Lesson 33 – Trials to first success: The language of Geometric distribution

So goes the legendary story: Bruce, Robert I, the King of Scotland, defeated the English armies on his seventh trial. He bore six successive defeats before that.

 

Giants are yet to win their first game of the season. Their record: L, L, … I wonder how many games until their first win.

 

The last deadly hurricane that hit New York City is Sandy in 2012. We are now five years through without such a dangerous event.

 

The common thread tying the three examples is the number of trials to the first success. This is the language of the Geometric distribution.

If we consider independent Bernoulli trials of 0s and 1s with some probability of occurrence p and assume X to be a random variable that measures the number of trials it takes to see the first success, then, X is said to be Geometrically distributed.

We can get the success on the first trial, in which case X will be 1. We can see the success on the second trial, in which case the sequence will be 01, and X will be 2. We can see the success on the third trial; the sequence will be 001 and X will be 3 and so forth. As you can guess, X = {1, 2, 3, … }, positive integers.

There is some probability that X can take any integer value. We should also figure out this probability, i.e., P(X = 1), P(X = 2), P(X = 3), and so on.

Let us take the example of a coin toss. The outcomes are head or tail. Binary outcomes → Bernoulli trial. The probability p of a head or tail is 0.5. In other words, if you toss a coin a large number of times, say 100, roughly 50 of them will be heads, and 50 of them will be tails. Let’s play. Heads you win, tails you lose.

 

Great, you win on the first trial. The probability of seeing head is 0.5. Hence, P(X = 1) = 0.5.

 

Let’s play again.

 

Ah, this time the first outcome is a tail and the second outcome is a head. You lose on the first trial but win on the second. It took two trials to wins. X = 2. P(X=2) is P(tail on the first toss)*P(head on the second toss) = 0.5*0.5 = 0.25. Why did we multiply? What is P(A and B) for independent events?

One more time.

  

Now it took three trials to win. X = 3, and P(X = 3) = P(tail on the first toss)*P(tail on the second toss)*P(head on the third toss) = 0.5*0.5*0.5 = 0.125.

For X = 4, it will be P(tail on the first toss)*P(tail on the second toss)*P(tail on the third toss)*P(head on the fourth toss) = 0.5*0.5*0.5*0.5 = 0.0625.

If we now plot X and P(X = k), k being 1, 2, 3, 4, …, we get a probability distribution like this.

The height of the line at X = 2 is 0.5 times the height of the line at X = 1. In the same way, the height of the line at X = 3 is 0.5 times the height of the line at X = 2 and so on. P(X=k) decreases in a geometric progression. Hence the name Geometric distribution.

We can generalize this for any probability p. In our game, we estimated P(X = 1) as 0.5, i.e., the probability of seeing a head p. P(X=2) is 0.5*0.5, i.e., (1-p)*p. P(X = 10) = (1-p)^9*p. First success on the tenth toss is nine tails followed by a head.

More generally,

We can derive the expected value and variance of X as: 

The expected value of a Geometric distribution relates to a special concept called return period → we will look at it next week.

Meanwhile, here are some more geometric probability distributions with different values of p.

p = 0.1

p = 0.3

p = 0.5

p = 0.7

p = 0.9

Notice how the shape changes with changing values of p. p is the parameter that controls the shape of the distribution. The greater the value for p, the steeper the fall.

If the probability of success is close to 1, the odds of winning in the first few trials is high → notice the height of the line for p = 0.9. If the probability of success is close to 0, it takes several trials to get to greater odds of winning overall.

Have you now conceptualized the idea of geometric distribution?

Let me challenge you to a bet then.

I have a coin toss game where I give you two times your bet if you win; you get nothing if you lose. Assume we have a fair coin, would you play the game with me and bet your money? If you will, then what is your strategy, assuming you are in it to win.

Since I challenged you to a bet, I also looked into some lottery games myself at nylottery.ny.gov.

First observation

The odds of winning first prize in any of the games is next to 0. So if you plan to keep buying the tickets until you win the first time, and then retire, you now know that you will keep buying forever.

The chances of winning per game get better for lower prize levels. For example, in the Mega Million, if you want to win the ninth prize, the odds are 1 in 21. Still low, and will take a long time to win.

But, who wants to win the ninth prize. It is like saying “America Ninth.”

Second observation

I keep wondering why on earth is New York Government running a lottery business … only to reconcile that “bread and circuses” have always been up the state’s sleeves to expand.

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