K Trips to the Fair Casino

This problem is a reworked version of a problem which has been all the rage on one corner of the Internet in the last couple of weeks. I have taken the liberty of recasting it in gambling terms, but will reveal the original context shortly.

A man goes once a day to the Fair Casino to play roulette. The Fair Casino is unusual in that its perfectly balanced roulette wheel has no zeros, there are no table limits and credit is unlimited. (Everyone would go, but it has no comps.) The man always bets on red. He bets one unit on every spin until he wins, then he leaves for the day. He does this for k days.

Now the questions:
(1) [Extremely easy] What is his expected total number of winning bets?
(2) [Pretty darn easy] What is his expected total number of losing bets?
(3) [The tough one] What is his expected ratio across all k visits of losing bets to total bets?

Some advice: Don't jump to conclusions on (3). Try solving it for k=1.

1- x
2- x
3- 1/2

By "x," I assume you mean k, so you got (1) and (2) right (told you they were easy!) But you're wrong on 3. (Told you not to jump to conclusions.)

1*K
1*K
1/3*K

OK. (1) and (2) are right, but (3) can't be right. Are you saying that if k=4 the expected ratio of losing bets to all bets is 4/3? I'm pretty sure that value has to be between 0 and 1!

Quote: jfalk

What is his expected ratio across all k visits of losing bets to total bets?

Thanks for posting the question. Agreed, the answer is not 1/2. I know why, but will give others a chance to figure out why.

Quote: Wizard

Thanks for posting the question. Agreed, the answer is not 1/2. I know why, but will give others a chance to figure out why.

Isn't this problem reducible to a coin-flipping scenario, where every day you flip a coin until the first instance of heads? So legal outcomes might be
TH
H
TTTTTTTH
TTTH
TTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTTH
etc.

If so, the partition into days (or trips) is irrelevant and you're left with one long series of coin flips. Assuming a fair coin, the ratio of tails to total flips is 1/2 almost by definition. Maybe I misread the problem, but where does this analogy break down? Is it the assumption of a fair coin or, back to the problem, the assumption that p(red) = 1/2?

The coinflip problem is indeed the same problem, as is the one I will reveal later. The (partly counterintutive) result is indeed the fact that although p(red)=0.5, and the number of days would appear to be irrelevant, they are not.

Quote: jfalk

The coinflip problem is indeed the same problem, as is the one I will reveal later. The (partly counterintutive) result is indeed the fact that although p(red)=0.5, and the number of days would appear to be irrelevant, they are not.

Agreed. If I may add a further hint, focus on what specifically was asked for, the "expected ratio."

Quote: jfalk

The coinflip problem is indeed the same problem, as is the one I will reveal later. The (partly counterintutive) result is indeed the fact that although p(red)=0.5, and the number of days would appear to be irrelevant, they are not.

Okay, but if p(red)=0.5, and then p(black)=0.5, then the expected number of losses on the first day is SUM(n*0.5^(n+1)), which converges to 1 as n->+inf. So if there's one average losing spin, and exactly one winning spin, and no other spins, then the total number of spins on day 1 averages to 2. An average of 1 over an average of 2 is an average of 1/2, no?

Is this one of those problems where infinity screws everything up?

"An average of 1 over an average of 2 is an average of 1/2, no?"

No, or at least not necessarily. And infinity has nothing to do with it, except that it will bail you out as k --> infinity.

It might help you to do a much easier problem, but I hesitate to give that problem until some more people have tried to get the answer.

Ahh, got it. I was misinterpreting your original statement, but the Wizard's hint plus your suggestion that the number of days wasn't irrelevant got me there.

The expected ratio is 1 - ln(2).

Edit: The misunderstanding I was having was that (3) is asking for EV(# losses / # spins) rather than EV(# losses) / EV (# spins).

Excellent, for k=1. Now the really hard math comes from k=2,3,4... So we can turn to a somewhat easier problem with the same counterintuitive feature which should be easier for people to solve (as an alternative). The man goes in to the casino and plays either once or twice. If he wins on the first spin he leaves, but he leaves after the second spin no matter what. It is easy to show that E(Win)=K and E(loss)=K, as before. But what about E(losses/(wins+losses))?

Did I miss something?

Was the answer provided?

MathExtremist gave us the correct answer (though not the derivation) for k=1. For k=1 the answer is 1-ln(2), or about 30.7 percent.

Oh. I didn't realize that "ln" is a math function. Excel cleared that up for me.

This is getting into areas of math that makes my head explode.

For those who have been following thus far, the internet issue was not roulette wheels, or coin flips, but male-female births under a rule in which parents stopped having kids once they had a boy, but stylized to 50 percent and an infinite number of possible births. This problem was apparently used by Google in prospective employee testing with the "correct" answer of 50 percent, which is wrong except as the number of families approaches infinity. Economist Steven Landsburg is the one who started the fuss here . And the best answer for the k-family problem is here . The latter, DJTeddyBear, you should not look at, because I don't want to be responsible for any exploding heads.

