Maths – probability question

[The Statman]

We obviously have a bunch of mathematicians in our midst, so can you help?

Peter and Quentin have a pub quiz book and, every so often, they meet up and quiz each other. There are 1000 questions in the book, and they decide that they should randomly pick the question that is asked (the book is split into different difficulty ratings and subjects, so they don't want to be able to choose which questions to have from knowledge of how the book is arranged).

So they get a pack of cards and get Ace to 10 of each suit in three separate piles (they discard the fourth suit). To pick a number, they pull a card from one of the piles which represents the unit, from another pile to represent the tens and from another pile to represent the hundreds. (The Ace is 1, the numbers 2-9 are 2-9, and 10 is zero.) They pick the digits in reverse order so as to keep in suspense which part of the book the quesition is from.

Thus, each question number has an equal chance of coming up - from 1 (when 0,0,1 are chosen) to 999 (where 9,9,9 are chosen) and 1000 (which is taken as 0,0,0).

However, as time goes on, they realise that the same numbers are coming up again and, frankly, the book is far too difficult and they're embarrassed at their apparent ignorance.

So they buy another book. Thing is, this book has 1500 questions rather that 1000.

So what they do is carry on with the three piles of cards A-10, and introduce a fourth pile containing just an Ace and a 10 of the fourth suit which before had been discarded.

Again they pick the units first, then the tens and then the hundreds. But if the three digit number produced is between 0,0,1 and 5,0,0, they pick one of the two in the fourth pile. (If 0,0,0 comes out this is taken as 1000; all other combinations are as they read.)

The question is – is there still an absolute equal chance of all the numbers having an equal chance of coming out?

[archicrooks]

before all the mathematicians start to 'rack' their brains mr statman do you know the answer ?

[cap]

I think numbers between 501-1000 are twice likely to be drawn than numbers 1-500 and 1001-1500. Is that correct?

[davis_greatest]

I think numbers between 501-1000 are twice likely to be drawn than numbers 1-500 and 1001-1500. Is that correct?

Yes, of course it is. The numbers 1 to 500 and 1001 to 1500 each have a chance of 1/2000. The numbers 501 to 1000 each have a chance of 1/1000.

[davis_greatest]

Incidentally, if they want a way of picking the questions so each has an equal (1/1500) chance of being selected, here is a simple way:

Step 1: Pick a card Ace or Ten from the 4th suit. Ten gives 0. Ace gives 1. This gives the first digit (the "thousands").

Step 2: Pick a card from Ace to Ten from the 3rd suit. Ten gives 0. Ace gives 1. The other cards give 2 to 9. This gives the second digit (the "hundreds").

Step 3: If the first two digits are 15 or higher (i.e. 15, 16, 17, 18 or 19), discard and start again at Step 1.

Step 4: Pick a card from Ace to Ten from the 2nd suit. Ten gives 0. Ace gives 1. The other cards give 2 to 9. This gives the 3rd digit (the "tens").

Step 5: Pick a card from Ace to Ten from the 1st suit. Ten gives 0. Ace gives 1. The other cards give 2 to 9. This gives the 4th digit (the "units").

Step 6: Put the 4 digits together. If the number is from 0001 to 1499, that is the question number to choose. If the number is 0000, choose question 1500.

[The Statman]

Incidentally, if they want a way of picking the questions so each has an equal (1/1500) chance of being selected, here is a simple way:

Step 1: Pick a card Ace or Ten from the 4th suit. Ten gives 0. Ace gives 1. This gives the first digit (the "thousands").

Step 2: Pick a card from Ace to Ten from the 3rd suit. Ten gives 0. Ace gives 1. The other cards give 2 to 9. This gives the second digit (the "hundreds").

Step 3: If the first two digits are 15 or higher (i.e. 15, 16, 17, 18 or 19), discard and start again at Step 1.

Step 4: Pick a card from Ace to Ten from the 2nd suit. Ten gives 0. Ace gives 1. The other cards give 2 to 9. This gives the 3rd digit (the "tens").

Step 5: Pick a card from Ace to Ten from the 1st suit. Ten gives 0. Ace gives 1. The other cards give 2 to 9. This gives the 4th digit (the "units").

Step 6: Put the 4 digits together. If the number is from 0001 to 1499, that is the question number to choose. If the number is 0000, choose question 1500.

