Monty Hall Revised
December 19th, 2008This riddle is my take on the Monty Hall problem. If you know the original version, this one should be very easy for you.
You are watching the TV series “Lets Make a Deal”.
You are very excited as you know your friend Heidi is participating today. You watch attentively throughout the show, only to find out that it ends without Heidi ever appearing in it.
You call Heidi, several times during the following day, but you keep getting her answering machine. Maybe something happened to her?
You recall the Rules of the Show:
The host presents to each candidate three closed doors. Behind one of these doors hides a big prize, behind the other two there is an empty bucket. The candidate chooses one of the doors (this choice is random as he has no information whatsoever on the location of the prize). Then the host opens one of the other doors behind which there is no prize. Then the candidate is given the option to remain with his initial choice or to switch doors and choose the other closed door. Then the remaining doors are opened, and the candidate is awarded whatever is behind the door finally chose. The goal of the candidate is (obviously) to choose the door with the prize.
Pondering about the rules, you develop the following theory:
Maybe the host does not know in advance which of the doors contains the prize. In that case, if the host happens to choose to reveal the door with the prize (thus ruining the show) the candidate is killed (finally, some proper riddle scenario) and his scene is cut-out during the editing of the show.
Maybe that is what happened to Heidi?
Assuming you have a lot of episodes (as many as you need), how can your theory be ruled out or strengthened?
December 19th, 2008 at 1:57 pm
An interesting (and deadlier) variation…
December 19th, 2008 at 5:52 pm
Interesting take on the original riddle. Here’s my two cents:
We will only consider games in which the candidate chose to switch doors (obviously this is also the correct strategy, but we don’t care, as we can simply ignore the other cases).
We need to distinguish between two hypotheses:
H0: The host never opens the prize door, all candidates get their 15 minutes of television fame.
H1: The host might open the prize door, in which case the candidate is executed.
WLOG we can assume the candidate’s original choice was door #1 (everything is symmetrical, so it shouldn’t really matter).
Now, there are 3 scenarios under H0, each w.p. 1/3:
1. The prize is behind door #1: so the host opens some random door 2 or 3, the candidate switches to the other one and loses.
2. The prize is behind door #2, so the host opens door #2 w.p. 1/2 and the candidate dies, or opens door #3 w.p. 1/2 and the candidate lives and wins.
3. the prize is behind door #3 - same as last.
All and all, under H0 we have:
candidate loses: 1/3
candidate wins: 1/3 * 1/2 * 2 = 1/3
candidate dies : 1/3 * 1/2 * 2 = 1/3
So under H0 and limiting ourselves to TV appearences only, we can expect to see about 1/2 of the candidates winning and the rest losing.
Under H1 we have also 3 scenarios, each w.p. 1/3:
1. The prize is behind door #1: so the host opens some random door 2 or 3, the candidate switches to the other one and loses.
2. The prize is behind door #2, so the host opens door #3 and and the candidate wins.
3. The prize is behind door #3 - same as last.
And all and all, under H1 we have:
candidate loses: 1/3
candidate wins: 1/3 * 2 = 2/3
So under H1 and limiting ourselves to TV appearences only, we can expect to see about 2/3 of the candidates winning and the rest losing.
Now we just need to apply the (weak) law of large numbers, compute the win proportion from the actual (virtually unlimited) television shows we have, and see which of H0 and H1 is more likely.
December 19th, 2008 at 5:56 pm
(Oops! Definitions should be the other way around, i.e. H0 - with deaths; H1 - no deaths).
December 20th, 2008 at 7:26 pm
You forgot to mention that they also occasionally kill participants even if the host didn’t ruin the show, just so the stats would be indistinguishable from the no-killing scenario.
December 20th, 2008 at 9:48 pm
Seb,
Good to hear from you!
Danny,
).
Very well analyzed. Note that in case the “killing theory” is correct (i.e. the host does not know where the prize is) switching is the best strategy only in a trivial sense - all the possible strategies are equivalent and give you a chance of exactly 0.5 of winning (given that you are not killed
Yair,
