of outcomes", then your "outcomes" must be EQUALLY LIKELY outcomes.
if not, then you can arrive at all sorts of absurd conclusions. for instance, i could conclude that my probability of winning the lottery is 1/2, because there are exactly two outcomes ("win" and "lose") and one of them is "win". the problem, of course, is that those 2 outcomes aren't equally likely; your argument has the same problem, albeit not on as ridiculous of a scale.

the correct 8 combinations are
HHH
HHT
HTH
THH
HTT
THT
TTH
TTT

[RonPurewal]

sorry i misread the question.

I thought 3 coins are used. Slip is right.

the analysis of this problem is EXACTLY the same in ALL of the following circumstances:
(a) three different coins are flipped simultaneously;
(b) three different coins are flipped in succession;
(c) the same coin is flipped three times in succession.

there is no difference at all.
in all three cases, there are three independent flips, so the probabilities are calculated in the same way each time.