Marilyn Vos Savant question in Parade a couple of weeks ago.

LCKurtz

LCKurtz

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Here's an image of the question:
Image (10).jpg

And here's her answer in spoiler tags if you want to think about it first:
answer.jpeg
My reason for posting is that I don't think her "solution" is valid although, surprisingly to me at least, the answer is apparently correct. I think the particular answer depends crucially on the fact that one set of coins is exactly 1 larger than the other, whereas I think her "solution" applies equally well where the numbers of coins differ by more than 1, and where she would get a wrong answer. It's wrong for 4 vs. 2 coins and dramatically worse for larger differences. And a proper solution looks to me to involve some identities with binomial coefficients that don't look trivial to me. What do you think?
 
Her answer does not seem to take into account the difference in the number of coins which makes her solution invalid. It seems clear that if one person has 1 coin and the other person has 100 coins that it is not true that 1/2 the times the person with 100 coins will toss more heads that the person is 1 coin. She does go onto say however that Brianna can't toss the same number of heads AND tails as Brian since she has an extra coin.

Now how to prove this? I would go like this. If both people had the same number of coins then on average they would get the same number of heads. However if one person then gets to flip one more coin then I would say that this person has a 50-50 chance of getting more heads then the other person. A bit of hand waving I know. Well maybe not on 2nd thought. Initially they both people have the same expectation in the number of heads tossed. Now with this extra toss the person has a .5 increase on the expected number of heads tossed.

Lets see if I can write this up better. Suppose both people toss a coin n times. Let X = the number of heads person 1 gets. Then E[X] = n/2. Let Y = the number of heads person 2 gets. Then E[Y] = n/2. If person 2 gets to flip an additional coin then E[Y]= (n+1)/2 and the results that were posted is correct.
 
LCKurtz said:
Here's an image of the question:
View attachment 23735

And here's her answer in spoiler tags if you want to think about it first:
View attachment 23736
My reason for posting is that I don't think her "solution" is valid although, surprisingly to me at least, the answer is apparently correct. I think the particular answer depends crucially on the fact that one set of coins is exactly 1 larger than the other, whereas I think her "solution" applies equally well where the numbers of coins differ by more than 1, and where she would get a wrong answer. It's wrong for 4 vs. 2 coins and dramatically worse for larger differences. And a proper solution looks to me to involve some identities with binomial coefficients that don't look trivial to me. What do you think?
Click to expand...
A coin can land 3 ways.

head-side up or

tail-side up or

on the edge (and roll under the sofa- never to be found again)

George Gammow told me so .......
 
Jomo said:
...and the answer to the problem is???
Click to expand...
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

​
Please share your work/thoughts about this problem.
 
Subhotosh Khan said:
Please show us what you have tried and exactly where you are stuck.
Click to expand...
I will tell you where I am stuck. I am stuck with you as my moderator. What have I tried? I pleaded and pleaded with Ted to reconsider your appointment as Super Moderator.
I hope that you have a fabulous day!
 
Jomo said:
I will tell you where I am stuck. I am stuck with you as my moderator. What have I tried? I pleaded and pleaded with Ted to reconsider your appointment as Super Moderator.
I hope that you have a fabulous day!
Click to expand...
I always do have a fabulous day.

You see I am not just a moderator - I am super moderator. That superlative is important to moderate you!!
 
Subhotosh Khan said:
I always do have a fabulous day.

You see I am not just a moderator - I am super moderator. That superlative is important to moderate you!!
Click to expand...
Is your job just to moderate me? It would not be the first time that happened to me.
 
Subhotosh Khan said:
A coin can land 3 ways.

head-side up or

tail-side up or

on the edge (and roll under the sofa- never to be found again)

George Gammow told me so .......
Click to expand...
Hard as it may be to believe, I actually experienced that. My brother and I were flipping a coin for the break on a pool table that had a typical cloth surface. We gave it good flip, it bounced on the table, spun a bit, and stopped on its edge. That was over 50 years ago and I still regard it as one of the most unlikely things I have ever witnessed.
 
I had the following experience playing pool. A ball was sooooo close to being in the pocket that when my friend went to pocket that ball the instant he struck the cue ball the other ball fell into the pocket. In other words the cue ball nor any other ball touched the ball that fell in. I literally thought that I was going to die laughing. I guess that happened due to vibration?? Funny funny incident! I continue to claim that it was a foul as his cue ball failed to hit the ball he called (to go into the pocket).
 
