Quickie Coin Probability Puzzle
There haven't been any new puzzles for a while so I thought I'd contribute this quickie.
There are 3 bags. One contains 2 gold coins, another has 2 silver coins and the 3rd has one of each.
You pick a bag at random, and without looking inside take out one coin. It's gold. What is the probability that the other coin in that bag is also gold?
Michael Josephson
Sunday, October 10, 2004
its 2/3
Saurabh Sharma
Monday, October 11, 2004
2/3rds??? Explain this...
Isn't it 1/2? You pick gold once, meaning you picked one of the two bags w/ gold. 50/50 that you have the gold/gold bag.
dave
Monday, October 11, 2004
There are, indeed, those two possibilitiesbut they aren't equally likely. The fact that you picked a gold coin makes it twice as likely that you picked from the gold/gold bag as the chance that you picked from the gold/silver bag.
Another way to think about this: There are six coins, and if you choose a coin and bag randomly, there's an equal chance you'll pick any of them:
Bag 1: Coin A (gold), Coin B (gold)
Bag 2: Coin C (silver), Coin D (silver)
Bag 3: Coin E (gold), Coin F (silver)
Now, you just picked a gold coin. That means you picked A, B, or E, and all three are equally likely (1/3 chance). So the chance that it was A or B (that is, that the bag holds another gold coin) is 2/3. The chance that you picked Coin E (that is, that the bag's other coin is silver) is just 1/3.
Avrom RoyFaderman
Monday, October 11, 2004
I was not sure about the answer, so I wrote a simulation program to test it.
Here is the result:
Total number of picks: 2000000
Picked silver: 1000960
Picked gold other also gold: 665914
Picked gold other not gold: 333126
I was some what surprised to see that the answer is indeed 2/3, as some of the other posters have already pointed out and explained.
JHY
Tuesday, October 12, 2004
Almost all of these seeminglycounterintuitive probability problems rely on our difficulty telling ordered pairs from unordered pairs(see, for example, "Painfully Easy"; I suspect that you could even explain the Monty Hall problem in these terms, although it would take some work). This one, too, gets its bite from this distinction. Compare the following, superficially similar, problem:
There are three bags. Each bag contains a pair of coins marked "1" and "2". In the first bag, both coins are gold. In the second bag, coin 1 is gold and coin 2 is silver. In the third bag, both coins are silver.
You pick a bag at random and draw a coin. You see that it is marked "1", and is gold. What's the chance that the other coin in the bag is gold?
Here, the answer really *is* 1/2. And the only difference is that, if the bag is the first bag, you don't just know that you drew a gold coin, but know *which* gold coin you drew. This makes all the difference.
Avrom RoyFaderman
Tuesday, October 12, 2004
i think its 2/3 as i first said ..
probability that you pick a gold coin is 1/3 + 1/3*1/2 say A
probability that you pick a bag with both gold coins is 1/3 say B
so probability that you pick a bag with both gold coins given that you have pick a gold coin is B/A
which is 2/3
tell me if I am wrong.
PS. I think many of you are neglecting the fact that probalitiy of getting a gold coin in a bag with 2 gold coin is twice more than the probablility with one gold coin.
Saurabh Sharma
Tuesday, November 16, 2004
Think of it this way:
The choice of the first coin is really the random choice of one of the six coins.
If you get a gold coin, two of the three come from the bag with the two gold coins. So 2/3 of the time, you get the bag with two gold coins.
This is a variation of the Monty Hall three door paradox.
...
Tuesday, November 16, 2004
Saurabh, I'm not sure if you meant to be replying to me. But if so, I think we're talking past each other. I agree with you that the probability is 2/3 in the original problem. But not in my modified problem. If you pick a coin marked "1", the fact that it's gold *doesn't* make it more likely that the coin game from the GG bag...since both the GG bag and the GS bag have one coin that's both gold and marked "1". My point is that the marking (or more specifically, the ability to distinguish members of pairs) makes a difference.
Avrom RoyFaderman
Wednesday, December 22, 2004
It is 5050. You pick it or you don't.
Aaron L Huddart
Saturday, December 25, 2004
Aaron, that's faulty reasoning.
You could imagine thinking "Well, either I will win the lottery or I won't. So I have a 50/50 chance of winning, yay!".
Ian McMeans
Wednesday, January 05, 2005
Oh, sorry. I didn't realize you were discussing your modified version of the problem.
Ian McMeans
Wednesday, January 05, 2005
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