For example, if the deck has 26 red cards on top, you'd end up dwindling your initial $1.00 stake to 0.000000134 before riding it back up to 9.08
If you start with $35,522.07 or less, you will lose it all after 26 incorrect cards.
⊢min←⌊⌿ratio←×⍀{1-d×0⍪¯1↓(+⍀d←¯1+2×⍵)÷⌽⍳≢⍵}52↑26⍴1
1.353223554704754E¯7
⌈÷min
7389762
⌊(⌈÷min)×⊃⌽ratio
67108864
Note: above code is Dyalog APL.
So another reading is “choose to give your children wealthy parents”.
This is what most people discover, you need to play like every toss of the coin(i.e tosses over a very long periods of time). In series, like the whole strategy for it to work as is. You can't miss a toss. If you do you basically are missing out on either series of profitable tosses, or that one toss where you make a good return. If you draw the price vs time chart, like a renko chart you pretty much see a how any chart for any instrument would look.
Here is the catch. In the real world stock/crypto/forex trading scenario that means you basically have to take nearly trade. Other wise the strategy doesn't work as good.
The deal about tossing coins to conduct this experiment is you don't change the coin during the experiment. You don't skip tosses, you don't change anything at all. While you are trading all this means- You can't change the stock that you are trading(Else you would be missing those phases where the instruments perform well, and will likely keep landing into situations with other instruments where its performing bad), you can't miss trades, and of course you have to keep at these for very long periods of time to work.
Needless to say this is not for insanely consistent. Doing this day after day can also be draining on your mental and physical health, where there is money there is stress. You can't do this for long basically.
You must always practice in real world conditions. Notice in the experiments conducted in programs, you are taking series of tosses as they come, even if they are in thousands in numbers, one after the other, without missing a single one. Unless you can repeat this in a live scenario. This is not a very useful strategy.
Kelly criterion is for people who are planning to take large number of trades over a long period of time, hence the idea is to ensure failures are not fatal(this is what ensures you can play for long). As it turns out if you play for really long, even with a small edge, small wins/profits tend to add to something big.
If you remove all the math behind it, its just this. If you have a small edge to win in a game of bets, find how much you can bet such that you don't lose your capital. If you play this game for long, like really really long, you are likely to make big wins.
Now, b) is false. You can change the code to extract 3 random numbers each time, discard the first 2 and only consider the third one, the results won't change.
Instead a) is generally true. In this case, the Kelly strategy is the best strategy to play a great number of repeated games. You could play some games with another strategy and win more money, but you'll find that you can't beat Kelly in the long term, ideally when the repetitions approach infinity.
Might be in theory. In practice, this is rarely true.
Take for example in trading. What happens(is about to happen), depends on what just happened. A stock could over bought/over sold, range bound, moving in a specific direction etc. This decides whats about to happen next. Reality is rarely ever random.
Im sure if you study a coin toss for example, you can find similar patterns, for eg- if you have tired thumb, Im pretty sure it effects the height of the toss, effecting results.
>>Instead a) is generally true. In this case, the Kelly strategy is the best strategy to play a great number of repeated games.
Indeed. But do make it a point to repeat exact sequences of events you practiced.
A person tosses a coin, so tosses are are connected to each other.
Ask yourself this question- Would your thumb hurt if you toss a coin 5000 times? If so, would that change the results?
https://www.stat.berkeley.edu/~aldous/157/Papers/diaconis_co...
The paper does discuss coins allowed to land on a hard surface; it is clear that this will affect the randomness, but not clear if it increases or decreases randomness, and suggests further research is needed.
The machine they use to toss the coin has a spring, and Im sure the spring tension varies through time effecting results.
I think the portfolio argument is an unnecessary detour though. There's a two-line proof by induction.
1. The payoff in the base case of (0,1) or (1,0) is 2.
2. If we are at (r,b), r >=b , have $X, and stake (r-b)/(r+b) on red, the payoff if we draw red and win is X * (1+(r-b)/(r+b)) * 2^(r+b-1) / (r+b-1 choose r-1) = X * 2^(r+b) * r / ((r+b) * (r+b-1 choose r-1)) = X * 2^(r+b) / (r+b choose r).
