# Deal or No Deal

## 13 posts in this topic

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Here's a twist on the Monty Hall problem.

Some of you might be familiar with a show called Deal or No Deal. For those you are not familiar, a simplified version of the show is as follows,

* There is a host, a player, and 40 suitcases of money.
* The host informs the player that there are 39 empty suitcases and 1 case with 1 million dollars. The suitcases are closed and the player has no idea how much each specific suitcase contains. At the beginning of the game, the player is requested to choose 1 suitcase at random. Let's call this the Chosen Suitcase
* At the beginning of each turn, the player is requested to choose 1 suitcase from the remaining Non-Chosen Suitcases, reveal the contents, and put this case and contents in a discarded pile. After this, the player has the choice of quitting the game and keep the content of his unopened Chosen suitcase, or switching his Chosen Case for one of the remaining Non-Chosen Cases and play on.

Let's say that the player goes through 38 turns, opening and discarding suitcases without ever switching his original Chosen Case. The 38 opened cases are all empty. There are now only 2 unopened cases left, and 1 of them is currently chosen by the player. Should he switch his original Chosen Case for the remaining Suitcase for the best chance to win 1 million dollars? Assuming optimal strategy, what is the player's probability of winning the 1 million dollars in this situation?

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Posted · Report post

I don't think it matters. From what I know of the Monty Hall problem, the option to switch doors/cases benefits the player only because the host has knowledge of which door/case gets eliminated, ensuring that the best option remains in the end.

With that being said, I would switch my choice anyway. In the event that I'm wrong on my assumption, the odds are still 50/50 at worst when I switch if my first paragraph assumption is correct, and 39/40 if I'm wrong in the first paragraph.

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Posted · Report post

i would tend to think so, that is switching is better than staying, but of course you want to know how much better.

my guess is that there is a 1/40 chance his case is the million dollar winner, with the other case being 39/40.

in truth however this seems like a trick question. for example, in the original Monty hall problem, it was the host opening the

other door; or the other suitcases in this scenario. and he knew which case has the million, and would only open the empty ones.

so in this scenario where the person opening the suitcases has no idea whether switching is better, i honestly don't know for sure.

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Posted · Report post

gonna go with the 50/50 crowd. "And now for something completely different," would have to see some Monte Carlo sim for this Monty Pythonish Monty Hall varient.

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Posted · Report post

Maybe we have to ask, "What is the probability we would see this situation if the Chosen case has the million?" vs. "What is the probability we would see this situation if the remaining case has the million?"

P( we would see 38 open cases | Million is in Chosen) = 1
P( we would see 38 open cases | Million is in remaining) = product of missing the million dollar case 38 times =38/39 * 37/38 * ...1/2 = 1/39

I think he should take the Chosen case, and I think the probability is the average of these two likelihoods = 20/39.

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Posted · Report post

This is a case of p(A|B). That is the probability that A is true GIVEN that B is true.

In this situation, it's A: having the winner GIVEN B:after 38 rounds of choosing a winner still remains.

The odds that the remaining case is a winner remains the same (this is crux of the Monty Python puzzle) as it would in the beginning: 1/40.

Thus, the probability of A (he has the winning case) is 39/40. He would be a fool to switch.

This isn't the case where the game show host purposely opened 38 of the empty cases. (in which case the probability would be switched). This is the case that B occurred (he himself picked 38 blanks).

Think about it: You have a 1/40 change that you pick the right case. If this happens the probability of B occurring is 1.

If the remaining case had the winner, the probability of B occurring would be (38/39 * 37*38 * ,,, * 1/2) which simplifies to 1/39.

This adds up to 1 + 1/39 = 40/39, with

p(B|A) = 1/( 40/39 ) = 39/40

p(B|!A) = ( 1/39 )/( 40/39 ) = 1/40

Thus, it is more likely that the events occurred as they did if you initially picked the winning case.

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Posted · Report post

It doesn't matter if he switches or not. The probability of having the winning case is 50%.

If he originally picked the winining case there are 39! ways he can open 38 other cases leaving one last case unopened.

If he had picked an empty case originally there are 38 empty cases and one winning case remaining among the 39 other cases. There are 38! ways to pick 38 empty cases, but there are 39 empty cases to begin with, so the total number of ways to start from an empty case and go through all remaining 38 empty cases is also 39!

So, there are a total of 2*39! possible scenarios that satisfy the conditions of the puzzle and in 50% of them the chosen case is the winner.

To make it easy to understand play it out with 3 or 4 cases and you'll see.

