I’m really not that interested in talking about this again, but I read a passage (about an entirely different topic) that made me think about this. Here’s the passage:

You can flip a coin 100 times and have it come up tails 60 times. It’s an unlikely result, but it happens. About 3 percent of the time, you’ll get 60 or more tails.

You can flip it again the next day, and it can come up tails 60 times again. This is possible, and no doubt has been done before. (Again. Don’t do this. It is so boring! Read more football instead. That’s a good use of your limited time on our doomed planet.)

If you flip the coin 100 times for 100 days, though, you’re not going to keep getting tails 60 percent of the time. Regression, the most powerful arm of probability, will tug the percentage back toward 50 percent. It just will. It doesn’t have a choice, and neither does the coin.

I’ll go over what stands out in the comment below.

3%? That seems really low for 60 of one side. I would think the chances that happens would be a lot higher. What about 59 of one side? 58? 57? I would think there must be some big jump in percentages at some point. The percentage of seeing 51/49 must be pretty high. Same with 52. What does the drop in percentages look like, getting to 60/40?

Let’s say the drop occurs between 54/46 and 55/45, going from 40% to 10%. If this were true, suppose you flipped the coin 95 times and you already had 54 of one side. Wouldn’t your chances of getting that side in the next 5 flips be a lot lower?

The other thing that stood out: the concept of regression to the mean seems like the one I’m using to argue my position.

I’ll address some of your other points later, but it’s important for you to understand that the limit on this condition makes 3% far more reasonable than some number close to 50%.

You’re specifying HEADS: 60 heads, not 60 heads or 60 tails. You can see how this makes a difference just with flipping two coins. If you flip 2 coins, the odds of them being the same are 50% and the odds of them being different are 50%. But the odds of them being BOTH HEADS are 1/4, or 25%, and the odds of them being BOTH TAILS are 1/4, or 25%. Just with two coins, you’ve cut the likelihood of getting your desired result in half, from 50% to 25%.

The math isn’t exactly parallel, because in the scenario you describe, each flip is independent of the other 99, and the 60+ heads can be in any order (they don’t have to be on specific flips), but you’re still talking about a LOT of flips (100) where you’re looking for a range of specific numbers (60 to 100) of specific outcomes (heads).

Rather than explain the math, I thought I’d just show you a demo. Here is a spreadsheet I made simulating 100 coin flips, 220 different times. I’m calling each set of 100 flips a “trial” for clarity’s sake. Each trial is numbered at the top, mostly for my own clarity. The smaller number beneath each trial number is the number of HEADS flipped in the trial below.

The red number in the upper left is the number of trials where 60 or more HEADS were flipped out of 100 tosses. As I’m looking right now, it says 7. The sheet is set to auto-recalculate, so if you refresh the page, when it loads, the flips will randomly re-simulate, and you’ll see different results (give it 10 seconds or so after the refresh; it takes a moment). I refreshed five times and the red number moved between 5 and 8. 220 is a very small sample size, so that kind of fluctuation is to be expected, but you’ll probably only see numbers pretty close to this range.

The blue number at the very top is the percentage (out of 220 trials) where 100 flips resulted in 60 or more heads.

Right now I have it set to let you look only. But if you want to play around with it (make it 61 or more heads, or 57 or more heads), I can give you editing privileges. You’d only have to change two cells to do it.

I’m not finished thinking about this, but something occurred to me, after thinking about what you said below:

I think this is helpful. Perhaps a better way to think about this is 60 heads

or60 tails–versus only one of these outcomes. This might not be clear, but let me explain how I arrived at this. While looking at your spreadsheet, I noticed that 39-38 heads seemed rare as well. 39 heads basically means 61 tails. I guess this isn’t a revelation so much something that made your example a lot clearer and more palpable.What would be the odds of getting either 60 heads

ortails? If this number is closer to the 3% than 50% that would be a little weird, too, right?I’m just guessing, but I would expect the odds of getting 60+ heads OR 60+ tails would be around 6%. Going to adjust the sheet and test it now.

Okay the sheet now shows the number of tails vs. the number of heads, and a total number of trials with either 60+ heads or 60+ tails. Looks like about 6% to me.

I admit if I were to just estimate what the probability would be, there’s no way I would have come up with anything near 3%. I might have guessed 25% or so, but when I thought about it 3% sounded much more likely, ‘though it still seems low to me too.

That makes sense. 6% still seems surprisingly low, right? Or is it just me?

I’m mildly interested in doing this, as long as it’s not too complicated. I’m interested in seeing if a big jump occurs or not between 50 and 60. If you don’t mind, check how many times 55 and 45 shows up.

I have it displaying the number of trials where there are 59+ heads or tails right now. It’s easy to adjust. Click the cell where the big red number is, change where it says “59” to whatever number you want, and hit enter. Then do the same thing for the cell where the big green number is. You don’t have to change the cells in the top row where they say 60; just keep in mind that they will display results for whatever number you’re changing to, not 60 anymore.

It didn’t allow me to do this.

By the way, any predictions about the way the percentages will change, as they get closer to 50 flips of heads or tails? What would you predict as the flips go farther from 60? For example, would 70 flips be closer to 0% or 3%?

Try it now. And yes, if the relationship isn’t exponential, I suspect it’s at least kind of exponential (I’m saying this because I haven’t worked out the exact math yet), the way the two-heads-with-two-coins thing is exponential, which would mean a steeper increase with every number beyond 50. 2 squared is 4, 3 squared is 9, 4 squared is 16, etc. But I don’t think this condition is that steep.

(OK, the problem was that I didn’t log in, I think. I will try to do this later.)

No; the problem was that I had editing turned off except for me.

Wait, I just realized that 59+ means “>=59.” Shouldn’t we just look at 59, and not all the numbers that exceed it?

I changed the number to “>=55” and using both heads and tails, it’s about 30% which seems like a huge jump.

I now adjusted it to “=55.” The numbers are lower–4.55% each or 9.1%, which doesn’t seem like that big of a jump.

At 51, the percentage is 4.09 and 4.55, which comes out to 8.6%, close enough. That’s seems low. What could be going on? My thought is that basically, from about 45-55 (heads or tails), the percentages are the same. My error is thinking that getting 50 heads or tails is significantly more likely than 55, 48, or any number between 45-55. (I didn’t really plug in 45, but I’m just assuming.)

It should be noted that 220 flips isn’t all that much, but I wouldn’t expect the percentages to be dramatically different. Or is this wrong? If we did a million flips, what’s the chances that the percentages would be dramatically different.

Still, should we think that a jump from 9% to 3% is not significant? 9% versus something closer to 0% does seem significant. If, on the first round of a 100 flips, you had 60 of either heads or tails, the odds of getting 65 (let’s say) would be closer to zero. So if you had 64 of either heads or tails and had a few more flips to go, wouldn’t we expect the the other side of the coin? That is, those last flips wouldn’t be 50/50 probability.