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Unit 7

Making Sense of Randomness

7.3 Simple Probability and Counting

A COIN TOSS

Probability theory enables us to use mathematics to characterize and predict the behavior of random events. By "random" we mean "unpredictable" in the sense that in a given specific situation, our knowledge of current conditions gives us no way to say what will happen next. It may seem pointless to try to predict the behavior of something that we fundamentally characterize as unpredictable, but this is exactly what makes the mathematics of probability so powerful. Let's think about a coin toss. There is no way to predict the outcome of a single coin toss. In this sense it is a random event.(Now, we have to be a bit careful here because a coin toss would not be random if we were able to know all of the initial conditions of the toss, but since we can't know all of the conditions that affect the outcome, we can treat it as random.) But we can say some definite things about it.

The first thing that we can say is that the outcome will definitely be either heads or tails. Putting this in mathematical terms, we say that the probability of the coin landing heads up or tails up is 1, or absolutely certain. An event of probability zero is effectively impossible. In the case of equally likely outcomes, such as in the dice example above, determining the probability that a particular outcome occurs basically involves counting. So, we do as Cardano did and compare the number of ways a specific outcome can happen to how many total outcomes are possible. The probability of the coin landing heads up would then be 1 out of 2, or 1/2. There is, of course, the same probability that it will land tails up. To see this we could start with the probability that the coin will be heads or tails, 1, and subtract the probability that it will be heads, 1/2. This leaves us with 1/2 as the probability that the coin will not land heads up, in other words, the probability that it will land tails up.

Determining the probability of the optional outcomes of a single coin flip may not seem that interesting, but it provides a good starting point for understanding the probabilities associated with any event or series of events that have only binary outcomes, (e.g., heads or tails, win or lose, on or off, left or right). We can abstract any such situation in the form of a simple machine known as a Galton board.

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THE GALTON BOARD

Imagine a peg fixed in the middle of an inclined board, with the base of the board divided into two bins of equal size. If we drop a marble towards the peg, it will hit it and deflect either into the right bin or the left bin. In terms of probability, this is just like our coin toss from before.

galton board

The two bins represent the possible outcomes for the dropped marble, and because there is only one way to get into either bin, the probability associated with a marble ending up in a particular bin is 1/2. The advantage of viewing binary systems in this way is that it is very easy to build complexity into the experiments by adding rows of pegs. Let's add a row of two pegs below the initial peg, one to the right and one to the left.

galton board

Notice that there are now three bins at the base. We can, as before, figure out the probability of the marble ending up in any particular bin. We may be tempted to think that all three bins are equally likely destinations, which would make the probability for any individual bin 1/3. This ignores the fact that if we look at the paths a marble can take through the machine, we find that there are two possible paths to the middle bin, whereas there is only one path leading to each of the side bins. This suggests to us that we need to count the paths to bins instead of the bins themselves. With such a simple system, enumerating the paths is straightforward: LL, LR, RL, RR . With four possible paths, two of which end up in the middle bin, the probability of the marble ending up in the middle bin is 2 out of 4, or 1/2. Because each side bin has only one path associated with it, the probability of the marble ending up in one particular side bin is 1 in 4 or 1/4. If we add all the probabilities together, we get the probability that the marble will end up in one of the three bins: 1/2 + 1/4 + 1/4 = 1.

This is what we would expect, because the marble cannot disappear and must, therefore, end up in one of the three bins. To represent more-involved binary outcome systems, we can continue adding rows to our machine:

galton board

Here we have shown all possible ways that the marble can traverse three rows of pegs. In each row the marble hits one peg and deflects either right or left. This is a good model for any collection of three binary decisions, such as our problem of points from the last section. If instead of left and right, you imagine each peg represents win or lose, you have a nice model of the three rounds that the two players might hypothetically play to finish their game. Let's call each deflection to the left a win for Player 1 and each deflection to the right a win for Player 2. As we said earlier, Player 2 needs three consecutive wins, so the only path that would lead him to victory would be the RR path. Because this is just one of eight possible paths, his chances of winning are 1 in 8. This means that if the game is interrupted at this point, Player 2, the one who is behind, should get 1/8 of the pot.

