Bertrand's Box variations


by Balfo
Tags: bertrand's box, monty hall, paradox, probability
Balfo
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#1
Apr30-12, 10:46 AM
P: 7
Hi, I have a question about the famous Bertrand's box paradox.

In this puzzle there are three boxes: a box containing two gold coins, a box with two silver coins, and a box with one of each. After choosing a box at random and withdrawing one coin at random that happens to be a gold coin, the question is what is the probability that the other coin is gold. As in the Monty Hall problem the probability is 2/3.
So my question is, what would the answer to the same question be if we eliminated the box with the two silver coins, or if we added two more boxes with mixed coins, or if for example we added 7 more boxes with two silver coins, so now we had 10 boxes altogether?

I think it would be easier to understand the paradox with answers to a few variations like these, thanks in advance,
Balfo
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arildno
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#2
Apr30-12, 01:17 PM
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1. "or if for example we added 7 more boxes with two silver coins, so now we had 10 boxes altogether?"
Would any of these come into consideration if you already knew the first coin is gold?
2. "or if we added two more boxes with mixed coins"
Knowing how many gold coins you have in total, how many are in the double coin box, and how many spread among the three one-gold boxes?
Diffy
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#3
Apr30-12, 01:37 PM
P: 443
If there are 4 boxes that have double gold coins, and 3 boxes that have gold and silver.

Then there are 3 ways to pick a gold coin that is losing, and there are 4x2 = 8 ways to pick a gold coin that is winning. there are 8 + 3 = 11 ways to pick a gold coin total. Thus there are 8/11 chances you picked a case where the other coin is gold.

So in general, L = number of boxes that have silver and gold. W = number of boxes that have 2 gold.

Your probability is going to be 2*W/ (2*W + L)

Does that help?

Balfo
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#4
Apr30-12, 06:01 PM
P: 7

Bertrand's Box variations


Thank you both for answering,
yes Diffy, it helps, thanks again :) , however I do have another question, how could we calculate the possibilities if we changed the number of coins in some boxes, let's say we have a box with two golden coins, a box with 3 silver coins , and two boxes that contain 3 golden and two silver coins each, so altogether 4 boxes?
thanks,
Balfo
Diffy
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#5
Apr30-12, 06:29 PM
P: 443
That is an excellent followup question.

Do this: try and answer that yourself. Try writing out all the possibilities. Start with two boxes and three coins.

I'll reply in another day if you still can't figure it out.
Balfo
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#6
Apr30-12, 07:03 PM
P: 7
will do, I'll try :)
Balfo
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#7
May1-12, 04:11 PM
P: 7
Quote Quote by Balfo View Post
how could we calculate the possibilities if we changed the number of coins in some boxes, let's say we have a box with two golden coins, a box with 3 silver coins , and two boxes that contain 3 golden and two silver coins each, so altogether 4 boxes?
thanks,
Balfo
So i did some thinking, I don't know if I'm right or not but I think the answer to my own question is 50% , but i did not know how to calculate the possibilities that we will get a silver coin after we withdrew the first coin and it was silver, the question goes for the same four boxes, thanks,
Balfo
Diffy
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#8
May1-12, 08:58 PM
P: 443
Sorry the question is a little bit different this time. The original was what is the probability that the other coin in the box was gold.

With these four boxes that you have now defined. We are assuming we picked a silver coin and the second coin we pick will also be silver?

If that is the case, and correct me if I am wrong. Than we can ignore the box with 2 gold coins as there is no way we can pull a silver from it.

Now that leaves 1 box with 3 silver. and two boxes that contain 3gold and 2 silver.

This changes the problem significantly. Because if I pick one of the silvers a box that contain 2 silver and 3 gold then I only have a 25% chance of picking another silver out of it.

So now there are 4 silver coins that I could pick and only have a 25% chance of picking a second, and 3 silver coins that I could pick where I will 100% definitely pick a second silver coin.

