Kind of a Monty Hall Problem

[Lestat]

I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

Oh, and for extra credit:



<font color="blue"> Prior to the time I retired, my wife used to drive me to the train station to go to work in the morning and pick me up at the train station the same time in the evening.

One day I got off work early and took an early train which got me to the train station one our early. My wife was shopping (of course) and her cell phone battery was dead. I could not reach her to come to pick me up one hour earlier than usual.

It was a nice day, so I decided to walk home. After a while I saw her driving to the station to pick me up and I stopped her, got in the car, and we went home. We got home 20 minutes earlier than the normal time we would have gotten home had I taken the usual train.

How long was I walking before she picked me up?

You don't need to know how fast she was driving, how far it is from home to the station, how fast I walked, etc. There is no trick. There is one correct answer.
</font>

I did (I'm pretty sure) get this one.

[vhawk01]

[ QUOTE ]
I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

[/ QUOTE ]

Hmmm...its simple to reduce your odds of winning. Put all the slips in one box and you now have a 25% chance of winning 1k.

EDIT: Which makes the solution obvious I guess. Put 1 1k slip in Box A, and the rest of the slips in Box B. You now win 2999/4000 times, right?

[Lestat]

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[ QUOTE ]
I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

[/ QUOTE ]

Hmmm...its simple to reduce your odds of winning. Put all the slips in one box and you now have a 25% chance of winning 1k.

[/ QUOTE ]\

Yeah, this much is obvious. So there should be a way to increase the chances. Btw- I added a problem to the OP, if you're interested.

[vhawk01]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

[/ QUOTE ]

Hmmm...its simple to reduce your odds of winning. Put all the slips in one box and you now have a 25% chance of winning 1k.

[/ QUOTE ]\

Yeah, this much is obvious. So there should be a way to increase the chances. Btw- I added a problem to the OP, if you're interested.

[/ QUOTE ]

Yeah, I edited my solution too.

[arahant]

[ QUOTE ]
[ QUOTE ]
I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

[/ QUOTE ]

Hmmm...its simple to reduce your odds of winning. Put all the slips in one box and you now have a 25% chance of winning 1k.

EDIT: Which makes the solution obvious I guess. Put 1 1k slip in Box A, and the rest of the slips in Box B. You now win 1999/3000 times, right?

[/ QUOTE ]

This problem has an ambiguity that also existed in the 'original' marilyn vos savant published monty hall problem...namely, does the host select the box at random or not?

[vhawk01]

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[ QUOTE ]
[ QUOTE ]
I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

[/ QUOTE ]

Hmmm...its simple to reduce your odds of winning. Put all the slips in one box and you now have a 25% chance of winning 1k.

EDIT: Which makes the solution obvious I guess. Put 1 1k slip in Box A, and the rest of the slips in Box B. You now win 1999/3000 times, right?

[/ QUOTE ]

This problem has an ambiguity that also existed in the 'original' marilyn vos savant published monty hall problem...namely, does the host select the box at random or not?

[/ QUOTE ]

Ugh, I made like 4 edits to my solution, 1 because I am dumb in general and 2 more because I am dumb specifically, and you guys managed to quote them all!

[Lestat]

<font color="blue">This problem has an ambiguity that also existed in the 'original' marilyn vos savant published monty hall problem...namely, does the host select the box at random or not? </font>

I agree, but I think we have to assume that he does (i.e. selects which box randomly).

[Neuge]

[ QUOTE ]
EDIT: Which makes the solution obvious I guess. Put 1 1k slip in Box A, and the rest of the slips in Box B. You now win 2999/4000 times, right?

[/ QUOTE ]
This depends on knowing which box has the 1k slips to begin with.

You have to choose one box at random, then move all but one to the other box. But you run the risk of choosing the box with $1 slips, so when the host picks that box you always win exactly $1.

The EV of doing this is:

0.5(0.5*1000+0.5*(49/99)*1000+0.5*(50/99)*1)+0.5(0.5*1+0.5*(49/99)*1+0.5*(50/99)*1000)= $500.47

which is less than the $500.50 of not doing anything to the boxes. I'd surmise that doing anything always reduces your odds below 50/50.

[Lestat]

<font color="blue"> This depends on knowing which box has the 1k slips to begin with. And if you know that, why do anything at all? Just pick the box with the 1k slips in it. </font>

Remember, you don't choose the box. The host does.

[Neuge]

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<font color="blue"> This depends on knowing which box has the 1k slips to begin with. And if you know that, why do anything at all? Just pick the box with the 1k slips in it. </font>

Remember, you don't choose the box. The host does.

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Yeah, I realized that and edited accordingly. But my point still stands, if you don't know which box has the 1k slips in it, you can't, with 100% confidence, put exactly one 1k slip in one box.

If you do know which box contains them, it becomes a trivially easy problem.

[dknightx]

i understood the first problem as you knowing which box has what, and you can arrange the boxs/slips any way you want. After that the host will pick a random box and then pick a random note from that box. is this not correct?

if so, solution seems very easy

[vhawk01]

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EDIT: Which makes the solution obvious I guess. Put 1 1k slip in Box A, and the rest of the slips in Box B. You now win 2999/4000 times, right?

[/ QUOTE ]
This depends on knowing which box has the 1k slips to begin with.

You have to choose one box at random, then move all but one to the other box. But you run the risk of choosing the box with $1 slips, so when the host picks that box you always win exactly $1.

The EV of doing this is:

0.5(0.5*1000+0.5*(49/99)*1000+0.5*(50/99)*1)+0.5(0.5*1+0.5*(49/99)*1+0.5*(50/99)*1000)= $500.47

which is less than the $500.50 of not doing anything to the boxes. I'd surmise that doing anything always reduces your odds below 50/50.

