Brute force probability

Earlier this week, a twitter bud of mine posted the following (WordPress is supposed to embed the tweet on the next line but it’s not working right now – let me know if you know why and I’ll fix it):
http://twitter.com/#!/samjshah/status/78898954753949696

Q: deck of cards with 6 suits (78 cards). Draw 2 cards without replacement. Probability that the sum of the cards is exactly 8?

Several people responded and came up with an answer around 4%. I put a tiny bit of thought into it and realized I’d have to do the usual bit of reminding myself how the various factorials etc go into such a calculation.

What I want to talk about in this post is the brute force way of finding the answer. I started thinking about how I could get Mathematica to simulate the act of drawing two cards and summing them over and over. It didn’t take long before I went from thinking about it to coding it. When I did, I got the same 4% answer but I realized I had the makings of much more.

First the code:

suit = Join[Range[10], {10, 10, 10}]
deck = Flatten[Table[suit, {6}]]
draw := RandomSample[deck, 2]

The first line defines a suit as 1-10 with 3 extra 10s for the face cards. The second line sets up a deck with 6 suits. The third line lets me randomly draw two cards. Here’s the code to do the draw 100,000 times and find out how many times it gets a sum of 8:

Total[Table[If[Total[draw]==8, 1.0, 0.0], {100000}]/100000

It uses an if/then construction to find out what we want and gives first the number of times it happens and then divides by the total number of times (the fundamental definition of the probability of an event).

That’s fine and all, but I think you learn more from looking at a histogram of all possible sums:

Histogram[Table[Total[draw], {n = 100000}], 20, "Probability"]

That gives this nice image (in Mathematica you can hover over the bars to find the probabilities):

Histogram of sum of 2 cards from 6 suit deck

The first bar is for 2 (drawing two aces) and the last bar is for 20 (lots of possibilities because of the face cards).

The reason I think this is a better answer to Sam’s original question is that it immediately lets you do some comparing and contrasting with other sums. What do you think?

Taking it further

What I really like about this particular approach is that you can explore other questions that might not be as easy to figure out using the typical probability tricks. For example, what are the chances that, when you draw 3 cards from that deck, the second card is larger than the square of the third card but smaller than the square of the first one?

f2[{d1_, d2_, d3_}] := If[d3^2 < d2 < d1^2, 1., 0.]
Total[Table[f2[d = RandomSample[deck, 3]], {n = 100000}]]/n

All I really had to do was change the function I was using from just adding and checking if it was eight to the new question. It turns out the answer is roughly 12% of the time, by the way (weird, huh?).