That Doesn’t Make Cents!

October 4, 2010 at 8:40 pm 2 comments

Two bad puns for today…

What’s the difference between an old penny and a new dime? Nine cents!

Why didn’t the quarter roll down the hill with the nickel? Because it had more cents.

And a math problem related to coins:

You have three coins — one is a typical coin with heads on one side and tails on the other. But the other two are strange — one has heads on both sides, and one has tails on both sides. One coin is chosen at random and placed flat on the table. It shows heads. What is the probability that the other side is tails?

Your first instinct may be to think that the probability is 1/2. But it’s not. I won’t give you the actual answer, but here’s a hint — create these “coins” using pieces of paper with H and T. Then try it several thousand times. You’ll see that if the coin shows heads when placed on the table, then the other side will not be heads and tails in equal proportions.

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Math Clocks Can You Digit?

2 Comments Add your own

  • 1. xander  |  October 9, 2010 at 6:19 am

    This is basically the Monty Hall problem, is it not?

    We have three coins: h has two heads, t has two tails, and s is a standard coin with one heads and one tails. There are two possible results: H is heads, and T is tails. Then the outcomes are as follows:

    Hh (we see heads, and it is the heads coin)
    Hh (we see the other heads on the heads coin)
    Hs (we see heads, and it is the standard coin)
    Ts (&c.)
    Tt
    Tt

    So if we see heads, there are three possibilities: we are seeing one side of the heads coin, we are seeing the other side of the heads coin, or we are seeing the heads side of the standard coin. Thus we are seeing one of three possible outcomes. Only one of these outcomes gives us tails on the opposite side, so the chance that the reverse is tails is 1/3.

    At least, I think that is the case. A degree in statistics, and I still have trouble with the Monty Hall problem.:\

    xander

    Reply
  • 2. venneblock  |  October 9, 2010 at 6:40 am

    Your argument is correct… but I’m not sure about your analogy to the Monty Hall problem. I think this is different. For them to be the same, I think we’d need to see what’s in front of and what’s behind one of the doors, as that woud be akin to a coin with H on both sides, no? I think the only similarity is the answer, which is 2/3 for both problems.

    Reply

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The Math Jokes 4 Mathy Folks blog is an online extension to the book Math Jokes 4 Mathy Folks. The blog contains jokes submitted by readers, new jokes discovered by the author, details about speaking appearances and workshops, and other random bits of information that might be interesting to the strange folks who like math jokes.

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