One famous pattern in Pascal's Triangle is that if you toss n coins, the chance of getting m heads is the mth number along in the nth row of Pascal's triangle. In this article, I...
One famous pattern in Pascal's triangle is that if you toss n coins, the chance of getting m heads is the mth number along in the nth row of Pascal's triangle. In this article, I have attempted to explain this intriguing link between Pascal's triangle and probability.
As an example to help us understand why there is a link between Pascals' triangle and probability, let's take the situation of tossing 10 coins and trying to get 6 heads. There are two ways that this can be done. We could toss the first 9 coins and get 5 heads. Then we would have to get another head from the 10th coin to give us 6 heads out of 10 tosses. A second option is that we could get 6 heads from our first 9 coin tosses, and then get a tail from our 10th coin, again giving us 6 heads out of 10 tosses. Amazingly, simple as this is, it is the essence of the proof of the link between Pascal's triangle and probability!
Think about what we have said so far: the number of ways of getting 6 heads out of 10 is the sum of the number of ways of getting 5 heads out of 9 and 6 heads out of 9...
All we need to do is think about where the situations discussed above appear in Pascal's triangle:
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1
The numbers corresponding to 5/9 heads and 6/9 heads are 126 and 84, and from the rules of Pascal's triangle. we know that these will equal 210, as they are (when the triangle is centralised) the two numbers directly above 210. We know that 6 out of 10 heads will be directly below 5 out of 9 and 6 out of 9 heads, because the link we are trying to prove states the total number of coin tosses shows you what row to look in, and the number of heads you're trying to get is one less than how many rows in you count. If you think about it, you get the 9th row, 6th number in, and the 9th row, 7th number in, which will be positioned directly above the 10th row, 7th number in if you centralise the triangle.
This argument is no different for getting any number of heads from any number of coin tosses. (Note: the only cases which are slightly different are when we are trying to get all heads or no heads. Then, there is only one way of doing this, not two. For example, when trying to get 0 heads from 5 coin tosses, the only way of achieving this is to get 0 heads from 4 coin tosses, so there is no addition of two different values. This is just like in Pascal's triangle, however, when you're at the edge calculating the 1s, and only adding a single number (a one) from above).
Essentially, what we have shown is that assuming this link between Pascal's triangle and probability works for the nth row, we can show it also works for the (n+1)th row. The rule clearly works for row 1 (you can check this), and therefore works for row 2, and as it works for row 2, we have shown it must work for row 3. We can carry on this reasoning indefinitely, showing the rule works for any row of Pascal's triangle (this method of proof is called proof by induction).
Therefore, we have shown our desired result, and it is (hopefully!) clear why Pascal's triangle is so inextricably linked the tossing of coins and the repetition of similar equal probabilities.
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