[Guest]

Dr. Brainard is inventing again. He invented an lazy susan with divided areas that can collapse and have the remaining areas expand across the entire area. Imagine a series of Japanese hand fans (an accordian folded piece of paper that looks like a stick but can stretch out to make a full circle) all connected at the center. He is tickled pink, just imagine one day you can have 2 compartments: 1 for chips and the other for dip and then the next day you can have wings divided into spicy, cajun, hot, bbq, and honey mustard. But no one is buying them so he invented a game to go with it.

The lazy susan starts with 9 compartments and a slip of paper goes in each one. As with all good games, all the slips but one say, "sorry, please try again" and the last one says, "WINNER". As a contestant, you pick a compartment and the host "closes" 3 adjacent compartments to one side or the other of the one you picked, saying/showing you that those 3 were not the winning compartments. Now you have a chance to pick a new compartment or stay after which the host will close 2 adjacent compartments to one side or the other the one you now have picked. This again repeats with the host closing 1 more compartment, leaving 3 compartments left. You have 1 last chance to change your pick.

So to help imagine them numbered 1,2,3,4,5,6,7,8,9 and if you pick 1 the host may close numbers 2,3,4 or 7,8,9.

so you end up with 1,5,6,7,8,9 or 1,2,3,4,5,6 and if you stay with 1 the host may close numbers 5,6 or 8,9 in the first case or 2,3, or 5,6 in the second case.

etc...

The 1st part is, "what is the best strategy? and what is the resulting win probability?"

People are still not buying them so Dr. Brainard decides to try and sell them to casinos. Now being for a casino, he had to come up with payments so being tired he came up with something easy: you pay $1 to play and win $8 if you pick the right answer, but you have to pay another $1 to have the host close compartments and make a new selection. So you pay end up paying $4 if you want to go to the end, but you can stop at any time. So does this change your strategy and what is your best ROI?

[superprismatic]

Dr. Brainard is inventing again. He invented an lazy susan with divided areas that can collapse and have the remaining areas expand across the entire area. Imagine a series of Japanese hand fans (an accordian folded piece of paper that looks like a stick but can stretch out to make a full circle) all connected at the center. He is tickled pink, just imagine one day you can have 2 compartments: 1 for chips and the other for dip and then the next day you can have wings divided into spicy, cajun, hot, bbq, and honey mustard. But no one is buying them so he invented a game to go with it.

The lazy susan starts with 9 compartments and a slip of paper goes in each one. As with all good games, all the slips but one say, "sorry, please try again" and the last one says, "WINNER". As a contestant, you pick a compartment and the host "closes" 3 adjacent compartments to one side or the other of the one you picked, saying/showing you that those 3 were not the winning compartments. Now you have a chance to pick a new compartment or stay after which the host will close 2 adjacent compartments to one side or the other the one you now have picked. This again repeats with the host closing 1 more compartment, leaving 3 compartments left. You have 1 last chance to change your pick.

So to help imagine them numbered 1,2,3,4,5,6,7,8,9 and if you pick 1 the host may close numbers 2,3,4 or 7,8,9.

so you end up with 1,5,6,7,8,9 or 1,2,3,4,5,6 and if you stay with 1 the host may close numbers 5,6 or 8,9 in the first case or 2,3, or 5,6 in the second case.

etc...

The 1st part is, "what is the best strategy? and what is the resulting win probability?"

People are still not buying them so Dr. Brainard decides to try and sell them to casinos. Now being for a casino, he had to come up with payments so being tired he came up with something easy: you pay $1 to play and win $8 if you pick the right answer, but you have to pay another $1 to have the host close compartments and make a new selection. So you pay end up paying $4 if you want to go to the end, but you can stop at any time. So does this change your strategy and what is your best ROI?

I have a question whose answer will affect the probability.

If the host can choose either of the two groups of compartments

on either side of your current choice, does he do so randomly?

