Let's Make A Deal 2/23 - Monty Hall Problem or No ?

Not sure if this should go here, in the Cafe, or elsewhere. Mods please move.


Wayne Brady just offered a contestant the classic Monty Hall Problem, Or did he?

Three “Boarding Passes” were offered to two contestants. Each contestant took one. They had to scan their pass to confirm that they had the trip.

Player one scanned their pass and Zonked, declining a Living Room group “sure thing”.

Player two was given a few options : Scan their pass, Switch their pass with the remaining pass, or to take the same Living room group.

Player took the Living Room, her pass was another Zonk ticket, leaving the remaining ticket as the only “good” ticket.

Does this qualify as a Monty Hall Problem?

I mean, I can see it argued either way. Given the outcome, I must assume that the player wasn’t aware of the Original Monty Hall Problem.

I mean, I Would have jumpped at the option to switch tickets. But the switch was to the only remaining ticket. I guess this changes the problem just enough, as to not qualify for a pure Monty Hall Problem.

That sounds like the Monty Hall problem to me. Contestant #2 originally had a choice from 3 tickets, one of which was the trip. She had a 2/3 chance of not having the trip in her hand. When one of the other two tickets was revealed to not be the trip, there was still a 2/3 chance that the other ticket she didn’t hold was it. She should have taken it.

Yeah, as presented it is not the Monty Hall Problem at all.

ETA: posted while Skammer was posting. I disagree with Skammer, here. Each ticket had a 1/3 chance, revealing the one leaves the others with a (1/3)/(2/3) chance, or 1/2 each.

That’s pretty much an exact definition of the Monty Hall Problem.

The Monty Hall game works because Monty Hall is restricted in what he can do by the rule that he never reveal the winning prize. If you choose incorrectly the first time, he can only open one other door. Maybe I’m misreading something, but in the boarding pass problem, the host doesn’t know which one pass wins the trip. Could the first contestant have won?

ETA: choosing incorrectly the first time restricts Monty, not correctly. If you choose correctly the first time, he can open either of the other doors at random.

I don’t think it’s unreasonable to assume that the host knew which passes were which. And regardless, if the second contestant had been holding the winning pass, the offer would have still been the same, with the same (improved) 50/50 odds.

No it isn’t. In the Monty Hall problem, because the host is restricted in his choice of which door to open, your chance of winning becomes 2/3 if you switch doors. In a random, non-Monty Hall problem where the host is unrestricted, your chance of winning is 1/2 if you switch doors.

The Monty Hall problem is this:

If “player 1” was identified as the first player to scan the boarding pass before the players chose their passes, then this was not a Monty Hall problem.
If Wayne chose which player was “player 1” randomly, then this was not a Monty Hall problem.
If Wayne chose which player was “player 1” because he knew that player had chosen a goat (and was therefore restricted in choosing which player could go first), then this was a Monty Hall problem.

Okay - that’s right. Sorry!

(The best way to remember (which I forgot to remember) is thinking of millions of doors, rather than 3. Once Monty boils the choices down between your 1-in-a-trillion lottery ticket and one other hand-picked lottery ticket sent from heaven, the odds are clearly not 50/50, but rather 999,999,999,999/1,000,000,000,000 if you switch).

This is still not quite right.

If you think about tit he scenario described by this condition both remaining players would be able to increase their chances by changing their choice, which is clearly paradoxical as the total probabilty would then add up to 4/3 and not unity.

You need an two additional conditions:

  1. One player must be pre-idenitified so that they will not be picked to try their boarding pass first whether they get a goat or not.

  2. Only players who know the player selected to fufil 1) are able to adopt the optimum strategy.

In this scenario the player selected for 1) increases their chances by switching (to 2/3), whereas the remaining player [i[reduces their chance by switching (1/3).

According to the OP, there are only two players total, so once player 1 is disposed of, there is only one remaining player. The fact that the third player is non-existent fulfills your two conditions.

Must read the OP properly. :smack:

In this case thoguh the condition that is needed is that “player 1” (the player who will beasked if he wishes to change his choice) is given a card out of the 3 at random, whereas “player 2”’ card is not completely random. If “player 1” selects a goat, then player 2 must be given the prize (which makes the problem seem quite trivial infact), if “player 1” selects the prize then it doesn’t matter which goat “player 2” selects.

Why?

There are three possibilities: Player A has the winner, player B has the winner, and neither has the winner. If A has the winner, host scans B, and if B has the winner, the host scans A. If neither has the winner, the host flips a coin. The possibilities and associated probabilities are

A picked the winner, B is scanned 1/3
B picked the winner, A was scanned 1/3
Neither picked the winner, A was scanned 1/6
Neither picked the winner, B was scanned 1/6

So if I’m A, and the host scans B (making B be player 1), half of those possibilities are eliminated, but 2/3s of the remaining times, I already picked the winner. So I shouldn’t switch, and it’s not the same as Monty. Kind of the opposite.

Oooh, right; I missed that. Wayne is constrained in a different way than Monty is, so the answer changes. That’s kinda neat.

Just adding here as well: I thought it was odd that you had to “scan” the boarding pass.

Jonathan Magnum, the Announcer, brought down a table that had a TV screen on it. Boarding passes were scanned, and the TV showed the result.

Given that the TV is a digital answer, and not an Analog answer, I am curious as to if the results were made up on the spot, or if the passes were locked in into a Zonk / Non-Zonk state before this game started.

I’d suspect the former as well. Human nature tries to validate any choice you’ve made. And most people think the answer to the Monty Hall problem is to not switch. And the actual true evaluation says that switching is the worst option. It would make sense to set it up where the card they didn’t pick was the winner. Or, if they really wanted to be unfair, always make it the last card scanned.