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## RE: [probability] 3 boxes with a coin in one!

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• Hi Veeresh, This looks like a the Monty Hall problem - however one critical item is missed - when you say C is opened - was it chosen (out of B and C)
Message 1 of 11 , Jun 1, 2001
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Hi Veeresh,

This looks like a the Monty Hall problem - however one critical item is
missed - when you say C is opened - was it chosen (out of B and C)
randomly? Or did someone who knew it was empty opened it? The situations
differ - the latter is the Monty Hall problem.

Suppose C is chosen randomly
----------------------------
Then, effectively we have a situation where out of three boxes, one is
randomly eliminated. The probability of coin being in either A or B is the
same - viz 1/2, so no point switching.

Suppose C is chosen the Monty Hall way
--------------------------------------
Now, C was chosen out of (B and C) by someone who knew that C is empty. As
someone who knew the coin is made this choice, the odds are better than even
that B has the coin. In fact, in this case probability works out to what
Reuben suggested - switching wins with probability 1/3.

Both the above situations can also be rigorously demonstrated with
simulation or proven with conditional probability arguments.

regards,
Rino.

-----Original Message-----
From: veereshad@...
To: probability@yahoogroups.com
Sent: 5/31/01 7:02 PM
Subject: [probability] 3 boxes with a coin in one!

Hello,
I have an example-riddle about probability. Please help me to find
answer for the riddle.

It goes:
There are 3 closed-similar-boxes, say A, B, C, one and only one of
them has a coin in it. Now I am asked to point the one having the
coin. Randomly I opt for one, say A. Then, when one of the other two
boxes is opened, say C, found to be not having the coin. Now I am
again have the chance of changing my selection of box.
Now do I stick to the old choice A or switch to B? Which one gives me
more chance of winning?
Any body, please answer me, which one? And Why? What's the
probability of winning? Not 50 %?

Regards,
Veeresh
• Typo in second situation below - switching wins with probability 2/3 as Reuben suggested.... ... From: Rino Raj(Taco) To: probability@yahoogroups.com Sent:
Message 2 of 11 , Jun 1, 2001
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Typo in second situation below - switching wins with probability 2/3 as
Reuben suggested....

-----Original Message-----
From: Rino Raj(Taco)
To: 'probability@yahoogroups.com '
Sent: 6/1/01 5:58 PM
Subject: RE: [probability] 3 boxes with a coin in one!

Hi Veeresh,

This looks like a the Monty Hall problem - however one critical item is
missed - when you say C is opened - was it chosen (out of B and C)
randomly? Or did someone who knew it was empty opened it? The
situations
differ - the latter is the Monty Hall problem.

Suppose C is chosen randomly
----------------------------
Then, effectively we have a situation where out of three boxes, one is
randomly eliminated. The probability of coin being in either A or B is
the
same - viz 1/2, so no point switching.

Suppose C is chosen the Monty Hall way
--------------------------------------
Now, C was chosen out of (B and C) by someone who knew that C is empty.
As
someone who knew the coin is made this choice, the odds are better than
even
that B has the coin. In fact, in this case probability works out to
what
Reuben suggested - switching wins with probability 1/3.

Both the above situations can also be rigorously demonstrated with
simulation or proven with conditional probability arguments.

regards,
Rino.
• ... He said it was found to be empty, so it doesn t matter who selected it, how they selected, what they knew, or why they selected it. If the box to reveal is
Message 3 of 11 , Jun 1, 2001
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>
> Hi Veeresh,
>
> This looks like a the Monty Hall problem - however one critical item is
> missed - when you say C is opened - was it chosen (out of B and C)
> randomly? Or did someone who knew it was empty opened it? The situations
> differ - the latter is the Monty Hall problem.
>

He said it was found to be empty, so it doesn't matter who selected it, how
they selected, what they knew, or why they selected it.

