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.
A very intriguing view of what the EV is for the lucky players who are drawn to pick a sealed envelope . I think the bigger aspect some folks are skeptical about is that they are not guaranteed often to being one of those lucky ten players for playing what they feel is a lot and there is no way of gauging how many entries are in the barrel to determine a player’s chance at getting selected.
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What bothers/confuses me is when the rules state each person can only get one envelope. I’ve been to drawings where the first person drawn has by far more tickets than anyone else, and he stands there on stage while his name is drawn again and again, and discarded, because he can only get one envelope. He “deserves” to get several of the envelopes, and all that extra gambling he did to get all those tickets is “wasted” as his well-earned tickets are being discarded. How do you calculate the diminishing return as you collect tickets, based on the worthless value of getting drawn more than once?
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It’s always a semi-educated guess — especially since you usually don’t know how many tickets are in the drum.
For “regular” drawings (once a week, once a month, every January, etc.) you assume there will be about the same number of tickets in the drum this time as last time. And you share information with other players who play similar amounts as you do. If, based on past drawings, that $10,000 coin-in gives you about a 20% chance of being drawn, that helps you figure out EV
As you indicate, you can play “too much,” as the marginal chance of being drawn keeps going down because you can only win once.
It’s not an exact science. If the crowd gets loud and angry about the same guy getting called again and again, the smart player will take some time off from entering drawings at that particular casino. If he doesn’t, eventually the casino will do something to mollify the loud and angry players — and the player won’t like the “solution” the casino comes up with.
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Interesting Math Puzzle. Intuitively, I guessed the 10th guy is worse off because in order to win the $10K 9 people have to miss it, but it turns out everyone is equally likely regardless of order.
Probability of 1st Guy Winning =1/10
Probability of 2nd Guy Winning =9/10*1/9=1/10
Probability of 3rd Guy Winning =9/10*8/9*1/8=1/10
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Probability of 10th Guy=9/10*8/9*7/8*6/7*5/6*4/5*3/4*2/3*1/2=1/10
In every case the numbers in the numerator cancel out the numbers in the denominator except for the original 1/10.
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Agreed with Mr. Dancer’s analysis. Of course there is the taxation issue.
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Being mathematically challenged, these types of questions always confuse me. An EV of $2500 just because the total sum has been divided? But, no one can ever win that mythical $2500–it’s not even one of the choices. Makes no sense to me unfortunately!
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Being mathematically challenged puts you at a disadvantage when trying to beat a casino. Casino games, and promotions, are frequently mathematically based.
I don’t know you and am certainly not criticizing you. But if you wish to succeed in the gambling world, you would probably be well served to take some beginning algebra classes — and later some beginning probability classes.
You certainly don’t have to do this — and gambling success may not be all that important to you anyway — but mathematical fluency is part of being a successful gambler.
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It seems to me that knowing, or estimating, EV is only an academic excercise. It doesn’t effect your chances of winning a larger prize. The only strategy is to play more to maximize your number of drawing tickets and to be present at the drawing.
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Estimating EV helps me decide on whether I should invest my time and dollars at Casino A or Casino B — and whether this particular promotion is worth playing for or not.
While you can’t cash EV at the grocery store, if you consistently base your playing decisions on playing for max EV, you’ll come out better over time than if you ignore EV considerations as an academic exercise only.
Do I calculate EV in the middle of a drawing? No. By the time the drawing takes place, pretty much you get what you get and it’s too late for any such calculation to be useful. My only purpose in using the concept of EV in this article was to argue that in this kind of drawing, it doesn’t matter mathematically which position you get selected.
A couple of nights ago, Bonnie was selected for the Silverton weekly seniors drawing — which is similar to what was described in the article only the prizes are smaller. It’s a “collect with 24 hours” drawing, and by the time we got there, the prizes remaining were $1,000; $500; $250; and $100. We didn’t do the exact EV calculation. She just tried to get the $1,000 envelope. I suggested she choose envelope #9 — can’t say exactly why — which turned out to be $100. Oh well. It was a blind guess and that was always a possibility. Slightly disappointed, sure, but overall glad she was selected to pick an envelope.
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Agreed. When you can participate in only one of simultaneous events, evaluating EV is the logical approach.
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Not to put down Mr. Dancer, but the passage “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!” is not accurate. The reason I say this may be seen as mere semantics, put there’s an important point: Ignorance does not constitute fact. Neither does partial knowledge. The EV for the 10th draw-er’s envelope is determined and fixed (not in-doubt or unknown) by the contents of the 9 envelopes that have already been drawn. If all the “goodies” have been drawn and only a $1,000 envelope remains, then the V of that envelope (not EV, because there is no multiple possibility) is $1,000. Period. The fact that Draw-er #10 hasn’t been watching or hearing the first 9 results doesn’t affect the V of the 10th envelope at all. It only affects his knowledge/ignorance.
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This is something I try to teach to my statistics students. I can’t access this site at my school because it falls under a gambling category. ( I used to be able to show it before when it stayed under LVA ) Am I allowed to copy and paste if I give both you and this site the proper citation?
This is a great example of why the expectation is the same for every person upfront even if it changed as events occurred in some formats. It clearly shows why it just does not matter under any format identified.
I love using some of your posts in my classes. My school system considers it community service,; a required component for graduation. I have convinced them that an educated gambler is less likely to become a problem gambler.
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Feel free to copy and distribute with attribution. Thanks.
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I have found Video poker for Winners to be very useful tool. Have you considered software that will be compatible with an iMac computer.
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