To see why, let's imagine that someone wishes to take Player 2's place in the game, even though he needs an unlikely three consecutive points to win. How much should this newcomer pay to get into the game (which is the same as asking how much Player 2 must accept to get out of the game)?

Using the language of the Galton board, the newcomer would need the sequence RR to win the entire pot-any other sequence results in a loss of however much she paid Player 2 to get into the game. Let's say that this newcomer is rather cautious and shrewd and, knowing that she is at a disadvantage, wishes to hedge her main wager (her payment to Player 2) with a series of side bets with onlookers at the contest. Because there are eight possible outcomes, and she can win on only one of them, she should place seven side bets to cover every possible outcome. Each side bet is a 50/50 bet on whether a certain sequence of events will happen. If the pot is $8, the newcomer should make the following side bets:

Onlooker A pays $1 if LLL happens and gets $1 if RR happens.
Onlooker B pays $1 if LLR happens and gets $1 if RR happens.
Onlooker C pays $1 if LRL happens and gets $1 if RR happens.
Onlooker D pays $1 if RLL happens and gets $1 if RR happens.
Onlooker E pays $1 if LRR happens and gets $1 if RR happens.
Onlooker F pays $1 if RLR happens and gets $1 if RR happens.
Onlooker G pays $1 if RR L happens and gets $1 if RR happens.

In the event that RR happens, the newcomer would win the $8 pot and owe a total of $7 on all of her side bets, resulting in a gain of $1. If any sequence other than RR happens, the newcomer would get $1 from one of her side-bets (the rest would be ties) and the pot would go to Player 1. No matter what happens, the newcomer ends up with $1, so to enter the contest she should pay Player 2 no more than $1.

The newcomer, of course, does not actually have to make all of the side bets. In fact, if she hopes to gain anything from a fair bet gamble, she shouldn't, because she would be guaranteed to break even. However, considering these bets, also known as hedges, helps in figuring out what the fair price is for entrance to the game. Besides, if someone offered her the opportunity to play for less than a dollar, then using these side bets she is guaranteed to make a profit. Such a guarantee of profit is called an arbitrage opportunity, and this view of probability as the hedge-able fair price plays a fundamental role in applying probability theory to finance. It also happens to be the way that early thinkers such as Fermat and Pascal viewed probability.

The newcomer should pay $1 to Player 2, which means that Player 2 could walk away from the game at this point with $1, or 1/8of the total pot. So, this is also the amount Player 2 should get if the game is interrupted, because in both cases he is leaving the unresolved game.

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ENTER PASCAL'S TRIANGLE

Let's now return to the Galton board and see what happens as we continue to add rows.

galton board

We notice that it quickly becomes unwieldy to enumerate every path. It would be nice to have some easy way to find the number of paths to each bin, given how many rows, or rounds, or decisions there are. Fortunately we can model this situation, as we did in our discussion of combinatorics in Unit 2, using Pascal's Triangle:

Pascal's Triangle

In Unit 2, we found the generalization that the number of paths from the top of Pascal's Triangle to the kth "bin" in the nth row is given by:

1/2

Let's verify that there are indeed six paths to the middle bin (k =2) of the fifth row (n=4).

4!/(2!(4-2)!)

Adding the path totals for each bin in the nth row gives the total number of paths available to the marble, 16. The probability that the marble will end up in the middle bin is, therefore, 6/16, or 3/8.

Our discussion up until now has been quite theoretical. We have used the power of combinatorics to enumerate all the paths available to a marble traveling through the Galton box, and we have calculated probabilities associated with those paths, assuming each individual path has an equal probability. If we would actually perform such an experiment, however, we would have no way of knowing in which bin a single marble will end up. We can speak only generally. However, even this general view has great power, as we shall see in the next section.

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Next: 7.4 Law of Large Numbers


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