If the box to reveal is chosen randomly, then 1/3rd of the time the box with
the coin is revealed, and there is no problem for those cases. You only
need to consider the cases where the box that's revealed is empty. Monty
Hall never revealed a box containing the prize - but the Monty Hall problem
is exactly the same as the problem where a box is selected randomly.
• Thnaks for a clear explanation. But in my view it s different. Here, once the box C opened and found not containing coin, an instance of probability ended.
Message 4 of 11 , Jun 1, 2001
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Thnaks for a clear explanation.

But in my view it's different.

Here, once the box C opened and found not containing coin, an
instance of probability ended. When again prompted with a chance of
selecting, it's altogether a different case, where in you have two
boxes and have to select a box that has coin in it. (May be 'cos of
the history of the record known, that's all.) Here do we need to link
the probability analysis of the first case to the second one. In fact
second one is as if refurnished case! Probability of the first case
(1/3) is valid till the first box opened. Once opened, before going
for the next step of selecting rules of probability need to be
applied again (where we have two equal possiblity! I mean 1/2
probability for each box!)

Am coming to say that chance is 50 % for both the box? Each box stand
50 % chance! So switch or not, 50 %!

Do you agree? Else, please tell me how do say that probablity of
first case extends to the second case? They are descrte cases.

--- In probability@y..., "Reuben Fries" <reuben@m...> wrote:
> You have a 1/3 (33%) chance of winning if you stay, and a 2/3 (66%)
chance
> of winning if switch.
>
> There is a 1/3 chance that the box you originally selected is
correct. 33%.
> This does not change, even after another box is revealed.
Therefor, staying
> with the same box means your chance of winning is 33%.
>
> There is a 2/3 chance that one of the two boxes you did not select
is
> correct. This also does not change after one of them is revealed
to have
> nothing. The chance of EITHER the one which you now know has
nothing in it
> OR the third (untouched) box being correct is 2/3. That is to say,
the sum
> of the probabilities of the revealed box and the last box is 2/3.
You know
> the probability of the revealed box is 0, because it has been
revealed to
> contain nothing. 2/3 - 0 = 2/3, so the probability of the third
box to have
> the coin is 2/3. Switching to this box is advised.
>
>
> ----- Original Message -----
> From: <veereshad@y...>
> To: <probability@y...>
> Sent: Thursday, May 31, 2001 6:32 AM
> Subject: [probability] 3 boxes with a coin in one!
>
>
> > Hello,
> > I have an example-riddle about probability. Please help me to find
> > answer for the riddle.
> >
> > It goes:
> > There are 3 closed-similar-boxes, say A, B, C, one and only one of
> > them has a coin in it. Now I am asked to point the one having the
> > coin. Randomly I opt for one, say A. Then, when one of the other
two
> > boxes is opened, say C, found to be not having the coin. Now I am
> > again have the chance of changing my selection of box.
> > Now do I stick to the old choice A or switch to B? Which one
gives me
> > more chance of winning?
> > Any body, please answer me, which one? And Why? What's the
> > probability of winning? Not 50 %?
> >
> > Regards,
> > Veeresh
> >
> >
> >
> >
> > Your use of Yahoo! Groups is subject to
http://docs.yahoo.com/info/terms/
> >
> >
> >
• I do not agree. The cases are not discreet. The fact that one is revealed afterwards does not change the exact probability we have already calculated. Our
Message 5 of 11 , Jun 1, 2001
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I do not agree. The cases are not discreet. The fact that one is revealed
afterwards does not change the exact probability we have already calculated.
Our calculation of 1/3 previous is not a guess. It is an exact probability.
One in three times, our box will be correct. Period. Nothing done after
you choose the box will change the fact that 1 in 3 times your first choice
will have been correct. Nothing. If you keep that in mind, it should be
easy to wrap your mind around my previous explanation.

Saying there is a 50% chance is simply an uninformed guess, as compared to
our exact calculations earlier. Just because the sum of probabilities A and
B equals 1 (100%), does not mean that A and B are both 50%. If we know
nothing of the nature of A and B, only that A and B sum to 100%, then the
GUESS which would minimize incorrectness is that A and B are both 50%.
However, we know the nature of A and B in our problem, and we can calculate
the exact probabilities of each, so we need not make such guesses.

