Probabilities

[NORiculous]

Goodness, I need a math person. Please.

Here is the picture:

Total of 10 balls lottery.
Team A : 4
Team B: 3
Team C: 2
Team D: 1

Are the probabilities the same if we pick the winner first (first ball out) or if the « last standing »( last ball out) is the winner.

This is a bit beyond my capabilities but its been haunting me for a while.

I can « feel » like the odds « must » be different but have no idea how to get the answer.

Anyone?

Click to read more...

 

[morehockeystats]

Take a college course?

 

[Doctor No]

Probabilities are the same.

 

[Canadiens1958]

Probabilities are the same.

Yes, just the order of attribution of prizes changes.

 

[NORiculous]

Probabilities are the same.

So the first ball has just as many chances of being team A then the last ball, before anything is picked. Right?

But when team A does get picked first, the probabilities of the last ball have to be changed, no?

Don’t the probabilities change once a ball comes out? And since team A had 40% chance on the first ball, it has the most chances of getting a reduced probability for the last ball... no?

 

[Doctor No]

(Try this - I'm a former college probability professor, for what it's worth).

Pull all ten balls out, then line them up in order without looking at them (cover them with a blanket if you must).

What's the probability that the first ball is Team D?

What's the probability that the last ball is Team D?

 

[Doctor No]

This problem may be easier to wrap your mind around if you consider a smaller (but almost as interesting) problem. You're right that the probabilities change as balls are drawn (the key to the above puzzle is that you must know nothing about the other balls' positions - otherwise it gets closer to the Monty Hall Problem, which if are interested in this stuff at all is a must-Google).

Suppose that...

Team A has 1 ball.
Team B has 2 balls.

What's the probability that Team A's ball is picked first? 1 in 3.

What's the probability that Team A's ball is picked last? (if I'm right above, then it sure as hell better be 1 in 3, but let's prove it a different way).

In order for Team A's ball to be picked last, the balls have to appear in the following order: B-B-A.

What's the probability that B is picked first? 2 in 3 (66.7%).

Given that B was picked first, what's the probability that B is picked second? Well, if B was picked first, then there are one A ball and one B ball remaining. So given that B was picked first, the probability that B is picked second is 1 in 2 (50%).

Given that B was picked first and second, what's the probability that A is picked last? If B was picked first and second, then there is one ball remaining - Team A's. So given that B went first and second, the probability that A is picked third is 100%.

Put these all together, since "and" probabilities are multiplied:

P (A is picked third) =

P (B is picked first) * P (B is picked second | B is picked first) * P (A is picked third | B is picked first and second) =

[The "|" notation is probabilistic slang for "given that" - so you read the second value as "Probability that B is picked second GIVEN THAT B is picked first", which fortunately we calculated above already.]

66.7% * 50% * 100% =

33.3% =

1 in 3.

So we took the long road there, but the probability that Team A's ball is picked last is 1 in 3, exactly the same probability that Team A's ball is picked first.

 

[Doctor No]

Once you get your head wrapped around that (and give it a bit to simmer - it's harder than it may look) - try adding a third team and playing with that. Probability is not intuitive for most people - when I taught college probability, I'd bring a craps table and a deck of cards into the class for the first week. I'd give everyone Monopoly money and then I'd take it back via gambling.

Once everyone was suitably interested ("okay, I can use this stuff"), then I had their attention for the rest of the semester.

 

[abo9]

So the first ball has just as many chances of being team A then the last ball, before anything is picked. Right?

But when team A does get picked first, the probabilities of the last ball have to be changed, no?

Don’t the probabilities change once a ball comes out? And since team A had 40% chance on the first ball, it has the most chances of getting a reduced probability for the last ball... no?

I'm pretty bad at counting problems so someone correct me if I'm wrong, it's a nice practice anyway.

I believe the bolded is true, but it also is a different problem than initially because you now have information in that you know which ball was picked first.

