Reeling In the Nimrod, March 2011 issue

[Tomas Blomberg]

In the March issue, Harry Anderson presents a way to use a classic probability puzzle to put a game in your favour. As far as I can see it will not work, so you'll end up losing money.

In short, if I understand it correctly, it was described as follows:

1. Two random cards face down.
2. You peek the color of one of them.
3. You name the peeked color and claim that the other card is of the opposite color. You offer to pay 3 dollars if you are right, and you get 2 dollars if you are wrong.

The article gives the example of the peeked card being Red but doesn't make it clear if you just don't play the game if the peeked card is Black or if you still play the game but claim that the other card is Red.

To be clear, either way of playing does not seem to give an edge.

/Tomas

[Richard Kaufman]

Nice to see you here again, Tomas. Perhaps someone will chime in.

[Tomas Blomberg]

Thank you very much, Richard!

I should mention that there is a _small_ edge, not because of what's mentioned in the article, but due to the fact that a deck has 26 cards of each color.

If you pull any two random cards from a deck, the probability of them being of different colors is 0.5098. That's enough to make you money in the long run if the bet is even.

But let's get back to the original probability puzzle. Can it be used at all for a scam?

Did I understand the description of it correctly?

/Tomas

[Pete McCabe]

I was going to post the exact same thing but I didn't want people to think I'm a nimrod.

The basic puzzle Harry's bit is based on is this one: A couple with two children has at least one boy. What are the odds the other kid is also a boy?

The answer is 1 in 3. This works because there are four possible combinations for two kids, all occurring in equal numbers:
BB
BG
GB
GG
And the wording of the puzzle (at least one boy) eliminates the GG pair, leaving only one third of the available combinations with two boys.

But Harry's Red/Black version doesn't workat least as I understand itbecause you do not eliminate the BB pair. Instead you take two cards, look at one, mix them, then announce that one of the cards is the color of the one you looked at. At this point the odds both are the same are 50-50 (technically 26/51, but close enough).

[Jonathan Townsend]

While I don't pretend to be intelligent - I also don't expect any audience to sit for such a game. Not even sure it would work for "special" audiences.

[Tomas Blomberg]

Jonathan, I'd sit for that game if you do it for me as written in the article. ;)

/Tomas

[Pete McCabe]

I pretend to be intelligent.

[Bob Gerdes]

I noticed this too.

But Harry's Red/Black version doesn't workat least as I understand itbecause you do not eliminate the BB pair.

I thought that peeking and seeing a red card (as in Harry's example) does eliminate the BB pair.

Of the four possibilities, when you peek you narrow them down to RB and RR.


The part I have a problem with is Harry's argument that simply mixing them so that you no longer know which card is the one you peeked at somehow gives you back the possibility of RB and BR.

It's not the positions of the cards that is important, but the fact that before you look at them they could be either red or black.

The card you peeked at--no matter where it is after the mixing-- will always be red. Still only one way to have a red and a black.

[Pete McCabe]

One thing that confused me was that Harry didn't specify that you peek at one card and only use that pair if the card you see is red. But this still wouldn't give you the result he suggests. If you pick a pair of cards and peek at one, and only use that pair if the peeked card is red, the odds of an RB pair are still 26/51.

Bob: there are 26 ways to have a red and a black, and 25 ways to have two reds.

[Bob Gerdes]

Hi Pete,
I agree with the odds you cite. I was just trying to see where my thinking differs from Harry's. He specifically mentions and refutes the idea that the odds are about even.

I have an issue with this bit (after peeking and mixing up the two cards):

"Your advantage lies in the fact that no one, not even you, knows which card is red. While you've eliminated the possibility of two black cards, there are still two ways of having one black card- red on the right, black on the left, or black on the right, red on the left. There is only one way to have two reds. You're hot."

[Pete McCabe]

Yes, I agree. There are two ways to have two reds: the first one you saw is on the left and another is on the right, or the first one you saw is on the right and another on the left.

[Tomas Blomberg]

One way this will work is if you somehow show the ability to detect when a Red card is present, but not how many there are.

I suggest using a third person:

The Mark hands the third person two random cards, which only he is allowed to look at. You ask "Is there at least one Red card there?" If "Yes" you play the game. If "No", both have to be Black so there is no reason to play the game.

At each new play you may randomly ask for a color, so you may sometimes ask if there is at least one Black. You can even ask the Mark to suggest which color to ask for.

Your bet is that there will be a card of the opposite color present. If you pay 3 dollars when you are wrong and get 2 dollars when you are right, you win 0.38 dollars a game on an average.

Any other ways to play the game?

/Tomas

[Tomas Blomberg]

A similar game:

Take six coins and glue them together to get one double tailed, one double headed and one with a tail on one side and head on the other.

Show the strange set of coins and get the Mark to agree with you that if you see a single of these coins and it shows a head, it has an equal probability of being heads or tails on the other side.

Have him jiggle the three thick coins between his palms and place one on the table without looking at the two other coins.

Whatever shows, you bet that the other side is the _same_. You win 2/3 of the time.

/Tomas