Monty Hall Game Revised?

In the classic Monty Hall game, a game contestant is shown 3 doors. 2 doors hide cheap prizes, like cleaning products, and 1 door hides the grand prize, maybe an all expenses paid trip to Bora Bora. The contestant gets to choose 1 door. After the door is chosen, the game host then [always] picks another door and opens up, [always] showing a cheap prize. The contestant is then offered a chance to switch to the other unopened door or stay put. But because this is now a much-studied, much-publicized game, and nearly every knowledgeable person knows now that it's far better to switch than to stay put, the game producers have decided to change the game format. They have kept the original game host, which we'll call A, and bring in 2 more, called B and C. After the contestant first makes a choice,

A always then picks another door that always hides a cheap prize
B picks another door with a flip of a coin, and if it reveals the grand prize, does the game over again repeatedly until he picks a door revealing a cheap prize
C picks another door always revealing a cheap prize, but only if the contestant has first picked a door hiding the grand prize

A contestant is invited to play the new format game, and knows about the different game hosts, A, B, C, but does not know who is which. However, he can pick any one of them to play the game. Moreover, after the contestant has made the 1st choice, and the game host picks [or does not pick] another door, and is given a chance to switch, the contestant is allowed to scrub the game and start over again, again with any game host. But only 2 retries are allowed.

Following the best possible strategy, what is the contestant's chances of winning the grand prize?

2011-05-13T09:56:13Z

Bonus: What if the contestant is allowed 3 retries?

2011-05-13T18:29:14Z

The location of the grand prize is always randomized, with any repeats by game host B, or re-tries by the contestant. Otherwise, the problem becomes too trivial.

2011-05-13T18:51:02Z

Anytime you see the words "game over again", or "start over again", the location of the grand prize is randomized for the contestant, and only game hosts A and C know exactly where it is. For game host B, it is irrelevant, as he decides with a flip of a coin.

2011-05-13T18:51:51Z

After randomization, the game starts over and the contestant is again given a choice to pick the first door.

2011-05-13T21:34:22Z

husoshi, once again, let's say that game host B throws a coin and opens and door and reveals the grand prize. He then does the whole thing over again from scratch, the grand prize being in a new location and the contestant asked to make his 1st choice. This is not a problem with game host A, because he always knows how to avoid opening the door to the grand prize. Ditto for game home C. And, yes, every time the contestant wants to retry a game, everything is started over again from scratch, and he can choose the same or different game host.

2011-05-14T23:16:53Z

All right, so this is a tough problem. Maybe I'll repost at another time.

2011-05-15T14:58:01Z

I'm continuing to get messages asking me to clarify the rules. I think this Y!A question is too bogged down with explaining the rules to go any further, I might have to close this out. Once again, the rules:

1) A game begins with the contestant's choice of game host and 1st door,
2) A retry is when the contestant voluntarily cancels a game and gets another start. The contestant is allowed up to 2 retries (or 3 retries in the bonus question).
3) When game host B opens a door revealing the grand prize, he does the game over again, but this does not count as a retry for the contestant.
4) In all the games, the contestant is given the chance to switch from his 1st choice to another unopened door, but does not have to switch. He can cancel the game and go for a retry.
5) Game host A will ALWAYS open another door that will ALWAYS reveal a cheap prize
6) Game host B will ALWAYS flip a coin to decide which door to open, which CAN reveal the grand prize, in which case he will do the g

2011-05-15T14:59:04Z

6) Game host B will ALWAYS flip a coin to decide which door to open, which CAN reveal the grand prize, in which case he will do the game over again, which does not count as a "retry", repeating until he reveals a cheap prize
7) Game host C will ONLY open another door if the contestant's 1st choice happens to be the one hiding the grand prize. Of course, if the contestant knew that the game host is C, and he does open another door, probability is 100% that the grand prize is behind the contestant's 1st choice.
8) At the beginning, the contestant is aware of the 3 different ways the game hosts play this game, but does not know who does what. That is for him to deduce, and to make the best choice based on what he's able to discern.

2011-05-15T15:00:47Z

See this wiki to grasp some of the complexities of this very controversial probability problem:

http://en.wikipedia.org/wiki/Monty_Hall_problem

Mugen is Strong2011-05-15T22:40:36Z

Favorite Answer

my strategy would be trying for 'normal' door-opening, like when A was the host because P(win with host A) = 2/3, and then decide to change for all cases with 1 exception, that is when C refused to open in 1st try but opened in the 2nd one, where i'll stick to my 1st choice. A would give normal opening in 2 retries (no changing host) by 100%, B by (2/3)^2 = 4/9 and C by 1/9.

let B' and C' denote B and C being recognized by contestant.

list of normal-looking opening
= {AA, BB, CC}
= {9/9, 4/9, 1/9}

P(get normal opening from A twice) = (3/3)^2 = 1
P(win from (2nd) A if i change my door) = 2/3

P(get normal opening from B twice) = (2/3)^2 = 4/9
P(win from (2nd) B if i change my door) = 1/2

P(get normal opening from C twice) = (1/3)^2 = 1/9
P(win from (2nd) C if i change my door) = 0

C'C is useful because i get to recognize C in the 1st try and deflect his tactic in the 2nd.
happens with probability of (2/3)*(1/3) = 2/9, but if it does happen, i'll stick for its 100% warranty.

