In a hypothetical scenario,
if Lebron James made 10 successful shots in the same spot consecutively, what is the probability that his 11 shot is successful? Is it the same probability as the 1st? or did any of his prior shots affect his 11th throw?
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What about in the instance of coin toss, similar scenario,
if i throw head for 10 times, what is the probability that i will get another head in the 11th time?
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Thanks
In: Mathematics
lebron making free throws and flipping a coin are INDEPENDENT events. ie the previous result does not affect the current attempt. so whatever the odds are, remains the odds.
now if you want to ask, what’s the probability that you get 11 heads in a row? then you take the probability of one heads and multiple by probability 11 times.
Depends on what assumptions you’re taking into the situation. We tend to assume coin tosses are “independent” meaning the outcome of one flip does not impact the outcome of another. So the probability of flipping heads on a fair coin on the 11th flip is 50%, regardless of the last ten flips. Those ten heads don’t give you any more information, assuming you know the coin is fair.
If you’re not sure the coin is fair, the chance of getting ten heads in a row is so low that you might want to change you assumption and question if the coin is truly fair.
You’d have a harder time arguing LeBron’s shots in a game are as independent. If LeBron is an 80% FT shooter in his career, then a decent first guess would be 80%, but if he made the last 10, I think it’s reasonable to argue he has a better than normal chance of making this. This is sometimes called “hot hands” in sports statistics, but it can be as simple as if he made the last ten, he’s really feeling good, he’s in the zone, he’s totally healthy. The 10 made FTs give you information you can use to adjust your guess. Similarly, if he missed the last ten, you probably wouldn’t guess he has an 80% chance on the eleventh, because the fact he missed ten in a row tells you he might be hurt or something.
IMO it depends on the activity, something like making free throws, you can get into a rhythm and keep feeling out each shot so its as close as possible. Coin toss though? You have much less, if any control over the out come of that so each time is a 50/50.
The question you have to ask yourself is, does the past affect the future in any way? A coin doesn’t care, and you can’t tell the difference between a coin that’s been last flipped heads and last flipped tails. So the odds are always 50/50 for each individual throw. If you get a lot of Heads in a row, the universe does not owe you a bunch of Tails to make that up.
In the case of Lebron James, you *can* tell the difference between a successful streak and a failing streak because he’ll either be smiling or frowning. For humans, things like that can influence success on a skill action like throwing a ball through a hoop. But if you’re just going to assume he has, say, a 70% chance of making the throw each time regardless and he keeps his cool, then his 11th attempt will still be a 70% chance and all the same stuff from the coin flip still apply.
Lucky streaks happen from time to time. They’re rare, but if it’s something that turns into like a news story, it’s not like they report “guy flips coin 10 times, gets a mixture of results”. You only hear about the unique and interesting events.
You are invoking something called the “Gambler’s Fallacy”, which holds that probability sort of gets stored up and that the longer an event goes without occurring, the more likely it becomes.
This is called a fallacy for a reason.
While there are random events whose probabilities do depend on previous events, that’s definitely not the case in a coin toss, and perhaps not the case for LeBron.
No matter how many heads you get in a row, the odds of the next heads will **always** be 50%. Yes, the odds of you have an event where you get 11 heads in a row is pretty unlikely (about 1 in 2000), but the coin has no memory. The coin doesn’t think about the fact that it’s only been flipping heads for a while, and isn’t that strange.
With LeBron James, if we say that each of his shots makes it with some probability that’s related to his inherent skill and traits and not the state of the game, that’s also true. In practice, other stuff can affect it. He can get tired. A minor injury he took early in the game could be affecting him, or he could be getting complacent, or maybe he’s trying to break a previous record. Humans are complicated systems^[citation ^needed] and it’s hard to isolate them from all influences.
But I think the spirit of your question is that, no, the past shots do not affect the current one.
If you want, you can play a game to convince yourself of this at home: flip a coin a whole bunch of times, and while you likely won’t get to 10 heads in a row, if you keep a record of each flip, you can easily show that the odds of heads or tails after 3 heads in a row are exactly the same as after 2 heads, or after 2 tails, or after 4 tails, or after heads-tails-heads, or whatever.
We know the coin has a 50/50 probability of landing on heads/tails. We don’t know what the probability is of LeBron making or missing this shot from this location. We can infer from 10/10 successful shots that the probability is very high, thus it would make sense to presume he’d make the 11th. The 11th shot is independent, but the difference between a coin and LeBron is that we have lots and lots and lots more data about coins than we have about LeBron shooting from this location.
We only know he took 10 shots, and succeeded all 10 times.
Coin tosses are independent; flipping 10 heads in a row doesn’t affect the outcome of the
If you’re falling into the trap “11 heads is very unlikely, so the 10th flip is probably a tail” then notice that tye combination of flips HHHHHHHHHHH (11 heads) and HHHHHHHHHHT (11 heads followed by 1 tail) have the same probability of occurring; that is 0.5^11. They are both unique outcomes out of 2^11 total Possible outcomes
But it might be different for Lebron shooting freethrows. I don’t know anything about basketball, but maybe after 10 throws he is getting tired or careless and so is less likely to Make his 11th throw than his 1st.
There’s an easy mathematical answer and there’s a deeper answer.
