You’re at your favorite casino. You’ve played a lot all month and are now there for the big drawing. Here’s the way it works:
Ten winners get called — they have a minute and a half to show up and identify themselves. If one or more spots are unclaimed after 90 seconds, more names are called. Eventually there are 10 contestants to “play the game.” Good news! You’re one of the chosen few — but I’m not going to tell you now whether you were first or last.
The way the game works is that 10 unmarked envelopes, in numbered spaces, are on a big board. Prizes total $25,000. The distribution of the prizes in the envelopes is:
First $10,000
Second $4,000
Third – Fifth $2,000 each
Sixth – Tenth $1,000 each
Any of the players may end up with any of the envelopes. The first player drawn has the biggest choice. The last player drawn has no choice at all, but clearly it’s better to have this “no choice” rather than not to have been called at all.
Here are the questions: What’s your EV (expected value) if you get the first choice? What’s your EV if you barely make it in and you end up taking the last envelope? (We’re assuming the envelopes are indistinguishable from one another. I’ve been at drawings where actual cash was in the envelopes and the envelope with 100 C-notes inside was quite a bit fatter than the ones with “only” 10 Benjamins. In that drawing, you definitely wanted to be first to pick because visual inspection of the envelopes contained valuable information.)
The answer, of course, is “it depends.” (I like questions where this is the answer. That gives me something to write about!)
For the first player to select, the EV is clearly $2,500. A total of $25,000 is being given away to 10 players, and $25,000 divided by 10 is $2,500. This is as simple as an EV calculation gets.
For the second player, his actual EV depends on what the first player chose. If the first player selected a $1,000 envelope, then the second player’s EV is $24,000 divided by nine, which is $2,667. If the first player selected the $10,000 envelope, then the second players EV drops to $15,000 divided by nine, which is $1,667.
By the time we get down to the last player, there will be one envelope left and the EV is whatever prize hasn’t been claimed — meaning $10,000; $4,000; $2,000; or $1,000.
How do you take a weighted average of that?
Before I answer that question, let’s change this discussion a little. Assume each of the players selected an envelope but didn’t open them until the very end when they opened them together. In that case, each of the players has an EV of $2,500. There is still $25,000 in the prize pool, so far as they know, and they each have one in 10 chances to get any of the prizes.
Now, change it again. Assume you are the last person in line but you put earphones and blinders on until it’s your turn. Based on the information you have, you now have the same $2,500 EV as you would if everybody opened the envelopes at the same time!
If you are watching what happens and you’re still last, and you do this many times, on average your EV will be $2,500 — with variance!
Mathematically, on average it doesn’t matter whether you pick first or last. It can matter psychologically however. You see the $10,000 and $4,000 envelopes opened by somebody else and it’s a real downer if you’re somebody who sweats your daily scores! But sometimes getting called last will mean you see all of the smaller envelopes being opened and you’re left with the big one! On average it doesn’t matter, but if you want to feel bad about it, knock yourself out.
Since there are five $1,000 envelopes out of 10 total, half the time the last guy will end up with $1,000. (Of course, half the time the first guy — with complete freedom to choose any of the envelopes — also gets $1,000.)
When the first guy picks $10,000 (which will happen 10% of the time), it LOOKS like having the first choice was a big advantage. But it really wasn’t. He just made a lucky pick.
