a problem

"If I start with $2 then I should finish with $5 if I win my bet and $0 if I lose. "


You do. The original message said you would get paid $3. Therefore, you are paid $3 for your $2 and walk away with $5.

Maybe I'm misunderstanding the odds notation...I've never encountered it before until reading about poker. (I'm only used to percentages.) In this case, it should be 1:2 (or .5:1) instead of 3:2, and 2:3 (or (2/3):1) instead of 5:3. Sorry for the confusion.

Except the trials AREN'T independent, in this case. Say you have a full deck, in order, As Ks Qs Js... to start with. You don't want any identical cards. The probability the first isn't is 51/52, true. But the probability the next one is identical is dependent on the first.


Let A = event the first card is not identical, and B = event the second card is not identical. The definition of "independent" means that


P(A intersect B) = P(A)*P(B),


(e.g. choosing suit vs. rank in a deck are independent because P(Qs) = P(Q)*P(s)).


Here, P(A) = P(B) = 51/52.


But what is P(A intersect B)? Well, how many permutations are there where the first two aren't identical? There are a number of cases. The first two cards could be completely different from the As and Ks. There are P(50,2) ways to get these cards, then. After that, there are 50 cards left, and they can be permutated in any way, so that's 50!, altogether, in this case, there are 50*49*50! ways. What if exactly one of those two cards falls among the first two? It's place is completely determined, because it can't fall on its own place, so here there are 2*50*50! ways (the 2 comes from the choice of 2 cards As and Ks, the 50 from choosing the other of the first 2 cards, and then 50! for the rest). And finally, they both could appear, but in reverse order. Here, there's 50! ways this could happen. Altogether, there are (50*49 + 2*50 + 1)*50! = 2551*50! ways. Dividing by 52!, we get the probability to be 2551/(52*51), about .9619155


The probability you get from multiplying P(A) by P(B), however, is (51/52)^2, which is 2601/(52*52), about .9619082, not the same.


Another way to see why simply assuming they're independent is wrong, is to imagine a deck with, say, 3 cards. If you assume independence, you get the probability of no identical card falling to be (2/3)^3 = 8/27. Using inclusion-exclusion (which IS known to give the correct answer), you get 1 - 1 + 1/2 - 1/6 = 1/3, which is not the same as 8/27. Or, better yet, use a deck with TWO cards...inclusion-exclusion gives you the (obviously correct, even without inclusion-exclusion) answer 1 - 1 + 1/2 = 1/2, whereas assuming independence gives you (1/2)^2 = 1/4.


This is an example of getting the right answer, but for the wrong reasons. It's true, that in the LIMIT, as n goes to infinity, the probability of a derangement on n elements (a permutation without fixed points) approaches the probability of taking n independent selections, with replacement, from a jar with n objects, and never selecting a specific fixed object. That doesn't mean the probabilities for FINITE n are identical.

I agree that the trials aren't independent in the strict sense. That's obvious when you consider that if all the cards up to the last one match, then a match on the last one is guaranteed.


The bigger question, IMO, is whether the trials can be treated as independent for the purpose of calculation. We have shown that the probability of no match is equal to the product of the individual probabilities of no match per card (51/52)^52 to within our desired accuracy of calculation. Therefore, for all practical purposes, the trials may be regarded as indpendent. This will be the case for a reasonbly large number of trials, in this case 52. You have shown yourself that even with as few as 3 trials, the approximation is quite close (8/27 vs 1/3). The inclusion-exclusion principle will generally also give an approximation since we will not generally bother to calculate all the terms for large n, and a calculation based on an assumption of independence may actually be just as accurate as well as much simpler to compute. So I disagree that it is "getting the right answer for the wrong reasons". It is getting the right answer for the right reason, namely the correct realization that the trials may be treated as independent.


In the other problem where we wanted the probability of a player holding a pair of jacks or better, the assumption of independence still held to a high degree of accuracy, though I would not necessarily have assumed that. Yet now that I know this result, I will continue to use it to simplify calculations where appropriate.

You had it right, you just have to make it clear that when I win you give me back my original $2 in addition to the $3 that I win. That is giving me 3:2 odds. I'm betting $2 in order to have a net win of $3.

Well, okay...you didn't say you were TREATING them as independent, you said they WERE independent. There's a big difference. When you say "the limit is 1/e, therefore the trials are independent", that's just false. If you mean to say, "for large n, the trials can be considered as independent", and then give a heuristic or intuitive justification for that, is one thing. But I interpreted what you said to mean that independent trials could be used to find the EXACT number of derangements, and that's not true. I'm sorry, if it seems like I'm incredibly pedantic and take things literally, but that's just my training. Anything mathematical, I interpret literally, unless I'm told otherwise.

I think i showed that the trials were okay to be considered at 1/52 each time in a very simple way

I said they were independent "in this problem" meaning that for the purpose of deciding whether or not to accept 3:2 odds, they could be regarded as behaving independently. Since the series solution represents 1/e in the limit, and since the first few terms of this series are sufficient to compute our probability, then surely [(n-1)/n)]^n which also approaches 1/e very rapidly will also yield the same probability to our required degree of accuracy, hence the trials behave as though they are independent for our purpose. That is all I was noting.


If P(AB)-P(A)P(B) = 0, A and B are independent. If P(AB)-P(A)P(B) = .0000001, to a mathematician they are not independent. However, to a physicist or engineer, they are still highly independent depending on what one is attempting to measure. Having been trained in all three disciplines, I use whatever interpretation gives me the results I'm after.