If you go blind (0 information) in both scenario:

1- First ball out is winner: Obviously you got A wins with 40% probabilities, B with 30% probabilities, C with 20% probabilities and D with 10% probabilities

2- Last ball out is winner: Remember, you have absolutely no idea which balls were taken out before. You could write out all the possible solutions and add up the probabilities but that would be time consuming. Instead, try imagining the whole draw as a bunch of possible paths, so each starting point must also be an end point because draw can always go both ways on a "path", so the probabilities must also reflect this and be the same right?. Eg one path is A-A-A-A-B-B-B-C-C-D obviously there is a draw which outcome is D-C-C-B-B-B-A-A-A-A.

Try it out with a smaller problem to familiarize yourself with the problem and create examples it should make it clearer (eg a problem with 2 teams and 3 balls)

Edit: I was so slow writing my response that Doctor No. answered you before with a similar answer haha. I'll leave it here anyway but just clarifying that I did not see the above answer before posting !

 

[Tweed]

Edit: I was so slow writing my response that Doctor No. answered you before with a similar answer haha.

Naw man... he's just that fast. Don't beat yourself up.

 

[Fourier]

Once you get your head wrapped around that (and give it a bit to simmer - it's harder than it may look) - try adding a third team and playing with that. Probability is not intuitive for most people - when I taught college probability, I'd bring a craps table and a deck of cards into the class for the first week. I'd give everyone Monopoly money and then I'd take it back via gambling.

Once everyone was suitably interested ("okay, I can use this stuff"), then I had their attention for the rest of the semester.

Probability is definitely not intuitive for most people. I've had conversations with colleagues about precisely the OP's question and they do not believe the answer. (Actually about the probability that the second card in a shuffled deck is an ace.) We're talking about people with Ph.D.'s in Mathematics here as well.

 

[morehockeystats]

One of the very few good things of growing up in the Soviet Union was the availability of books by E. Aleksandrova and M. Levshin which were "adventures into advanced math". One such book was dedicated to the early days of probability theory, to Chevalier de Mere's problems, and how Pascal and Fermat solved them. That gave me a wonderful introduction into the world of combinatorics and probabilities at a very early age (I was 11 or 12 when I read it, I think).

I am pretty sure these books were never translated, but it might just be that Martin Gardner produced something similar.

 

[Doctor No]

One of my classmates in grad school was from Russia (and would say things like "oh, we did measure theory in fifth grade" - not bragging, just matter of factly) has spoken to me about those books, and I've always wanted to see them.

And I agree that they don't appear to have been translated, and that Martin Gardner is nice too.

 

[Fourier]

One of my classmates in grad school was from Russia (and would say things like "oh, we did measure theory in fifth grade" - not bragging, just matter of factly) has spoken to me about those books, and I've always wanted to see them.

And I agree that they don't appear to have been translated, and that Martin Gardner is nice too.

Over the years I have experienced claims like the one above many times. Far more often than not it has been to the detriment of the student. Abstraction can be a problem if it happens too early. In the specific case of probability theory for example one could argue that probability theory is just a sub-branch of measure theory. For me this is not really true. I like to think of probability theory being measure theory with twists and attituded. If the first time you see probability theory is from a measure theory perspective there is a pretty good chance you can prove the Central Limit Theorem but not be able to correctly answer the OP's question.

 

[WarriorOfGandhi]

Probability is definitely not intuitive for most people. I've had conversations with colleagues about precisely the OP's question and they do not believe the answer. (Actually about the probability that the second card in a shuffled deck is an ace.) We're talking about people with Ph.D.'s in Mathematics here as well.

I haven't taken a math class in fifteen years. Is the probability that the second card is an ace something other than 1/13?

 

[abo9]

I haven't taken a math class in fifteen years. Is the probability that the second card is an ace something other than 1/13?

Ok so intuitively I thought "No way, what if the first card picked is an ace? The probability must be very close, but still a tad lower"

And so I did the calculation (prob of picking a on first draw X prob of picking ace on 2nd draw + prob of pick any other card on 1st draw X prob of pick ace on 2nd draw) = 1/13

Actually idk if I'm doing something wrong but using similar calculations for the picking an ace on the third draw is also 1/13. Could it be that without any information about previous draws, it does not matter where you pick, the probability of picking an ace will always be 1/13? Is that an example of a famous theorem?