[/ QUOTE ]

The way I read it, you can open both boxes, look at everything, switch them around fifty times if you want, its how you decide to set them and offer them to the host that matters.

[PairTheBoard]

You start with boxes A and B, each with 50 slips. It is unknown to you which box has the $1 slips. You can direct for the slips to be shuffled in such a way as now have in Box A, x slips from A and y slips from B. That will force B to have 50-x from A and 50-y from B.

If Box A had the $1 slips your EV is now,

.5[x/(x+y) + 1000y/(x+y)] +
+ .5[(50-x)/(100-x-y) + 1000(50-y)/(100-x-y)]

Call that expression F(x,y)

Then your total EV(x,y) = .5F(x,y) + .5F(y,x)

Algebraically simplify the expression you get for EV(x,y)

Then using Calculus if necessary, Fix x and maximize EV with y as the single variable. Get a function MaxEV(x). Then maximize that function wrt the variable x.

I don't want to do it though. Looks like too much work. Unless there's a shortcut.

PairTheBoard

[Neuge]

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EDIT: Which makes the solution obvious I guess. Put 1 1k slip in Box A, and the rest of the slips in Box B. You now win 2999/4000 times, right?

[/ QUOTE ]
This depends on knowing which box has the 1k slips to begin with.

You have to choose one box at random, then move all but one to the other box. But you run the risk of choosing the box with $1 slips, so when the host picks that box you always win exactly $1.

The EV of doing this is:

0.5(0.5*1000+0.5*(49/99)*1000+0.5*(50/99)*1)+0.5(0.5*1+0.5*(49/99)*1+0.5*(50/99)*1000)= $500.47

which is less than the $500.50 of not doing anything to the boxes. I'd surmise that doing anything always reduces your odds below 50/50.

[/ QUOTE ]

The way I read it, you can open both boxes, look at everything, switch them around fifty times if you want, its how you decide to set them and offer them to the host that matters.

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If that's the case, then yeah your answer is correct. It just seems stupid to pose a "problem" with such an easy solution.

[Rastadon]

Does anyone know the second problem?

[PairTheBoard]

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Does anyone know the second problem?

[/ QUOTE ]

I believe the answer is in white here &gt;&gt; <font color="white">50 minutes </font>

It requires the assumption that the wife travels along the route with the same local speed going either way. ie. It takes her the same time to travel any stretch of the route going one way as it does going the other. You only need to look at times traveled on the pertinent segments of the route.

PairTheBoard

[LCposter]

The train problem as stated does not seem correct. It seems trivial to come up with more than one answer:

Let's say the usual train arrives at the station at 5:00, the early train arrives at 4:00, the distance from home to station is 30 miles, and the wife drives at 60 mph. The normal pickup would occur at 5:30, and the normal home arrival would occur at 6:00. Today, to arrive home 20 minutes early, the pickup would occur at 5:20, and the home arrival would occur at 5:40. This would mean the pickup occurred after walking for 80 minutes.

If we keep the parameters the same but increase the distance from home to station to 60 miles, then the normal pickup would occur at 6:00 and the normal home arrival would occur at 7:00. Today, to arrive home 20 minutes early, the pickup would occur at 5:50, and the home arrival would occur at 6:40. This would mean the pickup occurred after walking for 110 minutes.

The amount of time spent walking does not seem constant. However, there are constants for this problem - the pickup always occurs 10 minutes earlier than normal, the time spent walking minus the normal driving time between home and station is always the same (it's the answer posted above).

If I'm wrong then someone please show the flaw in my counterexample.

[PairTheBoard]

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Let's say the usual train arrives at the station at 5:00, the early train arrives at 4:00, the distance from home to station is 30 miles, and the wife drives at 60 mph. The normal pickup would occur at 5:30,

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We assume the wife would leave at 4:30 so she can meet the train when it arrives at 5:00

PairTheBoard

[LCposter]

Thank you, of course you're right! I got confused assuming the wife was working and didn't get home until a set time each day (e.g. 5:00). Of course, the problem assumption is that the wife is home waiting and can leave at any time to ensure that she is at the station by 5:00.

See what the women's lib movement has done... I can't even properly think about a simple problem scenario where the wife isn't working along with the husband!

[Rastadon]

Thanks, does this mean the problem is impossible when the driving distance from home to train station is less than 10 minutes?

[PairTheBoard]

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Thanks, does this mean the problem is impossible when the driving distance from home to train station is less than 10 minutes?

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Yes.

PairTheBoard

[CallMeIshmael]

[ QUOTE ]
I've seen this before somewhere, maybe it was on here. But nothing turned up when I searched. I'm admitting defeat. I'm obviously being stupid somewhere and I can't figure out where. I know someone here will quickly be able to figure it out....



<font color="blue"> You are on a tv quiz show. The host has two boxes.

In one box there are 50 slips of paper that say "you have won $1"

In the other box there are 50 slips of paper that say "you have won $1,000".

You can leave the slips as they are, or mix them any way you wish. You can put them all in one box or split them any way between the two boxes. You cannot discard or hide any of the slips.

When you are done, the show host will select one of the two boxes and at random select one of the slips of paper from that box showing your prize.

Can you increase your odds of winning the $1,000 from 50/50? If so how and what will your odds be?

</font>

[/ QUOTE ]


Just to clarify, are we trying to maximize the chances that we win 1000, or maximize our expected payoff?

(they might be the same, I havnet done anything other than read the question, but Im just clarifying first)