D&D 5E Title / Subject - or probabilities are hard

[Ovinomancer]

What if to choose, I roll 1d6 assigning 1-2 to the site we were headed to, 3-4 to one of the other two sites, and 5-6 to the other?

The roll comes up 1-2.

Are my odds now 1:3 of being right, even though I'm still heading for the site I chose before Monte turned up? Are the odds for each of the two other sites now also 1:3?

No, and this:

Nope. Your dice roll added no new information to the problem, so it can't affect the odds. What CapnZapp posted before is a variation on the classic Monty Hall problem (or the game show problem).

Now, I'm making some assumptions, because I know that CapnZapp is a making a joke about the Monty Hall problem. If Monta Halle's appearance is truly random (say, a 1 in 4 chance that she appeared and told the party that their current destination is wrong), than my statement above would be wrong. The Monty Hall problem only applies if the outside participant (Monta Halle, in this example) is giving information based on knowledge of the overall state of the sites.

Again, I think this is true...I'm going of what I remember of the problem from reading something about it 20 some odd years ago.

Is pretty close.

The 3 doors version of this problem is the simplest, but it extends. I'll try to present the 4 sites version of it.

You are initially presented with 4 choices, only one of which is correct. You have no other information on the choices at this time. You chance of naively picking the correct choice is 1:4 (assuming a fair contest, as always).

Now it get interesting, because the Monty Haul stand in is going to provide more information before we see if our choice is correct, but, as 26 notes, that information is privileged information that is provided knowing the full truth of which choice is correct. Monte will now provide information about one of the choices you didn't pick, but will always, always, always show you a wrong choice. This is important -- Monte's information is not random at all, but must show an incorrect choice from those that you did not pick.

So, to back up a moment, your chance of picking the correct site at first is 1:4. This means that there's a 3:4 chance that you picked the incorrect site and the correct site is one of the 3 remaining sites. When Monte tells that one of the unpicked sites is wrong, he's providing much more information that is first apparent. In the likely event that you picked incorrectly, you now have one fewer incorrect other choices. In other words, the likelihood that the correct site is one of the choices you didn't pick has changes from 3:4 to 3:8 (as twosix points out). The why is more apparent if you break it down a bit.

Odds your first choice is correct (1 pick from 4) is 1:4. This never changes because it was picked when you had no information about the other doors. The reveal of one other incorrect site doesn't alter these odds because you already knew there were at least 2 wrong sites among the other picks (the case where you picked the correct site first and the case where you didn't), so this information doesn't change the odds of your first pick being correct. It does, however, change the odds that one of the 2 remaining unpicked sites is correct. In the pre-reveal information, the odds that each site is the correct site is 1:4. This adds up to a 4:4 total odds, or a probability of 1 that the correct site is among those four choices. But when Monte reveals one incorrect site, this changes. The odds you picked correctly to begin with are still 1:4 (see above for why), but now there are only two possible choices left instead of 3. The probabilities must still add up to 1, and we have the 1:4, so the remaining 2 choices have to split among the original 3:4 probabilities, so we get 3:8 chance that each of the remaining two choices are the correct site. 3:8 + 3:8 + 1:4 is 4:4, so we've satisfied the 1 probability requirement.

That's how the Monte Hall problem works, generally. By providing new information after the choice but before the reveal, the odds of success are shifted. The more choices available, the slighter the (at 3 choices it goes from 1:3 to 2:3, for instance).

To give a slightly different example, let's look at a more extreme reveal. If you have 15 sites to pick from, and, after you pick, Monte looks at the remaining 14 sites and then reveals 13 of them that are wrong, leaving only one site, and asks if you want to switch, it's really apparent that switching is far more likely to pick the correct site. While this is extreme, it does more clearly show the effect that the additional information has. Even if Monte only told you about 1 of the remaining 14 sites, the same effect, at a much smaller impact, would be there and you'd still, on average, be better off switching your guess.

 

[clearstream]

No, and this:


Is pretty close.

The 3 doors version of this problem is the simplest, but it extends. I'll try to present the 4 sites version of it.

