## Monty Hall – Bigger, Better, Badder!

This game is an extension of the old Monty Hall game.

There are four boxes on the stage, each with a stack of brand new £10 notes inside. Before the game starts, you don’t know how much money is in each, or even what the range of values is. You do know that no two boxes have the same size stack.

The host will let you open the boxes, one at a time, in whatever order you wish and examine the stack inside. You may then choose to take the visible stack or open another box.

Once passed, you may not return to a previous stack; e.g., if upon seeing the first stack you decide to open the second box, you may not then opt for the first stack. If you pass the first three boxes, the fourth stack will automatically be your choice.

Now here’s the catch: You only get to keep the money in the box you chose if it turns out to be, in fact, the largest cash value of all the four boxes…

Can you beat the odds with this game? Or is blind guessing just as good a system as any?

Assume that you can tell whether one stack is bigger than another just from observation.

September 4th, 2010 at 3:50 am

I think our chance is 5/12 to win the money?

September 4th, 2010 at 6:02 am

Well Slavy, as a random guess, the chances of winning are 1 in 4 or 3 in 12. You seem to have improved the odds with your answer, but what strategy are you using, and can you improve on your 5 in 12 odds?

September 4th, 2010 at 2:58 pm

The answer 5/12 should be the optimal one and I believe I can prove it, but the problem is good and it is too early to spoil the fun of the others. So I don’t want to reveal my strategy at least for two more days

September 5th, 2010 at 11:40 pm

Brand new £10 stacks may have serial numbers in order,

unless the notes have been shuffled for some reason.

This would not guarantee anything anyway.

And with the money box, one good reason not to pick

another box is to find a fully loaded box, so that there

could not possibly be more money in any other boxes.

Anyhow, humans are greedy. They tend to think the one to come

is better and the one you have already is not good enough.

September 6th, 2010 at 11:10 pm

You could lift the boxes individually and determine the heaviest that way you know the biggest stack without looking and loosing your chance to get the money.

September 7th, 2010 at 3:09 pm

I think that your best odds would be 11/24.

September 7th, 2010 at 7:10 pm

good one Karl

i don’t know,but:

i would say,because you don’t know the range,that you can’t stop at the first box-that’ll only give you 1/4 odds

if you open box 2 and see that it is less,you have to move on

same with box 3

if box 2 is more than box 1,i would stay

if you have to open box 3,and it is more than box 1,i really would stay

just wish i could tally the odds

September 8th, 2010 at 1:18 am

Correct idea, Knightmare! The only way to improve the odds involves a strategy of taking a bigger box than before. So we should always quit when we have reached a “local maximum” (i.e., have opened the biggest box so far) or we have exhausted all our options. The first local maximum, of course, is the first box, since it is bigger than all the boxes we have opened before it. But taking it is just pure guessing and the odds will be exactly 1/4. If we play, until we open the second local maximum then we will find the biggest box with probability 10/24, because this strategy covers all the cases, when the biggest box is the second one (6 out of 24), when the biggest box is the third one and the second box is smaller than the first one (it is easy to see that this is exactly one half of the cases when the biggest box is third, i.e. 3 out of 24) and when the biggest box is the last one and all the other three boxes are ordered in a descending order (which is exactly 1 case). Waiting for the third local maxima is a very bad idea (just compute the odds), so the above strategy is the optimal one.

P.S. This is the strategy that one should follow not only if he wants to exactly guess the biggest box, but also if he wants to earn as much money as possible on average (assuming that he always gets the quantity in the chosen box).

September 8th, 2010 at 9:26 am

11/25

September 8th, 2010 at 9:30 am

I meant 11/24. I.e. slightly better than 5/12.

September 11th, 2010 at 8:48 am

It is tempting to reason that since each box has an equal probability of having the fattest stack, your odds will always be one in four. Knowing what the first box contained doesn’t tell you a thing about whether the second will be larger or smaller.

You’d be right about the second part. The second box has a fifty-fifty chance of being larger or smaller than the first.

And once you’ve seen the first two boxes, there is only a fifty-fifty chance of the fattest stack being in one of those boxes.

But let’s say you knew it was- you could say with certainty which box it was. The larger of the two, of course. But this still doesn’t help you, because if the larger of the two boxes at this point turned out to be the first, it’s too late to go back.

Wait a minute, though: although it’s too late to go back, it isn’t too late to press ahead. If the second box is smaller than the first, you know with 100% certainty that the second box is a loser. And while the winner might have passed by, pushing on until you see a box larger than the first can only help.

If you employ the strategy of opening the first box and passing on it, then settling on the next box you see that isn’t a guaranteed loser (in other words, the first box that contains a stack larger than the first), you can improve your odds over one in four.

In fact, your odds of winning rise to almost 46%!

Milly scores. Another 45.8% chance winner at the game!

September 12th, 2010 at 3:22 am

Agree – I have missed one case at the end of my post, i.e., when the biggest box is last we just need the second biggest one to be first, so we will get it in 2 cases, not in only 1 and thus in total we have the above answer.

October 6th, 2010 at 4:22 am

I’m wondering if it would be possible to expand this problem to n boxes? And would the same strategy apply?

October 7th, 2010 at 2:41 pm

@Milly, I would say that a generalization is possible, but (so far only intuitively) the best strategy should be to take the [(n+1)/2] (integer part of (n+1)/2) time highest box, i.e., you open a box and it is the highest one so far, then you open another box and if it contains more money you count it as the second time highest box and so on. It is clear that the only good strategy is to take a box that is at least the biggest so far, and it is also clear that the strategy should be fixed from the beginning because you don’t gain any additional information throughout the process. So one way to prove the problem is to compute all the probabilities for winning if we take the k-th time highest box and see for which k between 1 and n the number is the largest. Unfortunately I don’t have the time to do it now – there are more important mathematical problems (in terms of my future) waiting for me – so I will try it in a month (probably). What is easy to compute is the generalization of the strategy for 4 boxes, i.e., the one of taking the second time highest box: the probability to get the biggest box among all is (1+1/2+1/3+…+1/(n-1))/n. When n=4 we derive the value (1+1/2+1/3)/4=(6+3+2)/24=11/24, which is the answer of the particular problem.