Jomo said:
Lets see if I can write this up better. Suppose both people toss a coin n times. Let X = the number of heads person 1 gets. Then E[X] = n/2. Let Y = the number of heads person 2 gets. Then E[Y] = n/2. If person 2 gets to flip an additional coin then E[Y]= (n+1)/2 and the results that were posted is correct.
Click to expand...
I don't see how that argument is any better. How do you get from E(Y) - E(X) = 1/2 to P(Y > X) = 1/2 ? How would you apply that reasoning to the case of 2 and 4 coins? Here [MATH]E(X) = np = 1[/MATH]. Now person 2 has two extra coins so [MATH]E(Y)=\frac{4}{2}= 2[/MATH]. How do you get from that to [MATH]P(Y>X) =\frac{21}{32}[/MATH], which is the correct answer?
 
LCKurtz said:
Here's an image of the question:
View attachment 23735

And here's her answer in spoiler tags if you want to think about it first:
View attachment 23736
My reason for posting is that I don't think her "solution" is valid although, surprisingly to me at least, the answer is apparently correct. I think the particular answer depends crucially on the fact that one set of coins is exactly 1 larger than the other, whereas I think her "solution" applies equally well where the numbers of coins differ by more than 1, and where she would get a wrong answer. It's wrong for 4 vs. 2 coins and dramatically worse for larger differences. And a proper solution looks to me to involve some identities with binomial coefficients that don't look trivial to me. What do you think?
Click to expand...
I'm trying to find an argument along the lines of hers, but without the nonsense of implying that when there are two things that can happen, they must be equally likely. Here's my idea:

I'm the one with n+1 coins, and you have n. I hold one coin aside (or flip it but hide the result), and we each flip n coins. Considering only those, it will be equally likely that you win or that I win, with some additional probability that we tie. So P(tie) = 1 - 2*P(I win with n).

Now I bring out my extra coin. If I won with the n coins, I win regardless of the outcome. If I tied with n coins, I win with probability 1/2 (that is, if this toss is heads). So my probability of winning with all n+1 is now P(I win with n) + P(I tied)/2 = ... [drum roll] ... 1/2.

The real justification for her conclusion, though, is symmetry (which I realized while thinking through my answer). If we posed the same problem, asking about the probability that I have more tails, the answer has to be the same as the probability that I have more heads, since heads and tails are interchangeable. Since, as she says, we can't tie, I must be equally likely to win or lose, and that probability is 1/2.
 
Dr.Peterson said:
The real justification for her conclusion, though, is symmetry (which I realized while thinking through my answer). If we posed the same problem, asking about the probability that I have more tails, the answer has to be the same as the probability that I have more heads, since heads and tails are interchangeable. Since, as she says, we can't tie, I must be equally likely to win or lose, and that probability is 1/2.
Click to expand...
I have wrestled with that argument when I read the puzzle. I don't see how it applies for 2 coins vs. 100 coins. It isn't whether you win or lose, it's whether you get more heads. You're going to get more of both heads and tails with a very high probability.
 
LCKurtz said:
I have wrestled with that argument when I read the puzzle. I don't see how it applies for 2 coins vs. 100 coins. It isn't whether you win or lose, it's whether you get more heads. You're going to get more of both heads and tails with a very high probability.
Click to expand...
What I meant by "win" was "get more heads". Can you clarify how the number of coins (say, my n) changes the argument?
 
Dr.Peterson said:
The real justification for her conclusion, though, is symmetry (which I realized while thinking through my answer). If we posed the same problem, asking about the probability that I have more tails, the answer has to be the same as the probability that I have more heads, since heads and tails are interchangeable. Since, as she says, we can't tie, I must be equally likely to win or lose, and that probability is 1/2.
Click to expand...
Dr.Peterson said:
What I meant by "win" was "get more heads". Can you clarify how the number of coins (say, my n) changes the argument?
Click to expand...