 

[Fourier]

Ok so intuitively I thought "No way, what if the first card picked is an ace? The probability must be very close, but still a tad lower"

And so I did the calculation (prob of picking a on first draw X prob of picking ace on 2nd draw + prob of pick any other card on 1st draw X prob of pick ace on 2nd draw) = 1/13

Actually idk if I'm doing something wrong but using similar calculations for the picking an ace on the third draw is also 1/13. Could it be that without any information about previous draws, it does not matter where you pick, the probability of picking an ace will always be 1/13? Is that an example of a famous theorem?

In fact, the probability that the second card in the deck is an ace is indeed 1/13 as your calculations show. However, your first thought "No way, what if the first card picked is an ace? The probability must be very close, but still a tad lower" is exactly what confuses many people. What is interesting though is that if you ask for the probability that the first card is an ace, almost everyone who is unsure of what happens with the second card will none the less say that the probability of the top card is 1/13.

On one hand the confusion has a positive aspect to it because it shows that we have a natural instinct concerning "conditional probability". Unfortunately, this instinct often results in mistakes, as it frequently does in this problem. You have correctly identified the two distinct possibilities that must be considered if you are to apply the normal mathematical rules to calculate the given probability. You have also correctly used both conditional probability and the more subtle notion of independence.

However, one can also think of this more intuitively. We seem to mostly agree that the probability of the first card being an ace, given no additional knowledge will be 1/13 because there are 52 possible cards that could be in that place, 4 are aces and each card theoretically has an equal likelihood of being the first card. But is it not also true that when the shuffling stops each card should have an equal likelihood of being in the second spot, or the third, or the 45th. So without any additional information there is really nothing special about the second spot in the deck. It could be any of the 52 cards and each is equally likely.

But what if the first card is an ace...It turns out that if we don't already know this to be the case, then this is technically irrelevant to our calculation. After all it has to be something but since its nature is unknown to us in the face of our own ignorance the first card is really no more of a concern than the nature of the 35th card. (In your original first thought should you not have been equally concerned about the nature of card 35?) But if we flip the first card over, then our game changes. Now we are essentially playing the original game with 51 cards all equally likely to be at the top of the deck but this time with only 3 aces left. So given that we know the first card is an ace, the probability that the second card is an ace would now be 3/51. This is conditional probability.

Similarly, if in our original deck we were told card 35 was an ace, then once again the second card in the deck would be one of 51 equally likely cards. with only three aces to choose from we are once more left with a probability of 3/51. So it turns out that learning the nature of the 35th card gives us as much information as learning the nature of the 1st card.

One way to resolve this once the deck is shuffled secretly mark the second card and then without your friend noticing place the cards in a big circle. Now ask your friend what the probability that the marked card is an ace.

 

[Fourier]

I haven't taken a math class in fifteen years. Is the probability that the second card is an ace something other than 1/13?

Yes!!

 

[LeHab]

I haven't taken a math class in fifteen years. Is the probability that the second card is an ace something other than 1/13?

If you are blindly drawing one card, position will not impact the probability. If you are picking a 2nd card while knowing the first (without replacement/reshuffeling) then your probabilities will change.

As already stated probabilities are rather contre-intuitive which is one of human flaws lotteries are exploiting. As humans we tend to look for patters/links between events - Gamblers Fallacy in case of lottery where events are independent.

 

[abo9]

In fact, the probability that the second card in the deck is an ace is indeed 1/13 as your calculations show. However, your first thought "No way, what if the first card picked is an ace? The probability must be very close, but still a tad lower" is exactly what confuses many people. What is interesting though is that if you ask for the probability that the first card is an ace, almost everyone who is unsure of what happens with the second card will none the less say that the probability of the top card is 1/13.

