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The Game Show Forum => The Big Board => Topic started by: davidhammett on March 29, 2011, 05:30:53 PM

Title: A Twist on the Monty Hall Problem
Post by: davidhammett on March 29, 2011, 05:30:53 PM
Today's LMAD featured a game where a couple could win a car by finding three of a kind from among six playing cards (by choosing only three of them).  The six cards consisted of three K's and three 3's.

After the couple chose their three cards, two of them were revealed.  Since the producers knew which cards were where, they revealed the pair which the couple was guaranteed to have under the circumstances.  The couple was then offered a sure thing to end the deal, rather than being given the chance to switch their final card to one of the other three unrevealed ones.

The questions for your consideration:
1.  What was the probability that the third card they chose did in fact make three of a kind?
2.  If they could have switched to one of the other three cards, would that probability change, and if so, to what?
And finally:
3.  Before revealing the third card the couple chose, the producers revealed a pair from the three unchosen cards (again, this was guaranteed to exist).  If the couple were allowed to switch at that point and take the one other unrevealed card, how (if at all) would that affect their probability of winning the car?

Enjoy.
Title: A Twist on the Monty Hall Problem
Post by: clemon79 on March 29, 2011, 06:11:27 PM
1.  What was the probability that the third card they chose did in fact make three of a kind?
3c6 = 20. Two combinations are winners. Ergo, 1 in 10.

Quote
2.  If they could have switched to one of the other three cards, would that probability change, and if so, to what?
Well, 1 in 4, ostensibly. Four cards left, one is right, three are wrong, they get to pick which one they want. As you said, they are shown their guaranteed pair, so everything up to that point is chrome.

Quote
3.  Before revealing the third card the couple chose, the producers revealed a pair from the three unchosen cards (again, this was guaranteed to exist).  If the couple were allowed to switch at that point and take the one other unrevealed card, how (if at all) would that affect their probability of winning the car?
This is exactly the Monty Hall problem, and as such I won't touch it with a ten-foot pole. Really, (2) was closer than I'm comfortable talking about.
Title: A Twist on the Monty Hall Problem
Post by: GameShowFan on March 29, 2011, 07:23:45 PM
I can't say if 1/4 is wrong, but here's my take on Question #2:

There is a 9/10 chance the contestant chooses to switch correctly. (The initial p = 1/10 is correct, so q = 9/10.) However, when they switch, there is only 1/3 chance the contestant is correct. Thus, wouldn't the probability here be 9/10 x 1/3 = 3/10?

I would agree that the last question fits the classic Monty Hall problem we are all familiar with.
Title: A Twist on the Monty Hall Problem
Post by: davidhammett on March 29, 2011, 07:54:52 PM
I can't say if 1/4 is wrong, but here's my take on Question #2:

There is a 9/10 chance the contestant chooses to switch correctly. (The initial p = 1/10 is correct, so q = 9/10.) However, when they switch, there is only 1/3 chance the contestant is correct. Thus, wouldn't the probability here be 9/10 x 1/3 = 3/10?

I would agree that the last question fits the classic Monty Hall problem we are all familiar with.
Your analysis of #2 is correct.  And #3 is an even more blatant Monty Hall problem... the couple is 9 times more likely to win if they switch!
Title: A Twist on the Monty Hall Problem
Post by: dale_grass on March 29, 2011, 08:57:13 PM
1.  What was the probability that the third card they chose did in fact make three of a kind?
3c6 = 20. Two combinations are winners. Ergo, 1 in 10.
Because I'm an alternate solution kind of guy, you could also go (1)(2/5)(1/4) = 1/10.  After picking the first card, you have a 2-in-5 chance of matching on the second and a 1-in-4 chance of matching the third.



