## Monty Hall-ish Problem

A forum for good logic/math puzzles.

Moderators: jestingrabbit, Moderators General, Prelates

Posts: 455
Joined: Fri Nov 28, 2014 11:30 pm UTC

### Monty Hall-ish Problem

You are playing a game with a standard deck of 52 cards. The dealer draws three cards. Looks at them, then places them face down before you. He then draws a fourth card and puts it face down without looking at it. You are given two choices: take first three cards or take just the fourth card.
You win if any of the cards you take are from a black suit. Otherwise, you lose.
However, there's an added twist! The dealer can reveal up to 2 cards after you make your decision. You are then given a chance to change your choice before the final reveal.

In a particular round of this game, you chose the first three cards and the dealer flips over two of them, both red suits. Should you stick with your decision or switch to the fourth card?
This is a block of text that can be added to posts you make. There is a 300 character limit.

Cauchy
Posts: 602
Joined: Wed Mar 28, 2007 1:43 pm UTC

### Re: Monty Hall-ish Problem

I don't think this problem is tractable, since the other player has agency. The lack of agency is what makes the Monty Hall Problem tractable, and its presence is what makes Let's Make a Deal interesting.

Anyway, switching yields a 52% chance of winning (26 black cards out of 50 cards not visible to you), so you should stay precisely if you believe you have greater than a 52% chance of winning. Is your friend revealing these in the hopes that you switch, or stay? Are they employing reverse psychology or double reverse psychology? If they take the strategy "always reveal cards 2 and 3 whenever all three cards are red, and never reveal otherwise", then you should switch. If they take the strategy "always reveal cards 2 and 3 whenever the cards are black, red, red, and never reveal otherwise", then you should stay. Without an insight into his psyche, it's hard to make any calls, and if we ignore his motivations completely, then we're left with the naive view of Monty Hall, where the two remaining cards are equivalent.
(∫|p|2)(∫|q|2) ≥ (∫|pq|)2
Thanks, skeptical scientist, for knowing symbols and giving them to me.

douglasm
Posts: 630
Joined: Mon Apr 21, 2008 4:53 am UTC

### Re: Monty Hall-ish Problem

Insufficient information - how does the dealer decide how many and which cards to reveal?

In the Monty Hall problem, the host is required by the rules to always reveal 1 non-prize door, and that information is a necessary part of the solution.

Xias
Posts: 363
Joined: Mon Jul 23, 2007 3:08 am UTC
Location: California
Contact:

### Re: Monty Hall-ish Problem

An easy fix is to say that the opponent must reveal two cards. Then, obviously, if there were two or three black cards in the first pile, you would win outright, but red-red-red and red-red-black are indistinguishable.

Spoiler:
At that point though it is easy to see that there is no benefit to switching. This is because the problem deviates from Monty Hall in a second way: What you choose in the beginning does not alter the choices available to the dealer. In Monty Hall, the door you select in the beginning cannot be opened by Monty. In this case, regardless of what you choose, the dealer still looks at the same cards and still may choose any of them to reveal.

Therefore, the game merely changes to "Is it more likely to get black-red-red over red-red-red than it is to get black over red?" and the odds are the same.

It might seem that red-red-red would be more likely after the reveal because there are three more ways to choose two red cards than in the black-red-red case. This sort of thinking would lead to a 50/50 solution to Monty Hall: you are half as likely to have picked the car than the goat, but if you chose the car there are two goats for him to show you, which evens it out.

mward
Posts: 123
Joined: Wed Jun 22, 2011 12:48 pm UTC

### Re: Monty Hall-ish Problem

Assuming that the dealer doesn't want you to win, we can be sure that the dealer will never turn over the single card or any black cards.

If the dealer always turns over two red cards when there are at least two red cards in the pile of three, then we have four possibilities which are (roughly) equally likely:

RRR
RRB
RBR
BRR

In three out of four cases, the third card is black. For the single card there are only two roughly equal possibilities R or B. So picking the three cards gives you a better chance of getting a black card.

So suppose the dealer only turns over two reds when there are three reds: to try and persuade you to stay with the three cards. In a long run of games you will be able to detect and take advantage of this strategy. In fact, any strategy used by the dealer will give you additional information about the cards (once you have worked out what the strategy is). Given that, in the absence of any information, you are much better off choosing three cards rather than one, it is hard to see how the dealer could gain an advantage by using one strategy to give you one set of information and then switching strategies to fool you into losing. I think you would win more during the "establishing the strategy" phase than you would lose in the "switching strategies to try and fool you" phase.

### Who is online

Users browsing this forum: No registered users and 8 guests