How much more common is it to flip 50 heads and 50 tails than 99 heads and 1 tail, and why?
Somebody was trying to tell me the odds are the same, and I need to explain why they aren’t.
I know there are only 100 ways to flip 99:1, but millions or more ways to flip 50:50.
I’m guessing there was some miscommunication between you and your friend. Any two unique combinations of heads and tails are equally likely. So these two outcomes have the same probability:
HTHTHT (50% heads)
HHHHHT (~83% heads)
However, as you noted there are more ways to have a run of 50% heads than 83% heads. This means it is more likely to a run of 50%.
That explains it right there. Toss a coin 100 times, and there are only 100 possible sequences that contain 1 head + 99 tails. But there are far more possible sequences that contain 50 heads + 50 tails. So some sequence of 50+50 is far more likely than some sequence of 99+1.
If you narrow the acceptable outcomes down to
-99 tails followed by 1 head, versus
-50 tails followed by 50 heads,
either of those two outcomes is unique and equally probable.
Thanks for the quick reply. I acknowledged that any 2 unique series were equally likely to him at the time, and then said my point about the 100- ways vs millions.
Is there a simpler way I can explain it to him? he was very intent on the 50/50 having the same odds as 99/1
He was making a point and explained it more or less like this:
“for example if you flip a coin 100 times. The chance of getting 99 tails and 1 heads is actually the same as getting exactly 50 heads and 50 tails.”
This sentence is incorrect for sure.
As far as intuitive explanations go, your last sentence in the OP is as good as anything. Illustrate why there are only 100 ways to flip 99:1, then from there, why there are so many more ways to flip 98:2. Explain that it continues from there until the 50:50 median.
Statistically speaking, the number of combinations of X things taken Y at a time is X!/(Y!(X-Y)!), and 100!/(99!(100-99)!) < 100!/(50!(100-50)!)…but that’s unlikely to convince anyone who doesn’t understand this in the first place.
I think what’s confusing your friend is a sort of reverse gambler’s fallacy. The crux of the gambler’s fallacy is that past results have no effect on future probabilities, and is often illustrated with a question such as “If five consecutive coin flips have come up heads, what is the probability that the next flip will be tails?”, the answer to which is 50 percent. He might have heard this sort of explanation, and decided it meant that a 5:1 result is as likely as 6:0 in all cases, which is not so; it’s true only when you’ve already gotten a 5:0 result. Maybe approaching it from that angle might help.
Explain it using 10 flips instead of 100. You can manually write out all 10 sequences of 9 heads 1 tails. Then you can write out 11 sequences of 5-5, write out more if you need to, but if you need to I’d advise you to give up. Probability is not intuitive. If he lacks the ability to grasp it, well, there isn’t much you can do.
Yes I will consider doing it with an example of 5 flips. Thank you.
An approach that might work for explaining the issue is this:
Suppose you start tossing the coin and the first two results are heads. There’s no way that this sequence can be the start of a sequence with 99 tails and one heads - but there are plenty of ways that two heads can be the start of a sequence that comes out 50/50.
Think up betting game based on the premise, say with 10 flips, and get him to agree to play it 10 times - maybe losing a bunch of money will convince him…
This is the best solution. Either he catches on and you win the argument, or you have a reliable new income source. For best results, offer him odds that would be favorable if he were right.
No, with 5 flips, the probability of getting equal numbers of heads and tails is much, much lower than the probability of only getting one heads.
Try 4 flips:
Here are all 16 possibilities:
HHHH
HHHT
HHTH
HHTT
HTHH
HTHT
HTTH
HTTT
THHH
THHT
THTH
THTT
TTHH
TTHT
TTTH
TTTT
As you can see, there are 6 cases with 2 heads and 2 tails. There are 4 cases with 3 heads and 1 tail. So 38% of the time you will see the same number of heads as tails, while only 25% of the time you will see all heads except one tail.
Now about about 6 flips.
Here are all 64 possibilities:
HHHHHH
HHHHHT
HHHHTH
HHHHTT
HHHTHH
HHHTHT
HHHTTH
HHHTTT
HHTHHH
HHTHHT
HHTHTH
HHTHTT
HHTTHH
HHTTHT
HHTTTH
HHTTTT
HTHHHH
HTHHHT
HTHHTH
HTHHTT
HTHTHH
HTHTHT
HTHTTH
HTHTTT
HTTHHH
HTTHHT
HTTHTH
HTTHTT
HTTTHH
HTTTHT
HTTTTH
HTTTTT
THHHHH
THHHHT
THHHTH
THHHTT
THHTHH
THHTHT
THHTTH
THHTTT
THTHHH
THTHHT
THTHTH
THTHTT
THTTHH
THTTHT
THTTTH
THTTTT
TTHHHH
TTHHHT
TTHHTH
TTHHTT
TTHTHH
TTHTHT
TTHTTH
TTHTTT
TTTHHH
TTTHHT
TTTHTH
TTTHTT
TTTTHH
TTTTHT
TTTTTH
TTTTTT
As you can see, there are 20 cases with 3 heads and 3 tails. There are 6 cases with 5 heads and 1 tail. So 31% of the time you will see the same number of heads as tails, while only 9% of the time you will see all heads except one tail.
This trend continues. For 100 coin flips, the probability of 50 heads/50 tails is 8%, while the probability of 99 heads/1 tail is 0.0000000000000000000000000008%.
This is easy to calculate. The probability for N flips and X tails, is “N choose X” divided by 2^N.
Seems like a money making opportunity to me.
In college I had a friend who was a math major who did not understand probability. We loved playing poker with him.
[nitpick] Not quite. After 5 flips that all come up heads, there is a small, but statistically calculable (but I can’t be arsed to do the calculation) probability that the coin is unfairly biased toward heads. That would be subtracted from the 50 percent. So the actual probability is some small but calculable amount less than 50 percent. ETA: If you preface it with “Assume an unbiased coin and flip method,” that small difference goes away. [/nitpick]
This is one of those things that seems kinda obvious once you think about it, but “COIN FLIPS ARE ALWAYS THE SAME PROBABILITY!!!” is one of those things most people think they know in the back of their brains.
You can’t calculate the “probability that the coin is unfairly biased towards heads” except relative to some assumed model of how exactly that bias would work, what other biases are possible and how they would work (e.g., coins “biased” to alternate between 5 heads in a row and then 5 tails in a row, or such things), and a priori probability distribution for the various biases. You certainly can’t give a meaningful universal calculation for “Given that a coin has come up heads 5 times, what is the probability that it is unfairly biased towards heads?”. Probabilities only exist relative to probability distributions.
Yes, I know all that. That’s why I couldn’t be arsed to figure out all those details and do a calculation. Nevertheless, after 5 straight heads, a biased coin starts to become a suspicion. One who’s probability is calculable, once you figure out all the possibilities, and that very small probability would be subtracted from the 50%. I WAG it to be somewhere in the 49.9999 area for tails, give or take a few 9’s. You’d need a lot more straight heads before I started to consider it a realistic possibility, worth the trouble of calculating.
As I said, preface the thought experiment with “assume an unbiased coin and flip method” and the issue goes away, which is the main point I was intending to make.
I’m curious, in practice, what calculation you would carry out if you could be arsed to do so.