Suppose the winning number is 9 but you chose 4. The host

closes 1,2,3. Now, you change your guess to 6. The host may

wish to close 7,8 rather than 4,5 to make you think that the

4 that you just abandoned may actually be the winner. This

may entice you to go back to 4 later. If he's allowed this

kind of shenanigans rather than pick randomly, the

probabilities will be affected.

[Guest]

Dr. Brainard is inventing again. He invented an lazy susan with divided areas that can collapse and have the remaining areas expand across the entire area. Imagine a series of Japanese hand fans (an accordian folded piece of paper that looks like a stick but can stretch out to make a full circle) all connected at the center. He is tickled pink, just imagine one day you can have 2 compartments: 1 for chips and the other for dip and then the next day you can have wings divided into spicy, cajun, hot, bbq, and honey mustard. But no one is buying them so he invented a game to go with it.

The lazy susan starts with 9 compartments and a slip of paper goes in each one. As with all good games, all the slips but one say, "sorry, please try again" and the last one says, "WINNER". As a contestant, you pick a compartment and the host "closes" 3 adjacent compartments to one side or the other of the one you picked, saying/showing you that those 3 were not the winning compartments. Now you have a chance to pick a new compartment or stay after which the host will close 2 adjacent compartments to one side or the other the one you now have picked. This again repeats with the host closing 1 more compartment, leaving 3 compartments left. You have 1 last chance to change your pick.

So to help imagine them numbered 1,2,3,4,5,6,7,8,9 and if you pick 1 the host may close numbers 2,3,4 or 7,8,9.

so you end up with 1,5,6,7,8,9 or 1,2,3,4,5,6 and if you stay with 1 the host may close numbers 5,6 or 8,9 in the first case or 2,3, or 5,6 in the second case.

etc...

The 1st part is, "what is the best strategy? and what is the resulting win probability?"

People are still not buying them so Dr. Brainard decides to try and sell them to casinos. Now being for a casino, he had to come up with payments so being tired he came up with something easy: you pay $1 to play and win $8 if you pick the right answer, but you have to pay another $1 to have the host close compartments and make a new selection. So you pay end up paying $4 if you want to go to the end, but you can stop at any time. So does this change your strategy and what is your best ROI?

For a)

As in the monty hall problem, it is always better to be switching. By not switching throughout, you will have a 11.1% chance of winning.

If you switch after 3 are closed, I believe your chances improve to 17.8%.

If you switch after 3 are closed, then after 2 are closed, I believe your chances improve to 27.4%

If you switch after 3 are closed, then after 2 are closed, then after 1 is closed, I believe your chances improve to 36.3%

However, if you switch after 3 are closed, keep the choice, then switch again after 1 is closed, I believe your chances will be 41.1%

Not switching after 3 are closed, but switching after 2 are closed improves your chances to 29.6%

Switching after 2 then after 1 gives you a 35.2% chance

The best scenario though, is switching only after all 6 are closed. This will give you a 44.4% chance of winning.

For b) I believe your odds are as follows if you are forced to change after compartments are closed:

No closing: $1 to win 8 for 11.1%

closing 3: $2 to win 8 for 17.8%

closing 5: $3 to win 8 for 27.4%

closing 6: $4 to win 8 for 36.3%

Even in the best case scenario where you are not forced to change after compartments are closed, you will have the same odds per dollar won as if you didn't close anything.

[Guest]

I have a question whose answer will affect the probability.

If the host can choose either of the two groups of compartments

on either side of your current choice, does he do so randomly?

Suppose the winning number is 9 but you chose 4. The host

closes 1,2,3. Now, you change your guess to 6. The host may

wish to close 7,8 rather than 4,5 to make you think that the

4 that you just abandoned may actually be the winner. This

may entice you to go back to 4 later. If he's allowed this

kind of shenanigans rather than pick randomly, the

probabilities will be affected.

SP - Yes, if the correct one doesn't dictate which side then it is random

Ogden - I agree with some of your probabilities and disagree with others and you didn't get the best, see the spoiler for my comments:

Also, just to clarify, you are not forced to 'switch' after paying, you can always pick the same compartment you had before. My use of the word 'new' may have been misleading, sorry.

