Becky's T-Blog

Friday, August 11, 2006

Friday Mind-Bender

One of my pet interests is logic puzzles, particularly ones relating to odds and statistics. Here are two classic logic problems, both of which have answers that go against what you might consider "common sense":

You are on a game show and the host shows you three doors. The host tells you that behind one of the doors is a pot of gold, and the other two are empty. The host knows which door holds the gold. He gets you to choose one, then he chooses another one and opens it to show it's empty. He then offers you to keep the door you chose, or swap to the third, unopened door. Is it better to stay with your door or swap?

A married couple have two children. You know that one of the children is a girl, what are the rough odds that the other one is a boy?

Have a think about it, I'll give you the answers later. Then have a think about the fact that often juries are asked to make decisions based on a "common sense" understanding of statistics, and get a bit worried.

Jane do you want us to answer these in the comments or send you an email, will there be a prize for the especially selected winner?

Say trip to King's Lynn, Thai Green Curry and Long Island Iced Tea chez Bex?

Becky You can answer in the comments. There'll be no prize except the pride of everyone knowing what a clever clogs you are.

Siobhan Curran I've seen Jessica get the first one wrong elsewhere on the net ;-P

Becky Have you? I've had a niggling feeling I've blogged about this before. :-S

Siobhan Curran It was on a different blog.

There are many ways to explain the first one - I like to think about it as the odds of "getting things wrong", rather than the odds of which box.

But yes, I too worry about juries...

Valerie S This is fun! I'll give my bets, a bit shady so I don't ruin anybodys fun:
1) depends on things but if the presenter had the gold there wouldn't be anymore suspense and he'd kind of ruined the show..?
2) I suppose here we are looking at all the families, not an individual child..

Lauren Teo 1. At the start you have an ~66% chance of picking an empty door. Assuming you do (and the odds are on your side) the host is forced to pick the other empty door.
You should change; I don't know how to calculate the odds they're higher that the gold is behind the unpicked door.

2. I daren't answer this one as it all still seems so obvious, and must therefore not be. (Ooohh, it's just like QI.)

Alan Davies The answer to question 2 is 50/50

Steven Fry BONG! Ohohohoho so very wrong, Alan! Minus 50 points, I'm afraid.

Bo!

Anonymous Hi,

If you really like puzzles, you might like this site -

http://perplexus.info/

If you have seen the american show, Numbers, (the math wiz with the FBI brother) you first problem was used in one of the shows. The second one I am not sure, but I think it is also 66% in favor of the other being a boy. Here is how I think of it.

If not knowing either child's gender ( and who really does...but I digress) the options are
girl - girl
girl - boy
boy - girl
boy - boy

and it is 3/4 or 75% chance of at least one boy. Once you say you know one is a girl, you take area the last option. With three options, two have a boy, or 66%.

Or I am missing something.

Anyway, have fun and check out that puzzle site.

Dawn

Siobhan Curran > "the options are girl - girl, girl - boy, boy - girl, boy - boy"

Sounds like my porn collection...

jessica_sweet_tv If answers go against common sense
then
Problem number one, I am too tired to think, plus if you made a choice why change??
Problem number two, well the second children might look like a girl in that pretty dress, but its actually a boy (oops, that wicked mind on me)

Cathii Scott There are lies, damned lies and statistics. Since these problems are both based around probablity, and probability IS statistics, then both answers are lies, no matter what they are.

First problem.... three choose one, then two choose one. Mathmatically the probablity is 1 divide by (3 x 2) that you choose the right door each time. However the reality is that you have 50/50 of chosing the right door the second time. (human factors not considered) there for the odds are far better to choose the unopened door.

Second problem.... Combination, there are four possible combinations or outcomes for 2 children assuming the binary model is selected (ie no transexual or intersexed children) as already pointed out above. One of the possible combinations is boy-boy and can be discounted. One of the remaining three options is girl-girl there for the odds are 2 in 3.