\Hey mods, how 'bout some mutha-lovin' LaTeX up in here?
Title: A Twist on the Monty Hall Problem
Post by: Johnissoevil on March 29, 2011, 09:08:42 PM
All this talk about the Monty Hall Problem has me hoping I'll find the LMAD commercial on GSN from 2001 where someone sounding like Daffy Duck is trying to explain it. :-)
Title: A Twist on the Monty Hall Problem
Post by: TLEberle on March 29, 2011, 09:31:15 PM
If you have a game that has a one-in-ten chance of winning a car, why would you increase those odds to one in four and one in two?  

Some of my favorite games from the old days were the stank-assed-luck, like finding $7 out of four envelopes with $1 and $2 bills, or Beat the Dealer, or the Cash Register, and on and on. Not a one of them ever had Monty reveal some part of the game and then play it over, and I thought all the better for it.

They're playing some really good games from the sample size I've seen; so I can't understand why they'd water it down that way. (I like Panic Button as much as I loathe the game where a player has to roll 21 in 5d6.)
Title: A Twist on the Monty Hall Problem
Post by: chris319 on March 29, 2011, 10:24:14 PM
2.  If they could have switched to one of the other three cards, would that probability change, and if so, to what?
Well, 1 in 4, ostensibly. Four cards left, one is right, three are wrong, they get to pick which one they want. As you said, they are shown their guaranteed pair, so everything up to that point is chrome.
This is the conclusion I always arrive at with the Monty Hall Problem. The problem is, most people approach the Monty Hall Problem as if it were a casino game where the player has an unlimited number of trials. If that were the case then it is 100% true that the player should switch every time, giving himself a 66.6% chance of winning. However, if it is a true Monty Hall Problem within the context of a TV game show, the player has exactly one trial and the odds are as you stated above. The number of trials is less than the number of possible outcomes.
Title: A Twist on the Monty Hall Problem
Post by: Matt Ottinger on March 29, 2011, 10:53:02 PM
This is the conclusion I always arrive at with the Monty Hall Problem. The problem is, most people approach the Monty Hall Problem as if it were a casino game where the player has an unlimited number of trials. If that were the case then it is 100% true that the player should switch every time, giving himself a 66.6% chance of winning. However, if it is a true Monty Hall Problem within the context of a TV game show, the player has exactly one trial and the odds are as you stated above. The number of trials is less than the number of possible outcomes.
No matter how many times you say that, it still doesn't make any sense.  If I only have one opportunity to play a game of chance, I am going to make the decisions that give me the best odds of winning that one game.  Changing always improves your odds.

Pick a number between one and ten.  You've got a 10 percent chance of being right.  Now I'm going to give you the opportunity to swap your one number for all nine of the other numbers.  The correct answer isn't "this is my only chance to play, so it doesn't matter."  The Monty Hall problem is essentially the same thing, only with less extreme -- but still very real -- odds.
Title: A Twist on the Monty Hall Problem
Post by: dale_grass on March 29, 2011, 11:07:38 PM
This is the conclusion I always arrive at with the Monty Hall Problem. The problem is, most people approach the Monty Hall Problem as if it were a casino game where the player has an unlimited number of trials. If that were the case then it is 100% true that the player should switch every time, giving himself a 66.6% chance of winning. However, if it is a true Monty Hall Problem within the context of a TV game show, the player has exactly one trial and the odds are as you stated above. The number of trials is less than the number of possible outcomes.
I'm not following what you're saying.  Do you mean to say that the (theoretical) probability of an outcome in a single trial shouldn't affect decision-making?
Title: A Twist on the Monty Hall Problem
Post by: davidhammett on March 29, 2011, 11:15:10 PM
If you have a game that has a one-in-ten chance of winning a car, why would you increase those odds to one in four and one in two?  