For a)

As in the monty hall problem, it is always better to be switching. By not switching throughout, you will have a 11.1% chance of winning. I agree

If you switch after 3 are closed, I believe your chances improve to 17.8%. I agree

If you switch after 3 are closed, then after 2 are closed, I believe your chances improve to 27.4% I disagree

If you switch after 3 are closed, then after 2 are closed, then after 1 is closed, I believe your chances improve to 36.3% I disagree

However, if you switch after 3 are closed, keep the choice, then switch again after 1 is closed, I believe your chances will be 41.1% I disagree

Not switching after 3 are closed, but switching after 2 are closed improves your chances to 29.6% I disagree

Switching after 2 then after 1 gives you a 35.2% chance I disagree

The best scenario though, is switching only after all 6 are closed. This will give you a 44.4% chance of winning. I agree with the % but disagree that it is the best you can do

For b) I believe your odds are as follows if you are forced to change after compartments are closed:

No closing: $1 to win 8 for 11.1%

closing 3: $2 to win 8 for 17.8%

closing 5: $3 to win 8 for 27.4%

closing 6: $4 to win 8 for 36.3%

Even in the best case scenario where you are not forced to change after compartments are closed, you will have the same odds per dollar won as if you didn't close anything.

[Guest]

For a)

As in the monty hall problem, it is always better to be switching. By not switching throughout, you will have a 11.1% chance of winning. straightforward - 1 in 9

If you switch after 3 are closed, I believe your chances improve to 17.8%. while the 1/9 chance for your original does not improve, the remaining 8/9 chance is now split equally amongst the remaining 5

If you switch after 3 are closed, then after 2 are closed, I believe your chances improve to 27.4% following the same logic as above, if your 17.8% chance does not improve, and the remaining 3 have an equal probability

If you switch after 3 are closed, then after 2 are closed, then after 1 is closed, I believe your chances improve to 36.3% again, if your 27.4% chance does not improve, and the remaining 2 have an equal probability

However, if you switch after 3 are closed, keep the choice, then switch again after 1 is closed, I believe your chances will be 41.1% if your 17.8% chance does not improve, and the remaining 2 have an equal probability

Not switching after 3 are closed, but switching after 2 are closed improves your chances to 29.6% here your 11.1% chance doesn't change, but the remaining 3 have an equal probability

Switching after 2 then after 1 gives you a 35.2% chance same logic

The best scenario though, is switching only after all 6 are closed. This will give you a 44.4% chance of winning. same logic

For b) I believe your odds are as follows if you are not forced to change after compartments are closed:

No closing: $1 to win 8 for 11.1%

closing 3: $2 to win 8 for 17.8%

closing 5: $3 to win 8 for 29.6%

closing 6: $4 to win 8 for 44.4%

Even in the best case scenario where you are not forced to change after compartments are closed, you will have the same odds per dollar won as if you didn't close anything.

Edited October 9, 2009 by ogden_tbsa

[Guest]

I suppose if at the end one of the compartments you picked originally but switched away from is still left, the probability may be different.

[Guest]

I suppose if at the end one of the compartments you picked originally but switched away from is still left, the probability may be different.

If you choose compartment 1, then choose the compartment that is 3 away from it, meaning 7 in the first case, or 4 in the second case, then compartment 1 can not be closed. This will allow you to improve the probability.

You would still have 11.1% for the first pick. After you change you would still have 17.8%. However, changing again would further improve your odds to 35% by picking a new compartment.

If you stay with 1 after 3 are closed, but change to the middle compartment of the 3 remaining for your next choice, this may optimize your chances. You would then switch to the remaining 1 of the 3 that was not closed after the last compartment is closed (sorry if this sounds confusing). Imagine you pick 1, and 2,3,4 are closed. Then, you stay with 1, and 5,6 are closed. Now, change to 8. If 7 is closed, choose 9, and vice versa. This should improve the chances. When you chose 8, I believe your chances were 29.6%. Switching to 7 or 9 as above should increase your chances to 59.3%. This should also yield the best best at the casino and would definitely be worth closing them all (if my math is correct).

