Randomness, Part 2
Before I begin, let me clarify something from last week, regarding how you multiply two permutations.
A permutation like (14235) means that 1 gets mapped to 4, 4 gets mapped to 2, etc. If you perform the same permutation twice, then in the first step 1 will get mapped to 4, and 4 will get mapped to 2, so the resulting permutation will map 1 to 2, so we start by writing “(12”. Next, we’ll see w hat 2 gets mapped to. 2 in the first permutation gets mapped to 3, and 3 gets mapped to 5, so we continue the permutation “(125”. Then we see what 5 gets mapped to, etc. I hope that makes things a bit more clear. If ever a number gets “left out”, we start another set of parentheses for that number and see what it gets mapped to. If it gets mapped to itself, we just leave it out.
Now, on to more randomness. Last time I said that the chances of drawing a card in your opening hand that you are playing four of is 4/60+4/60+..+4/60 = 7/15. But that’s not entirely true, because we’re assuming that each of the card draws has the same probability of drawing that card. But what happens if your first four draws are that card? Then suddenly the probability that the fifth card is that card drops to zero, so clearly something can’t be quite right.
The reason for this is that the probabilities for drawing a given card in each of the first seven draws are not independent, so we need to think about how we can come up with the actual number. There is a tool that we use quite often in probability theory, namely the fact that sometimes it’s easier to look at the probability that something doesn’t happen than it is too look at how likely something is to happen.
Let’s look at how likely it is that, given seven card draws, how likely is it that we will not draw at least one of the card we are looking for. The first card we draw has a 56/60 chance of not being that card. Now, if we draw one of the cards with the first draw, we already know that the “event” we are looking at doesn’t match the criteria, so we can assume the first card draw is not the card we want. Which means that there are 59 cards left, 55 of which are not the card. So we now multiply the probability that the first card we draw isn’t the card with the probability that the second card isn’t it, so we have
We go on like this for the first seven cards, ending up with
P’=56/60*55/59*54/58*53/57*52/56*51/55*50/54 ≈ 0.6 =60%
To come up with the probability we do draw it, we subtract from one (or 100%) the probability that we won’t draw it, so
P=1-P’ ≈ 0.4 = 40%
So, in your first hand you have about a 40%, or two in five, chance of drawing any given one of your four-ofs.
Looking at the complementary event can come in very handy when dealing with a complex situation. If we wanted to look at the probability of drawing at least one copy, we’d have to add the probability of drawing one copy, plus the probability of drawing two copies, and so on.
That will do it for today. I forgot to add exercises last week, so I’ll make sure to add some this time.
Next week we will talk some more about shuffling. Noticing a pattern yet?
1. What is the probability of drawing all four copies of a card in your first hand?
2How does the probability change if you are on the draw.
3Some card games have a minimum deck size of forty, but only allow three copies of a card. Are you more or less likely to draw a three-of in your opening hand if you follow those construction rules?