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Monty Hall, we have a PROBLEM

Re: Monty Hall, we have a PROBLEM

If you've seen this one before, my apologies. The first time I heard it, it bedeviled me for hours, until the "aha" moment came.

A person walks into a hardware store, looking for a common item that is used in most houses.

"Ah, I see that the item is $1.50 for each one," he says.
-- that's right.

"So, then, 57 would be $3.00?"
-- right again.

"Okay, then, I'll take 132."
-- That will be $4.50.


What did s/he buy?

Numbers to put on the front of their house.
 
Re: Monty Hall, we have a PROBLEM

Totally concur. That would have been completely clear, and lend itself to a solution. Then your math in post # 106 would have provided an answer, and the difference between the two problems would be that a less-than-infinitely fast search rate would suffice in the second problem, compared to requiring an instantaneous 2nd lap in the first problem.

in 15 minutes he searched 10 pockets, so in the next 15 minutes he'd need to search 34 pockets or 136 pockets per hour. Perhaps desperation would provide the adrenaline he needed to search faster.

Thank ye - that's what I was hoping for :). Although one can think of "pockets" like links at a racetrack, Larry didn't burn half the distance, he burned half the time. He can indeed finish off his half-hour search with an average pockets-per-hour more than twice the rate he achieved in the first 15 minutes (and find the ticket in time).

I apologize for the original q. not being more specific.
 
Re: Monty Hall, we have a PROBLEM

You're in an auction with one other person. Up for bids is a crisp $100 bill. There's a catch, however: the losing bidder must pay his bid as well and gets nothing.

Bidding starts at $1.

What is the optimal bid?
There's a silent auction going on at work where each bid increments by $1 right now. The winner gets $200 towards an iPhone 5S purchase, to be made in the presence of the person offering the auction "item". Since the losing bid does not pay, unlike your story, I'm curious to see how this one plays out. I immediately thought of this thread when I read the auction description.
 
Re: Monty Hall, we have a PROBLEM

Assume everyone works for the government.

What is the minimum income tax rate required so that everyone gets paid?

Ah, you say, I did not provide enough information / specifications.

For example, is government borrowing allowed? If so, would anyone lend to such a government in this situation?

What other sources of income might the government have besides income tax revenues?


I don't want this to come across as some political snark. I'm genuinely intrigued by how the math would work out once one specifies the ancillary assumptions needed to attack the problem. Is it as intractable as it seems after five minutes of thought? or is there a possible solution?

Example: suppose no borrowing / lending, and non-tax revenues = R and Tax revenues = T, while total payroll = P and non-payroll spending = S.

Then R + T = P + S. we also know T = taxrate*P. So R+taxrate*P = P + S, or R - S = P (1-taxrate), (R-S)/P = 1-taxrate, or taxrate = 1-(R-S)/P. Therefore non-tax revenues must exceed non-payroll spending or else there is no possible solution.
 
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Re: Monty Hall, we have a PROBLEM

Assume everyone works for the government.

What is the minimum income tax rate required so that everyone gets paid?

Ah, you say, I did not provide enough information / specifications.

For example, is government borrowing allowed? If so, would anyone lend to such a government in this situation?

What other sources of income might the government have besides income tax revenues?

$Texas
 
Re: Monty Hall, we have a PROBLEM

Dumb question for you math geeks who have time and effort (and motivation, which I have none):

Using a 1-16 confidence factor, as some sports bars use, put a "1" on the team you think is going to win but have no confidence in, and "16" on the team you think WILL win hands down (Example: 1 on MN vs CAR this past week, 16 on DEN over JAX).

How's the math on betting $10 on your "1," increase bet in $10 increments up to $160 for your "16?" Profitable strategy in the long run? Money line only, no spread bets.
 
Dumb question for you math geeks who have time and effort (and motivation, which I have none):

Using a 1-16 confidence factor, as some sports bars use, put a "1" on the team you think is going to win but have no confidence in, and "16" on the team you think WILL win hands down (Example: 1 on MN vs CAR this past week, 16 on DEN over JAX).

How's the math on betting $10 on your "1," increase bet in $10 increments up to $160 for your "16?" Profitable strategy in the long run? Money line only, no spread bets.

Totally depends. How accurate are your picks? Is a 16 always going to win? What are the odds on the money line? Even money? Or Vegas style where favorites can command 1:5 or worse odds for true mismatches?

Simply not enough information to tell you how that strategy would work.
 
Re: Monty Hall, we have a PROBLEM

Totally depends. How accurate are your picks? Is a 16 always going to win? What are the odds on the money line? Even money? Or Vegas style where favorites can command 1:5 or worse odds for true mismatches?

Simply not enough information to tell you how that strategy would work.

There's the rub, and obviously many a variable in play. It's more of a "win consistently" instead of a "win big" strategy. I am running at about at 70% clip this season, in picking the winners, with most of my big (above the average of 8) numbers winning. So for THIS year, I would be profitable. And that is what spurned the question.

There has to be some sort of graph that says "if you are winning x%, your average point-win needs to be at y-point." I guess that is my question.
 
There's the rub, and obviously many a variable in play. It's more of a "win consistently" instead of a "win big" strategy. I am running at about at 70% clip this season, in picking the winners, with most of my big (above the average of 8) numbers winning. So for THIS year, I would be profitable. And that is what spurned the question.


There has to be some sort of graph that says "if you are winning x%, your average point-win needs to be at y-point." I guess that is my question.

