The following is a very interesting problem copied from p.132 of "Elements of Information Theory" by Cover and Thomas. I've been reading information theory recently, and I'm finding it very fascinating.
Example 6.3.1 (Red and Black):
In this example, cards replace horses and the outcomes become more predictable as time goes on.
Consider the case of betting on the color of the next card in a deck of
26 red and 26 black cards. Bets are placed on whether the next card will
be red or black, as we go through the deck. We also assume the game
pays a-for-l, that is, the gambler gets back twice what he bets on the
right color. These are fair odds if red and black are equally probable.
We consider two alternative betting schemes:
1. If we bet sequentially, we can calculate the conditional probability of the next card and bet proportionally. Thus we should beton (red, black) for the first card, and
for the second card, if the first card is black, etc.
2. Alternatively, we can bet on the entire sequence of 52 cards at once. There arepossible sequences of 26 red and 26 black cards, all of them equally likely. Thus proportional betting implies that we put
of our money on each of these sequences and let each bet “ride.”
We will argue that these procedures are equivalent. For example, half the sequences of 52 cards start with red, and so the proportion of money bet on sequences that start with red in scheme 2 is also one half, agreeing with the proportion used in the first scheme. In general, we can verify that bettingof the money on each of the possible outcomes will at each stage give bets that are proportional to the probability of red and black at that stage. Since we bet
of the wealth on each possible outtut sequence, and a bet on a sequence increases wealth by a factor of 2^52 on the observed sequence and 0 on all the others, the resulting wealth is:
Rather interestingly, the return does not depend on the actual sequence. This is like the AEP in that the return is the same for all sequences. All sequences are typical in this sense.
It's amazing that the strategy which bets everything at the start can beat one which bets after seeing which cards are no longer in the deck. The calculation of resulting wealth of the 2nd betting scheme, which results in 9.08, is transparent but I didn't see how it could be exactly equal to the resulting wealth of the other betting scheme.
Here is an excel spreadsheet I made to simulate the first betting scheme: excel 2007 | older 2003, (functions in 2003 version may not work). On the first sheet, it randomly generates a sequence of draws from the deck, according to actual probabilities, and calculates the wealth you would have after the deck runs out. You can see the resulting wealth at the bottom (cell H54) of the "wealth" column- it's always 9.08. It's very surprising to me and I don't fully understand how the two results can match. I'm new to the field though.
Tell me if you have any insight or a similar example that makes the paradox less paradoxical.





6 comments:
Hi Max,
"The calculation of resulting wealth of the 2nd betting scheme, which results in 9.08, is transparent but I didn't see how it could be exactly equal to the resulting wealth of the other betting scheme."
The reason that the two schemes produce the same results is because they are really one and the same strategy. In the second betting scheme, half of the C(52,26) possible sequences start with red cards, so we are effectively betting 1/2 on the first card being red. Suppose that the first card we draw is black, then we lose on all the sequences that start with red, but we double our money on those sequences that start with black. So we double the 1/2 and end up with a total of 1 after the first card. There are now C(51, 26) sequences left, each with the same probability of happening. Among these sequences there are 26*C(50,25) with second card being red. A quick calculation shows that 26*C(50,25)/C(51,26) = 26/51. So we are effectively betting 26/51 on the second card being red (so 25/51 on black). Similarly for the third card and on, there is really no difference between the first scheme and the second scheme.
In fact, one can bet 1 on one particular sequence. The expectation of the resulting wealth is still 2^52/C(52,26) = 9.08. The two betting schemes you described are special in that every possible outcome equals the expectation.
"It's amazing that the strategy which bets everything at the start can beat one which bets after seeing which cards are no longer in the deck."
The two schemes both use the useful information that we are drawing (without replacement) from a deck of 52 cards with 26 red/black. That's why we end up making money on our bets.
How come we couldn't beat the second scheme after seeing which cards are no longer in the deck? My explanation is that knowing which cards are out is useless information to us. Consider the situation that we have $k and there are r red cards and b black cards left in the deck. To maximize our expectation of final outcome, should we bet on the next card being red or black? When either r or b equals 0, there is only one viable strategy. When r and b are both different from 0, it turns out it doesn’t matter what we do on the next card - we can bet all $k on the next card being black, or all $k on the next card being red, or any combination – the expectation of the final outcome is the same. This can be proved using induction on r+b.
Xun
Xun,
"My explanation is that knowing which cards are out is useless information to us. ... When r and b are both different from 0, it turns out it doesn’t matter what we do on the next card - we can bet all $k on the next card being black, or all $k on the next card being red, or any combination – the expectation of the final outcome is the same. This can be proved using induction on r+b."
So if there is one red card left and two black, and you bet all on red it has the same expectation as if you bet all on black? (1/3)*2+(2/3)*0 ?= (2/3)*2+(1/3)*0
Regards,
Max
Max,
"So if there is one red card left and two black, and you bet all on red it has the same expectation as if you bet all on black? (1/3)*2+(2/3)*0 ?= (2/3)*2+(1/3)*0"
Let E(r,b) = expectation with r red cards and b black cards. Then it should be
(1/3)*2*E(0,2)+(2/3)*0 = (2/3)*2*E(1,1)+(1/3)*0
The equality holds because E(0,2)=4 and E(1,1)=2.
Xun
Xun,
I see that the expectations are the same; I misinterpreted your original claim.
The "paradox" has become clearer to me as I've been thinking about the definition of (x choose k) over the past few days. The factorial division (x!)/((x-k)!(k!)) is exactly what you get if you write it in terms of sequential bets using conditional probabilities (i.e. like P(A|B)).
I guess I only saw it as a paradox because I hadn't been taught the combination formula, (n choose k), as a parallel to the sequential conditional probability approach. Mathematically they're equivalent though.
Regards,
Max
you may enjoy Ernest Chen's working out the kelly approximation based on the Sharpe formula as an input. He uses a log approximation and of course assumes stability in vol, but a nice starting place. http://nickgogerty.typepad.com/designing_better_futures/2009/03/quantitative-trading-ernest-p-chans-new-book-on-trading-is-extremely-useful.html
Thanks
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