Similarly, if we draw black and lose, the payoff is X * (1-(r-b)/(r+b)) * 2^(r+b-1) / (r+b-1 choose r) = X * 2^(r+b) * b / ((r+b) * (r+b-1 choose r)) = X * 2^(r+b) / (r+b choose r). QED
In probability theory, Proebsting's paradox is an argument that appears to show that the Kelly criterion can lead to ruin. Although it can be resolved mathematically, it raises some interesting issues about the practical application of Kelly, especially in investing. It was named and first discussed by Edward O. Thorp in 2008.[1] The paradox was named for Todd Proebsting, its creator.
That's good to know. Kelly is good if you know the probabilities AND they don't change.
If you don't know or if they can change, I expect the right approach has to be more complex than the Kelly one.
For instance, casinos have different payout schedules for Blackjack based on minimum bet size and number of decks in the shoe. Payouts for single deck Blackjack are very small in comparison to multi-deck games, as well as requiring larger minimums (and they shuffle the deck after only a few hands).
As for the actual probability being different from the expected probability, that's not too difficult to account for. Just set up a distribution (more or less generous depending on your risk tolerance) for where you believe the actual probability may lie, and work out the integrals as necessary, recalling that you want to maximize expected log-value. It's not the trivial Kelly formula, but it's exactly the same principle in the end.
Almost all people can achieve at least 20 point accuracy with a few hours of training. Unknown probabilities are not so problematic as people make them out to be.
Probabilities are literally measures of uncertainty. It's okay for them to be uncertain. They always are.
Also, your estimate of the true probability doesn't have to be that exact, if your edge is big enough to begin with. Once I made great profits (betting with fake internet points) just by naively taking the sample proportion from a small sample. In fact, the event turned out to be more 'streaky' than a weighted coin flip, but it didn't matter, since everyone else there was betting on vibes.
In any case, it's not like there's just the trivial Kelly formula, and woe to you if its toy model doesn't apply precisely to your situation. It's a general principle that can be adapted to all sorts of scenarios.
Only quibble i have is that black should be +$1 and red -$1 to follow standard finance conventions, i.e. be in the "black" or "red".
Incorrect. https://entropicthoughts.com/the-misunderstood-kelly-criteri...
The Kelly criterion generalises just fine to continuous, simultaneous, complicated allocations.
All it takes is a list of actions which we are choosing from (and these can be compound actions with continuous outcomes) and the joint probability distribution of wealth outcomes after each action.
The Kelly criterion seems excellent for many forms of gambling, but poker seems like it could be an exception: in poker, you’re playing against other players, so the utility of a given distribution of chips seems like it ought to be more complicated than just the number of chips you have.
(I’m not a poker player.)
So, if one's skill would turn your session probability to +EV, by limiting your losses and using the fact that in poker the strongest hands or better tourney positions would give you a huge ROI, it would be just a matter of time and discipline to get to a good bankroll.
Just remember that for the better part of this challenge he was averaging US$ 0.14/hour, and it took more than 9 months.
[1] https://www.thehendonmob.com/poker_tips/starting_from_zero_b...
But consider the rate of return! He turned $1 into $10,000 in 9 months. Could he then turn that $10k into $100M in another 9 months?
Or if he'd started with $100 instead of $1, could he have turned that into $1M in 9 months? That would still be incredibly impressive.
Certainly the game changes as the bets and buy-ins get higher, but even if he couldn't swing the same rate of return with a higher starting point and larger bets (though still only betting that same certain low percent of his bankroll), presumably he could do things like turning $5k into $1M. Even $100k into $1M would be fantastic.
Actually, the main assumption that leads to the Kelly criterion is that you will have future opportunities to bet with the same edge, not constrained by the amount.
For example, if you knew this was your last profitable betting opportunity, to maximise your expected value you should bet your entire stake.
I'm slightly surprised it leads to such a nice result for this game - I don't see a claim that this is the optimal strategy for maximizing EV zero variance is great, but having more money is also great.
Of course you are right about roulette and, if you are playing standard casino roulette against the house, the optimal strategy is not to play. But that's not because bets are uncorrelated, it's because they are all negative value.
Not the same edge -- any edge! And this condition of new betting opportunities arriving every now and then is a fairly accurate description of life, even if you walk out of the casino.
https://en.wikipedia.org/wiki/TRS-80_Model_100
to simulate it and it never failed. Recently I thought about it again and wrote a Python script that tried it 30 million times and... it never failed.
I've been thinking about what to do with it and came up with the options of (i) a prop bet and (ii) a magic trick, neither of which seemed that promising.
