Conundrum: A Fair Deal
I often like to come up with games of chance. There have been times in my life when this has been profitable, but mostly I’m just interested in questions of statistics and probability.
I had considered the math behind putting together my own Deal or No Deal style game, but with greatly reduced suitcase amounts and with a cost to play. Determining a fair cost (one which I would agree to if I were the player or the banker) at first seems like a hopelessly difficult problem, but the math is actually quite simple. The player has the option of keeping the initial suitcase until the end, and the banker has the option of offering whatever small amount he wants. At any given time the chosen suitcase is worth the average of all unopened cases. The banker certainly isn’t going to offer more, and if the player accepts less it’s just because he’s hedging his bets. The cost to play should be the average of all of the cases, whatever they may be.
A couple of months ago, while discussing the Two Envelopes problem, we briefly discussed what’s known as the Monty Hall problem, after the host of Let’s Make A Deal. Thinking of that problem has inspired another gambling proposition which is this week’s Conundrum.
Let’s continue to call our two gamblers the banker and the player. The banker has three boxes and hides a $10 bill in one of the boxes and a $1 bill in each of the other two. The player pays a set amount to the banker and chooses one of the three boxes. The banker must then open one of the other two boxes and show the player a $1 bill. Then the player can decide whether to keep the contents of the box he chose or switch to the other unopened box.
What would be the fair amount for the player to pay the banker to play this game?
UPDATE: Question solved by David. See comments for the answer.
June 5th, 2007 at 1:49 pm
$3. The players earns dollars by correctly determing the price of, say, varoius household items. In this way, the player has the option of risking what is earned for the possibilty of gettting more. If, of course, the price is right. And the game should have plinko and the showcase showdown.
June 5th, 2007 at 2:36 pm
All good ideas, but unfortunately, the price is wrong.
$3 is incorrect.
October 21st, 2007 at 12:14 am
This is the Monty Hall problem disguised. You have a 2/3 chance of winning $10, and a 1/3 chance of winning $1. Your expected winnings are thus (2/3)$10 + (1/3)$1 = $7. That’s the fair price you should pay to play this game.
October 21st, 2007 at 6:52 am
$7 is correct. David, you are on a roll!
It is a version of the Monty Hall problem, and I think that’s what made it so difficult. The math behind the answer to Monty Hall is easy to understand, but so counterintuitive that many have trouble grasping it.
So when you add in the pyschological element of actually having to pay the money, it just feels wrong. $7 just feels like way too much money to have to pay to play this game, even though that is mathematically the exact amount that’s fair to both players.
Which means that you might be able to con someone into let you play it for $5, and let them think they were conning you. That is, if you were that sort of person.