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Posted · Report post

[spoiler=Nice puzzle.]The beauty of puzzles of this type is the presence of a simple and compelling analysis that leads to the wrong answer. I should say the beauty, for me, lies in the presence of a simple, and equally compelling analysis that leads to the right answer. And when considered side by side, the correct analysis identifies and explains the fallacy of the incorrect approach.

The beauty is particularly enhanced when it obviates the need to crank through the appropriate Bayesian formulas. Gah!

Here, the [erroneous] analysis springs from assigning forever a 1/40 probability to the first suitcase holding the money. How on earth could that value change, simply because we inspected 38 other suitcases? "It can't," is the compelling thought, so the other 39/40 probability certainly must lie with the remaining cases, finally devolving upon the 40th case. Switch! You're a fool not to. Just like in Monte Hall.

But then one remembers that in the former puzzle Monte knew the intermediate cases [doors] were losing choices. Moreover, Monte was able to open a losing door for every distribution of hidden contents. Here, one does not know a priori they are losing choices. It just turns out that they are. And, suitcase 1 was chosen from a sample population of cases in which 2-38 [numbered by the order chosen] were losers. This was true even though we did not know it. An assignment of 1/40 is valid only when a choice is made from an unbiased distribution of equal a priori likelihoods. How do we see [intuitively] that the conditions of the problem make 1/40 the wrong measure of probability for the initial suitcase? Read on.

I will change the problem slightly. In a way that [a] still adheres to the conditions of the puzzle and does not change the answer.

Since events are independent, order does not matter. Defer the choice of the "initial" case until after 38 "intermediate" cases have been inspected. In ninety-five percent of the trials, money will be found in one of the "intermediate" suitcases. These instances do not fit the conditions of the puzzle, so when that happens we begin again. Eventually we will open 38 empty suitcases; now the puzzle conditions have been realized. Next, we choose an "initial" suitcase. Finally, we choose between the "initial" case and the 40th suitcase. Shall we switch, or not?

Now intuition leads clearly to the correct answer.

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Posted · Report post

The original Monty Hall puzzle is somewhat counterintuitive, where 50/50 chance is the intuitive answer, but not the correct one.

In this puzzle, both doors enjoying the 50% chance seems both intuitive and the correct answer. Was it meant specifically for those people who are familiar with Monty Hall puzzle?
Although, the OP does not imply that \$1 million is the highest prize available. It’s just the one we must pick up. Either way it’s 50/50.
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The problem is skillfully analyzed and solved by k-man and bonanova. Kudos, everyone.

Was it meant specifically for those people who are familiar with Monty Hall puzzle?

Yes, the Monty Hall reference was meant as a red herring, but apparently that false clue isn't enough around these parts.

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Posted · Report post

1. We form all possible sequences of 40 cases (39 empty and 1 with the treasure).

2. Then we remove all sequences, where treasure is in the middle (1 < case# < 40). In other words we retain only sequences with treasure in case #1 or case #40.
3. Finally we ask what kind of sequences are more frequent among retained: those with treasure in case #1 or those with treasure in case #40.

It should be obvious that both types of sequences are equally frequent, since there is an obvious 1-1 correspondence between those two sets.

(k-man nicely proved the same by counting cardinality of both sets and showing their equality.)
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Posted · Report post

I thought of a better way to show that the probability to win is the same whether the player chooses to keep the chosen case or chooses to switch.

Let's consider 2 players with 2 different strategies. Player A keeps his chosen case till the end while Player B sticks with his case until there is one last case remaining and then switches. Both players always follow their strategy, so if we find the probability of winning by each player we can determine which strategy is better or if they are the same.

For Player A to win he must pick the winning case from the start, so he obviously has 1/40 chance of winning the game.

For Player B to win he must sequentially pick 39 empty cases leaving the winning case as the last case, so that he wins by switching to it. Probability of picking 39 empty cases is 39/40 * 38/39 * 37/38 * ... * 2/3 * 1/2, which equals 1/40.

Since both strategies have exactly the same winning probability it doesn't matter which strategy to follow.

Nice puzzle, Bushindo.

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If [with knowledge of their contents] Monte had opened 38 empty suitcases after my initial choice, then I should switch.

What makes the difference? For every distribution of which suitcase holds the money, there will be 38 empty suitcases for Monte to disclose. Thus no distributions have been ruled out a priori. My initial choice in this case therefore has a true 1/40 probability. The remaining suitcases comprise the remaining 39/40 probability of having the money. By exposing 38 empty ones, Monte has directed me to the single suitcase that has that 39/40 chance.

I switch; and by doing so I multiply my winning chances by 39.
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