Care to test your intuition on a probability problem?

½ – there is a fifty fifty chance of that happening.

If you had worded it differently, the asnwer would be different.

I am coming out at 5%, but that is assuming that each person has a chance to get their hat, which is impossible, because with each wrong hat handed out, one more of the subsequent people has a 100% probability of not getting their hat.
I think until the halfway point, the probability of #10 getting the wrong hat is at 50%, the rest have a 100% chance of getting the wrong hat, so overall the chance of none of them getting their hat is at 50%. I think.

I don’t think this problem is a very good example of people’s intuition about probability, because the scenario and the language make it more of a logic puzzle than of intuition about probability.

A better example would be a common one such as: Paul thinks he has a 50% chance Sue will give him a ride, and a 50% chance Mark will give him a ride. Sue and Mark aren’t coordinating at all; the just might or might not show up. What are Paul’s chances that at least one of them will show up?

I have seen a smart teenager who had had probability lessons in math at least for the last two years, not only give the wrong % chance, but adamantly insist, believe and trust his wrong conclusion even after it was repeatedly explained to him in detail using non-math explanations.

@elbanditoroso, I do not see what you are saying. How could the problem have been reworded?

The answer is about ⅓, .36 to two places and rapidly approaching 1/e as the number of people increases, where e is the natural log base.

When I first came across the problem, I thought that the probability would be fairly low that nobody would get the right hat. It seemed to me that the probability should keep getting smaller as the number of hats increased.

The problem relates to a general mathematical concept. If you take a sequence of objects and reorder them so that none of the objects occupy the same position, the new sequence is called a derangement, where, in this case, derange means to disorder.

I’d think at least half, as even distributed randomly, the one handing out would surely hand womens hats to the ladies and vice versa helping with the accuracy. I mean, if you see a middle aged man surely you won’t hand him a bonnet with pink flowers.

Math is not where my intuition works, but if I read the problem correctly, I would say 1/1000. What I mean to say is that I think there’s a very strong likelihood that at least one person will get the right hat.

Horrible wording. The chance not a single person gets the correct hat, when more typically the question would be what’s the probability of someone getting their correct hat.

My guess is 9/10 or 90% chance people don’t get the correct hat. Guessing on my own I might have said 75%, but ⅔ is the next choice and seemed too low.

@LostInParadise So, ⅔rds chance people do get the correct hat?

I don’t see how this is an intuition problem. Seems like it’s straightforward statistics. (which I haven’t done in garbled-mumbled years.)

@JLeslie , Yes ⅔ chance that at least one person gets the correct hat.

@LostInParadise Ok, so I STILL interpreted the Q wrong, but that makes sense to me that it’s around ⅔. I was thinking how many will get the correct hat, and I thought it would be between 10–25%, so my guess was 90% but you’re saying the odds of at least one person.

I still say it’s the type of Q people get wrong because of how it’s written even if they do the math right.

I had thought the problem was similar to the birthday problem and that it was very likely at least someone would get the correct hat. Like @Jeruba I thought the odds were about a thousand to one that no one would get the right hat.

So much for intuition. I don’t know how much a hat check gets paid but I don’t think it is enough.

Anyone interested in the math behind the problem, you can check here The good thing about it is that it is relatively short, but it could have been done better toward the end.

@LostInParadise, I don’t follow this ^^. Could you just please tell me if my answer is right or wrong? My answer in words, that is; I’m none too sure of my answer expressed as odds. It seems to me that @flutherother said the opposite of what I said, but I’m doubtful of that too.

Your guess is the same as my initial guess, that the chances of nobody getting their hat is very low. As I interpret it, @flutherother had the same feeling. What he meant was that he thought the chances are 1000 to 1 against no one getting the right hat, which works out to a likelihood of 1/1000.

In fact, the chances are about ⅓ (over 36%) that not a single person gets their own hat. Other than by showing the math, I can’t think of any way of explaining why it works out the way it does.

1/10.

Here is a really simple case where our intuition fails us.

Given 4 cards numbered 1 to 4, choose 2 of them. What are the chances that they add up to an even number? 50%, right? Well, no, ⅓. Here is a way of looking at it. Suppose the first card chosen is even (2 or 4). To get an even sum, the next card must be even, but there is only one even number remaining and two odd numbers, so the chances are 1 out of 3. The same logic holds if the first card chosen is an odd number.

That is counterintuitive, but logically there are two ways to get an even total (1,3) and (2,4) and four ways to get an odd number (1,2) (1,4) (2,3) and (3,4) so you are twice as likely to get an odd total.

The way I now look at the hat problem is that every time you add another gentleman with another hat you create one more possibility of matching a man with his hat. However, there is less chance of any one man being matched with his hat as there are now more hats to choose from.

The advantage of the approach I used for the card number problem is that it generalizes easily. If the numbers go from 1 to 2n then, if either an even number or an odd number is chosen for the first card, there will be n-1 ways to form an even sum and n ways to form an odd sum, so the probability of an even sum is (n-1)/(2n-1), which must also be the proportion of all pairs of cards that have an even sum. This probability goes to ½ as n gets larger, which is why I chose a small value to give a more dramatic result.

Here is yet another way of looking at the card selection problem in which our intuition, with a little guidance, can find the right answer.

Suppose that after choosing the first number, the card is placed back in the pile, which is then shuffled before selecting the second card. With two even numbers and two odd numbers, our intuition again tells us, correctly this time, that there is a ½ chance of getting an even sum. So what’s the difference between the two cases? The only difference is that by returning the first card it is now possible to choose the same number twice. In each of these results the sum is even. That means that in the first case, by eliminating the same number possibilities, there must be more odd sums than even sums.