A Fistful of Brain Teasers

Don Atkinson posted:

Oh dear.........

Let’s stick with Monty Hall for the moment.

The odds on winning the car are most definitely 2/3 if you change your mind, ie you accept Monty’s offer to open the other door. If you stick with your original door, the odds on winning are 1/3.

........yes, I know it isn’t blatantly obvious, and I also know that many eminent mathematicians disputed these probabilities when first presented, but..........

Suppose the contestant was part of a husband and wife team, but the wife was delayed at the hair dressers. She dashes onto the stage at the point that one door has been opened revealing a goat. Husband says to wife, there’s a car behind one of those two doors, we’ve got to chose between door 1 and door 2, you choose.

She’s got a 50/50 chance of winning, whichever door she chooses. The fact that he has the same identical choice between door I and 2, must mean his chances of winning are also 50/50 whichever door he chooses.

fatcat posted:

Don Atkinson posted:

Oh dear.........

Let’s stick with Monty Hall for the moment.

The odds on winning the car are most definitely 2/3 if you change your mind, ie you accept Monty’s offer to open the other door. If you stick with your original door, the odds on winning are 1/3.

........yes, I know it isn’t blatantly obvious, and I also know that many eminent mathematicians disputed these probabilities when first presented, but..........

Suppose the contestant was part of a husband and wife team, but the wife was delayed at the hair dressers. She dashes onto the stage at the point that one door has been opened revealing a goat. Husband says to wife, there’s a car behind one of those two doors, we’ve got to chose between door 1 and door 2, you choose.

She’s got a 50/50 chance of winning, whichever door she chooses. The fact that he has the same identical choice between door I and 2, must mean his chances of winning are also 50/50 whichever door he chooses.

Of course, it all changes if the TV company is unscrupulous and moves the caras soon as the door os chosen...

Oh dear, oh dear...............

Frank, you are creating a different set of circumstances.

I think i’m With you down to, but not including, the last sentence.

Remember, husband has more information than his wife.

Innocent Bystander posted:

fatcat posted:

Don Atkinson posted:

Oh dear.........

Let’s stick with Monty Hall for the moment.

The odds on winning the car are most definitely 2/3 if you change your mind, ie you accept Monty’s offer to open the other door. If you stick with your original door, the odds on winning are 1/3.

........yes, I know it isn’t blatantly obvious, and I also know that many eminent mathematicians disputed these probabilities when first presented, but..........

Suppose the contestant was part of a husband and wife team, but the wife was delayed at the hair dressers. She dashes onto the stage at the point that one door has been opened revealing a goat. Husband says to wife, there’s a car behind one of those two doors, we’ve got to chose between door 1 and door 2, you choose.

She’s got a 50/50 chance of winning, whichever door she chooses. The fact that he has the same identical choice between door I and 2, must mean his chances of winning are also 50/50 whichever door he chooses.

Of course, it all changes if the TV company is unscrupulous and moves the caras soon as the door os chosen...

They didn’t. This isn’t Brexit 

dave marshall posted:

The explanations out there for the Monty Hall conundrum suggest that one's chance of winning increases from one in three to two in three if one switches from the original choice to the other door remaining.These suggestions that you refer to are correct. The chance of winning increases from 1/3 to 2/3.

This seems to be based on the premise that by sticking with one's original choice, when the odds were one in three, the odds have not improved, and that by changing doors, one's odds shorten to two in three. Again, that is correct. It just needs to be explained concisely and clearly.

Seems to me that this overlooks the fact that there are now only two doors from which to choose, so that the often dismissed view that the odds are now 50 / 50, is, in fact correct. This is where the logic you are using, breaks down. The fact that there is a choice between these two doors does not mean that the odds are 50/50. This aspect is the crux of the conundrum !

The popular explanation overlooks this,The real explanation doesn't overlook this point. it simply recognises that it is false. and puts forward a theory based on the original choice of three doors ............. completely overlooking the fact that the choice has started all over again, The choice hasn't started all over again - you have retained knowledge from your earlier choices now that the first door has been removed, i.e. 50 /50?

dave, I have inserted comments to help identify where the 50/50  probability is wrong.

the Black/White card senario is similar, but let's stick with Monty hall first, until yourself, David H and Fatcat are converted. It;s not easy to grasp as i'm sure IB will affirm. A year or so ago It took him about three pages of forum activity to accept that he had been wrong !

If nobody else comes up with a clear, concise explanation, I will provide one that I hope will satisfy you.

But meanwhile, you could try an experiment at home by getting your other half to set up game with (say) three matchboxes (to represent the doors), a £1 coin to represent the car and 2 x20p pieces to represent the goats. run it for say 100 times, and consistently "change your mind".  You will win the £1 about 2/3 of the runs. If you "stick with your first coice" you will only win about 1/3 of the runs. Give it a try - but you do need to do a reasonable number of runs. Mrs D wouldn't co-operate, so I set up a simulation in excel with 1,000 runs. It works !