As a prop bet I can offer $1000 to somebody's $10 which is not the route to great prop bet profits, also I worry that if I make a mistake or get cheated somehow I could be out a lot of money. (Now that I think of it maybe it is better if I re-organize it as a parlay bet)
As a magic trick it is just too slow paced. I developed a patter to the effect that "Parapsychologists were never able to reliably demonstrate precognition with their fancy Zener cards, but I just developed a protocol where you can prove it every time!" but came to the conclusion that it was not entertaining enough. It takes a while to go through a deck which doesn't seem like a miracle, you will have to do it 7 times in a row to exclude the null hypothesis at p=0.01. Maybe somebody with more showmanship could do it but I gave up.
https://en.m.wikipedia.org/wiki/Boyer%E2%80%93Moore_majority...
The one flaw in the magic is that "2nd pass to verify" is a significant cost, transforming the algorithm from online streaming O(1) space to O(n) collection-storage space.
There are enough possible card sequences that you might be concerned about your source of pseudorandomness failing to exhaust the space. Simulations can give you very misleading results when that happens.
Even if you do have enough entropy, 30 million trials is definitely not enough.
Turn 1 r = b so no bet
Turn 2 bet 1/3 on whichever card wasn't revealed in turn 1.
Turn 3 either you were wrong on turn 2 and you now have 2/3 of your stake but you know the colour of the next two cards so you can double your stake each time to end up with 4/3 after turn 3 or you were right and you have 4/3 of your stake but have one of each red or black left so you don't bet this turn.
Turn 4 you know the colour of the final card so you double your money to 8/3 of your original stake.
And then the exercise to the reader is to prove optimality (which is fairly straightforward but I don't believe there is a short proof)
I was thinking these are very different strategies, but they’re not exactly. The Kelly strategy does the same thing when there’s only one color left. The difference is this strategy does nothing before that point.
Still, they feel like limit cases. Betting it all with only one color left is the only right move, so it’s what you do before that. Nothing and Kelly seem like the only good strategies.
It would be interesting to do the math and show why they're equal. It seems like you should be able to make the same sort of portfolio probability argument.
I guess the number of possible arrangements of cards with N of one color remaining is... The number of permutations of N times 2 times the number of permutations of 52 minus N times 26 choose N?
Ah, yes this works, you can see it here: https://www.wolframalpha.com/input?i=%28summation+of+N%21+*+....
That is: (summation of N! * (52 - N)!* (26 choose N) * 2^N/52! from N=0 to 26 (for some reason the * 2 for different suits was over counting, so I removed it. Not sure why? Also it seems like it should be from 1 to 26, but that also doesn't give the right answer, so something is whack)
Do we know if the Kelly strategy is optimal here?
If you are willing to take some risk in exchange for possibility of higher payout just bet a bit more then Kelly recommends. That's your "optimal" strategy for the amount of risk you are willing to take. I imagine it's expected return is the same as Kelly and calculating it's variance is left as the exercise for the reader.
I imagine it's expected return is the same as Kelly
Accepting higher variance with no increase in expected return has a name: gambling.
I don’t recall this specific example, but I learned about the Kelly criterion in a class that Thomas Cover taught. He was one of my favorite teachers, and any discussion with him was guaranteed to be interesting and worthwhile. RIP.
It shouldn't be a problem because the RNG is advanced each run. Might save someone a check though.
After making a bet, you gain information about the contents of the rest of the deck of cards. I could see it being possible to do better by pricing in that information into your bet.
Original paper by Thomas Cover: https://isl.stanford.edu/~cover/papers/paper93.pdf
A good breakdown with links to code examples by Andy Jones: https://andrewcharlesjones.github.io/journal/universal-portf...
What is your bankroll? Cash on hand? Total net worth? Liquid net work? Future earned income?
Depending on the size of your bankroll, a number of factors come in to play. For example, if your bankroll is $100 and you lose it all it's typically not a big deal. If you have a $1 million bankroll, then you are likely more adverse to risking it.
What is the expected value? Is it known? Is it stationary? Is the game honest?
Depending on the statistical profile of your expected value, you are going to have to make significant adjustments to how you approach bet sizing. In domains where you can only estimate your EV, and which are rife with cheats (e.g. poker), you need to size your wagers under significant uncertainty.
What bet sizes are available?
