Ask Cool Daddy, Not Marilyn
Dear Cool Daddy: Every Sunday I read that “Ask Marilyn” column in “Parade” magazine, but sometimes she drives me nuts. That “4 envelopes” question still has me scratching my head. I figured if you couldn’t suss it out, nobody could. Go! …from Wayne in West Dullsville
Dear Wayne: I know exactly what you mean about Marilyn vos Savant. BTW, her father’s last name is Mach, but she chooses today to use her mother’s maiden name, which really is vos Savant, in case you were skeptical. In all fairness, she’s right most of the time, altho sometimes she simplifies things, so technically she’ll be right, but there’s more to the story, so in another sense, she’s isn’t. That could very well be a result of the limited space she’s allotted in the magazine, I dunno. But her biggest fault is she isn’t good at explaining things clearly…and this is especially true of relatively complicated mathematical puzzles.
I should give the devil his due…I recently reread an analysis of this problem here: Stolf’s Blog. This a daily deal written by a guy I used to do a radio show with…in fact, I taught him everything he knows…altho not everything I know! Anywho, I believe he really nails it, and thinking about it, I’ve come up with 4 further scenarios to illustrate why her answer to this puzzle is wrong.
And for those who haven’t heard it, here’s how it goes…
There are on a table 4 envelopes. There is a $100 bill in one of them, and the other 3 are empty. You get to choose one envelope. I take 2 of the remaining envelopes and open them, showing them to be empty. I then say you may trade the envelope you chose for the remaining envelope on the table. The question is, should you switch?
This is thought to be one of those “anti-intuitive” questions, that is, the immediately obvious answer is purportedly wrong, and the seemingly illogical solution is actually the right one. Most people think in this case, there are 2 envelopes unopened…yours and the one left on the table. One has the $100, the other doesn’t, so it’s 50/50…switch or don’t, it makes no difference.
Her argument is you should switch, because the odds of your having chosen the right one are 1/4, and the odds of the money being in one you didn’t choose are 3/4. And here’s the key: What’s really happening here is you are being asked whether you want what’s in your envelope, or everything that’s in the other 3 envelopes. Showing you the 2 empties was just misdirection aimed at confusing you. Put that way, it now looks like a no-brainer that you should switch. But why was the 50/50 assumption wrong?
Well, many people think 50/50 was right after all, and so does Stolf…and me also. Here are 4 observations to help you see why…
(1) Suppose the game is played with 2 people instead of one, say you and your sister. At a signal, you each reach out and pick an envelope. If you pick the same one, you go again until you take different ones. I take the 2 remaining envelopes and open them, revealing them to be empty. I then ask the both of you if you would like to trade envelopes with the other. Now using Marilyn’s reasoning, it would be in your interest to trade, 3/4 if you do, 1/4 if you don’t…but the same thing applies to your sister! How can it be in the interest of both of you to switch? Only if the odds are the same for both of you, and that has to make them 1/2 or 50/50. It’s mathematically impossible that both of you, over time, could win more than the other by switching rather than not switching, because winning for one has to mean losing for the other.
(2) Now apply this reasoning to the original 1-person game. You have your envelope in you hand, and you’re staring at that one remaining envelope on the table, wondering if you should switch. Just ask yourself this: What if I had chosen that envelope still on the table, instead of the one I did? In that case, by Marilyn’s reasoning, you should be anxious to get rid of that envelope on the table, the same one, by her reasoning, you are now anxious to have! How can that be? Only if the odds are indeed 50/50!
(3) Or try it this way. Suppose you are allowed to choose 3 envelopes, not one. You do, and I take 2 of your envelopes, open them, and they are empty. Now you are offered the switch. Well, by Marilyn’s reasoning, having everything contained in 3 envelopes, once you have chosen them, gives you a 3/4 chance of winning. And even if you did end up winning, 2 of your envelopes would have to be empty anyway….my showing you which ones they were has no impact on the odds. So it’s 3/4 versus 1/4 that you had the right one among your 3…that’s 3/4 odds against switching. Trouble is, the odds of winning are 3/4 if you can keep the contents of any 3 envelopes…and that includes your 2 empty ones and the one on the table! So it’s 3/4 to switch. the same as not switching! And even odds can only mean odds of 1/2.
(4) Finally, let’s go back and start the game in original way: you pick one envelope. I ask you to mark it with with an X and put it back on the table. I then open 2 envelopes and they are empty. You may now pick one envelope from the table and keep its contents…
Your odds are clearly 50/50. It’s the same as if I had said, before the game even started: What are your odds of winning if I remove 2 empty envelopes before you pick? The answer would have been 50/50 then, as it is now. The fact that you once had one of the 2 now on the table in your hand, and marked it with an X, makes no difference. As the weak-armed traveling salesman said, I rest my case…
But let me say this: while I believe all 4 of my examples refute Marilyn’s answer, I welcome your input…if you can show that one of them, or even all 4 of them, do not…please give it a shot. I’m pretty sure that I’m right. But I’m also pretty sure that I’m not always right!
shameless plugs, choose any 5…
Other Daily blog at http://stolf.wordpress.com (the legendary Stolf’s Blog)
Audio samples http://stolfspots.podbean.com/