| Coin Flip Portfolio |
I was explaining the dozens of different Buy & Sell schemes over our morning coffee and my wife grunted: "Why not just flip a coin."
It was a rhetorical question, but it sounded like an interesting idea, so I ...
>Aw, c'mon You're not going to flip a coin.
That's exactly what I'm going to do.
I'll download ten years worth of monthly stock prices and see what happens if I start with, say $100K in Cash and flip a coin each month to decide whether I BUY or SELL.
If I SELL, I'm 100% Cash. When my coin says "BUY", I buy ... provided I have Cash on hand. If ...
>So you're either 100% in the stock or 100% Cash, right?
Exactly ... except you can stick your Cash into Money Market and give yourself some small interest rate.
The spreadsheet looks like this:
>Huh? Coin-Flip wins 76% of the time? You kidding?
Okay, this is what happens:
Each time you press F9 you get another bunch of 120 coin flips. ... because there are 120 months in 10 years.
BUT, if you click the Who Wins? button you'll get 100 sets of 120 coin flips and the number of times (out of 100 trials) that the Coin-Flip Portfolio wins.
That's where I get: Coin-Flip wins 76% of the time. See?
>Do you expect anybuddy to actually use that system?
It's cheaper than paying a manager, right? 
Actually, I think that's really funny ... and my wife is still laughing!
>But what about during a bull market?
Speak up! I can't hear you!
>And what about transaction fees?
Huh? What's that? 
| d'Alembert's Scheme |
Here's an interesting modification of the coin-toss-all-Cash-all-Stock ritual ... it's a betting scheme attributed to d'Alembert
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>d'Alem-who?
A French mathematician, physicist and philosopher. His betting scheme goes like this:
It says you should increase your bet when you lose, hoping to win the next time to cover your loss ... then decrease your bet when you win so you get to keep some of your winnings. |  d'Alembert Betting (starting at $20): Lose, Win, Lose, Lose, Win ... etc. |
Okay, here are some examples where X = +1 means a Win and X = -1 means a Loss. 
The POT is what you end up with after each bet, starting with a $20 bet and playing 8 times: See how it goes? To identify a win or a loss at the kth bet, we assume that X(k) = +1 if we Win and X(k) = -1 if we Lose. Then, starting with a first bet of B(1), our Bet#2 would be B(2) = B(1) - X(1) ... so we're decreasing (or increasing) our bet by X(1) when we Win (or Lose). Then Bet#3 would be: B(3) = B(2) - X(2) = B(1) - X(1) - X(2). After 2 Wins, we'd be betting B(1) - 2. After a Win and a Loss we'd be betting B(1) again. After 2 Losses we'd be betting B(1) + 2. Our winnings after 3Bets is:
For 4Bets. We have:
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Observations:
- The final value of our winnings does NOT depend of the order in which the wins and losses occur.
That is, Win, Win, Lose gives the same final POT as Win, Lose, Win or Lose, Win, Win. - For the 3Bets example, there are 4 possibilities: 3 Wins, 2 Wins, 1 Win and 0 Wins.
- The values of X(1), X(2) and X(3) which give these 4 possibilities are 1,1,1 and 1,1, -1 and 1, -1, -1 and -1, -1, -1:
- 3 Wins gives POT3 = 3 B(1) -3
2 Wins gives POT2 = B(1) + 1
1 Wins gives POT1 = -B(1) + 1
0 Wins gives POT0 = -3 B(1) - 3
The possibilities are illustrated below:
Note: By "POT" we're referring to our winnings after umpteen bets. If POT>0 we've made money.X Bet Win
LosePOT 1 20 20 20 1 19 19 39 1 18 18 57 X Bet Win
LosePOT 1 20 20 20 1 19 19 39 -1 18 -18 21 X Bet Win
LosePOT 1 20 20 20 -1 19 -19 1 -1 20 -20 -19 X Bet Win
LosePOT -1 20 -20 -20 -1 21 -21 -41 -1 22 -22 -63 - Note, however, the possibe POTs are not equally likely.
In a jillion games (each with 3 bets), 2 Wins and 1 Win will each occur 3/8 of the time whereas 3 Wins or 0 Wins will each occur 1/8 of the time. - In general, the frequency with which each Win/Loss sequence occurs is given by the Binomial Coefficients.
For 3Bets, they're: 1, 3, 3, 1 and 4Bets they're 1, 4, 6, 4, 1 and 5Bets they're 1, 5, 10, 10, 5, 1 - The "Expected" value of W(3) (in the 3Bet game) is then weighted by the frequency of occurrence and is proportional to:
3C3 POT3 + 3C2 POT2 + 3C1 POT1 + 3C0 POT0 where mCn are the Binomial Coefficients: mCn = m! / n! (m-n)! - Now look again at 4Bets. We have:
W(4) = B(1) [ X(1)+X(2)+X(3)+X(4)] - [ X(1) X(2) + X(1) X(3) + X(1) X(4) + X(2) X(3) + X(2) X(4) + X(3) X(4)] - 4 Wins gives POT4 = 4 B(1) - 6
3 Wins gives POT3 = 2B(1)
2 Wins gives POT2 = 2
1 Wins gives POT1 = -2B(1)
0 Wins gives POT0 = -4 B(1) - 6
The possibilities are illustrated below:X Bet Win
LosePOT 1 20 20 20 1 19 19 39 1 18 18 57 1 17 17 74 X Bet Win
LosePOT 1 20 20 20 1 19 19 39 1 18 18 57 -1 17 -17 40 X Bet Win
LosePOT 1 20 20 20 1 19 19 39 -1 18 -18 21 -1 19 -19 2 X Bet Win
LosePOT 1 20 20 20 -1 19 -19 1 -1 20 -20 -19 -1 21 -21 -40 X Bet Win
LosePOT -1 20 -20 -20 -1 21 -21 -41 -1 22 -22 -63 -1 23 -23 -86 - Note that, although the Binomial Coefficients are symmetrical, the final POTs are NOT!!
- In the 3Bets example, out of a jillion trials, we'd expect to get POT3, POT2, POT1 and POT0 with a relative frequency of
1, 3, 3, 1.
That'd give an "Expected" POT of (1/8) [3 B(1) -3] + (3/8) [B(1) + 1] + (3/8) [-B(1) + 1] + (1/8) [-3 B(1) - 3] = 0. - In the 4Bets example, we'd expect to get POT4, POT3, POT2, POT1 and POT0 with a relative frequency of
1, 4, 6, 4, 1.
That'd give an "Expected" POT of (1/16) [4 B(1) - 6] + (4/16) [2B(1)] + (6/16) [2] + (4/16) [-2B(1)] + (1/16) [-4 B(1) - 6] = 0.
Don't you see? In d'Alembert, you're expected to break even.
>Huh? Then why would anybuddy use that scheme?
To enjoy the game without losing a fortune, eh? And who knows? For short periods you could win. 
You can try the spreadsheet:
I might mention the following:
d'Alembert was a great mathematician who turned his back on Theology.
He is also known for incorrectly arguing that the probability of a coin landing heads increased every time that it came up tails.
I imagine the following discussion:
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