Edited October 9, 2009 by ogden_tbsa

[Guest]

sorry spoiler fxn didn't work....let me try again.

Edited October 9, 2009 by gastropod

[Guest]

Since there is no negative consequence in this game until the end, there is no probability that you will lose until the very last decision point. As the final decision is only between 2 choices, it is a 50/50 choice, your original choices have no relevance. $4 bet pays $8. Casino would not like this payout against these odds.

[roolstar]

Is this related??

It is isn't it?

[roolstar]

Since there is no negative consequence in this game until the end, there is no probability that you will lose until the very last decision point. As the final decision is only between 2 choices, it is a 50/50 choice, your original choices have no relevance. $4 bet pays $8. Casino would not like this payout against these odds.

Research "MONTY HALL"...

A bit confusing but makes a lot of sense... eventually....

Edited October 10, 2009 by roolstar

[Guest]

I'd like some confirmation that this answer is correct before putting up my reasoning. I'm lazy I guess. Not enough time to make it reader friendly. Anyhoo:

By going Switch Switch Switch or Stay Stay Switch (and knowing which ones to switch to) I can get an average chance of 55.56%.

Also, By going Switch Switch I can get an average winnings of 56 cents per game.

As for the Monty Hall Problem, this riddle actually highlights the weakness in the common explanation of why the Monty Hall Problem works the way it does.

Edited October 11, 2009 by Tuckleton

[superprismatic]

I wrote a simulator to try all ways to stay/switch, each way being tried 10,000,000 times. The best I could do was win 38.4% of the time. Of course, I could have a bug even though the program is pretty simple.

[Guest]

I wrote a simulator to try all ways to stay/switch, each way being tried 10,000,000 times. The best I could do was win 38.4% of the time. Of course, I could have a bug even though the program is pretty simple.

Well that's discouraging. I probably did the probabilities wrong somewhere along the line. Perhaps you could enter my method into your simulator and let me know what it comes out as to help me locate my error.

So pick any door, then, once doors are opened, follow the line of opened doors from the picked door and switch to the first unopened door. Do this again after doors are opened a second time. After the last door is opened, switch to the door that could have been opened but was not. For example, you pick door number 1. The 3 doors on it's right (234) are opened so you go right from 1 to the first unopened door, which is 5. Then 2 doors on the left (19) of 5 are opened so you go left to the first unopened door, which is 8. Finally, the door on it's right (5) is opened so you pick the door on it's left, door 7, and cross your fingers. Let me know if that made sense. I'm kind of tired

[superprismatic]

Well that's discouraging. I probably did the probabilities wrong somewhere along the line. Perhaps you could enter my method into your simulator and let me know what it comes out as to help me locate my error.

So pick any door, then, once doors are opened, follow the line of opened doors from the picked door and switch to the first unopened door. Do this again after doors are opened a second time. After the last door is opened, switch to the door that could have been opened but was not. For example, you pick door number 1. The 3 doors on it's right (234) are opened so you go right from 1 to the first unopened door, which is 5. Then 2 doors on the left (19) of 5 are opened so you go left to the first unopened door, which is 8. Finally, the door on it's right (5) is opened so you pick the door on it's left, door 7, and cross your fingers. Let me know if that made sense. I'm kind of tired

Yes, it made sense. I'm also tired. This will have to wait for 18 hours or so for my attempt to run this thru the program.

But I will.

[Guest]

Yes, it made sense. I'm also tired. This will have to wait for 18 hours or so for my attempt to run this thru the program.

But I will.

Super, any chance you could run my method outlined in post dated 09 October 2009 - 10:18 PM through your program?

59.3%

[Guest]

Super, that doesn't sound right, I suspect an error in the program.

I want to say that I have enjoyed this puzzle because my initial intention was to use the basic concept in the monty problem multiple times to further improve the odds, but I in the process i have not only gained greater insight into these problems, but also realize that this problem has factors that the monty problem doesn't. In the spoiler I have a hint of what to think about.

In the monty program the host will pick out any of the other doors.

