I was not able to find a similar question when searching, but if I've missed one please feel free to point me to it. Unfortunately the closest example in the textbook was not terribly helpful either.

I am working on a problem that asks me to make use of the binomial model for discrete time option valuation with the objective of finding the probability that a call option finishes "in the money," which is to say $S - K > 0$ where $S$ is the security value and $K$ is the strike price of the option. The problem doesn't say anything else, so I believe I want a general probability in terms of $p$ and $q$.

We have primarily used the binomial model to determine the value of securities and options up to this point, so I'm not sure exactly where to start to find the probability.

The formulation we have used was of this form:

$V_0 = (1 + i)^{-N}E(F(S_n))$

$ = (1 + i)^{-N} \sum_{j=0}^N {N \choose j} \times (S_0 \times u^j \times d^{N-j}) \times p^j \times q^{N-j}$

Where $u$ and $d$ are the factors by which a security increases or decreases in value each period.

I started by taking this model and replacing $F(S_n)$ with $F(S_n - K)$. But thinking about it that seems to only get me the expected value. If I just focus on the probability component, which is just binomial we have $p^j * q^{N-j}$ for any given event, but with knowing the other parameters I dont know how I am supposed to find out when the option is in the money or not (for example, this would depend on the values of $u$ and $d$).

Any and all help would be much appreciated!


2 Answers 2


Let's just deal with the aspect of probabilities. The answer is easier than you think.

Consider the simplest one-step model. At the end, the stock will either be up (to $Su$) or down (to $Sd$). It will move up with a probability p or down with a probability (1-p). Here's how to calculate p:

$ p = \frac {e^{rT} - d}{u - d} $

For example, consider a stock (from Hull), where $ u = 1.1, d=0.9, r=0.12, T $ is 3 months.

$ p = \frac {e^{0.12 \times 0.25} - 0.9}{1.1 - 0.9} = 0.6523 $

Now consider a two-step model. The stock will end up in one of three states.

  1. uu
  2. ud or du
  3. dd

Note that because multiplication is commutative, $S \times d \times u = S \times u \times d$, and we got to state 2 through 1 down and 1 up, not caring whether the down or the up came first. (This is a recombining tree. If we have dividends, the tree does not recombine and it becomes more complex.)

The probabilities that each of the states will occur will correspondingly be:

  1. $1 \times p^2$
  2. $2 \times p \times (1-p)$
  3. $1 \times (1-p)^2$

Why the factors 1, 2, 1 at the beginning? These correspond to the different paths to get there, namely {uu}, {ud, du}, {dd} having a set size of 1, 2, and 1.

The three-step model will have four final states.

  1. 3 u / 0 d: $1 \times p^3$
  2. 2 u / 1 d: $3 \times p^2 \times (1-p)$
  3. 1 u / 2 d: $3 \times p \times (1-p)^2$
  4. 0 u / 3 d: $1 \times (1-p)^3$

The factors 1, 3, 3, 1 correspond to the different paths to get to the final states, namely

  1. {uuu}
  2. {uud, udu, duu}
  3. {udd, dud, ddu}
  4. {ddd}

The multipliers are the choose function that you described in your question. Another way to think of this is through Pascal's Triangle.

   1 1
  1 2 1
 1 3 3 1
1 4 6 4 1

Once you understand the pattern, it should be fairly easy to extend.

There's a much better graph than I can draw at http://en.wikipedia.org/wiki/File:Arbre_Binomial_Options_Reelles.png, and no discussion of binomial option trees would be complete without it. While there are 3 paths to get to each of the middle boxes at the far right, there is only 1 path to get to the top box.


First find minimum value of j that assures the option is in the money. This would be function of u,d,n,S,K,p,q

Any particular value of j has a probability associated with it. You gave formula above. Then you need sum probabilities from j=min j needed to n. See wikipedia binomial distribution to find formula for sum. Sum is based on cdf of binomial distribution. cdf is not closed form.


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