### Hey Kids, Spot the Math Mistake!

(Wabulon, please give someone else a chance before posting the answer.)

A while ago I sent my brother Peter Bernstein's book Capital Ideas.* I was trying to get him interested in becoming a "financial engineer," i.e. one of the Wall Street whiz kids who can tell you how safe mortgage-backed securities are. (Now you know why I cannot charge for my advice.)

Anyway, he was looking at it again and noticed a pretty dumb mistake. I reproduce the relevant excerpt below. The author is referring to Cowles (whose first name I forget):

[Cowles] must have been a fiendish bridge player. Here is one passage from his notes on the game:

"If each of 50 million bridge players in the US plays 200 sessions of 40 deals each, this adds up to 50 million*200*40 = 400 billion hands dealt each in US (sic). The probabilities on any given hand being dealt with 13 cards of one suit are .00000000000156. The chances of a hand with 13 cards of one suit being dealt in the US in any given year, therefore are 400 billion times .00000000000156=.624."

I will post the first comment to this thread, giving a hint from my brother.

* Isn't it odd that in that sentence, it seems for a second that my brother is Peter Bernstein? Cuz he's not.

1. Slightly paraphrasing, here was the reaction of my brother (that is also a hint):

"Wow, it's a good thing Cowles didn't run the numbers for China, otherwise he would've gotten a probability greater than 1!"

2. I don't think the multiplication is meaningful.

I'll give a stab at the correct solution. If the probability of a 13 card hand being dealt is 1.56E-12, then the probability of a 13 card hand not being dealt is (1-1.56E-12). So then, over 4E11 hands, the probability that none are 13 card hands is (1-1.56E-12)^4E11. So the chances that at least one 13 card hand is dealt is 1 - (1-1.56E-12)^4E11)

3. Jacob,

It is meaningful: it's the expectation of the number of such occurrences. But obviously that's not the same thing as the probability of it happening (at least once).

Your approach seems right; I am not bothering to verify the numbers. (Wabulon, feel free to tell us the exact answer now if you want... :))

4. It seems that the author uses the multiplication table as a substitute for the binomial distribution

5. Anonymous11:50 AM

Does it have to do with the fact that cards are going into the other players hands as well, instead of the probability of just laying down 13 cards straight from the deck?

6. If you have a 1-in-2 chance of getting heads on a coin toss, you are not guaranteed to get at least one head if you flip the coin twice.

7. David,

Right... But nobody said that, right? Not the author or me.

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