Some of my favorite games from the old days were the stank-assed-luck, like finding $7 out of four envelopes with $1 and $2 bills, or Beat the Dealer, or the Cash Register, and on and on. Not a one of them ever had Monty reveal some part of the game and then play it over, and I thought all the better for it.
To clarify, the switching questions I posed do NOT reflect what happened on today's show; they're simply hypothetical questions I posed similar to those in the original Monty Hall Problem.
Title: A Twist on the Monty Hall Problem
Post by: davidhammett on March 29, 2011, 11:23:09 PM
2.  If they could have switched to one of the other three cards, would that probability change, and if so, to what?
Well, 1 in 4, ostensibly. Four cards left, one is right, three are wrong, they get to pick which one they want. As you said, they are shown their guaranteed pair, so everything up to that point is chrome.
This is the conclusion I always arrive at with the Monty Hall Problem. The problem is, most people approach the Monty Hall Problem as if it were a casino game where the player has an unlimited number of trials. If that were the case then it is 100% true that the player should switch every time, giving himself a 66.6% chance of winning. However, if it is a true Monty Hall Problem within the context of a TV game show, the player has exactly one trial and the odds are as you stated above. The number of trials is less than the number of possible outcomes.
Chris, I agree with you that there is a practical difference between playing the game once and playing it a large number of times.  Indeed, it's not that different from DoND, where even though the bank offer may be significantly less than the average of what's left, this is the one time the contestant is playing, and it's real money, so that affects their decision.  However, it does not mean that the probabilities are any different.  There may not be any long-term considerations if the player plays only once, but the probabilities don't change because of that.
Title: A Twist on the Monty Hall Problem
Post by: vtown7 on March 29, 2011, 11:30:08 PM
They're playing some really good games from the sample size I've seen; so I can't understand why they'd water it down that way. (I like Panic Button as much as I loathe the game where a player has to roll 21 in 5d6.)

Slightly OT: Are there clips of "Panic Button" anywhere?  I watch LMAD from time to time but I've never seen this game show up.

Ryan :)
Title: A Twist on the Monty Hall Problem
Post by: Matt Ottinger on March 29, 2011, 11:35:25 PM
Chris, I agree with you that there is a practical difference between playing the game once and playing it a large number of times.  Indeed, it's not that different from DoND, where even though the bank offer may be significantly less than the average of what's left, this is the one time the contestant is playing, and it's real money, so that affects their decision.  
To be clear, I agree with this as well.  I'm simply saying (as everyone else is) that it doesn't change the underlying odds affecting a single trial.
Title: A Twist on the Monty Hall Problem
Post by: TLEberle on March 30, 2011, 12:59:23 AM
Slightly OT: Are there clips of "Panic Button" anywhere?  I watch LMAD from time to time but I've never seen this game show up.
Beats me, I don't really seek out clips or episodes.

The contestant/couple is situated at a Starship Enterprise console sort of thing with six big arcade dome buttons. Three of them won't do anything, but three will close one curtain each. The contestant has to push three buttons, and any prizes that are still visible are won by the player. The contestant can then bet those prizes on a fifty-fifty final pair of buttons, for all the prizes including a fourth jackpot prize, or nothing at all.
Title: A Twist on the Monty Hall Problem
Post by: chris319 on March 30, 2011, 02:58:50 AM
My conclusion makes all kinds of sense if you are familiar with the Law of Large Numbers, viz.:

Quote
In probability theory, the law of large numbers (LLN) is a theorem that describes the result of performing the same experiment a large number of times. According to the law, the average of the results obtained from a large number of trials should be close to the expected value, and will tend to become closer as more trials are performed.

The LLN is important because it "guarantees" stable long-term results for random events. For example, while a casino may lose money in a single spin of the roulette wheel, its earnings will tend towards a predictable percentage over a large number of spins.
http://en.wikipedia.org/wiki/Law_of_large_numbers

In this case the expected value is 66.6% wins for switching. One independent trial, as on a TV game show, is an insufficient number of trials for the Law of Large Numbers to give the player an edge. If the player had a larger number of opportunities to play, the odds would approach 66.6% for switching.