You are initially presented with 4 choices, only one of which is correct. You have no other information on the choices at this time. You chance of naively picking the correct choice is 1:4 (assuming a fair contest, as always).

Now it get interesting, because the Monty Haul stand in is going to provide more information before we see if our choice is correct, but, as 26 notes, that information is privileged information that is provided knowing the full truth of which choice is correct. Monte will now provide information about one of the choices you didn't pick, but will always, always, always show you a wrong choice. This is important -- Monte's information is not random at all, but must show an incorrect choice from those that you did not pick.

So, to back up a moment, your chance of picking the correct site at first is 1:4. This means that there's a 3:4 chance that you picked the incorrect site and the correct site is one of the 3 remaining sites. When Monte tells that one of the unpicked sites is wrong, he's providing much more information that is first apparent. In the likely event that you picked incorrectly, you now have one fewer incorrect other choices. In other words, the likelihood that the correct site is one of the choices you didn't pick has changes from 3:4 to 3:8 (as twosix points out). The why is more apparent if you break it down a bit.

Odds your first choice is correct (1 pick from 4) is 1:4. This never changes because it was picked when you had no information about the other doors. The reveal of one other incorrect site doesn't alter these odds because you already knew there were at least 2 wrong sites among the other picks (the case where you picked the correct site first and the case where you didn't), so this information doesn't change the odds of your first pick being correct. It does, however, change the odds that one of the 2 remaining unpicked sites is correct. In the pre-reveal information, the odds that each site is the correct site is 1:4. This adds up to a 4:4 total odds, or a probability of 1 that the correct site is among those four choices. But when Monte reveals one incorrect site, this changes. The odds you picked correctly to begin with are still 1:4 (see above for why), but now there are only two possible choices left instead of 3. The probabilities must still add up to 1, and we have the 1:4, so the remaining 2 choices have to split among the original 3:4 probabilities, so we get 3:8 chance that each of the remaining two choices are the correct site. 3:8 + 3:8 + 1:4 is 4:4, so we've satisfied the 1 probability requirement.

That's how the Monte Hall problem works, generally. By providing new information after the choice but before the reveal, the odds of success are shifted. The more choices available, the slighter the (at 3 choices it goes from 1:3 to 2:3, for instance).

To give a slightly different example, let's look at a more extreme reveal. If you have 15 sites to pick from, and, after you pick, Monte looks at the remaining 14 sites and then reveals 13 of them that are wrong, leaving only one site, and asks if you want to switch, it's really apparent that switching is far more likely to pick the correct site. While this is extreme, it does more clearly show the effect that the additional information has. Even if Monte only told you about 1 of the remaining 14 sites, the same effect, at a much smaller impact, would be there and you'd still, on average, be better off switching your guess.

My thought is that the die enters the scene with no information. From its perspective, there are three sites to pick from, and one of those three contains the idol.

So for the die, the odds are different. The odds of the rolling the number on the die that corresponds to the right site is 1:3 (discarding my knowledge). I think what we're dealing with is no longer our odds of choosing the right site, but rather our odds of seeing the corresponding numbers come up on the die.

Think of it this way. A rogue hides an idol under one of five cups. We lift a cup and see the idol is not there. We then nominate one of the remaining four to lift. The rogue lifts one we did not nominate and lo, the idol is not there either. At this point, there are three cups and - being a fair game - the idol is under one of them. A paladin passing by, having no information at all about the game, is asked to choose. What are the odds from the paladin's perspective? A cleric, passing by, takes a wager on the paladin choosing correctly: the odds are 1 in 3. Rolling the die makes me that cleric: I've switched to betting on what I will roll on the die.

It's an inferior strategy, but seems to have improved my chances if I end up back on the site I already chose. As that became a 1 in 3.

 

[Ovinomancer]

My thought is that the die enters the scene with no information. From its perspective, there are three sites to pick from, and one of those three contains the idol.

So for the die, the odds are different.

Think of it this way. A rogue hides an idol under one of five cups. We lift a cup and see the idol is not there. We then nominate one of the remaining four to lift. The rogue lifts one we did not nominate and lo, the idol is not there either. At this point, there are three cups and - being a fair game - the idol is under one of them. A paladin passing by, having no information at all about the game, is asked to choose. What are the odds from the paladin's perspective?

