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I'm the host, you're the player.
I shuffle 3 cards, 2 of which have the word "LOSE" on them, one has "WIN".
You randomly select a card but you're not allowed to turn it over and I do not turn over my 2 cards.
AT THIS POINT, WHO IS MORE LIKELY TO HOLD THE WINNING CARD?
I look at my cards and reveal a losing card.
NOW, WHO IS MORE LIKELY TO HOLD THE WINNING CARD!
I ALLOW YOU TO SWITCH TO THE REMAINING FACE DOWN CARD. SHOULD YOU?
I WOULD!
Hey, I figured I'd try my "hand" at this classic too! An important point here is whether my model of the original puzzle is equivalent.
Your thoughts?
Sent from my Verizon Wireless 4GLTE Phone
I shuffle 3 cards, 2 of which have the word "LOSE" on them, one has "WIN".
You randomly select a card but you're not allowed to turn it over and I do not turn over my 2 cards.
AT THIS POINT, WHO IS MORE LIKELY TO HOLD THE WINNING CARD?
I look at my cards and reveal a losing card.
NOW, WHO IS MORE LIKELY TO HOLD THE WINNING CARD!
I ALLOW YOU TO SWITCH TO THE REMAINING FACE DOWN CARD. SHOULD YOU?
I WOULD!
Hey, I figured I'd try my "hand" at this classic too! An important point here is whether my model of the original puzzle is equivalent.
Your thoughts?
Sent from my Verizon Wireless 4GLTE Phone
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12 comments:
Yes, I think this is the same as the Monty Hall Problem. Though you're right, it's easy to tweak the Monty Hall Problem in small, nuanced ways, so that it's no longer equivalent.
"You randomly select a card but you're not allowed to turn it over and I do not turn over my 2 cards.
AT THIS POINT, WHO IS MORE LIKELY TO HOLD THE WINNING CARD?"
You are more likely to hold the winning card. The probability of MY card being the right card is 1/3. The probability of YOU having the winning card is 2/3.
"I look at my cards and reveal a losing card.
NOW, WHO IS MORE LIKELY TO HOLD THE WINNING CARD!"
Still the same. No new information. The probability of my card being the right card is still 1/3. The probability of one of your two cards being the right card is still 2/3. That hasn't changed. We can point to your face-down card and now say that IT is more likely to be the winning card, so...
"I ALLOW YOU TO SWITCH TO THE REMAINING FACE DOWN CARD. SHOULD YOU?"
...absolutely, I should switch. The original probability that MY card was right was 1/3. The probability that YOUR face-down card is the winning card is 2/3.
Thanks for your support of my argument!
I've read and viewed almost every explanation and there always seems to be something missing that would convince me and students. I remember coming up with a mathematical approach a few years ago that really made sense to me but unfortunately I left it in the margins of some book I was reading!
We would need a little more information about the set up to be truly monty hall.
We need to know that you will ALWAYS reveal a losing card and that you will ALWAYS give us the chance to switch.
Without this information then in your posted scenerio it could be that you only give us the chance to switch if we have the winning card. Which means switching will always be a bad move.
Thank you, Mr. B.
Yes, I should have explicitly stated those rules rather than imply them. That's why I asked for suggestions!
With that said, does this version of the problem make it more or less transparent?
Again, all the videos and explanations I've seen online and in print do not give a satisfactory explanation to dispel the myth that the contestant's chances of winning do NOT become 1/2 when Monty Hall reveals the goat. After all, the contestant gets to make a SECOND DECISION so why shouldn't the odds change.
Here's how I see it. Instead of looking at it from the contestant's point of view, consider Monty's chances of keeping the car (a win for him!).
If contestant DOES NOT SWAP, Monty will win 2 out of 3 times (if you chose Goat 1 or Goat 2, he wins), which means your chances of winning are 1/3. BUT IF YOU SWAP, YOUR CHANCES OF WINNING BECOME 2/3!
Darn it! Five minutes after I post a comment, I realize I missed the point I had intended to make!
