Calculating the number of possible outcomes of a Coin Toss

In the previous posts, we've been concentrating a lot on coin tosses.

I thought it might be useful for us to look at how we can calculate the possible number of outcomes.  

In order to calculate the probability of an event to occur mathematically (or to be able to effectively analyze what happened, we need to be able to calculate all possible outcomes).

So in the case of a coin toss.   There are always two possible outcomes in a coin toss.   You will either flip heads or tails.   So let's look at how this breaks down for multiple coin tosses.

1 coin = 2 outcomes =  H,T
2 coins = 4 outcomes = HH,HT,
                       TH,TT
3 coins = 8 outcomes = HHH,HHT,HTH,HTT,
                       THH,THT,TTH,TTT
4 coins = 16 outcomes = HHHH,HHHT,HHTH,HHTT,HTHT,HTTH,HTTT,
                        THHH,THHT,THTH,THTT,TTHT,TTTH,TTTT

And so on and so forth.   And it grows quite quickly.   Here's how the growth looks charted in Excel.

Hey doesn't the graph look like the streaky graph we did the other day.  Sure it's a typical binomial distribution (we'll discuss that another day).  For now we'll concentrate on how to calculate on the number of outcomes for higher number ranges.

In our run of 39 coin tosses, it would probably take us a long time to figure out how many outcomes there were if we were to calculate each and every outcome.   So we wanna take a shortcut.

Inferring the number of outcomes for a coin toss

Just by looking at the coin toss data, we can probably infer that the number of outcomes doubles every time we add a coin.   2 outcomes, 4 outcomes, 8 outcomes, 16 outcomes...

And you'll probably notice the pattern as well.   If you look at 3 coin tosses,  you can see that the first line is the exact same sequence as 2 coin tosses except the first coin toss is a head.   And the second line is again the same sequence as 2 coin tosses but with a tail as the first coin toss.    So sure, it makes sense that it doubles up all the time.

So we are really number of combinations (heads or tails) * number of combinations for previous number of coins

1 coin = 2 outcomes

2 coins = 4 outcomes = 2 outcomes * 2 combinations

3 coins = 8 outcomes = 4 outcomes * 2 combinations

4 coins = 16 outcomes = 8 outcomes * 2 combinations

What would happen if we had a 3 sided coin

Let's say for a second our coin has 3 sides, (Heads, Tails and errrrr Body).  What would happen?

1 coin = 3 outcomes =    H, T, B

2 coins = 9 outcomes =   HH, HT, HB, 

                         TH, TT, TB, 

                         BH, BT, BB

3 coins = 27 outcomes =  HHH,HHT,HHB,HTH,HTT,HTB,HBH,HBT,HBB,

                         THH,THT,THB,TTH,TTT,TTB,TBH,TBT,TBB,

                         BHH,BHT,BHB,BTH,BTT,BTB,BBH,BBT,BBB

Okay, so now we seem to be tripling up from the previous set of outcomes.

And we're following the same pattern as before.

First line is the exact same sequence as 2 coin tosses except the first coin toss is a head.   And the second line is again the same sequence as 2 coin tosses but with a tail as the first coin toss.    And the third line is again the same sequence as 2 coin tosses but with a body as the first coin toss.  So sure, it makes sense that it triples up all the time.

So we are still number of combinations (heads, tails, body) * number of combinations for previous number of coins

1 coin = 3 outcomes

2 coins = 9 outcomes = 3 outcomes * 3 combinations

3 coins = 27 outcomes = 9 outcomes * 3 combinations

4 coins = 81 outcomes = 27 outcomes * 3 combinations

This is cool but would still be a pain to calculate 39 combinations.   What's the formula for doing this?

Quick way for calculating number of outcomes

The quick way of doing this is calculating this is 

number of combinations ^ (number of tosses)

otherwise known as 

number of combinations to the power of number of tosses.

So for a 3 sided coin with 2 tosses, we would calculate it as 

1 coin = 3^1 = 3 = 3 outcomes

2 coins = 3^2 = 3 * 3 = 9 outcomes

3 coins = 3^3 = 3 * 3 * 3 = 27 outcomes

4 coins = 3^4 = 3 * 3 * 3 * 3 = 81 outcomes

And for our regular 2 sided coin

1 coin = 2^1 = 2 = 2 outcomes

2 coins = 2^2 = 2 * 2 = 4 outcomes

3 coins = 2^3 = 2 * 2 * 2 = 8 outcomes

4 coins = 3^4 = 2 * 2 * 2 * 2 = 16 outcomes

How would we do this in Excel (or google docs)?

Pretty easy really.  Lets generate a table of coin tosses (2 sided coin) with their number of outcomes.  Something like this

To generate this in excel, we just do the following

1) In column A, just put the number of coin tosses to generate.  

2) In column B, just put in the following formula

=POWER(2,A2)

3) Do the cell draggy thing to generate the table.

And if you want, you can generate a chart like I did.

BTW, the answer is there is 549,755,813,888 (around 550 billion) possible outcomes for 39 coin tosses.

And if you don't believe me, try it yourself, you know how to check it now