Just to make sure we are on the same page, I'm questioning Marilyn's argument and your symmetry argument here.
Let's say we have two coins versus 100 coins. The 100 coin tosser will get in the neighborhood of 50 heads, plus or minus, and virtually certainly more than 2 heads. Probability greater than .99+. Ditto for more than 2 tails. What I don't understand is why your symmetry argument above (and Marilyn's argument) don't apply to this case just as well as the n+1 case and give the incorrect answer of 1/2. The difference of just 1 coin must be critical to the argument because that's apparently the only case where 1/2 is correct. Maybe it's just too late at night and I'm being dense...
 
Suppose one person has n coins and the other person has n + k coins where k is some positive integer.
Suppose they both toss n coins. Now I agree that P(tie) = 1 - 2*P( I win with n coins). Now this person brings out the extra k coins. If they won with the n coins they will still win after tossing the k coins. If they tied then this person will win if they toss at least one head with the k coins. The probability of tossing at least one head = 1 - .5^k. So the prob of winning with the n+k coins = p( win with n coins) + (1-.5^k)p(tie) = p(win with n coins) + (1-.5^k)(1-2*p(win with n coins) = 1 - .5^k+ (2*.5^k - 1)p(win with n coins)
 
LCKurtz said:
Just to make sure we are on the same page, I'm questioning Marilyn's argument and your symmetry argument here.
Let's say we have two coins versus 100 coins. The 100 coin tosser will get in the neighborhood of 50 heads, plus or minus, and virtually certainly more than 2 heads. Probability greater than .99+. Ditto for more than 2 tails. What I don't understand is why your symmetry argument above (and Marilyn's argument) don't apply to this case just as well as the n+1 case and give the incorrect answer of 1/2. The difference of just 1 coin must be critical to the argument because that's apparently the only case where 1/2 is correct. Maybe it's just too late at night and I'm being dense...
Click to expand...
Please follow my general case where k = 1.
 
Jomo said:
Suppose one person has n coins and the other person has n + k coins where k is some positive integer.
Suppose they both toss n coins. Now I agree that P(tie) = 1 - 2*P( I win with n coins). Now this person brings out the extra k coins. If they won with the n coins they will still win after tossing the k coins. If they tied then this person will win if they toss at least one head with the k coins. The probability of tossing at least one head = 1 - .5^k. So the prob of winning with the n+k coins = p( win with n coins) + (1-.5^k)p(tie) = p(win with n coins) + (1-.5^k)(1-2*p(win with n coins) = 1 - .5^k+ (2*.5^k - 1)p(win with n coins)
Click to expand...
I think you're missing the case where I had fewer heads among the n, but my extra k make up for it.

That also is where 1 comes into my argument: To have more heads with n+1 coins, I have to have either tied or better with n. That isn't true for n+k.
 
LCKurtz said:
Just to make sure we are on the same page, I'm questioning Marilyn's argument and your symmetry argument here.
Let's say we have two coins versus 100 coins. The 100 coin tosser will get in the neighborhood of 50 heads, plus or minus, and virtually certainly more than 2 heads. Probability greater than .99+. Ditto for more than 2 tails. What I don't understand is why your symmetry argument above (and Marilyn's argument) don't apply to this case just as well as the n+1 case and give the incorrect answer of 1/2. The difference of just 1 coin must be critical to the argument because that's apparently the only case where 1/2 is correct. Maybe it's just too late at night and I'm being dense...
Click to expand...
Yes, the 1 is critical, right here (in the argument you are not talking about at the moment):

Dr.Peterson said:
Now I bring out my extra coin. If I won with the n coins, I win regardless of the outcome. If I tied with n coins, I win with probability 1/2 (that is, if this toss is heads). So my probability of winning with all n+1 is now P(I win with n) + P(I tied)/2 = ... [drum roll] ... 1/2.
Click to expand...
The unstated line: But if I lost with the n coins, I have no chance of winning with n+1, because I must have had 2 fewer heads, and 1 more can't overcome that.

In the symmetry argument, I agree it's a lot more subtle:
Dr.Peterson said:
If we posed the same problem, asking about the probability that I have more tails, the answer has to be the same as the probability that I have more heads, since heads and tails are interchangeable. Since, as she says, we can't tie, I must be equally likely to win or lose, and that probability is 1/2.
Click to expand...
... In fact, it's too late for me to take the time to think it through enough, so I'll look into it tomorrow.
 
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