Below are the results of 10 Monte Carlo simulations. The results are anomalous until we get up to five trials. One million trials give the expected 2:1 odds:

Trials: 1

No switching
Wins: 0
Zonk: 1
0% wins

Switching
Wins: 1
Zonk: 0
100% wins

Trials: 2

No switching
Wins: 1
Zonk: 1
50% wins

Switching
Wins: 1
Zonk: 1
50% wins

Trials: 3

No switching
Wins: 0
Zonk: 3
0% wins

Switching
Wins: 3
Zonk: 0
100% wins

Trials: 4

No switching
Wins: 2
Zonk: 2
50% wins

Switching
Wins: 2
Zonk: 2
50% wins

Trials: 5

No switching
Wins: 1
Zonk: 4
20% wins

Switching
Wins: 4
Zonk: 1
80% wins

Trials: 6

No switching
Wins: 2
Zonk: 4
33% wins

Switching
Wins: 4
Zonk: 2
67% wins

Trials: 7

No switching
Wins: 1
Zonk: 6
14% wins

Switching
Wins: 6
Zonk: 1
86% wins

Trials: 8

No switching
Wins: 2
Zonk: 6
25% wins

Switching
Wins: 6
Zonk: 2
75% wins

Trials: 9

No switching
Wins: 2
Zonk: 7
22% wins

Switching
Wins: 7
Zonk: 2
78% wins

Trials: 9

No switching
Wins: 2
Zonk: 7
22% wins

Switching
Wins: 7
Zonk: 2
78% wins

Trials: 10

No switching
Wins: 2
Zonk: 8
20% wins

Switching
Wins: 8
Zonk: 2
80% wins

Trials: 1000000

No switching
Wins: 334281
Zonk: 665719
33% wins

Switching
Wins: 665719
Zonk: 334281
67% wins

Title: A Twist on the Monty Hall Problem
Post by: dale_grass on March 30, 2011, 08:45:32 AM
My conclusion makes all kinds of sense if you are familiar with the Law of Large Numbers ...

I'm familiar with the law of large numbers.  I'm confused as to your using the law of large numbers and a simulation of thousands of trials to defend your position of "it's a one time shot."  Maybe I'm mistaken.

Better yet, let me just ask this: If you were on the show looking at the guaranteed pair and Wayne asked if you wanted to switch, would you?
Title: A Twist on the Monty Hall Problem
Post by: Neumms on March 30, 2011, 09:12:24 AM
Quote
2.  If they could have switched to one of the other three cards, would that probability change, and if so, to what?
Well, 1 in 4, ostensibly. Four cards left, one is right, three are wrong, they get to pick which one they want. As you said, they are shown their guaranteed pair, so everything up to that point is chrome.

That said, one chance in four to win a car is no better than your chances in many TPIR games for a car. The little games for a car that offered little chance to win (guess the suggested retail price of a can of paint on the West Coast within 50 cents, for example) got old even on Monty's version.
Title: A Twist on the Monty Hall Problem
Post by: davidhammett on March 30, 2011, 12:19:33 PM
My conclusion makes all kinds of sense if you are familiar with the Law of Large Numbers ...

I'm familiar with the law of large numbers.  I'm confused as to your using the law of large numbers and a simulation of thousands of trials to defend your position of "it's a one time shot."  Maybe I'm mistaken.
Same here.  Certainly things will average out over time, and will not (in fact, cannot) exactly mirror the theoretical probabilities for the first case or cases.  However, that does not change what the probability is... it only changes how you choose to interpret and use it.
Title: A Twist on the Monty Hall Problem
Post by: Dbacksfan12 on March 30, 2011, 03:16:03 PM
The little games for a car that offered little chance to win (guess the suggested retail price of a can of paint on the West Coast within 50 cents, for example) got old even on Monty's version.
Yes, but in those situations, people could make a somewhat educated guess.
Title: A Twist on the Monty Hall Problem
Post by: Mr. Armadillo on March 31, 2011, 09:17:59 AM
Even if it's still a one-time shot, you're going to want to do everything you can to increase your chances to win.

If I told you tomorrow that you'd have one shot to nail a half-court shot at halftime of the Butler-VCU game for a million dollars, you'd hit up a local basketball court and take a few practice shots beforehand, right?  Same principle here.  Anything you can do to get a few extra percentage points is usually worth it.