On one hand the confusion has a positive aspect to it because it shows that we have a natural instinct concerning "conditional probability". Unfortunately, this instinct often results in mistakes, as it frequently does in this problem. You have correctly identified the two distinct possibilities that must be considered if you are to apply the normal mathematical rules to calculate the given probability. You have also correctly used both conditional probability and the more subtle notion of independence.

However, one can also think of this more intuitively. We seem to mostly agree that the probability of the first card being an ace, given no additional knowledge will be 1/13 because there are 52 possible cards that could be in that place, 4 are aces and each card theoretically has an equal likelihood of being the first card. But is it not also true that when the shuffling stops each card should have an equal likelihood of being in the second spot, or the third, or the 45th. So without any additional information there is really nothing special about the second spot in the deck. It could be any of the 52 cards and each is equally likely.

But what if the first card is an ace...It turns out that if we don't already know this to be the case, then this is technically irrelevant to our calculation. After all it has to be something but since its nature is unknown to us in the face of our own ignorance the first card is really no more of a concern than the nature of the 35th card. (In your original first thought should you not have been equally concerned about the nature of card 35?) But if we flip the first card over, then our game changes. Now we are essentially playing the original game with 51 cards all equally likely to be at the top of the deck but this time with only 3 aces left. So given that we know the first card is an ace, the probability that the second card is an ace would now be 3/51. This is conditional probability.

Similarly, if in our original deck we were told card 35 was an ace, then once again the second card in the deck would be one of 51 equally likely cards. with only three aces to choose from we are once more left with a probability of 3/51. So it turns out that learning the nature of the 35th card gives us as much information as learning the nature of the 1st card.

One way to resolve this once the deck is shuffled secretly mark the second card and then without your friend noticing place the cards in a big circle. Now ask your friend what the probability that the marked card is an ace.

Thank you for your detailed explanations! It's a bit of a shame to admit but I'm actually doing an undergrad in Stats. However we don't see pure probability problems like that since 1st year and I've never been able to clearly understand the intricacies of some of these "short and sweet" problems.

I like how you point out that we have a "natural" instinct about conditional probabilities. And I realize now that in the initial problem from the OP I was able to make abstraction of the information, probably because it was first/last balls, and also because it's not a situation I'd encounter a lot. With cards though, my brain automatically assumes that if I am to pick the 2nd card, I would have picked the first card and known it's value, which is false in our brainsteaser.

I will try to recognize when my brain tries to create information that's not there!

 

[lomiller1]

This problem may be easier to wrap your mind around if you consider a smaller (but almost as interesting) problem. You're right that the probabilities change as balls are drawn (the key to the above puzzle is that you must know nothing about the other balls' positions - otherwise it gets closer to the Monty Hall Problem, which if are interested in this stuff at all is a must-Google).

Suppose that...

Team A has 1 ball.
Team B has 2 balls.

What's the probability that Team A's ball is picked first? 1 in 3.

What's the probability that Team A's ball is picked last? (if I'm right above, then it sure as hell better be 1 in 3, but let's prove it a different way).

In order for Team A's ball to be picked last, the balls have to appear in the following order: B-B-A.

What's the probability that B is picked first? 2 in 3 (66.7%).

Given that B was picked first, what's the probability that B is picked second? Well, if B was picked first, then there are one A ball and one B ball remaining. So given that B was picked first, the probability that B is picked second is 1 in 2 (50%).

Given that B was picked first and second, what's the probability that A is picked last? If B was picked first and second, then there is one ball remaining - Team A's. So given that B went first and second, the probability that A is picked third is 100%.

Put these all together, since "and" probabilities are multiplied:

P (A is picked third) =

P (B is picked first) * P (B is picked second | B is picked first) * P (A is picked third | B is picked first and second) =

[The "|" notation is probabilistic slang for "given that" - so you read the second value as "Probability that B is picked second GIVEN THAT B is picked first", which fortunately we calculated above already.]

66.7% * 50% * 100% =

33.3% =

1 in 3.

