Math Puzzles!

I had dinner with my high school friends, all of whom are extremely bright and talented people. Naturally, we started to talk about politics and work, etc, and someone brought up the average intelligence of his colleagues, how they are are extremely bright and it is interesting to tease out this fact in interviews. We were discussing off the wall interview questions we’ve been hearing about lately.

Another friend of mine posed the following math puzzle:

You are given a standard deck of cards. The ace has a value of 1, jack is 11, queen 12 and king is 13.

You are asked to assemble a magic square - the sum of the values in every row = the sum of values in every column = the sum of values in both diagnols - of dimension 3x3. This sum, by the way, is called the magic sum or magic constant.

What is the maximum ‘magic sum’ possible such that every card is unique? (again, dimension 3x3)

I have another, simpler probability word problem. The above problem, by the way, I don’t think is on the internet. I have checked.

EDIT: Yo Momma is right, but I did specify ‘of dimension 3x3’

52 cards don’t make a perfect square.

He says it’s a 3x3 square.

27

6 11 10
13 9 5
8 7 12

I got 27, but I don’t think I did it right. I got 27 without re using any values, but we actually CAN use a value more than once since there are 4 suits, right?

[quote]malonetd wrote:
I got 27, but I don’t think I did it right. I got 27 without re using any values, but we actually CAN use a value more than once since there are 4 suits, right?[/quote]

Damn! There’s the rub.

[quote]malonetd wrote:
I got 27, but I don’t think I did it right. I got 27 without re using any values, but we actually CAN use a value more than once since there are 4 suits, right?[/quote]

Nope! Good catch, though.

All cards are unique regardless of the fact that there are four suits. That is a good question, though, and a perfectly legit constraint.

Here’s another one:

You are given a revolver. Two of the six chambers are loaded consecutively. The trigger is pulled once and no shot is fired.

Now, handed the gun, you are asked to choose between:

A) Putting it to your head and immediately pulling the trigger a second time
B) Spinning the chamber before putting it to your head and then pulling the trigger

In which scenario are you more likely to survive and why?

Spinning the chamber gives you a 4/6 chance of the chamber being empty. If you chose A you would only get a 3/6 chance of the chamber being empty.

[quote]PonceDeLeon wrote:
A) Putting it to your head and immediately pulling the trigger a second time
B) Spinning the chamber before putting it to your head and then pulling the trigger

In which scenario are you more likely to survive and why?[/quote]

Answer is “A”. If you’ve hit an empty chamber, then you only have 1/4 chances that the next one will contain a bullet.

Choice B) gives you 2/6 = 1/3 chances of randomly finding a bullet.

So A gives you a 75% chance of survival vs. B’s 66.6%.

Scenario A. Since you can exclude two chambers, the one with a bullet and the one just fired, that leaves you with a 1/4 chance of getting shot.

If you were to just spin the chamber, you’d be giving yourself a 2/6, or 1/3, chance of getting shot.

That was an interview question at Facebook. A friend’s colleague applied there and they asked him this revolver question.

[quote]Hellfrost wrote:
Spinning the chamber gives you a 4/6 chance of the chamber being empty. If you chose A you would only get a 3/6 chance of the chamber being empty.[/quote]

You’re not taking into account the fact that the bullets are consecutive.

We agree that a random spin gives you 4/6 = 2/3 = 66.6% chance of survival.

But option A is better, since if you look at the gun:

1 2 3 4 5 6
E E E E B B (E=empty, B=bullet)

The fact that you’ve hit an empty chamber means that you’re somewhere between position 1 and 4. Only position 4 will kill you on the next pull; positions 1 to 3 let you live. Hence, 3/4 chance to live = 75%.

Scenario A leaves you with a 4/6 chance of survival/ 2/6 chance of death whereas scenario b leaves you with a 3/5 chance of survival/ 2/5 chance of death, so Scenario A, despite it only being a 6.6% difference.
Edit: Never mind, Pookie’s answer is the right one.

What is interesting is the difference in thought process when you can do these at your leisure and when you are put on the spot (such as during an interview or an exam).

Ahh ok, now I understand. English is my third language so I din’t take the word consecutive into account. Thanks for the explanation

[quote]PonceDeLeon wrote:
What is interesting is the difference in thought process when you can do these at your leisure and when you are put on the spot (such as during an interview or an exam).[/quote]

If it’s asked orally, then it can be tough, since you’re even more likely to miss the little details.

You’re on a game show and you have to choose between 3 doors, 1 of which contains a car, the others, goats.

You choose a door. The host opens one of the remaining doors to reveal a goat and then gives you the option to change your choice.

Should you switch?

No. By the sound of it, the host didn’t open the door that was chosen.

[quote]NickRageSkursky wrote:
You’re on a game show and you have to choose between 3 doors, 1 of which contains a car, the others, goats.

You choose a door. The host opens one of the remaining doors to reveal a goat and then gives you the option to change your choice.

Should you switch?[/quote]