Why is Deal or No Deal not the Monte Carlo problem?

[Lottery Larry]

10 cards face down, Ace through T. You pick one card without exposing it, I randomly expose the other cards

We're down to two possible cards, the Ace and the 5. I offer you the choice to switch. It is a 50/50 proposition that you choose the Ace.

How do I explain that this 2.22% scenario is NOT the same as a Monte Hall version of DoND, where I am forced to expose only losing cards... making the Ace 90% to be the card remaining from the big pile that I whittled down?

What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

Or, in other words, how does the fact that I randomly exposed 8 goats differ from when I intentionally exposed 8 goats?

[Lottery Larry]

here's one stumbling block in the attempt to clear this up:

In the Monte Carlo version, it's 90% chance that one particular card that you choose to "shoot for" is in the bigger pile, rather than the smaller pile.

There is a 90% chance that the Ace is in the big pile.... and, independently, it's a 90% chance that the 5 is in the big pile.

When you get down to two cards, how is it that you can determine that it is the Ace, not the 5, that is 90% likely to be the card you haven't chosen?

I think this argument changes the conditions of the Monte Carlo problem, but I can't explain it properly

[SheetWise]

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What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

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If you're exposing losers only -- you're adding information.

[WordWhiz]

And it's the Monty Hall problem, not the Monte Carlo problem.

[Lottery Larry]

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What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

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If you're exposing losers only -- you're adding information.

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I know, but saying that isn't going to convince them in the discussion

[ShaneP]

[ QUOTE ]
[ QUOTE ]
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What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

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If you're exposing losers only -- you're adding information.

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I know, but saying that isn't going to convince them in the discussion

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Here's one approach (how to solve it mathematically)...chance you pick the A to begin with: 1/10.

Chance it is in the remaining pile: 9/10. But we have to pick 8 of the 9 AT RANDOM that aren't the ace to get to the final situation, so the chances of that are (9/10) * (1/9) = 1/10. So there's a 1/10 chance that the ace was in the pile of 9 and we didn't pick it. So the probability either of the two remaining cards is the ace is 50/50, since each occurs 10% of the time.

BTW, the 1/9 can be seen easily if you note that you're flipping all but one, so you could actually just pick the one you aren't going to flip instead of trying to sum the probabilities for each individual flip.

The difference between that is the cards in this case are picked at random, without knowledge. In the Monte Hall problem, if we always have to show 8 of the 9 cards (and none can be the ace) then the last 1/9 probability becomes a probability of 1 (since we can always flip over 8 cards without an ace turning over if we know which one is the ace). Thus if the 'host' flips over 8 of the 9 cards and tells us he isn't flipping an ace, then the probability the unflipped card is an ace is (9/10) * 1 = 9/10, and we should switch.

Shane

[Lottery Larry]

I got to the same place as you did, Shane, up to here.

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The difference between that is the cards in this case are picked at random, without knowledge. In the Monte Hall problem, if we always have to show 8 of the 9 cards (and none can be the ace) then the last 1/9 probability becomes a probability of 1 (since we can always flip over 8 cards without an ace turning over if we know which one is the ace). Thus if the 'host' flips over 8 of the 9 cards and tells us he isn't flipping an ace, then the probability the unflipped card is an ace is (9/10) * 1 = 9/10, and we should switch.

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This is where I'm running into problems trying to translate for the person. The person's argument was, if it was 90% for the Ace to be available in the big pile... why isn't it 90% for the 5 to be in the big pile instead?

I know we're not lowering the percentage chance of a card being left from the big pile, because we're intentionally not lowering the initial probability (we're only exposing bad cards). My friend sort of understands that.

But their argument is that I am arbitrarily picking the Ace to be the 90% card, rather than the 5, to say that you should always switch in a Monty Hall version of Deal or No Deal.

I can't figure out a way to explain this sticking point.... especially when you compare end results of the Monty Hall version (8 unwanted cards exposed, down to choosing two, when we want the higher value one) and the DoND version (8 unwanted cards exposed, leaving 2 of which we want to pick the higher value one).

It's hard to look at that and say "Oh yeah, one should be 90% and the other 50%"... and even harder to think of a way to explain it.