The easy mathematical answer is it doesn’t matter. Both coin tosses and free throws are “independent events”. That means there isn’t anything that changes the next outcome based on the previous outcome. Let’s talk about free throws later, I already hear objections. For a coin toss, there’s nothing cosmic that ever makes one side more likely than the other. If you have flipped 99 heads the probability of the next coin being tails is still 50%. (I will have an interesting footnote about this.)
The deeper answer, and the objections I hear, is that some things aren’t really as “independent” as the math dictates.
If I try to make a free throw and fall short, I’m going to throw the ball harder next time. If I overshoot, I’ll use less force next time. If I make the first shot, I’ll try to do it the same way. So for an inexperienced person, to some extent we kind of expect if they take 4 or 5 shots the last few shots are the most likely to make it, with some falloff as they get tired.
But Lebron James is a professional who has practiced tens of thousands of free throws. He knows how to sink one and is making fewer adjustments between throws. If he has sunk 10 shots in a row without moving, all we can go by is that it must be very likely his 11th will make it. But if we look at his lifetime free throw percentage, we should use that instead. I did a quick search and it says he has a 73.5% rate. That doesn’t mean he’s never made 10 in a row. It means if he’s made 1,000 shots, 735 of them were successful. That’s way better than my percentage would be. So the problem here is, “How likely is a free throw to be made?” is very unique to each person. That’s why coin tosses are better for probability. 10 coins should behave the same way unless we know they are manufactured very differently.
That said, even coin tosses are weird. A recent paper proved that due to some interesting Physics, a coin is ever-so-slightly more likely to land in the same orientation it was when you flip it. So if you held it heads-upwards every single toss, they found that human coin tosses have a teeny-tiny bias, slightly less than 1%, towards heads. So if your life ever depends on a coin toss and you get to make the toss, be careful to bet on which side is up. If you call it in the air and couldn’t see how it was held, it sort of balances out.
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The only way this is not true is for “dependent” events, which are events where something about the previous attempts changes how future attempts work. A Bingo game is a good example. The game starts with balls 1 through 100 in a bin. If you take one ball out and it is 35, now there are only 99 balls. At the start, each ball only had a 1% chance of being selected. Now each ball has a 1 in 99 chance, which is slightly more than 1%. So if you are betting on one particular number being pulled, the odds it is pulled first are very low. But each number after that it gets more likely, and you will never pull all 100 balls without seeing your number unless someone cheated.
That is not so with coin tosses. Nothing in the universe prevents a coin from landing heads 1,000,000 times in a row. The key part to understand here is nothing “decides” how the coin lands. There is no cosmic force tracking every coin toss and trying to balance them. It is an act that can end in one of 2 ways, and ideally nothing about the last flip changes the coin, so it will always end one of those two ways with equal probability.
The hard part of this is if you’re gambling, if you choose the side that doesn’t land you think you “chose wrong”. What gamblers have to realize is “bad beats” exist. There have been poker plays where a person had 95% or better odds of winning and *lost*. That doesn’t mean they shouldn’t have bet on their own win. It just means they got unlucky. Being “good” at gambling means you know it’s smart to always bet on the highest odds of winning and to never let the times you were unlucky change what you know to be true about probability. People who are “good” at gambling understand it is possible to do everything “right” and lose. That’s why it’s called “gambling”.
The odds of flipping a coin and getting heads are 50/50.
The odds of flipping a coin 10 times in a row and getting heads 10 times in a row are 0.1%.
The odds of getting heads on the 11th flip are 50/50. Because the odds of flipping a coin and getting heads are 50/50, right?
The odds of flipping a coin 11 times in a row and getting heads 11 times in a row are 0.05%.
The odds of flipping *once* have nothing to do with the odds of flipping a specific result over and over and over.
Consider flipping a coin three times, there are eight possible outcomes: TTT, TTH, THT, THH, HTT, HTH, HHT, HHH.
If you look at any individual coin flip you see that Tails occurs as often as Heads. T** happens four times, and H** happens four times; * T * happens four times, and * H * happens four times; ** T happens four times, and ** H happens four times. So clearly, any individual coin toss has 4/8=50% odds of being Heads or Tails.
But what about the odds of getting HHH? Well, that only happens 1/8=12.5% of the time.
Now what about the odds of getting HHH if we already have HH*? Well, if we already have HH* then that means that we are in a world where TT*, HT*, and TH* didn’t happen. So, despite those accounting for 6/8 of the total possible “three coin flips” outcomes, we’re only looking at what fraction of the time HH* (which is the remaining 2/8 possible outcomes) becomes HHH (1/8 outcomes). And what fraction of 2/8 is 1/8? Half. 50%.
Your hypothetical is *similar* in that each shot would be *independent* just as the coin flips were independent – so the odds of hitting the last shot is the same as the odds of hitting the first shot. However, your hypothetical will be different because some more complicated math is needed when the odds of each shot aren’t *exactly* 50:50. (Basically, if the odds are 75% he makes it, you’d have to *weight* out the possibilities differently; the first shot’s outcomes would look more like like (H,H,H,T)=(3H,1T) since that gives the proper 1/4 odds of hitting tails – then the second shots outcomes would compound on top of that like (H(H,H,H,T),H(H,H,H,T),H(H,H,H,T),T(H,H,H,T)) which would give (9HH,3HT,3TH,1TT) and so on.)
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