A Twist on the Monty Hall Problem

[davidhammett]

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.

[clemon79]

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.

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.

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.

[GameShowFan]

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.

[davidhammett]

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!

[dale_grass]

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?

[Johnissoevil]

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. :-)

[TLEberle]

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.)

[chris319]

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.

[Matt Ottinger]

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.

[dale_grass]

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?

[davidhammett]

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.

[davidhammett]

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.

[vtown7]

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

[Matt Ottinger]

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.

[TLEberle]

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.