How to Win a Fifty-Fifty Scenario - Bertrand's Box

Say you’re in a club with a friend, and see a very attractive girl sitting in the next table. Both of you wanna ask her out, so your friend offers to resolve that issue by a coin toss winner takes the girl. Well, there's a variation of Bertrand's Box that might help you here...
You’ll need 3 napkins, 3 beer bottles, and a pen. Ask your friend to mark the napkins this way:
First napkin marked with an “X” on both sides.
Second napkin marked with an “O” on both sides.
Third napkin marked with an “X” on one side, and an “O” on the other.
Don’t look while he’s marking the napkins, shuffling them around, and placing the beer bottles on top of them. Now, pick one napkin at random and remove the beer bottle to reveal the top mark on that napkin. Let’s say you see an “X” on the top side. At this point you must say out loud something like “Well, it can either be the XX napkin or the XO napkin” (which is true), and convince your friend that if you’ll manage to guess the mark on the other side of the napkin, you’ll win the girl. After you’re sure he’s convinced it’s a simple 50-50 chance scenario (after all, it can be the XX or the XO), you can act like you’re not sure, change your mind a few times, ask other people for advice, use an ouija-board, or whatever, but eventually - bet there’s also an “X” on the other side. It’s true that it can either be the XX napkin or the XO napkin, but it’s not a 50-50 scenario. The chance for it being the XX napkin is actually 2/3 and not 1/2 as you might think. It becomes clear if you won’t try to guess what napkin it is, and instead try to guess which mark it is. There are 3 possibilities: It can be the “X” from the XO napkin, the FIRST “X” from the XX napkin, or the SECOND “X” from the XX napkin. Now you see why it’s much better than a fair coin toss. Of course, you should always bet the bottom mark is the same as the top mark (so if you see an “X” – bet there’s an “X” on the other side, and if you see an “O” – bet there’s an “O” on the other side). Also, after you're done, don't forget to check all the napkins to make sure your friend marked everything correctly. You won’t win all the time, but don’t be too greedy – winning two third of the girls should be enough.

One day, many years from now, it will happen. By then, you’ll have a gorgeous wife whom you’ve probably met in that club, three beautiful kids and a big dog. On that day, your old friend, still single, will sit in a bar, dozens of marked napkins scattered on the table, when suddenly, he’ll get it. He’ll understand how you tricked him all these years. He’ll probably be very angry, and will come looking for you. It’s a good thing you got that big dog…

10 comments:

Hater Von G said...

And chances are while all this takes place, her ginormously muscular boyfriend sits next to her and they start making out.

:)

Lise A said...

Hm. Is this related to the Monty Haul thing? 3 doors, choose 1, 1 is opened, now choose the 3rd? Or are they just awfully close to each, but different in some significant way?

Bob said...

I love a good sucker bet.

M said...

There is only one way to solve this problem; you introduce the girl to the friend. Help your buddy get the girl. Then you will come off as cool and the cute girl will introduce you to all her cute friends. Lets face it, good looking girls tend to "flock" together. This is how to solve the problem.

EA said...

I talked to hot girl once while huge boyfriend was in the same room; don't be intimidated. Move on to next girl if you know for sure she is taken, otherwise what do you have to lose?

Study in USA said...

Same opinion like EA..:d

dennis hodgson said...

Ah! I see that you have tackled what I called "the three boxes problem" Uri (and an interesting practical application too). My apologies for thinking that you might not have encountered it before.

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