> Thnaks for a clear explanation.
>
> But in my view it's different.
>
> Here, once the box C opened and found not containing coin, an
> instance of probability ended. When again prompted with a chance of
> selecting, it's altogether a different case, where in you have two
> boxes and have to select a box that has coin in it. (May be 'cos of
> the history of the record known, that's all.) Here do we need to link
> the probability analysis of the first case to the second one. In fact
> second one is as if refurnished case! Probability of the first case
> (1/3) is valid till the first box opened. Once opened, before going
> for the next step of selecting rules of probability need to be
> applied again (where we have two equal possiblity! I mean 1/2
> probability for each box!)
>
> Am coming to say that chance is 50 % for both the box? Each box stand
> 50 % chance! So switch or not, 50 %!
>
> Do you agree? Else, please tell me how do say that probablity of
> first case extends to the second case? They are descrte cases.
• Hello. I will try to explain this problems formally, which I guess is the correct way. I will show that 1.- If the box is selected by an observer who knows
Message 6 of 11 , Jun 1, 2001
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Hello.

I will try to explain this problems formally, which I guess is the correct
way. I will show that

1.- If the box is selected by an observer who knows where the coin is, I
agree that switching boxes increases chances.
2.- But if the box is selected at random there are various possibilities,
depending on the exact rules of the game. The probability of winning is also
2/3, but the strategy is not necessarily changing boxes.

1.-This is the most interesting game. Lets call A the box we choose, B the
box shown and C the other one. Then:

P(coin in A| B is shown) = P(coin in A and B is shown)/P(B is shown)
(Bayes Theorem)

P(coin in A and B is shown) = 1/6 (If the coin is in A, the observer
may choose with P =1/2 between boxes B, C)
P(B is shown) To calculate this one, we need the formula for total
probability

P(B is shown) = P(B is shown|coin in A)P(A) + P(B is shown|coin in B)P(B) +
P(B is shown|coin in C)P(C)
P(A), P(B), P(C) are the a priori probs. that the coin is in any of the
boxes (1/3 for each).For the rest:

P(B is shown|coin in A) = 1/2 (the observer may choose between B and C
with prob 1/2)
P(B is shown|coin in B) = 0 (the observe will never show the box
where the coin is)
P(B is shown|coin in C) = 1 (same as above)

Then P(B is shown) = 1/2*1/3 + 1*1/3 = 1/2

Therefore P(coin in A| B is shown) = 1/3

Working the same for P(coin in C| B is shown) = 2/3, which shows that
switching boxes is the good strategy.

2.-Now, if the box is selected at random this is not the same. To work the
problem we should know:

- If my box can be chosen (which I find goes against the game's spirit)
- If I can change to an opened box (in case it shows the coin)

A real random election should take into account this two situations. If it
does, then.

Prob of showing the box with the coin = 1/3. I sure win if I change to that
one.
Prob of showing an empty box is 2/3. Changing boxes or not does not matter.
I win either case woth prob. 1/3
In the random problem then I win with prob 2/3. The probability is the same
as in the case above (win with P=2/3), but the strategy is NOT necessarily
changing boxes (unless it shows the coin!).

Hope this helps.

Pablo
• ... the ... spirit) ... Both of these cases can be eliminated with separate strategies. Obviously, you only change to the open box (assuming that s allowed)
Message 7 of 11 , Jun 1, 2001
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> 2.-Now, if the box is selected at random this is not the same. To work
the
> problem we should know:
>
> - If my box can be chosen (which I find goes against the game's
spirit)
> - If I can change to an opened box (in case it shows the coin)

Both of these cases can be eliminated with separate strategies. Obviously,
you only change to the open box (assuming that's allowed) if the coin is
there. Also, you should obviously switch if your box is revealed to have no
coin, and stay if your box is revealed to have it. These two cases are now
eliminated, and the only game which is left is the one where only an empty
box is revealed, and that box is not your original choice. We have already
solved the remaining game. We always switch, and we win 2/3rds of the time.

