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The Magic Cafe Forum Index » » Magical equations » » The Monty Hall problem. (0 Likes) Printer Friendly Version

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Kevinh5
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Wow that is very visual. Thanks.
S2000magician
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Quote:
On 2012-11-17 04:55, Scott Cram wrote:
So, if you chose the door with a car behind it, the host can show you either of two doors, and has a 50% chance of choosing either of those.

This assumes that the host is unbiased when he has a choice. This is, in fact, the host's best strategy. If the host shows any bias when he has a choice (say, choosing the lower-numbered door more frequently than the higher-numbered door), the player can use that to his advantage to improve his probability of making the correct decision (stay or switch) to a number higher than 2/3.
Scott Cram
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That's true, but since there's nothing in the original puzzle to suggest that he has a door bias, I just went with 50%.

It would be interesting to run through the numbers, though, with a, say, 90% chance of the host picking the lower-numbered door.
Scott Cram
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Quote:
On 2012-11-19 00:27, S2000magician wrote:
Quote:
On 2012-11-17 04:55, Scott Cram wrote:
So, if you chose the door with a car behind it, the host can show you either of two doors, and has a 50% chance of choosing either of those.

This assumes that the host is unbiased when he has a choice. This is, in fact, the host's best strategy. If the host shows any bias when he has a choice (say, choosing the lower-numbered door more frequently than the higher-numbered door), the player can use that to his advantage to improve his probability of making the correct decision (stay or switch) to a number higher than 2/3.


I thought this might be a fun thing to ponder. So, let's re-run this whole thing with a new assumption: When the host has a choice of doors, he'll choose the lower-numbered door 9 out 10 times, with a 1 in 10 chance of picking the higher door.

I'll also assume that, having watched this game played over and over again, the player knows this.

First, let's re-do the previous diagram to reflect this.

I'll start with the same example as before - The player chooses door #1, and the host shows door #2.

Pruning the tree with the same reasoning as before, we now see that a joint probability of 19/90 is still in play.

Given that the player chose door #1, and that the host chose door #2, the probability of the car being behind door #1 is now (1/10)/(19/90), or 9/19.

By comparison, given the same facts, the probability of the car being behind door #3 is 10/19.

Interesting - the probabilities are closer, but still favor switching. However, since it seems it will affect the probabilities, what happens if the player chooses door #1 and the host shows door #3?

Pruning the tree for this examination, we see that a joint probability of 11/90 remains.

Under these circumstances, the car will be behing door #1 1 out of 11 times, and behind door #2 10 out of 11 times.

Either way, the strategy still favors switching.

With a player knowing of the host's tendency to choose the lower-numbered remaining door 90% of the time, when the player sees that the higher of the two remaining doors is chosen, the player can be very confident that it was a smart move.

However, when the host chooses the lower-numbered remaining door and then the player decides to switch, it's more of a nail biter in the original puzzle, as the probability is just over .5 (about 52.6%) this way.
S2000magician
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Scott: That's just what I meant: in that case, the (overall) probability of winning the car by switching is (10/11)x(9/10) + (10/19)x(1/10) = 182/209, or about 87.1%. That's higher than 66.7% if the host is unbiased. Thus, the host's best strategy (in the sense of minimizing the probability of the contestant winning the car and costing the show more money) is to be unbiased, but it is still a losing game (for the show).

And you're correct: no matter what strategy the host follows, the contestant's best strategy is always to switch.
Scott Cram
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Wait...When he goes with the lower-numbered door, which he does 90% of the time, you only have a 52.6% probability. Only 10% of the time do you wind up with the 91% probability, so wouldn't it be (10/19)x(9/10) + (10/11)x(1/10), which i......t 56.45% overall?
S2000magician
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Quote:
On 2012-11-20 02:37, Scott Cram wrote:
Wait...When he goes with the lower-numbered door, which he does 90% of the time . . . .

He does that 90% of the time when he has a choice, and only 50% of the time when he doesn't.

I'll recheck the calculations when I'm back at home tomorrow.
Therealmagician
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Didn't Derren Brown use this principle for his shoe trick as the opener for his Svengali show? It worked really well with the audience because it started off with full audience participation then a couple (which had a lot of laughs) then one person who had the chance to change her mind but didn't. He explained the theory with graphics mid way though. I use a prop which allows you to move the goal posts as it were and always win, or lose, which means the audience can vote on whether or not they want the car or goat to be behind the door and it will alway match which allows a new level of patter and participation. I don't know if that makes sense but basically I can always decide what's behind the door, a goat or a car, so I can play with the audience even more, rather than leaving it up to probability. And as a kicker make all 3 goats or cars depending on which way I want the trick to go.
Andy Moss
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A further idea comes to mind as to how to increase the odds of Monty Hall. Have someone hand you three cards from ANY playing card deck. They are instructed to select any duplicate pair of black spot cards and any red picture card.

You simply subtly crimp the corner of the queen whilst laying the three cards face down in a row.

Then apply the routine as outlined earlier in the thread making sure that you add the variety in your approach so that no methodology can be discerned.

You could of course use a crimp for the Hummer routine as well.

Present as a feat of intuition.
Gordon the discombobulator
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This is not a trick but it helps to easily visualise the Monty Hall problem.
The aim is for them to have the Ace of Hearts. If they have that card they win.

stage 1: Start with a full pack and ask them to select one at random but don't look at it. I look through all the remaining cards. I must discard one card (any one except the ace of hearts). should they now swap their one unseen card for the 50 cards in my hand to increase their chance of winning ?... Yes
stage 2: start with all the red cards. ask them to select one at random but don't look at it. I look through the remaining cards. I must discard one card (any one except the ace of hearts). Should they swap their one unseen card for the 24 cards in my hand... yes
stage 3: start with only the hearts... repeat... should they swap for the 11 in my hand.. yes
stage 4: remove the number cards (leaving the A K Q and J (of hearts)).. repeat.. should they swap for the 2 in my hand.. yes
stage 5: finally only the Q, K and A.. repeat... should they swap for the one in my hand... YES (this final stage is the Monty Hall problem)

no matter how many cards, swapping always gives them the best odds.
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