What's really going on here? Going back to our roulette table, the decision to stay until you get a win slightly biases the ratio of wins to losses, since it removes all the cases in which you get only losses. It doesn't make you any money of course, and I don't want to encourage martingale strategies. But if someone tells you: "I have a fair bet, and I guarantee that if you follow my strategy you can expect winning bets almost 70 percent of the time" you might be tempted to pay him something for this strategy. But it doesn't work because while it does have a high ratio of winning bets to all bets, it also stops betting when it wins, leaving you no better off than you were before, on average.

Note that there is nothing magical about the "bet until you win once strategy." The same phenomenon applies if you use the strategy: bet until you win or have lost x times in a row. The numbers change slightly, and it is easy to program an Excel recursive algorithm to yield the expected ratio of losses to bets.

Quote: jfalk

A man goes once a day to the Fair Casino to play roulette. The Fair Casino is unusual in that its perfectly balanced roulette wheel has no zeros, there are no table limits and credit is unlimited. (Everyone would go, but it has no comps.)

I'm sure glad I printed this out. I've been hitchiking my way to Vegas to visit The Fair Casino but in about the fourth or fifth bar it happened: when I went in and put all my accumulated coins on the bar top so as to convert people's spare change to Guinness, someone saw the printout at told me the Fair Casino only exists in Mathland, not in Las Vegas. I should have known. No one ever has a goal of only winning one bet. And if he does have that goal, he never leaves after attaining it!

We here in Mathland realize that you people in "Las Vegas" (if it really exists) have trouble understanding our ways. Believe me, we have trouble with yours as well, what with your "Field bets" and "Keno." I promise not to make fun of Wayne Newton if you promise not to make fun of Danica McKellar, who performs nightly at the Fair Casino, located, by the way, on the famed Moebius Strip here in Mathland.

Quote: jfalk

what with your "Field bets" and "Keno."

Field bets? Sure. Its 2.76 isn't it or something like that? Roulette Bets too. Those are 5.28 or something house edge. Keno?? No way!!! some of those at 80 percent house edge. Even the video keno is way too much of a house edge for me to risk.

I tell you though. All these young guys who buy four yard-long drinks and walk the entire strip. I tried walking the Moebius Strip and it just kept going and going. I just couldn't find an end to it. Couldn't find any crosswalks to its other side either.

Amusing... you actually got me to look and see that field payoffs have actually changed since I learned to ignore them. OK. Substitute "Any 7." The main problem with the Moebius strip is that it's only half as edgy as the one in Vegas.

Quote: MathExtremist

Edit: The misunderstanding I was having was that (3) is asking for EV(# losses / # spins) rather than EV(# losses) / EV (# spins).

Having reviewed the other references, I think it's pretty clear that most of the argumentation on the topic has to do with precisely this misunderstanding. The vagaries of English are often sufficiently imprecise to lead to this sort of ambiguity. Similarly, using "and" in English tends to mean implication rather than conjunction; e.g. "I'm thirsty and I'm going skydiving" seems a non-sequitur if you expect skydiving to slake thirst.

(As for derivation, I just computed the first 50 terms or so in Excel. I know that doesn't get me anywhere toward k>1, but I'm far, far faster with Excel than with a pencil.)

Quote: MathExtremist

The misunderstanding I was having was that (3) is asking for EV(# losses / # spins) rather than EV(# losses) / EV (# spins).

If I remember correctly, the two will be the same if the variables are independently distributed.

Well, in this case #losses and #spins can't be independently distributed, since#spins explicitly depends on #losses (indeed, in this example it"s #losses+k). Even if they were independent, the two are not the same in general (It's impossible in this example, but suppose X=1 or 2 with prob 50 percent each and Y=0 or 1 with prob 50 percent each, indepedently. E(X)/E(Y) = 3, but E(X/Y) isn't even defined. Change Y to be .1 or 1.1 with prob 50 percent each, independently, and E(X)/E(Y) is still 3, but E(X/Y)= (10+.0909...+20+.1818...)/4=30.2727227.../4=7.568

To respond to MathExtremist (and to recapitulate a discussion I had offline with the Wizard) The English language has ambiguity, but if it's a choice between E(loss)/E(spins) and E(loss/spins), isn't the latter clearly more sensible? The only reasn, it seems to me, that we'd ever bother to calculate E(loss)/E(spins) is that it's much easier to calculate. Once we realize that the, because of nonindependence the two ought not be the same, why would anyone ever be interested in E(loss)/E(spins)?

To go back to the original birth example, what fraction of born children are expected to be female is about as unambiguous a question as can be asked, I think. And people who answer it by answering the question with the answer to the expected number of girls divided by the expected size of the population of children are just making an error in both math and the understanding of the English language. But I admit that there are few proofs where English language understanding is involved, so I have no evidence other than introspection to back up this assertion.

Quote: MathExtremist

...The expected ratio is 1 - ln(2)...

That's the answer for k =1, and 2 ln(2) - 1 is the answer for k =2.