Any more help, Shane? Or anyone else?

[ShaneP]

Well, I might not be understanding something totally, but I'll put things in a different way to show why the 50/50 versus 90/10. I'm not sure why your friend is fixated on a specific card, so I might be missing something. Actually it IS 90% that ANY card is in the big pile...which would include the ace or any other card. Then it's 10% that any particular card is the one you picked initially, and 10% chance that it was in the 9 pile and survived the random cuts. (if your friend knew it, I'd say then you just Bayesian update...but that probably doesn't help)

I'll assume you're trying to pick the ace.

The first case can be seen as: pick one card. OK, pick another card (that's the one of the pile of nine that wasn't chosen). Now, here's all the cards you didn't pick. Gee, there's an Ace and a 5 (or a 6 and a 3, or whatever) that haven't been seen. You've got no new information about which card is which. It's just you've picked the two cards in a slightly different way than above, but which results in exactly the same distribution. So if you know the Ace and 5 are still out there, you don't have any way of differentiating between the cards.

The Monte Hall problem is : pick one card. OK, now I'll look through the cards you didn't pick, and show you eight of the nine. Now, do you want to switch. Since we can't reveal the ace, given there was a 90% chance it was in the pile of nine, and it can't be revealed, you now have a 90% chance that the ace is the card that was left after the other cards were revealed to be not aces.

One other thing to remember, is that yes, the ace has a 90% chance of being in the pile of 9, but if it is, it has a 89% chance of being revealed, so that really cuts down the probability of the card being in the pile of nine and being the card left unrevealed at the end.

I think the difference can be summed up as follows: you don't have any information (since they are essentially chosen in the same way) to differentiate between the two cards in the first case, but in the Monte Hall version, the cards are chosen quite differently...one was chosen by the participant, and one was chosen by Monte by taking away bad cards.

A final comment: in the DonD version with 10 cards, there's only a 20% that the ace isn't revealed until there are only two cards left...it probably doesn't affect the understanding of the problem, but in the monte hall problem there is a 100% chance the ace isn't revealed until the final two. That should be another indication that the final two cards are chosen differently.

Hope this helps...as I said, I dont' think I'm quite parsing the initial question totally, but I think this is on the right path...

Shane

[Pokerlogist]

In the Monte Hall door scenario, if Monte picked the door randomly like you did in the card game then the player would lose the game 2/3 and win 1/3 of the time when switching. Why? The player can lose because of the the first pick or if not, then by swicthing. The player would instantly lose the 1/3 of the time Monte randomly picked the prize door first. When Monte didn't pick it first then the player now has a one of two chances when switching. So overall, 2/3 of the time the player would still have a 1/2 chance of loss=1/3 for a total of 1/3 + 1/3=2/3 total chance of loss.

When Monte specifically picks a non-prize door first (as in the non-random orginal version classic problem) it eliminates the "first pick" loss chance and leaves the player with only the 1/3 chance of loss when switching.

In your card game the player would instantly lose if you happened to pick an Ace. You left the player the chance of the "first pick" loss along with the potential loss from switching so it is equivalent to the Monte Hall random pick scenario.This is NOT the original version of the problem. Under your card game scenario, the chances of loss at the end must be 50%. You could re-configure the card game so there was no immediate player loss from you picking an Ace. So if you found an Ace, you would replace it without showing. Then when it got down to one card, the player would have 90% chance of winning by switching.

[Sevenfold]

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What is it about one condition- being forced to exposed losers only- that makes the probabilities different in the end result?

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Player needs ace to win.

Player picks a card.

You randomly expose all but one.

Just ace and 5 are left.

This is 50/50.

What we are forgetting is 80% of the time the game ends when we expose the ace.

So---of the 20% of the time we randomly expose cards and no ace appears, we are 50/50 by switching.

If we look at the cards, the game is always played out, and the ace is on our side 90%.

So, by not looking at the cards, game ends 80% of the time.

So from this point ---the 20% played out with no ace---10% he has ace, 10% the other card, we are 50/50 (10% of 20%).

If you want to make more conditions, the ace and 5 both play say, then the game ends every time those cards are exposed.

So of the times that only the 5 and ace are left , it is 50/50.