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Author Topic: Urgent help needed with bar puzzle  (Read 4768 times)
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I don't think you connected the grounds, Dave.
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Of course, it's only a "win" if you didn't want a goat in the first place.

A bit like the old joke about the church raffle - first prize is a weeks holiday in Blackpool, second prize is two weeks holiday in Blackpool.
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" second prize is two weeks holiday in Blackpool."

Loved that one !
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It certainly goes against instinctive logic

I think it is a really interesting problem that gives an insight as to how our minds work. Firstly, having made a decision, we are psychologically inclined to stick with that decision. Maybe there are good survival reasons for this to avoid dithering in a dangerous situation. Next we are not really that logical, even when told that it is always better to swap people remain blind to the reality.

I bet there are more than a few instances where this flawed logic has found its way into computer programs.
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It certainly goes against instinctive logic
Next we are not really that logical, even when told that it is always better to swap people remain blind to the reality.

I bet there are more than a few instances where this flawed logic has found its way into computer programs.

I have know about this for years, and it still gets me. I can't help but think "but what if I chose right the first time?", even though the odds are against me.
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I can't help but think "but what if I chose right the first time?",

It does seem odd that if you choose right, you throw it away.     But its only once in 3 times, the other 2 you win !
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I would have to say that the odds change when you know the result of the ash tray that was removed.  Originally, you have a 1 in 3 chance.  Because you know that an ash tray has been removed that has nothing under it, you odds have changed to 50:50.  Think of it this way.  You're sitting at the bar and have the 1:3 chance and so on.  The bar tender removes the one ash tray and it has nothing under it.  You walk away.  Someone else steps up to take your place.  There are two ash trays sitting there.  One is close to him, the other is pushed back a little.  There is no 3rd ash tray because the bar tender took it away.  What are the odds?  2 in 3, right?  WRONG!!! They are 50:50.
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"There is no 3rd ash tray because the bar tender took it away.  What are the odds?  2 in 3, right?  WRONG!!! They are 50:50."

That's what it might seem, if there was no history.

If the barman tells you what has happened, and which one was originally chosen, the odds are 2 in 3 of a win if you change the original choice.

Look at it this way,  if you choose one on the left, and move the others to the right, the chances of the prize being on your right are 2 in 3, and on your left 1 in 3.  OK?

So if the barman removes one on the right, it doesn't alter the odds of the one you chose on the left, which is still 1 in 3.

There is a 2 in 3 chance of the prize being on the right still, and the barman has kindly removed one of the options, so there is only one choice for the 2 in 3 option on the right.

Look at the table I posted and it shows the results for the various choice and prize positions and you can see how it works.
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But all of that is only true so long as you don't know whether anything was under the tray that was removed by the bar tender.  As soon as you know the condition of the tray, the odds change.
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What if the barman removes the empty ashtray behind a screen so you cant see what he is doing,  you reckon the chances of your choice go from 3:1 to 2:1 the second he removes it, or only when he tells you what he has done ?

There is still a 2:1 chance that one of the other 2 has the prize, and it only helps you if he shows you the empty one.
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As long as you don't know where the coin is, including the possibility of it being under the tray that was removed by the bar tender, yes, you have a 1:3 chance.  As soon as you know that the coin isn't under that tray, you now have 1:2.  There's no difference between the 3 trays with 1 coin, remove 1 tray with no coin under it---and the two trays with 1 coin under 1 of them.  In the end of both cases you have two trays with a coin under one of them.  You still have a choice.  1:2
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As long as you don't know where the coin is, including the possibility of it being under the tray that was removed by the bar tender, yes, you have a 1:3 chance.  As soon as you know that the coin isn't under that tray, you now have 1:2.  There's no difference between the 3 trays with 1 coin, remove 1 tray with no coin under it---and the two trays with 1 coin under 1 of them.  In the end of both cases you have two trays with a coin under one of them.  You still have a choice.  1:2

But the third tray has not been removed from the game, it still exists, and you can still choose it (though why would you, now that you know its not the right one?).
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Suppose you choose ashtray 1 ( you can try it with the others its he same ) here are the possibilities :-

ashtay 1        ashtray 2        ashtray 3           result if switching                result if staying

coin                nothing        nothing                  lose                                        win
nothing            coin            nothing                  win                                        lose
nothing           nothing         coin                      win                                        lose

If you switch you win in 2 out of the 3 combinations

If you dont switch you win in 1 of the 3 combinations

It only works if the barman knows which you chose, and where the coin is, when he removes an empty ashtay.


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... Someone else steps up to take your place.  There are two ash trays sitting there.  One is close to him, the other is pushed back a little.  There is no 3rd ash tray because the bar tender took it away.  What are the odds?  2 in 3, right?  WRONG!!! They are 50:50.

@flyboy you are correct. That is what makes this problem so mindbending.

If a person steps up to bar, sees only two ashtrays (does not know the history), and makes a choice his odds of picking the ashtray with the coin under it are 1:2.

On the other I, who made a choice while there were three astrays, have a 1:3 chance of getting the coin if I stick with my choice but a 2:3 chance if I change my choice.

The critical thing is that I make my choice before it is revealed that the coin is not under one of the two ashtrays I did not choose. That act of revelation gives me additional information and means I can double my odds by changing my choice.

Look at the "truth tables" as the name implies they don't tell lies. Every possible combination is explored in the tables in these posts - you have a 2:3 chance of winning (much better than 1:2).

If you still don't believe me then perhaps we could play this game for real a couple of hundred times  smiley-money  - when its your turn you always pick and stick with your choice, when its my turn I'll always pick then change. I'll take you to the cleaners.
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If you still don't believe me then perhaps we could play this game for real a couple of hundred times  smiley-money  - when its your turn you always pick and stick with your choice, when its my turn I'll always pick then change.

Why not write a small simulation with counters and random generator on Arduino to run it a few million times should converge to result ...
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Hey Radman,   have you any more of these conundrums ?

I have really enjoyed getting my ancient head round this ( I thought it was a 50/50 at first ).
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