Now we have three simple strategies for three simple games which can be
combined to make an optimal strategy for any kind of host, with or without
the rules you've proposed could be allowed. This is simpler, easier, and
just as correct as providing a complete strategy each time the rules are
changed slightly, or the host becomes malicious.
• Well, looks like the case where an empty box is knowingly selected seems to meet agreement. In the other case, perhaps I can use a formal argument like
Message 8 of 11 , Jun 1, 2001
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Well, looks like the case where an empty box is knowingly selected seems to
meet agreement. In the other case, perhaps I can use a formal argument like
Pablo...

One of boxes A, B, C contain a coin. A is chosen by player. Let 1 denote
coin and 0 empty state.

Now, for X = A, B, C
p(X=1) = 1/3 and p(X=0) = 2/3

alternately, each of the following cases have equal probability
A | B C
========
1: 1 | 0 0
2: 0 | 1 0
3: 0 | 0 1

Random situation
----------------
C is randomly opened and found to be empty.
Now p(A=1 | C is randomly opened and found empty)

= p(A = 1 and C is randomly opened and found empty)/ p(C is randomly opened
and found empty)

p(C is randomly opened and found empty) = (1/2)*(2/3) = 1/3
p(A = 1 and C is randomly opened and found empty) = (1/3)*(1/2)*(1) = 1/6

Thus p(A=1 | C is randomly opened and found empty) = (1/6)/(1/3) = 1/2

Hope Reuben agrees that additional information may often influence/ improve
the chances of locating something. There is nothing sacrosanct about the
first box chosen and its probability of 1/3. You open, say a box B, and see
the coin, that probability definitely changes. Now if you don't see a coin
also, the probability may change, as shown above.
• This is is the situation I described in the random experiment as My box is not chosen You have make very clear that the probability here is 1/2 Of course
Message 9 of 11 , Jun 2, 2001
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This is is the situation I described in the random experiment as "My box is
not chosen"You have make very clear that the probability here is 1/2
Of course information given by random esperiment may change our initial
assignments of probability. That is the whole point of conditional probability
and expectation, aside from Bayesian decision.

Please, take a look at the problems I sent yesterday. I really need them
solved.

Pablo

"Rino Raj(Taco)" wrote:

>
> Well, looks like the case where an empty box is knowingly selected seems to
> meet agreement. In the other case, perhaps I can use a formal argument like
> Pablo...
>
> One of boxes A, B, C contain a coin. A is chosen by player. Let 1 denote
> coin and 0 empty state.
>
> Now, for X = A, B, C
> p(X=1) = 1/3 and p(X=0) = 2/3
>
> alternately, each of the following cases have equal probability
> A | B C
> ========
> 1: 1 | 0 0
> 2: 0 | 1 0
> 3: 0 | 0 1
>
> Random situation
> ----------------
> C is randomly opened and found to be empty.
> Now p(A=1 | C is randomly opened and found empty)
>
> = p(A = 1 and C is randomly opened and found empty)/ p(C is randomly opened
> and found empty)
>
> p(C is randomly opened and found empty) = (1/2)*(2/3) = 1/3
> p(A = 1 and C is randomly opened and found empty) = (1/3)*(1/2)*(1) = 1/6
>
> Thus p(A=1 | C is randomly opened and found empty) = (1/6)/(1/3) = 1/2
>
> Hope Reuben agrees that additional information may often influence/ improve
> the chances of locating something. There is nothing sacrosanct about the
> first box chosen and its probability of 1/3. You open, say a box B, and see
> the coin, that probability definitely changes. Now if you don't see a coin
> also, the probability may change, as shown above.
>
>
>
>
>
> Your use of Yahoo! Groups is subject to http://docs.yahoo.com/info/terms/
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