I have 100 origami swans. Written on the inside of one origami swan are the words “I owe you $100.” Written on the inside of 99 others are the words “you picked the wrong swan, dude.”

You get to select 1 origami swan, which you expect to be the one with the $100 note inside. You can’t look at it yet, as that would spoil the fun.

After you make your selection, I remove a loser swan from the 99 you passed over. I then realize that I had not yet asked if your selection was final.

Irked about my error, I offer you the option to swap your current swan for one of the 98 swans left.

**Question:**

Should you swap your swan for one of the 98 remaining?

Photo by: Izzie Whizzie

{ 13 comments… read them below or add one }

I am assuming that you don’t forget each and every time until I find the one with $100. I am also assuming that I didn’t get to look at my first pick either. Sounds to me like you are trying to have some fun with Bayes Theorem. Dude, it’s 6:30 in the morning, and I haven’t had breakfast yet. Too much to ponder.

Your assumptions are correct, but Bayes’ Theorem is overthinking it.

I don’t think it matters. Your odds of selecting the right swan initially would be 1/100. After I remove one (assuming I don’t look at it), you remove a non-winning swan. The odds of my swan or any of the swans on the table being the right one are 1/99. Assuming I randomly selected mine, it’s not any different from a swan on the table and the option to swap shouldn’t make a difference.

You’re right. Err…maybe you’re wrong!

I’m going to post the answer on Thursday, but thanks for your answer. Hopefully more people will jump in to play along. 😉

Fair enough! Looking forward to checking my answer…. We’ll see if I need to brush up on statistics.

Ok, it is now just past 10:30 and I have had a cup of coffee. I re-read the post and the question. To answer your question, yes I would switch. I will not explain why as to allow other contestants to play. By the way, what is your costume choice?

I would switch. In step 1, you have a 1% chance of guessing right and a 99% chance of guessing wrong. In step 2, you can either choose your original swan (with its 1% probability) or you can choose a new one, which now has (99% chance you didn’t guess right in step 1) * (1/98 chance the new guess is correct) = 1.0102%. Not great odds, of course…

Yes I would switch. Won’t spoil the reason for everyone. I believe this is version a Monty Hall problem, and as cashflowmantra said earlier, Bayes Theorem can be applied to this problem.

When I first read it at 6:30, I thought I was being expected to calculate the actual probability and come up with the numeric solution. I thought it was a little much to ask of a simple blog reader at that hour. But just a generalized idea is much easier to deal with. Odds are pretty decent when there are only 3 choices, but 100 and the benefit of having a negative choice removed is much smaller.

I’ll go against the grain and say I would keep my original swan. Why? Because I’ve an avid Deal or No Deal watcher, and I fully believe in my ability to choose the winning case (or swan, in this situation). I know probability is against me and switching is the “right” answer, but I’m not that worried about it since I don’t lose anything if I’m wrong.

Switch for exactly the same reason as Jonathon.

I know the answer, since it was in the movie 21 and I just watched that a couple weeks ago. I’d switch.

Ok, I know I’m late to the game but I haven’t looked at the answer or anyone else’s comments, I promise. If we know the swan that was taken away was a losing swan then you should switch because the swan you are holding in your hand has a 1/100 chance of winning and the ones on the table have a 1/99 chance of winning, now that one has been removed.

If we don’t know if the swan that was removed is a winner or not then it doesn’t matter if we switch.