So we took the long road there, but the probability that Team A's ball is picked last is 1 in 3, exactly the same probability that Team A's ball is picked first.

What if the host reveals one of the Team B balls that isn’t in the first spot, so you are left with 1 ball from team A and 1 ball from team B.

 

[Doctor No]

What if the host reveals one of the Team B balls that isn’t in the first spot, so you are left with 1 ball from team A and 1 ball from team B.

So in your hypothesis, the host shows (say) that ball 2 is Team B's?

Let's see if I handle this right, since these start to get tricky. Hopefully Fourier peer reviews my work

Let's call the three balls A, B1 and B2. Without your hypothesis, these are the possible orderings:

A B1 B2
A B2 B1
B1 A B2
B1 B2 A
B2 A B1
B2 B1 A

These are six equally likely outcomes (which also shows that since 2 of the 6 outcomes have A first and 2 of the 6 have A last, the 1/3 probability holds).

If the host shows that ball 2 is Team B's, then we can eliminate the third and fifth options above (because we know that they aren't possible), leaving us with four equally likely options. Two of those have A first (so 50% chance of A being first) and two of these have A last (so 50% chance of A being last).

Bonus question: suppose B1 and B2 are different colors so you can tell them apart. Does knowing which of B's balls is in the second spot change the final answer?

 

[lomiller1]

So in your hypothesis, the host shows (say) that ball 2 is Team B's?
Let's see if I handle this right, since these start to get tricky. Hopefully Fourier peer reviews my work
Let's call the three balls A, B1 and B2. Without your hypothesis, these are the possible orderings:
A B1 B2
A B2 B1
B1 A B2
B1 B2 A
B2 A B1
B2 B1 A
These are six equally likely outcomes (which also shows that since 2 of the 6 outcomes have A first and 2 of the 6 have A last, the 1/3 probability holds).

If I’ve asked the question properly (far from a given ) I think it should be as follows

If the host shows that ball 2 is Team B's, then we can eliminate the third and fifth options above (because we know that they aren't possible), leaving us with four equally likely options. Two of those have A first (so 50% chance of A being first) and two of these have A last (so 50% chance of A being last).

In this case the host can show the third ball instead. In fact in all 6 of the original options the host can eliminate at least 1 B ball without showing exactly where the A ball is so none of the original 6 possibilities can be eliminated. Since all 6 of the original possibilities are still possible and A is only in the fist spot in 2 of them, the chances of it being in spot 1 (of 2) is 33 1/3% while one of the 2 B balls being in spot 1 (of 2) 66.2/3 %

 

[Doctor No]

Gotcha - so a pretty strict formulation of the Monty Hall problem (one of my favorites, and deliciously hard to gather intuitively) other than that Team A can't "switch" spots once they have the information about B.

So working from our original six equally-likely outcomes:

A B1 B2
A B2 B1
B1 A B2
B1 B2 A
B2 A B1
B2 B1 A

You're right that the new information given about B doesn't eliminate any of the options (this assumes that the host does know where the teams' balls are and would not reveal an A ball), and the odds that A is in the first position would be 1 in 3.

 

[lomiller1]

Gotcha - so a pretty strict formulation of the Monty Hall problem (one of my favorites, and deliciously hard to gather intuitively)

Pretty much what I was going for.

other than that Team A can't "switch" spots once they have the information about B.

Ok, lets say the balls represent a new lottery system the NHL has implemented for the 2020 draft. Whoever has the ball that lands in the first slot gets the #1 OA pick in the draft, the other team will get the #2 pick.
Ottawa has finished last overall but have inexplicitly traded it’s first round pick to Winnipeg earlier that season so the Jets have 2 balls in the lottery while the Oilers have one.

In order to increase the drama Bill Daly has revealed a non-winning ball from the Jets, so we are down to 1 ball for the Jets and 1 for the Oilers. Jets management looks at the situation and says to themselves "Hey this is the oilers, of course their ball is going to win, we should offer a straight up trade", our ball for their ball.

The question is should the Oilers trade their ball for the Jets remaining ball?