[Guest]

I'm fairly certain you can achieve 16/27, or 59.3%.

[superprismatic]

Super, any chance you could run my method outlined in post dated 09 October 2009 - 10:18 PM through your program?

59.3%

I'd be glad to try. I'm not exactly sure of your algorithm, though. Please explain in more detail.

Edited October 12, 2009 by superprismatic

[superprismatic]

Yes, it made sense. I'm also tired. This will have to wait for 18 hours or so for my attempt to run this thru the program.

But I will.

I fixed the bug and ran with your strategy 100,000,000 times and got 55.561676%. A very good agreement with your analysis, Tuckleton.

Edited October 12, 2009 by superprismatic

[Guest]

I'd be glad to try. I'm not exactly sure of your algorithm, though. Please explain in more detail.

This approach is stick-switch-switch. By sticking to your original pick after the original offering, other than the compartment chosen originally, there will be 3 remaining compartments left, all next to each other. The next choice is to remove your original compartment selection from consideration, and force it to not be closed by choosing the middle compartment of the 3 remaining. You are now faced with the original Monty Hall problem, except for the 1/9 chance that the first compartment you selected was correct. A compartment next to your second choice will be closed, and you then switch to the 3rd of the 3 contiguous compartments, giving you a 2/3 * 8/9 chance.

If that's not clear, I tried to make it clear in my earlier post using compartment numbers with an example.

Edited October 12, 2009 by ogden_tbsa

[superprismatic]

This approach is stick-switch-switch. By sticking to your original pick after the original offering, other than the compartment chosen originally, there will be 3 remaining compartments left, all next to each other. The next choice is to remove your original compartment selection from consideration, and force it to not be closed by choosing the middle compartment of the 3 remaining. You are now faced with the original Monty Hall problem, except for the 1/9 chance that the first compartment you selected was correct. A compartment next to your second choice will be closed, and you then switch to the 3rd of the 3 contiguous compartments, giving you a 2/3 * 8/9 chance.

If that's not clear, I tried to make it clear in my earlier post using compartment numbers with an example.

Give me a few examples, please. I'm confused about the middle compartment business because I think, at that point, there are 2 things in every compartment. Examples will clear things up.

[Guest]

Give me a few examples, please. I'm confused about the middle compartment business because I think, at that point, there are 2 things in every compartment. Examples will clear things up.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 2,3,and 4 are closed.

You don't switch.

Now, compartments 5 and 6 are closed, or compartments 8 and 9 are closed. For this example, let's say 5 and 6 are closed.

At this point, you originally chose compartment 1. You may now switch to 7, 8, or 9. You switch to 8.

Now, compartment 7 is closed, or compartment 9 is closed. For this example, let's say 7 is closed.

You switch to 9, with 16/27 chance of being correct.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 7,8,and 9 are closed.

You don't switch.

Now, compartments 2 and 3 are closed, or compartments 5 and 6 are closed. For this example, let's say 2 and 3 are closed.

At this point, you originally chose compartment 1. You may now switch to 4, 5, or 6. You switch to 5.

Now, compartment 4 is closed, or compartment 6 is closed. For this example, let's say 6 is closed.

You switch to 4, with 16/27 chance of being correct.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 7,8,and 9 are closed.

You don't switch.

Now, compartments 2 and 3 are closed, or compartments 5 and 6 are closed. For this example, let's say 5 and 6 are closed.

At this point, you originally chose compartment 1. You may now switch to 2, 3, or 4. You switch to 3.

Now, compartment 2 is closed, or compartment 4 is closed. For this example, let's say 4 is closed.

You switch to 2, with 16/27 chance of being correct.

Edited October 12, 2009 by ogden_tbsa

[superprismatic]

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 2,3,and 4 are closed.

You don't switch.

Now, compartments 5 and 6 are closed, or compartments 8 and 9 are closed. For this example, let's say 5 and 6 are closed.

At this point, you originally chose compartment 1. You may now switch to 7, 8, or 9. You switch to 8.

Now, compartment 7 is closed, or compartment 9 is closed. For this example, let's say 7 is closed.

