# Coin flip probability. At least 2 tails from 3 flips?

Ok, so I'm a parent helping my 12yo with a 7th grade math question. We couldn't figure out the right way to get the answer for this one...

Q: A coin is flipped 3 times. What is the probability of getting at least 2 tails?

I thought the answer would be 1/2 x 1/2 which would equal 1/4 with the third flip not mattering, but that's not correct. Listing the outcomes (H being heads and T being tails... HHH, HHT, HTH, HTT, THH, THT, TTH, TTT), it's clear that 1/2 the outcomes result in at least 2 tails. So, is there a way to figure this out mathematically as a function of fractions, with each coin toss being a 1/2 probability? Is listing the outcomes and counting from there the only way?

• You could take a look at the Binomial distribution thought I am not sure if it's appropriate for a 7th grade task – RScrlli Oct 2 at 20:30
• In seventh grade, it is probably expected that the student manually creates the sample space (list of possible outcomes) and counts the favourable ones. I would approach this using a tree diagram as it helps introduce scenarios where the probabilities are not 1:1. – Andrew Chin Oct 2 at 20:32
• For higher levels of mathematics, I would refer you to an answer that I had posted a while ago which talks about the probability of throwing at least $n$ tails in $2n-1$ coins -- I hope it's an interesting read for you. – Andrew Chin Oct 2 at 20:37
• @AndrewChin I prefer the more elegant (and intuitive) method in the answer by Kevin P Costello: Realize that you can exactly match up each sequence with more tails than heads against a sequence with more heads than tails, and this listing gives you all possible sequences. – Robert Shore Oct 2 at 21:10
• Thanks all! It's been awhile for me since 7th grade! :) – AChez Oct 3 at 0:49

One option is to, as you did here, just list out all the possibilities. Another way of thinking about things: The condition "at least two tails" is equivalent to "more tails than heads". If you flip three coins, you'll be in exactly one of two situations:

Which of these two situations (if any) is more likely? What does that imply about the probability of each situation?

A follow up question you and your 12yo might want to think about: If I flip $$4$$ coins, the probability of getting more tails than heads is not $$\frac{1}{2}$$. Why can't I just use the same argument as I did for $$3$$ coins?

Let's take case of two tails:

First flip gets you tail then in second it doesn't but in third it does. P = 0.5(tails)*0.5(heads)*0.5(tails)

Or there can be an another case

Where first flip gets you tail then second flip also gets you tail but third doesn't . P=0.5(tails)*0.5(tails)*0.5(heads)

Or

First flip gets you head second gets you tail and third also gets you tail. P=0.5(heads)*0.5(tails)*0.5(tails)

Now take the case of three tails:

First gets you tails and second and third also gets you the same. P=0.5(tails)*0.5(tails)*0.5(tails)

Now add them all P= 0.125+0.125+0.125+0.125=0.5

So the probability is 0.5

You could have use combinations to count how many ways are there to arrange but for a 12 year child that would be too much ( if he is not Sheldon Cooper :P)