Cathii Scott Talking of answers to problems... what were the top 20 trannie movies of all time?????

Becky The dog ate it. :-S

Jessica I spent a long time on that first one and I still can't quite get my head around it, as for number 2, I'll go with 50/50

NH Problem 1: If I'm on a game show that asks me to choose a door for a prize with a bearded host goading me, then I have forfeited the right to life and thus don't deserve the pot of gold.

Problem 2: I have two choices; either I mathematically work out the chances that the 2nd child is a boy, or I instead ask the couple what the gender of it is. I might not win any logic awards, but my social skills will improve.

Stephanie Delacey Thank you, Cathii, for asking about the Top 20 tranny films! I've been dying to ask that myself - but I was afraid that Becky would snort derisively and point to a post about it back in April or something and I'd be left looking old and forgetful...

Daisy 1- Either is door or the hosts door has the gold so it must be 50-50. Therefore he has not greater chance that either door has the gold, so it makes no difference to the odds.

2- The second child can either be a boy or a girl (unless we are counting Hermaphrodites. But as it's just a logic problem I'll asume not), so the odds are 50-50.

Miss K here are the answers:

1. Yes
2. No
3. Maybe or all of the above

Miss K Oh, and

4. South Carolina

Becky Well done, you win the speedboat.

Anonymous Re juries not understanding statistics. There were several examples of deliberate attempts to mislead the jury with spurious probability arguments in the OJ Simpson case.

Simpson's defense lawyers argued:

1)That evidence that Simpson physically abused his wife be dismissed on the grounds that only 1 in a 1000 men who beat their wives also murder them.

2)The chance that a sample of DNA matched that of OJ Simpson was put at 1 in 4 000 000. Since there are 20 000 000 people in the Los Angeles area his lawyers argued it could not be conclusively proved that the DNA was his.

Neither argument, of course, is remotely convincing.

Stephanie Delacey That was me, not Miss Anonymous. Now what are the chances of that happening?

Miss K Dunno, but I already won the speeboat. You can have the cuddly toy.

Miss K > speeboat

Must. Use. Preview. Button. Provided.

VB-W If I were the game show host, and you were the contestant, I'd offer you the option to switch only if you initially chose the correct door. In this case, the first door has a 100% chance of winning, the second door has a 0% chance, and switching would be a sure loser.

Unless you understand the motives and behavior of the game show host, all the mathematics in the world won't help you answer this question

Savannah Hemingway 1. It doesn't matter which door you choose. Each has a 50% chance of holding the gold.

2. Each child has a 50% chance of being a boy. The gender of the other has no effect on the odds.

Kate Weston A couple have five children - half of them are boys ...

1) You have a better chance of winning by changing your door for the reasons described by others above. If the host had said choose either door 1 or both doors 2 and 3 together you would have picked the 2 wouldnt you - that is what you are doing when you swap. (er i think).

2) Gut reaction says 50-50 since the odds of having any particular sex are even (disregarding the weight of x chromosome carrying sperm) but the G-G B-G G-B B-B argument seems good too - hmmm don't know.

Natalie I'm sure it's all been covered already but it's 50/50 on both. Once one door is eliminated in the first question your odds raise to 50/50 regardless of your original pick. Same with the second one. The sex of the first child has no effect on the odds for the sex of the second and so it is still 50/50.

I see you've added the word verification again...

Genette The answer to the first one is that you should swap, as you will lose in the one third of cases where you picked the correct door in the first place, but you will win in the two thirds of cases where you picked the wrong door to start with.
The answer to the second one is 50/50. GG, BG, GB, BB cover all the possibilities, and we know we can rule out BB. But we can also rule out either BG or GB. (We don't know which one, but the girl is either older, in which case we rule out BG, or younger ruling out GB.) That leaves us with GG and one of the GB/BG options and an equal chance that the other child is boy or girl.

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