Probability Paradox
December 14, 2006 am31 9:40 am
In Combinatorics we are studying some combinatorial probability. Tomorrow the kiddies will have to deal with this annoying bit of coin flipology:
1. The more times a coin is flipped, the more likely the results are close to 50/50 heads/tails
2. The more times a coin is flipped, the less likely the results are exactly 50/50 heads/tails
So, how upset will my students be? (I am guessing: fairly)
And, do you have a neat way of explaining this seeming paradox?
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JD2718 
It’s easiest to understand the “paradox,” I think, if you express it as the contrast between a single quantity (the heads-tails difference) and a ratio (the heads-tails difference divided by the total number of flips). This distinction is currently implicit in your language: point 1 is true only if “close” applies to the ratio.
The key insight for your students to make is that both the expected heads-tails difference and the number of flips grow, but that the number of flips grows faster.
You could use the idea of a random walk–that the heads-tails difference has no “memory” of its initial value–to illustrate why the expected heads-tails difference grows as it does.
Lastly, I might avoid the wording of point 2 if I were you, given that the probability of an exact heads-tails split is zero for all odd N, no matter how large. :)
Well, yes. Half of 5 is two and a half, and it is hard to get two and a half heads!
They will be able to calculate, say, the probability that 2 of 4 flips are heads is 6/16, that the probability of 3 of 6 flips being heads is 20/64, and that 4 of 8 being heads is 70/256 (about 38%, 31%, and 27%, respectively.)
I should emphasize the binomial-ness of all this, of course. But I am thinking there is a visual representation, ie, the axis of symmetry of Pascal’s triangle, that they may hang onto best. I do this in 2 hours, and that is how I am leaning.
I will post a follow-up puzzle, or two, tomorrow.
Just to add another layer of complexity, I was reading a couple weeks ago about coin tosses, and the 1/2 probability of getting H or T is only true if you have a perfect coin-tosser. Most real human beings are not perfect coin-tossers, and will have a slightly higher chance of tossing the same side as they started out with (if it starts heads up, it will be slightly more likely to land heads up than tails up–if you’re really dextrous, and you practice, you can get this to happen every time, and with enough of a coin wobble that no one will notice).
Now if only I could find that article again… I really need to clean my office.
Gah, I have to remember not to use the less-than signs in your comments.
Taking Lsquared’s comment and going even further from the intended topic: can you figure out a way to take an unfair coin (or an unfair flipper; so long as the probability of heads is consistently some value p between 0 and 1) and create a system that gives you a way of making a random (50:50) choice? (In other words, how do you use an unfair coin to simulate a fair coin?)
My personal resolution of the seeming conflict: after many flips, there are many more values “close” to 1/2, and since there are many more situations which are close, the fact that each of them individually is of decreasing weight is not too suprising.
I asked them to calculate P(.5) for n = 2,4,…,18,20
I asked them to calculate P(.4 less than or equal to P less than or equal to .6) for the same values.
I asked them to make several observations about the data.
I will follow up (I am still thinking) by asking them to find n such that the P(.5) = 0% when rounded to the nearest whole percent. I don’t think it is a great exercise, but it has the possibility of being memorable.
Btw, if you want to jump in, I don’t know the answer yet.
It’s a big number — we want P(n) = C(2n, n)/2^(2n), the probability of getting exactly n heads on 2n tosses, to be less than 1/200 (so it rounds to 0 percent). Using Stirling’s Approximation (which gets very good for binomial coefficients, since the (small anyhow) errors tend to cancel out), we have P(n) is about one over the square root of pi*n, which tells us that n should be at least 40000/pi, which is about 12732.4. Indeed, with n = 12732, we get the probability of 12732 heads in 25464 flips is 0.005003…, while the probability of 12733 heads in 25466 flips is 0.00499983…
(All calculations courtesy of my computer.)
If you only ask that the probability be less than 1% (instead of less that 0.5%), it only takes 6366 flips.
I don’t talk Mathematics talk, but here’s how a Social Studies teacher would think about, as an issue of logic.
Since the odds of landing heads and landing tails is both 1 in 2, the longer one tosses the coin, the more likely that abberant patterns, such as five heads in a row, will cancel each other out, and come closer to 50%/50%. However, for the percentages to remain exactly 50%/50% much of the time, one would have to keep fairly even sequences, such as “one head, one tails” or “two heads, two tails.” The longer one flips the coin, the less likely it would be to maintain such sequences. In the first case, randomness brings the odds closer to 50%/50%; in the second case, randomness makes it harder for the odds to be exactly 50%/50%.
Now I will let you mathematicians get back to your numbers.