YES, AFTER MONTY SHOWS HIS CARD OR OPENS A DOOR TO REVEAL A GOAT, THERE ARE TWO POSSIBLE OUTCOMES REMAINING --- BUT THEY ARE NOT EQUALLY LIKELY! THAT IS THE KEY.
IF CONTESTANT DOES NOT SWITCH, THE OUTCOME THAT MONTY HOLDS THE WINNING CARD OR REMAINING DOOR REVEALS THE CAR IS 2/3 AS I EXPLAINED IN PREVIOUS COMMENT. THE ONLY WAY THE CONTESTANT CAN IMPROVE HIS ODDS IS TO SWITCH.
SO, FIVE MIN FROM NOW, I'LL HAVE ANOTHER EXPLANATION!
1. You're more likely to get the wrong card than the right card in the beginning, correct? (Since there are 2 wrong cards and 1 right card, so you have a 1/3 chance of getting the right card and 2/3 chance of getting the wrong card)
2. This doesnt change if 1 of the 2 wrong cards is revealed, correct? (Because your decision is still based off the circumstances described in #1)
3. Important Question: Would it change if you switched your decision?
Yes, because you know there is 1 right card and 1 wrong card now. Your odds have increased from 1/3 to 1/2 of getting the right card. The circumstances have officially changed.
I think the hardest part to understand is WHY should the person switch. I'm would ask, why not just go with eeny, meeny, miny, moe and pick one? And even if it does land on the same card you picked in the beginning, wouldn't it still have been 50/50 chance?
I guess one can explain it by saying the first card you picked is "tainted" with the increased likelihood of being the wrong choice (since again, you picked it under the circumstances where you would be more likely to lose rather than win). Since that card is "tainted", your best call is to switch your decision to the remaining card? I hope I got the logic correct.
Oh, and long time no see Mr. Marain! You are not forgotten!
Hi YzW731! I remember you too.
Read over my previous comment re why it would not be 1/2. Of course I may be wrong in my logic but I keep trying!
Indeed this is difficult to understand fully... do you think that my "tainted" explanation could be valid somehow?
Hi YzW731!
'Tainted' is an interesting idea but how do you define it quantitatively? I keep coming back to staying gives a 1/3 chance whereas switching gives you twice the chance.
This is late, but I thought I'd join in.
First, as is mentioned, it must be made VERY clear that you will ALWAYS turn over a losing card.
Second, there's actually nothing wrong with the logic that says you originally had a 1 in 3 chance of picking the right card, and that now you have a 1 in 2 chance. This aspect is much clearer when you consider the following scenario:
Suppose there were 10 cards, 9 with LOSE, 1 with WIN. The player chooses a card. At that point, you flip over 8 cards, all of which say LOSE (again, they *must* all say LOSE). There are now two cards left unturned, the player's original pick and one of your cards.
Would you switch? I believe in that case, most people would realize that their odds went from 1 in 10 to 1 in 2, and that the odds are much better for switching.
So that's what I call the "1 in 2" explanation.
However, if you prefer the "2 in 3" explanation (using the original scenario), here's a good way to explain it:
The player is offered a chance to switch or stay. If the player had perfect knowledge, when would he stay? He'd stay when he had the WIN card.
So what is the probability of him first choosing the WIN card? 1 in 3.
So, given imperfect knowledge, it is more likely that he did *not* choose the WIN card, and thus he should always switch. Again, this is much clearer if you use 10 cards.
I'm assuming you are going to give them the cards to try this experimentally?
Dear educationrealist,
Thank you for your carefully reasoned comments and, yes, of course, I would have students working in pairs perform this experimentally and then play me!
I find going to extreme cases as you recommended to be highly effective. Why not start with an ordinary deck of 52 cards and tell the students that the winning card is the Ace of Spades. It should be evident to most that the Ace is far more likely to be in the remaining cards! If the Ace is among those, then you have NO CHANCE of winning unless you switch!
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