Yes I was thinking about it further. Essentially what your doing here is pick a random number between 1 and 2000 and, if >1500 occurs, start again – although you know it's going to be >1500 before you've done the whole thing.

The best way would be to have 15 cards in the first pile, and agree which cards mean which hundred.

[davis_greatest]

Yes I was thinking about it further. Essentially what your doing here is pick a random number between 1 and 2000 and, if >1500 occurs, start again – although you know it's going to be >1500 before you've done the whole thing.

Exactly.

The best way would be to have 15 cards in the first pile, and agree which cards mean which hundred.

Lots of ways of doing it. Another way would be to take a pile of 30 cards (and number them 0 to 29) for the "fifties" and then take a pile of 50 cards (and number then 1 to 50).

So you take (50 x value of first card) + value of 2nd card.

If you're concerned that you only have 52 cards and not 80, you can always use the 30 cards from the first pile again as part of the second!

[cap]

Here is another interesting probability problem which was featured in a well-known game show.

Behind one of the three closed doors there is a prize, say a car. The presenter knows behind which door the car is. The contestant picks a door. Without revealing whether this is the correct choice or not, the presenter reveals one of the remaining two doors that does not have the car. Now the presenter offers the contestant the chance to change their choice. Should the contestant change their choice?

[davis_greatest]

A well-known problem, popularly known as The Monty Hall Problem. Suffice to say that the answer depends on the presenter's rules for opening a door - fundamentally, whether he will always open one of the 2 remaining doors, and if so, ensure that it is one without a car. If yes, then of course the contestant should switch, to double his chance of winning the car from 1/3 to 2/3.

[cap]

Indeed that's the Monty Hall problem. A colleague of mine was asked this question in one of his job interviews. I have to say my first impression was that it would make no difference at all, but after turning my brain around I came to the same conclusion.

[davis_greatest]

Indeed that's the Monty Hall problem. A colleague of mine was asked this question in one of his job interviews. I have to say my first impression was that it would make no difference at all, but after turning my brain around I came to the same conclusion.

The probability of choosing the car at outset is 1/3. If the presenter knows where the car is, he can always choose to open another door without the car (either one or both the other doors won't have the car). Thus, his opening the door does not affect the chance that the door you chose at outset has a 1/3 chance of having the car. Once he has shown a door with no car, there is only one final door - this must then have a probability of 2/3 of having the car, which is why you switch to improve your chances!

[The Statman]

The probability of choosing the car at outset is 1/3. If the presenter knows where the car is, he can always choose to open another door without the car (either one or both the other doors won't have the car). Thus, his opening the door does not affect the chance that the door you chose at outset has a 1/3 chance of having the car. Once he has shown a door with no car, there is only one final door - this must then have a probability of 2/3 of having the car, which is why you switch to improve your chances!

Surely 1/2, not 2/3? Surely, once one door is opened that does NOT contain the car, the two remaining doors each have a 1 in 2 chance of having the car.

?

[davis_greatest]

Surely 1/2, not 2/3? Surely, once one door is opened that does NOT contain the car, the two remaining doors each have a 1 in 2 chance of having the car.

?

That would be true only if the presenter had not known where the car was. But if he knows where the car is, he can always open a door with no car. That doesn't change the probability that the original door you chose had (and still has) a 1/3 chance of having the car.

Imagine that, instead of 3 doors, there were 1 million doors. You choose a door and the presenter, who knows where the car is, goes down the line, opening 999998 of the remaining 999999 doors, always opening one with no car. Then, if you were given the chance, would you keep the door you chose originally, or go for the one remaining door that he happened to leave unopened?

If you keep your original choice, you have a 1/1000000 chance of winning. If you switch, then if the car were originally behind any of the other 999999 doors, you will win it - so if you switch to the last remaining door, you will have a 999999/1000000 chance of winning.

The same principle applies with 3 doors: keep your original choice and you have a 1/3 chance of the car; switch, and you have a 2/3 chance.


(Note - this would not apply if the presenter DID NOT know where the car was, and merely happened by chance not to reveal the car. In such a case, your chance of getting the car would be 1/2 if you keep the original door and 1/2 if you switch - so there would be neither advantage nor disadvantage in switching. The argument also only applies if we know in advance that the presenter will ALWAYS open a door (or 999998 doors in my 1 million door example) with no car, and give you a chance to switch.)