In practice, you won't have a continuous range of bet sizes you can make. You will typically have discrete bet sizes within a fixed range, say $5-$500 in increments of $5 or $25. If your bankroll falls to low you will be shut out of the game. If your bankroll gets too high, you will no longer be able to maximize your returns.
At the end of the day, professional gamblers are often wagering at half-kelly, or even at quarter-kelly, due in large part to all these complexities and others.
You may also be required to pay for the privilege of placing a bet (spread and commissions in trading; the rake at a casino table).
9.08 ~
52/52 × 52/51 × 50/50 ÷ 50/49 × ... 2/2 × 2/1
= 52/51 × 50/49 × ... × 2/1
= 2^52 × 26!² / 52!
= (52/52 × 50/51 × ... × 2/27) × (52/26 × 50/25 × ... × 2/1)
This also points to a non-"many worlds"/portfolio version of the prod of zero-variance.
Every bet is e/d, where e is current edge and d is current deck size. So every outcome multiplies the stack by (d + e × (-1)^i)/d, where is ±1, depending on win or lose.
Note that the product of all the values of d is constant, so we can ignore the denominator.
Since we know (from the OP proof) that the product of these numbers is constant for all shuffles of the deck, we can split a shuffled deck anywhere such that both parts are balanced red=blue, and the total (multiplicative) return over each part of the deck is constant across all shuffling of that part of the deck. (There are at least two ways to prove this part!)
This is gives a further hint toward another fascinating fact: over any span of the deck between points where the deck is balanced, the numerators of the bet results double-cover all the even numbers between the starting and ending deck size.
To see why:
* A loss after a loss has a numerator (deck minus edge) of 2 less than the previous bet, as the deck size decreased by 1 and the edge has inccreased by 1.
* A win after a win also has a numerator (deck plus edge) of 2 less than the previous bet, as the deck size decreased by 1 and the edge has decreased by 1.
* A win after a loss, causes a big swing in the numerator, exactly back to the largest not yet double-covered numerator that started the streak that just ended. Then the new win streak continues making the second cover of even numerators, until... a loss after a win jumps the numerator back to continuing the sequence of decreasing even numberators, which will get their second cover later when the later wins come.
Since the deck is balanced, the number of wins always equals the number of losses, as long as we consider the 0 wager on a balanced subdeck to be a loss, since it increases the edge like non-degenerate losses do.
(When the deck is balanced, edge is 0, so the return of no-bet is same as a win is same as a loss)
You can visualize the numerator changes like so: a crane is driving from 52 to 0. Its arm is pointing either forward or backward, and there is a counterweight of the same length pointing in the opposite direction. At each step, the crane arm is either pointing toward 0 and stretches another step toward 0, or points backward to 52 and shrinks (toward 0 milestone and toward 0 arm length), or it swings to the other direction. Whenever the crane stretches toward 0, the counterweight stretches backward, its end not moving relative to the ground.
Because the deck is balanced at start and empty deck is balanced, the crane starts and ends with a 0-stretch arm. The front side is either the frame arm stepping 2 steps forward at a time relative to the ground, or holding still while the backside crane arm shrinks closer, and the crane arm occasionally flips back and forth pointing forward or ackward. And vice versa for the counterweight.
Over the course of the drive, the crane arm end reaches every even milestone once pointing forward and once again pointing backward.
When the deck has d cards left, it is sensible to make d bets of 1/d your stack, where each bet is that one specific card is next. If there are r reds and b=r+e blues, r of these bets simply cancel out r other bets, leaving e (times 1/d) remaining to be a nontrivial bet.
Eg. it's probably pretty easy to convince them that with 15 cards in a deck, out of which 5 are red and 10 are black, chances are bigger (and in particular 10/15 or ~67%) that they'll pull out a black card, and that you should bet more on this happening. If you happen to miss, you should only bet even more on black since the chances grow further — to be able to maintain this strategy, you only need to never bet too much so you have enough "funds" to bet all the way through (eg. in the worst case where the least likely thing happens: in my example, that would be 5 red cards coming up first).
Putting all this reasoning into formulae is what math is, and I do believe some struggle with abstracting these more than others (which is why the divide does exist and why many people believe those good at math are "smart", which is very much not so — seen plenty of "stupid" mathematicians, even professors). Does not make them "dumb", but might make them "modern math dumb". A signal that someone can be good at math today is that they are unfazed with more-than-3-dimensional spaces (you need to stop tying things to physical world).