Even Cooler Problem Involving e

[David Sklansky]

There is a bunch of woman at a party. You are not invited. But you have some enticements in your pocket that is guaranteed to get any one of them to go home with you.

Your plan is to approach one as she leaves. You know that they will leave one by one. You know how many ladies are at the party. You hope to pick the prettiest. Your strategy is to let a certain number of girls walk past you (all of whom you see perfectly) and then pick the first girl you consider prettier than any of the ones you let go by. Of course if the prettiest girl at the party is in the first group, you are out of luck, and are stuck with the last girl. (Once any girl passes she is gone.)

The question is "if the number of girls at the party is n, what are the chances you go home with the prettiest one?"

But before the mathmeticians here answer that question, please let the non mathmeticians ESTIMATE the answer to this second question:

If you were told the answer to the question when the number of girls at the party was 100, and you were asked to answer the question if the number of girls was 10,000, you would take the solution for 100 and divide it by about what? In other words how much more unlikely would it be that you would succeed with 10,000 vs 100 at the party?

[borisp]

So you are assuming that the first woman is rejected, and you are allowing for the possibility that your method will not select a woman? (For instance, they might leave in reverse order of attractiveness, in which case this event contributes zero to the overall probability.) Further, you are assuming that all of the women have distinct attractiveness levels?

So, in the case n=2, you get 1/2, since the second woman is either more or less attractive than the first. Is this correct? You are evaluating the women one by one? Or do you first pick some # that "must" go by, after which you pick the first one that exceeds the attractiveness of this group?

wiki link...beware this is a spoiler!

[NNNNOOOOONAN]

[ QUOTE ]
There is a bunch of woman at a party. You are not invited.

[/ QUOTE ]

haha, i never read this forum, but i guarantee this would NEVER happen.

silly DS. [img]/images/graemlins/cool.gif[/img]

[mbillie1]

As a non-mathematician, this depends highly on the average level of hotness. Because if the girls average about a 5/10, I'm letting a lot more go by. If the first 2 girls I see are knockouts, I'm probably taking the next hot piece of ass that walks out.

Also, how many beers have I had? Can I get some of these magic DS enticements? What Would Jesus Do? Is it acceptable to pick an ugly girl just to argue about religion with, but save the enticements for a hottie?

Mathematically I have no answer... I'd let 5-10 go by either way I suppose, because then my patience will have worn out and I'll want some ass already.

[bluesbassman]

I peeked at the solution to this stopping problem on wiki.

I assume that for both Mr. Sklanky's formulation and the "secretary" variant described on wiki, there is no known maximum theoretical upper bound on the value (in this case attractiveness) of a given candidate. That is, no matter how pretty a given woman is, the next one could be prettier. In other words, the domain of possible values upon which the candidates are judged is an open set. (Is this assumption correct?)

Suppose, however, we can define a finite upper bound? In Sklanky's example, suppose it is known the attractiveness of each woman can be assigned a value on the closed interval [1, 10]. (Non-integer values are allowed.) Thus, if a perfect 10 walks through the door, we can immediately stop.

Does this change the stopping problem solution? If so, how?

Edit: Actually, a bounded domain is not sufficient; it must also be closed. For example, if the attractiveness of the women can be assigned values on the open set (1, 10), then (I think) the published solution applies. (In that case a perfect 10 is excluded.) So modify my preceding question to consider domains which are both bounded and closed.

[jay_shark]

Cool problem .

We've had this discussion a few times in the probability forum and the answer is very neat .

[luckyme]

I'm guessing about 30% of the time and it doesn't matter how many girls. If you let 20-50% of them go by then take the 1st better-then-the-group one.

The times you haven't let the best slip through, you are protected by chances of the 2nd best being in the 1st group , or the 3rd best, etc which drags you into the winning 30% range.

very rough in-my-head calcs 'guess'.

luckyme

[PairTheBoard]

[ QUOTE ]
do you first pick some # that "must" go by, after which you pick the first one that exceeds the attractiveness of this group?


[/ QUOTE ]


[ QUOTE ]
DS -
Your strategy is to let a certain number of girls walk past you (all of whom you see perfectly) and then pick the first girl you consider prettier than any of the ones you let go by.

[/ QUOTE ]

I'm sure I've seen this before but I can't remember how it goes or what common sense type insight might cut through quickly to a solution. My first idea is to look at what happens if you let half the group go by before choosing the first prettiest. You then have half a chance of having a shot at the prettiest, but may still miss her. If I was ambitious I would grind through some math to calculate the probability exactly for this case. See how it goes, then do the same for a general cuttoff point and optimize. I'm not that ambitious though and it would be much more interesting if there is an intuitive approach that cuts quickly to a solution without a lot of math.

I wonder if there's a way to view this as somehow equivalent to the matching hats problem.

PairTheBoard

[joes28]

e= ecstasy?

[madnak]

[ QUOTE ]
I'm guessing about 30% of the time and it doesn't matter how many girls. If you let 20-50% of them go by then take the 1st better-then-the-group one.

The times you haven't let the best slip through, you are protected by chances of the 2nd best being in the 1st group , or the 3rd best, etc which drags you into the winning 30% range.

very rough in-my-head calcs 'guess'.

luckyme

[/ QUOTE ]

If you cut at the 50% mark (I think this is most efficient, but I can't prove it), then there's a 50% chance that the hot one is in the first 50% and you lose. But in the other 50% of cases where the hot one is in the second half, this means there's a >50% that the second-hottest chick is in the first half. And if the second-hottest is in the first half while the hottest is in the second half, then the hottest will be chosen.

Thus, there is a bare minimum 50%*50%=25% chance that you'll hit. But you'll also hit in many cases where the hottest and second-hottest are both in the latter group. So the chance is considerably higher than 25% but considerably lower than 50%. My answer in white below:

<font color="white">The title of this thread suggests that the answer has something to do with e. Since 1/e (~36% range) fits my observations, I'm guessing that as n approaches infinity, the success rate approaches 1/e.</font>