Two Plus Two Older Archives Expectation, Independence, and Tipping (Not Poker)
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#1
01-08-2003, 06:52 PM
 RocketManJames Senior Member Join Date: Nov 2002 Posts: 118
Expectation, Independence, and Tipping (Not Poker)

I have a question about expectation and independence with regards to tipping (perhaps to a BJ dealer).

I have a friend who is a card counter. Let's assume for the sake of my question that he's a successful/profitable card counter. Let's assume that his expected return is 2% per dollar wagered.

From any session to the next his future expected return should be independent of the events from previous sessions. So, if he lost \$500 in one session, his expected return is still 2% per dollar wagered.

Now, let's say he plays through \$5000 and he nets \$200. He is \$100 ahead of schedule. So, what if he tips the dealer \$50 (say he's feeling ultra-generous, or it was the holidays and he felt like giving). In my opinion (again, I am NOT a mathematician, but I like to think about this stuff from time to time), it makes no real difference that he does that this one time. But what if he repeatedly gives these large tips a lot more than once. This is where my logic confuses me.

How will this affect his long run? Assume that he will give up 50% of anything in excess of his normal 2%. If he is at the expected 2% or less (win or lose), he tips nothing. My questions:

1) How does this affect his overall expectation?
2) Will he go broke, or better yet is there some tipping strategy such that he will go broke so long as he does not tip more than his expected return for any session?

Interested to hear your analyses.

Much appreciated,
RMJ
#2
01-08-2003, 09:18 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Expectation, Independence, and Tipping (Not Poker)

Your friend's 2% expectation is an average of two numbers: the amount above 2% that he wins on average when he wins over 2%, and the amount below 2% that he wins on average when he wins less than 2%. Note this last number is probably a negative number; that is he loses money on average if you only look at the times he wins less than 2%. These two numbers are equal distance from 2% in opposite directions. The size of these two numbers depends on his standard deviation (sigma) for each session. If you assume his session results follow a normal distribution, then of the times he wins over 2%, the average amount that he wins turns out to be 2% + .7734*sigma. The amount that he wins when he wins less than 2% is 2% - .7734*sigma which is probably negative or a loss. Now if he tips half of his winnings over 2% when he wins over 2%, then his average win when he wins will become 2% + .7734/2 sigma. So his overall expectation will be (2% -.7734 sigma + 2% + .7734/2 sigma)/2 = 2% -.7734/4 sigma. This is almost certainly a negative expectation as the following example shows:

Say he bets an average of \$1250/hr for a 4 hour session, for a total of \$5000/session. He expects to win 2% of this or \$25/hr or \$100/session. Let's give him an hourly standard deviation of \$500 which would be as low as you could ever expect; it could be twice this much. Then his standard deviation for the session would be sqrt(4 hours)*500 = \$1000. So when he wins more than \$100, on average he would actually win 100 + .7734*(1000) = \$873, and on average when he wins less than \$100 he would actually win \$100 - .7734*(1000) = -\$673, that is he would lose \$673. Now if he is going to tip (873-100)/2 when he wins, then his overall session expectation would be (100 + 773/2 - 673)/2 = -\$93.25. So instead of winning \$100 for the session, he loses almost that same amount! Note again that this will become an even larger negative expectation if his standard deviation is larger which it most likely is.

In order to avoid a negative expectation, your average tip cannot exceed your average win. If you only tip when you win an above average amount, then you can tip twice as much since you are only tipping half as often. This doesn't guarantee you won't go broke; you can go broke even if you don't tip at all depending on the size of your bankroll. This is just how to avoid a negative expectation. If you have a negative expectation you will go broke for sure.

You probably don't want to tip so much that you have a zero expectation; you probably want to make some money. Tipping 10% of your expected win would be more than generous (it's like a tithe) and you could accomplish this by tipping 20% of your AVERAGE SESSION EXPECTATION only when you win over average, NOT 20% OF YOUR WIN FOR THE SESSION. The problem is that if you only tip this amount when you win big, many dealers will be more insulted than if you didn't didn't tip at all. They only see that you won big, and they expect to be tipped like a waiter 15-20% of your win for the session or something. If that was the only blackjack you played all year this might make some sense, but they don't understand that you are counting on that big win to make up for all the smaller wins and losses that they don't see.
#3
01-08-2003, 10:24 PM
 Ray Zee Senior Member Join Date: Aug 2002 Location: montana usa Posts: 2,043
Re: Expectation, Independence, and Tipping (Not Poker)

first if you play well figure 1% not two at best. and if you tip enough that you may go broke from tipping you should do it as you dont deserve to have any money at all.

i have never seen a dealer offer a broke player money for a bowl of soup.
#4
01-08-2003, 11:52 PM
 BruceZ Senior Member Join Date: Sep 2002 Posts: 1,636
Re: Expectation, Independence, and Tipping (Not Poker)

Concerning the number I quoted above as .7734*sigma. I said this was how far above the average you are on average when you are above the average, or how far below the average you are when you are below the average. This is not quite accurate but it is very close. Actually this number is the "median" amount above or below the average. That is, when you are above the average, you will be at least this amount above exactly half the time. For some distributions the median and the average are the same, but in this case they are very slightly different. The actual average distance from the average turns out to be sigma*sqrt(2/pi) = .7979*sigma. This will make the tipping results a little worse. Remember awhile back we also talked about another type of average distance from the mean called the rms average which is exactly 1 sigma.

To illustrate why you will look like cheapskate even though you're not, consider in the above example that you could tip 10% of your expecation by tipping \$20 per session only when you win over 2%. Since half the time that you tip you will win over \$873, this will be a tip of less than 3% of your net win for the session, and you will likely hear "gee thanks" or some such thing.
#5
01-09-2003, 01:02 AM
 RocketManJames Senior Member Join Date: Nov 2002 Posts: 118
Re: Expectation, Independence, and Tipping (Not Poker)

Good point, Ray.

In reality, I have no such card counter friend and I'm not too interested in being a card counter. I just wanted to see how all the math worked out, and it made most sense to put this all in a "familiar" context.

Bruce:

Thanks for your lucid response. I think I absorbed what you said fully. The main thing that got me confused was the when I thought about it after isolating a single case.

For example, what if I tipped huge just once. In the long run, my overall expectation should not change given that I am a +EV player. Then I figured if I did it twice and only twice, again, it shouldn't change either. I guess if you attach a frequency of occurrence to the oversized-tipping, then you will ultimately affect your expectation. Please correct this view if you believe it is flawed.

So for any K times of oversized-tipping (where K is some constant), your long run expectation shouldn't change, right?

Thanks again. I apologize in advance if I simply did not get what you said.

-RMJ

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