CC' and C'C' not so much. with C' in the 2nd round i only know the winner is NOT my 1st choice door. only 50% of the time i win if door is chosen randomly. both collectively happens = 1 - 1/9 - 2/9 = 2/3.

any B' would be the same, because knowing the host identity won't give me any clue which door to choose nor to avoid, as the host themselves have no prior info of it. i could change host if i got B' in the 1st round / try, but it did not produce a better result than by just sticking with B (same result actually). maybe if it was 3 retries changing from B' will give better result with the higher disparity of A : C = 9 : 1 in contrast to 3 : 1. so BB' + B'B' + B'B = 1 - 4/9 = 5/9. winning (by changing) in these cases is 50%.

i get 31/54 as final result, which is much less than 2/3 in the original game with host A.
= 1/3 * [(1 * 2/3 + 4/9 * 1/2 + 1/9 * 0) + (2/9 * 1) + (2/3 * 1/2) + (5/9 * 1/2)]
= 1/3 * [8/9 + 2/9 + 1/3 + 5/18]
= 31/54

edit
P(it was AAA) = 27/27 , P(win by changing if it was AAA) = 2/3
P(it was BBB) = 8/27 , P(win by changing if it was BBB) = 1/2
P(it was CCC) = 1/27 , P(win by changing if it was CCC) = 0

P(it was C'C) = 2/9 , P(win by sticking if it was C'C) = 1
P(it was CC'C) = 2/27 , P(win by sticking if it was CC'C) = 1
P(it was C'C'C) = 4/27 , P(win by sticking if it was C'C'C) = 1

P(CCC' or CC'C' or C'C'C') = 1 - 1/27 - 2/9 - 2/27 - 4/27 = 14/27 , P(win by changing) = 1/2

if i get normal looking opening in 1st try i'd just stick with that host. i would only change if i get B' in 1st round.
P(get normal looking opening in 1st try but turn out to be host B later) = 2/3 - 8/27 = 10/27
P(win by changing with those) = 1/2

P(it was B'AA) = 1/6 , P(win by changing if it was B'AA) = 2/3
P(it was B'CC) = 1/54 , P(win by changing if it was B'CC) = 0
P(it was B'C'C) = 1/27 , P(win by sticking if it was B'C'C) = 1
P(it was B'CC' or B'C'C') = 1/9 , P(win by changing if it was B'CC' or B'C'C') = 1/2

answer,
= 1/3 * [(2/3 + 4/27 + 0) + (2/9 + 2/27 + 4/27) + (7/27) + (5/27) + (1/9 + 0 + 1/27 + 1/18)]
= 1/3 * [22/27 + 12/27 + 7/27 + 5/27 + 11/27]
= 1/3 * 57/27
= 19/27

better than original game of 2/3.

50

husoski2011-05-14T04:25:05Z

It's not clear to me what happens when B happens to choose the grand prize. Does the reshuffle of the prizes count as one of the players retries? If not, is the player aware of this reshuffle?

I'm not sure it affects the player's strategy, though. I don't see how anything beats:

The player should reroll the game, if allowed, any time a door is opened. If no door is opened, he should stay with his latest choice. If a door is opened on the final reroll, he should switch.

Assuming an invisible reshuffle that doesn't count as a retry, then B is indistinguishable from A, provided that A doesn't have a known bias such as picking the lower numbered door when two zonkers are available to show. All that's known is that a random door, known to be a zonker by the opener, is opened. At that point, the prize is behind the unpicked, unshown door with probability 2/3, as in the standard version of the problem.

Only C can refuse to open a door, and that's a certainty that the chosen door is the winner. That happens with probability 1/9 on any trial. So, with k total tries (k-1 rerolls maximum) the player sees a door opened on the final try with probability (8/9)^k and loses only on 1/3 of those cases. The overall probability of a win is then 1 - (1/3)*(8/9)^k. Even without a retry (k=1), the player wins more than in the classic game, just because of C.

Edit: After a reread, it sounds like the odds might be better for the player. 1/9 of the time, (1/8 of the cases where the player didn't win outright by C failing to open), the player will know who B is and can avoid him for the rest of the game. That improves the odds of winning outright to 1/6 for any future games. Further, he should switch hosts after a reveal, since that host is less likely to be C.

I don't have the energy to work out the exact probabilities tonight, but it's do-able...if messy...on this basis for specific values of k. I don't know if a readable closed form in terms of k is there.

20

torquestomp2011-05-14T01:07:46Z

This seems like a pretty straightforward application of probability laws, but first, I'll ask for some clarification:

- What does B 'doing the game over again repeatedly' mean? He flips the coin again? Shuffles the doors and flips a coin? Simply chooses not to reveal a cheap prize? Or something else?

- If/when the contestant chooses to scrub the game, are the prizes shuffled again?

I think the clarification on B is most significant. If B is unable to not reveal a door, then C is the only one who can, and thus if our host chooses not to open any door, we have 100% certainty that our chosen door contains the grand prize.

20

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