Don Atkinson posted:

dave marshall posted:

The explanations out there for the Monty Hall conundrum suggest that one's chance of winning increases from one in three to two in three if one switches from the original choice to the other door remaining.These suggestions that you refer to are correct. The chance of winning increases from 1/3 to 2/3.

This seems to be based on the premise that by sticking with one's original choice, when the odds were one in three, the odds have not improved, and that by changing doors, one's odds shorten to two in three. Again, that is correct. It just needs to be explained concisely and clearly.

Seems to me that this overlooks the fact that there are now only two doors from which to choose, so that the often dismissed view that the odds are now 50 / 50, is, in fact correct. This is where the logic you are using, breaks down. The fact that there is a choice between these two doors does not mean that the odds are 50/50. This aspect is the crux of the conundrum !

The popular explanation overlooks this,The real explanation doesn't overlook this point. it simply recognises that it is false. and puts forward a theory based on the original choice of three doors ............. completely overlooking the fact that the choice has started all over again, The choice hasn't started all over again - you have retained knowledge from your earlier choices now that the first door has been removed, i.e. 50 /50?

dave, I have inserted comments to help identify where the 50/50  probability is wrong.

the Black/White card senario is similar, but let's stick with Monty hall first, until yourself, David H and Fatcat are converted. It;s not easy to grasp as i'm sure IB will affirm. A year or so ago It took him about three pages of forum activity to accept that he had been wrong ! 

Let's not be upsetting certain forum members by getting into three or more forum pages of discussion, inevitably involving numerous quotes and re-quotes!

If nobody else comes up with a clear, concise explanation, I will provide one that I hope will satisfy you.

But meanwhile, you could try an experiment at home by getting your other half to set up game with (say) three matchboxes (to represent the doors), a £1 coin to represent the car and 2 x20p pieces to represent the goats. run it for say 100 times, and consistently "change your mind".  You will win the £1 about 2/3 of the runs. If you "stick with your first coice" you will only win about 1/3 of the runs. Give it a try - but you do need to do a reasonable number of runs. Mrs D wouldn't co-operate, so I set up a simulation in excel with 1,000 runs. It works !

 I can't possibly ask my better half to indulge me in this, since she already suspects that I am completely bonkers, and I'd be reluctant to provide her with proof positive!

Best draw a veil over this, Don, as my unshakeable sense of logic prevents me from seeing the theory behind this, and no matter how much thought I have given to the problem, the penny still resolutely refuses to drop. 

dave marshall posted:

Don Atkinson posted:

dave marshall posted:

The explanations out there for the Monty Hall conundrum suggest that one's chance of winning increases from one in three to two in three if one switches from the original choice to the other door remaining.These suggestions that you refer to are correct. The chance of winning increases from 1/3 to 2/3.

This seems to be based on the premise that by sticking with one's original choice, when the odds were one in three, the odds have not improved, and that by changing doors, one's odds shorten to two in three. Again, that is correct. It just needs to be explained concisely and clearly.

Seems to me that this overlooks the fact that there are now only two doors from which to choose, so that the often dismissed view that the odds are now 50 / 50, is, in fact correct. This is where the logic you are using, breaks down. The fact that there is a choice between these two doors does not mean that the odds are 50/50. This aspect is the crux of the conundrum !

The popular explanation overlooks this,The real explanation doesn't overlook this point. it simply recognises that it is false. and puts forward a theory based on the original choice of three doors ............. completely overlooking the fact that the choice has started all over again, The choice hasn't started all over again - you have retained knowledge from your earlier choices now that the first door has been removed, i.e. 50 /50?

dave, I have inserted comments to help identify where the 50/50  probability is wrong.

the Black/White card senario is similar, but let's stick with Monty hall first, until yourself, David H and Fatcat are converted. It;s not easy to grasp as i'm sure IB will affirm. A year or so ago It took him about three pages of forum activity to accept that he had been wrong ! 

Let's not be upsetting certain forum members by getting into three or more forum pages of discussion, inevitably involving numerous quotes and re-quotes!

If nobody else comes up with a clear, concise explanation, I will provide one that I hope will satisfy you.

But meanwhile, you could try an experiment at home by getting your other half to set up game with (say) three matchboxes (to represent the doors), a £1 coin to represent the car and 2 x20p pieces to represent the goats. run it for say 100 times, and consistently "change your mind".  You will win the £1 about 2/3 of the runs. If you "stick with your first coice" you will only win about 1/3 of the runs. Give it a try - but you do need to do a reasonable number of runs. Mrs D wouldn't co-operate, so I set up a simulation in excel with 1,000 runs. It works !

 I can't possibly ask my better half to indulge me in this, since she already suspects that I am completely bonkers, and I'd be reluctant to provide her with proof positive!

Best draw a veil over this, Don, as my unshakeable sense of logic prevents me from seeing the theory behind this, and no matter how much thought I have given to the problem, the penny still resolutely refuses to drop. 

Rest assured dave that you are in good company with respect to the penny not dropping, as I said above, many emminent maths professors had difiiculty at first !