I am not sure I solved it correctly but I counted 10 cases where I could pick a silver coin and the second coin I pick could be silver, and 22 cases all together. Thus making the probability 45.4545454....%

Here it is all written out. Hopefully someone who is better at probability can verify.

Box 1
s1 s2 1
s1 s3 1
s2 s1 1
s2 s3 1
s3 s1 1
s3 s2 1
Box 2
s4 g1 0
s4 g2 0
s4 g3 0
s4 s5 1
s5 g1 0
s5 g2 0
s5 g3 0
s5 s4 1
Box 3
s6 g4 0
s6 g5 0
s6 g6 0
s6 s7 1
s7 g4 0
s7 g5 0
s7 g6 0
s7 s6 1
Balfo
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#9
May2-12, 10:16 AM
P: 7
Thanks, seems that you wrote out all the outcomes, but something is still not clear to me, in this video: youtube.com/watch?v=BqXToEU5eDY , it's said that for the original problem the result would be 2/3 even if we put a 100 gold coins in the first box, but if we tried to write down every possibility we would not get that result, what about that?
Balfo
arildno
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#10
May3-12, 07:12 AM
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Diffy:
I believe you are wrong in this.
We are after:
Chance that the second coin is silver, given that the first was silver.

This must equal:
P(chance having picked second box|given first coin silver)*P(getting second silver| given picked second box)+P(chance having picked third box|given first coin silver)*P(getting second silver| given picked third box)+P(chance having picked fourth box|given first coin silver)*P(getting second silver| given picked fourth box).

Now, we have, for example:

P(chance of picking first silver coin out of second box)
=P(chance having picked second box|given first coin silver)*P(getting first silver)
=P(chance of picking first silver|given picked second box)*P(chance picking second box).

We have :
P(getting first silver)=1/4*0+1/4*1+1/4*2/5+1/4*2/5=1/4*9/5.
Since P(chance picking second box)=1/4 (and P(chance of picking first silver|given picked second box)=1), we get that:

P(chance having picked second box|given first coin silver)=5/9.

(Note that this is wholly independent of the actual number of silver coins in the second box.)


Thus, we find:
P(second silver|first silver)=5/9*1+2/9*1/4+2/9*1/4=6/9=2/3.

-----------------------------------------------------------
Your flaw is not to take into consideration how the knowledge that that the first was silver actually redistributes the probabilities among "all possible outcomes", regarding them as "equally likely", rather than weighting them properly.

Knowing all possible outcomes doesn't give you any probabilities at all, you need the probability distribution between those possibilities to compute..probabilities.
Balfo
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#11
May4-12, 12:16 PM
P: 7
Quote Quote by arildno View Post
we have :
P(getting first silver)=1/4*0+1/4*1+1/4*2/5+1/4*2/5=1/4*9/5.
Since p(chance picking second box)=1/4 (and p(chance of picking first silver|given picked second box)=1), we get that:

P(chance having picked second box|given first coin silver)=5/9.
Thanks for the effort arildno, if you could just explain again the part which I quoted, i do understand the math you used to get 5/9 but I don't really understand the logic behind it,
Balfo
arildno
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#12
May4-12, 02:07 PM
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Quote Quote by Balfo View Post
Thanks for the effort arildno, if you could just explain again the part which I quoted, i do understand the math you used to get 5/9 but I don't really understand the logic behind it,
Balfo
Now, the event (getting first silver out of second box) can be written as the double event:

(picking the second box AND getting a first silver)

Agreed?

A)Now, we can rewrite the probability of that double event occuring in terms of a product like this:
P(picking the second box AND getting a first silver)=P(picking the second box| given getting a first silver)*P(getting first silver),

B) But, we could equally well rewrite the probability of that double event as:
P(picking the second box AND getting a first silver)=P(getting first silver|given picked second box)*P(picking second box).

Thus, the products in A) and B) must be equal to each other.
Balfo
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#13
May4-12, 07:08 PM
P: 7
Thanks for your help


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