The paladin is playing a different game, though. Essentially, the paladin (our the die) is overweighting the chance that the first pick is correct and underweighting the others as potentially correct.

This is clearer in the extreme example. If there are 10 picks, you have a 1:10 chance initially. That's a 9:10 chance of being wrong. If, of the unpicked 9, I reveal 8 that are wrong and offer you the chance to switch, you have a 90% chance that the switch will be correct compared to 10% for your initial. The paladin walks by and naively picks between 2 available choices, 50/50. He overweights the initial pick too much and underweights the switch pick by too much, but, for his information, this is the best he can do.

The Monte problem hinges on the fact that the offer to switch is no longer made under niave conditions. Substituting a new naive choice, even though more constrained, does not accurately reflect the probabities. Even in the original, the d3 roll overestimates the probability of the first pick being correct (1:4 to 1:3), and underestimates the switch pick (3:8 to 1:3). The difference in the switch is slight, which is where some of the confusion can set in. Few look at the change in probabilities of the first pick as relevant, but they very much are.

 

[clearstream]

The paladin is playing a different game, though. Essentially, the paladin (our the die) is overweighting the chance that the first pick is correct and underweighting the others as potentially correct.

Indeed, but going back to the die: what are the odds of the number rolled on the die corresponding to the right site?

3 sites remaining. 1-2 is 1, 2-3 is 2, 4-5 is 3.

I think what is happening is we change the game from betting on which door it is, to betting on whether the die will roll that door.

 

[Ovinomancer]

Indeed, but going back to the die: what are the odds of the number rolled on the die corresponding to the right site?

3 sites remaining. 1-2 is 1, 2-3 is 2, 4-5 is 3.

I think what is happening is we change the game from betting on which door it is, to betting on whether the die will roll that door.

Yes, as I said, the paladin and the die are playing a different game. I guess I'm not following where you're going with this, as I don't see what the different game the die plays has to do with the Monte Hall question. Using the die, or the paladin, is in all ways worse -- it can make you overconfident that your original choice is better than it is and it hides the improvement in odds from switching your choice. The die does worse in all cases because it is a naive choice vs a partially informed choice.

 

[TwoSix]

Indeed, but going back to the die: what are the odds of the number rolled on the die corresponding to the right site?

3 sites remaining. 1-2 is 1, 2-3 is 2, 4-5 is 3.

I think what is happening is we change the game from betting on which door it is, to betting on whether the die will roll that door.

Yes, you are. It's an inferior strategy (as opposed to choosing to move to a new site, which has 3/8 odds), as you've willing sacrificed information which gave you a clue as to the choice to make and left it up to random chance. But it does impact your odds as you've stated.

 

[Harzel]

Typo, but a potentially confusing one...

In other words, the likelihood that the correct site is one of the choices you didn't pick has changes from [-]3[/-]1:4 to 3:8 (as twosix points out). The why is more apparent if you break it down a bit.

 

[Harzel]

Indeed, but going back to the die: what are the odds of the number rolled on the die corresponding to the right site?

3 sites remaining. 1-2 is 1, 2-3 is 2, 4-5 is 3.

I think what is happening is we change the game from betting on which door it is, to betting on whether the die will roll that door.

Assuming you meant 1-2 is 1, 3-4 is 2, 5-6 is 3.

Yes, you are. It's an inferior strategy (as opposed to choosing to move to a new site, which has 3/8 odds), as you've willing sacrificed information which gave you a clue as to the choice to make and left it up to random chance. But it does impact your odds as you've stated.

Sort of interesting. Uniform random choice among the three sites is a better strategy than always sticking with the one you picked originally, but worse than always switching to one of the others. (6:24, 8:24, 9:24)

 

[Harzel]

The 3 doors version of this problem is the simplest, but it extends. I'll try to present the 4 sites version of it.
<snip>

Hey, you folks are being stingy with the XP; that was a good explanation.

 

[CapnZapp]

Die, Paladin, Die!

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