As for "the better half suspecting you are bonkers"......................join the club ! They are probably right

OK it's Saturday evening, i've finished work early because of poor weather, Strictly is an hour or two away, Mrs D is out babysitting grandchildren........

Let’s see if I can take this Monty Hall thingy one step at a time.

1.         If I am given 3 guesses for the car, I am certain to win it
2.         If I am given 2 guesses my chances are 2/3
3.         With one guess, my chances are 1/3 and
4.         With zero guesses my chance of winning the car is zero

Hopefully everybody agrees with the above statements ? (1 and 4 are dead certs. 2 and 3 are probabilities)

1.        Just to reinforce the above statements, assume I decide to pick Doors A & B and Monty opens both Door A & B,
2.        my chance of winning that car is 2/3.

Does everybody agree so far ? (this is a genuine question. No sleight of hand. Just plain probabilities)

Don Atkinson posted:

1.        Just to reinforce the above statements, assume I decide to pick Doors A & B and Monty opens both Door A & B,

If you pick A&B, Monty must open C plus A or B.

Hi Fatcat, thanks for seeking clarity,

In my explanation so far, I am just outlining the basic probabilities. I should have made that situation more clear. When I wrote :- ......

"Just to reinforce the above statements, assume I decide to pick doors A & B and Monty opens both Door A & B"

.....this was just to emphasis that, not withstanding the format of the game, if I was allowed to simply pick any two Doors, say Doors A & B, my chances of winning would be 2/3. I appreciate that the format of the game doesn't involve the contestant asking Monty to open two Doors (A & B for example). But at this stage of my explanation, i am hoping that we can all agree thatif  the contestant was allowed to pick two doors his chances of winning will be 2/3.

We really need to ensure these basic concepts are agreed, even though the Monty Hall game doesn't involve the contestant actually saying "Monty, i'd like you open both Door A and Door B !"

I hope the above is reasonably clear, if not, just say so and if possible outline what is unclear.

Once we have sorted out these first, relatively simple concepts, we can move on to the next part of the explanation.

Was the goat alive or dead ? Or perhaps both ?

Two goats and both alive and kicking !

of course, the assumption is that the contestant wants to win the car and doesn’ t really want to win a goat, dead or alive !

You can do a lot with a goat.

I'm assuming that Dave M, David H and Fatcat (plus anybody else who is following this teaser) are all satisfied with the basics of the probabilities now, especially the element that IF (and only IF) the contestant can select two doors, his probability of winning the car is 2/3.

So, to summarize - the probability of it being gold is 2/3, so the probability of it being silver (the question asked) is 1/3, as in my original post.

Dozey posted:

So, to summarize - the probability of it being gold is 2/3, so the probability of it being silver (the question asked) is 1/3, as in my original post.

Hi Dozey,

The wording of the original question on Page 20 was as follows :-

I select one box at random.

I open one lid of said box and reveal a gold coin.

What is the probability that this box is the one with the gold and silver coins

So not exactly the wording you have used above. Mind you, that is not to say that your answer to the question on Page 20 is wrong !

I don’t pretend that my explanation is any better than the explanation given by others but I set it out below nonetheless. I have already alluded to some aspects of it anyway.

Doors are labelled A, B and C

A car lies behind one door and a goat behind each of the other two

Given the choice of zero doors, the probability of getting the car is 0

Given a choice of only one door, the probability of getting the car is 1/3

Given a choice of all three doors, the probability of getting the car is 1 (it’s a dead cert !)

Given the choice of any two doors, the probability of getting the car is 2/3

That last option is crucial and………..it’s yours for the taking !

Instead of choosing C and sticking with it (1 in 3 chance)…..

…………..choose A and B

Yes, you heard right…..choose A and B

All you need now is to persuade the Show-Host to open both doors A and B

….and you stand a 2/3 chance of winning that car !

And what’s so good about this is, you already know how to do it !

You tell the Show-Host that you have “Chosen” Door C.

He then voluntarily opens either A or B (let’s say A)

You then tell him to open the other one ie B

Checking above, I note that if I can get Monty to open both Door A and Door B my chance of winning the car is 2/3.

Now, fatcat’s  business of “the Wife” joining us half way through the game.

If the wife has no knowledge of the door that the husband initially selected, she only stands a 50/50 chance of winning the car. For her, in that situation, the game is different.

If, however, she is told which door her husband initially selected, she is now playing the original game and would be wise to follow her husband and let him “change his mind”, or do it for him !

Birthdays.

What’s the smallest number of people you need in a group such that it’s more probable than not that two of them share the same birthday ?

Just a few “rules”….

It’s a random group of people, but no twins

Birthday means the day and month but can be different years eg 22^nd July 1965 and 22 July 2002 are the same birthday

365 days in a year ie ignore Leap Years

“More probable than not” means the probability has to be > 0.5

24. 

Any "advance" on 24 ?

23

23 is the point at which the probability is 0.5, or 50/50. So, assuming you can’t have half a person (which is a jolly good Smiths song by the way) the answer to Don’s question must be 24. 

I think it’s 23...

No, it’s not. It’s 24.