You switch to 9, with 16/27 chance of being correct.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 7,8,and 9 are closed.

You don't switch.

Now, compartments 2 and 3 are closed, or compartments 5 and 6 are closed. For this example, let's say 2 and 3 are closed.

At this point, you originally chose compartment 1. You may now switch to 4, 5, or 6. You switch to 5.

Now, compartment 4 is closed, or compartment 6 is closed. For this example, let's say 6 is closed.

You switch to 4, with 16/27 chance of being correct.

Let the compartments be numbered 1 through 9, in order. You begin by selecting compartment 1.

Now, compartments 2,3,and 4 are closed, or compartments 7,8,and 9 are closed. For this example, let's say 7,8,and 9 are closed.

You don't switch.

Now, compartments 2 and 3 are closed, or compartments 5 and 6 are closed. For this example, let's say 5 and 6 are closed.

At this point, you originally chose compartment 1. You may now switch to 2, 3, or 4. You switch to 3.

Now, compartment 2 is closed, or compartment 4 is closed. For this example, let's say 4 is closed.

You switch to 2, with 16/27 chance of being correct.

I printed out the selection array at intermediate steps after I changed the program to do your algorithm. The selection array does exactly what your examples do. When I run the program, I get something very close to 55.56%. Oddly enough, it looks like the chance of being correct is 15/27, not 16/27. I suspect you made a small mistake in your calculations.

[Guest]

I printed out the selection array at intermediate steps after I changed the program to do your algorithm. The selection array does exactly what your examples do. When I run the program, I get something very close to 55.56%. Oddly enough, it looks like the chance of being correct is 15/27, not 16/27. I suspect you made a small mistake in your calculations.

1) Select a door (call it D1) (1/9 chance of being right, 8/9 chance of being wrong).

2) Allow 5 doors to close, leaving 3 unselected, and the 1 that was selected with 1/9 chance of being right.

3) Apply traditional Monty Hall problem to 3 unselected doors which have 8/9 chance of being right:

a) Choose a door (call it D2) (middle door of remaining 3).

b) Another door is closed - NOT the original selected door D1.

c) Switch to final door, which is NOT D1 or D2 (call it D3).

Original Monty Hall problem: of 3 doors, you pick door 1. Door 2 is opened. By switching to door 3 rather than keeping door 1, you have a 2/3 chance of being correct.

I believe that the math here is simply 8/9 (chance of one of the three unselected doors in step 3 being correct) multiplied by 2/3 (chance of D3 being correct over D2).

edit: My error may come from not adjusting the probability of D1 being correct when the final door is closed, and instead leaving it at 1/9...

Edited October 12, 2009 by ogden_tbsa

[superprismatic]

1) Select a door (call it D1) (1/9 chance of being right, 8/9 chance of being wrong).

2) Allow 5 doors to close, leaving 3 unselected, and the 1 that was selected with 1/9 chance of being right.

3) Apply traditional Monty Hall problem to 3 unselected doors which have 8/9 chance of being right:

a) Choose a door (call it D2) (middle door of remaining 3).

b) Another door is closed - NOT the original selected door D1.

c) Switch to final door, which is NOT D1 or D2 (call it D3).

Original Monty Hall problem: of 3 doors, you pick door 1. Door 2 is opened. By switching to door 3 rather than keeping door 1, you have a 2/3 chance of being correct.

I believe that the math here is simply 8/9 (chance of one of the three unselected doors in step 3 being correct) multiplied by 2/3 (chance of D3 being correct over D2).

edit: My error may come from not adjusting the probability of D1 being correct when the final door is closed, and instead leaving it at 1/9...

Your logic seems iron clad to me. I'm really interested in getting to the bottom of this. At the moment I'm in programming mode and it's hard to switch my mind to probabilities quickly. I'll mull it all over and get back to you if I get any ideas. This turned out to be a pretty interesting problem. Oddly enough, Tuckleton computed his (different from your) algorithm's success at precisely 15/27. I hope he (or someone else) jumps in and tries to explain this stuff.