This time last year, you were no doubt glued to your computer screens each morning to see what lay behind the door on Mrs C's Advent Calender. As we got a bit giddy with all the excitement we've decided to give it a miss and calm ourselves down this year. Instead here's three puzzles for you. First correct answer in the comments box for all three wins er... something good.
1) We are at a Xmas Party exchanging merry banter and you happen to mention that at the last party you attended, it turned out that two of the guests had the same bithday.
Quick as a flash, I offer to bet you £100 that two people in this very room have the same birthday. We quickly count them and see that there are 48 guests. Assume that I have never met any of the guests before; should you accept my bet? (+explain why?)
2) Ignoring your reluctance to engage me in any further conversation, I produce a deck of cards and you notice they all have a number on one side and a letter on the other. I announce that the cards obey a simple rule: if a card has a number three on one side then it must have a letter 'H' on the other.
I throw down four cards onto a convenient table '3' '8' 'T' and 'H'. You have 20 seconds to tell me which of them you need to turn over to prove or disprove my rule.
If you have not timed yourself ruthlessly and submit an answer, I shall know.
3) Even though you are now sidling away from me, I refuse to take the hint and regain your attention by producing a large diamond and three cups. I place the diamond under one of them and shuffle the cups around so that you lose track of where it is. I announce that if you guess correctly then you can keep the diamond. After a brief pause you place your finger on one of the cups. Rather than turning it over, I actually turn over one of the two other cups revealing it to have nothing underneath.
Now I offer you a choice: you can either stick with the cup you have chosen or swap to the other untouched cup. What should you do and why?
The last puzzle is quite famous, so you can have a bonus point if you know its name.
Merry Christmas
Frank
15 comments:
1. No, as it's more likely than not.
2. 3 & H as the others don't prove anything.
3. Monty Hall. I'd switch.
1. No, I shouldn't take your bet. If we simplify the calculation to assume uniform distribution of birthdates (and forget about leap years for the time being), there are 365^48 possible combinations of birtdates, and 365C48 ("365 choose 48") ways for people to not have a birtdate in common. Thus, the probability at least two people have the same birthdate is given by
1 - ( 365C48 /
365^48 ) = really quite close to 1 (ie. you're exceedingly likely to win).
One could fart around with a slightly more involved calculation using leap years, but it still leads to your winning the bet with a strong likelihood.
2. I turn over '3' and 'T'. If I turn '3' over and it doesn't have an 'H' on the other side, this disproves your rule. If I turn 'T' over and it has a '3' on it, then this also disproves your rule. It doesn't matter what's on the other sides of the remaining two cards; 'H' may have any number it likes, '8' may have any letter it likes.
However, it should be noted that if your rule hasn't been disproved after turning over these cards, I have not actually proven your rule unless the pack consists of four cards only.
3. I should swap cups. There was 1/3 chance the cup I chose first was correct, which means there is a 2/3 chance the diamond was in one of the other cups; this does not change merely because you've jettisoned one which definitely doesn't have the diamond. So there's a 2/3 chance the remaining cup has it.
I believe this is better known as the Monty Hall problem, although the lecturer who first taught it to my year called it (inaccurately) the "Three Card problem".
Rats ... beaten to it.
One is the birthday paradox, once you get 26 people (I think, I haven't looked up the exact number) you have a 50/50 chance of having two people with the same birthdate. So any bet after that isn't in your favour. As Mark says once you get as far as 48 people you're pretty much going to get a match.
Two I screwed up, I thought you just needed to check 3 but Mark is right you need to flip T too. But I was trying to keep to your time limit ;)
Three is, as everyone said, the Monty Hall problem.
You should switch, if the cups are A, B and C. You choose A, and switch after the reveal. Either diamond is in A (you lose), B (C is flipped, you switch, you win), C (B is flipped, you switch, you win. So you've got a 2/3 chance of winning if you switch and a 1/3 chance if you don't.
Interestingly that seems to mean you should always switch on the last box on Deal or No Deal.
3. I don't know the name but didn't the original problem involve goats?!
okay, probabilty gurus, if i have a pack of 72 cards, and deal five cards, then shuffle, what are the odds of drawing the same five cards again (but not necessarily in the same order as before?).
assume each card in the 72 card deck is unique.
assume the decks are shuffled randomly.
Anonymous,
1 in 1473109704
Not exactly a Merry Christmas but fun anyway. Yes. Fun.
rich
Number of different ways of picking five objects from seventy two is 5/72*4/71*3/70*2/69*1/68
So the odds of picking a certain five are one in 13 991 544
Maths teachers - what a sad bunch of buggers! I bet the Media Studies teachers aren't doing this sort of nonsense at Christmas time...
Tefl Tradesman you're right. The Media Studies teachers will be watching daytime TV.
It works out that the birthday bet should only be taken if there are fewer than 23 people present. If there are 23 or more there is a better than 50% chance that two or more will share a birthday.
The Monty Hall problem only has the solution suggested if the person running the show knows where the diamond is. In effect, you are swapping the choice of someone who doesn't know where the prize is for the choice of someone who does. I don't know whether this applies in 'Deal or No deal'.
I liked it better when all you had to do was guess whether it was going to be a Christmas Pudding or a Robin.
The three cup trick I think is this.
You keep the cup that you have your finger on. The assumtion is that the dealer knows where the diamond is. If you had your finger on an empty cup, then all he has to do is lift that cup and say "You lose." By giving you a choice, assuming you have the right cup, he ups his chances from 100% loser to 50-50.
For problem 1, no one has mentioned whether there are any twins in the room!
At least it would avoid all this geeky maths stuff....I just have to look around the room!
You been reading Derren Brown?
Its a sad reflection on British teaching that Derren Brown's parents couldnt spell 'Darren' correctly.
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