It still depends on the payout. If they all pay even money, I'd ask why you're even betting on the tossups rather than the one or two certainties.

Presuming different payouts and the accuracy of your own odds, your best bet in the long haul is to load-up on the games where the payouts are the most different from your expectations. This includes underdogs whom you think have a better shot than the rest of the gambling market does, because even though you'll lose more individual bets, your overall payouts will cover it and then some.

It doesn't matter where in your personal weekly certainty ratings they fall. Some weeks you'll presumably have lots of tough calls, other weeks there will be lots of easy ones. But there's no point in loading up on Denver over Jacksonville if it pays crappily if your #15 game pays 2 or 3 times as much but is nearly as certain.
 
Re: Monty Hall, we have a PROBLEM

There's the rub, and obviously many a variable in play. It's more of a "win consistently" instead of a "win big" strategy. I am running at about at 70% clip this season, in picking the winners, with most of my big (above the average of 8) numbers winning. So for THIS year, I would be profitable. And that is what spurned the question.

There has to be some sort of graph that says "if you are winning x%, your average point-win needs to be at y-point." I guess that is my question.

Yeah but do you have a 70% win rate in each separate category across the board or does your winning % vary depending upon the category?


Reminds me of someone I knew who was successful at betting on horse races. He never picked a winner, he made his money sometimes on the "place" but mostly on the "show." He'd only bet a horse to "place" about 15% of the time. He didn't bet every race either. Maybe a bit more than half if that much.
 
Re: Monty Hall, we have a PROBLEM

Fish: 70 on the straight up winner.

And it sounds like this possible strategy has way too many variables to make it work. Thanks guys; guess I'll chalk it up to another crazy idea.
 
Re: Monty Hall, we have a PROBLEM

Fish: 70 on the straight up winner.

And it sounds like this possible strategy has way too many variables to make it work. Thanks guys; guess I'll chalk it up to another crazy idea.
Track your figures. If you have a solid record for your, say 13-16 confidences, then only bet those or go even lower if you're doing well lower. If your whole intent is to simply maximize your payout, then pay attention to the payoff rates and bet more often on those that have a higher payout AND those that you have a good history predicting.
 
Re: Monty Hall, we have a PROBLEM

Track your figures. If you have a solid record for your, say 13-16 confidences, then only bet those or go even lower if you're doing well lower. If your whole intent is to simply maximize your payout, then pay attention to the payoff rates and bet more often on those that have a higher payout AND those that you have a good history predicting.

Oh, I'm not going to actually DO this. I just thought of it while filling out the free confidence forms for Old Chicago (you beat the house, you win a free appetizer; hey, free food!).
 
Re: Monty Hall, we have a PROBLEM

I'll spare you the back story.

You have twelve coins, each identical in appearance. Eleven of them have exactly the same weight, one of them weighs either a little bit more or a little bit less but you don't know which.

You have a pan balance (like the scales held by Lady Justice). You are allowed to use it three times.

Identify which coin is not like the others.

No fair looking up the answer on the internet.
 
Re: Monty Hall, we have a PROBLEM

I'll spare you the back story.

You have twelve coins, each identical in appearance. Eleven of them have exactly the same weight, one of them weighs either a little bit more or a little bit less but you don't know which.

You have a pan balance (like the scales held by Lady Justice). You are allowed to use it three times.

Identify which coin is not like the others.

No fair looking up the answer on the internet.
I would start by weighing 6 vs. 6. Then I would take the 6 from the heavier side and weigh 3 vs. 3. Then I would take the 3 coins from the heavier side and weigh two of them. If they balance, the third coin is the odd sized one. If they don't balance, the heavier coin is obviously the odd sized one.

Edit: Now I see that you say it may be lighter as well, in which case my idea wouldn't work.
 
Re: Monty Hall, we have a PROBLEM

I would start by weighing 6 vs. 6. Then I would take the 6 from the heavier side and weigh 3 vs. 3. Then I would take the 3 coins from the heavier side and weigh two of them. If they balance, the third coin is the odd sized one. If they don't balance, the heavier coin is obviously the odd sized one.

Edit: Now I see that you say it may be lighter as well, in which case my idea wouldn't work.
First weighing should just be 3 vs 3. If they're the same, do 3 vs. 3 with the 6 that weren't included, and so on.
 
Re: Monty Hall, we have a PROBLEM

One I was young and doing this kind of puzzle, I am pretty sure that the first weighing was four coins vs. four coins. There are then two possibilities, ...
 
Re: Monty Hall, we have a PROBLEM

One I was young and doing this kind of puzzle, I am pretty sure that the first weighing was four coins vs. four coins. There are then two possibilities, ...

Any of the mentioned possibilities would work.

The trick is there are three states the scale can be in (side A heavier, side B heavier, both sides equal) and they can all yield information about where the counterfeit coin is.
 
Re: Monty Hall, we have a PROBLEM

Any of the mentioned possibilities would work.

The trick is there are three states the scale can be in (side A heavier, side B heavier, both sides equal) and they can all yield information about where the counterfeit coin is.



Yes, but "can you work out all the permutations?" is the question. I doubt starting with 3 vs 3 would work. Even if you start with 4 vs 4 as Papa Baer suggests, there is still a subtle twist required on the second or third step in order to get the solution in only two more tries.

e.g, "start with 4 vs 4" if it balances, then....

If it does not balance, then....

The tricky part is that you don't know whether the odd coin is heavier or lighter than all the others.
 
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