Optimal Mean Reversion Betting Strategy?

A coin is flipped 1000 times, resulting in 500 Heads and 500 Tails (for the sake of this question). What would the optimal betting strategy be for this scenario? I assume Kelly, but the ROR is so low. I'm wondering if a more aggressive betting approach with a higher risk of ruin would be more optimal.

Obviously there is no ROR for any betting system over the long-term, but I am stumped for the best answer to this problem. Size of bankroll and table mins/max are secondary factors and can be whatever one chooses.

Quote: hawkcatcher

A coin is flipped 1000 times, resulting in 500 Heads and 500 Tails (for the sake of this question). What would the optimal betting strategy be for this scenario? I assume Kelly, but the ROR is so low. I'm wondering if a more aggressive betting approach with a higher risk of ruin would be more optimal.

Obviously there is no ROR for any betting system over the long-term, but I am stumped for the best answer to this problem. Size of bankroll and table mins/max are secondary factors and can be whatever one chooses.

Is your goal to break even?

The goal is to earn the highest rate of return without busting by utilizing the knowledge that the coin flips will revert perfectly to the mean.

Quote: hawkcatcher

The goal is to earn the highest rate of return without busting by utilizing the knowledge that the coin flips will revert perfectly to the mean.

Is it guaranteed to happen within 1000 flips?

Yes. Knowing that the coin will end up 50-50 over 1000 flips, what would be the best way to take advantage of it?

What kind of coin is this? A magical one which counts how often it has landed each way and frantically plays catchup as it approaches 1000?

Here's the absolute optimal strategy and it has nothing to do with Kelly and the risk of ruin is absolutely zero....
Sit out and watch the first 999 tosses and keep very careful note of the result, meanwhile arrange to sell or mortgage all your possessions. Then stake everything you own in the world on that last toss. You are absolutely guaranteed to double your wealth :)
You may not have to wait for the last coin toss in the situation where maybe it has landed 490 times heads and 500 times tails, where the prudent strategy is to bet everything you own in the world on those last 10 tosses, doubling up each time!!!! Because, again we know it will have to end with 500 heads and 500 tails. When you come away with the biggest fortune ever won in history, send me a personal thank you message and I'll let you know where to send my thank you gift.

Seriously. Reversion to the mean does not work like that. It works like this:-
https://wizardofvegas.com/forum/questions-and-answers/math/2709-roulette-question/2/#post376699

Or to explain more in context of your original post:

Let's say for the sake of interest that after 100 tosses we have seen 47 heads and 53 tails: Not unreasonable.
That's 47% heads.

Lets say that the next 400 tosses behave perfectly with 200 heads and 200 tails. Rather unreasonably optimistic but still...
That's a total of 247 heads to 253 tails or 247/500=49.4% Heads. Hey we are reverting to the mean of 50%. Just 3 tails too many still.

Now let's say that by some quirk of fate the next 500 tosses are again perfectly even: 250 Heads and 250 Tails.
That's a total of 497 Heads and 503 Tails (Actually pretty unlikely) = 49.7% Heads: Getting still closer to the 50% mean. Still 3 tails too many. Dammit.

That 49.7% is far closer to 50% than the 49.4% when we were 500 tosses elapsed and much much closer than the 47% that we had at 100 tosses. We have been steadily reverting to the mean while always exactly 3 heads short of expectation.

Now let's say for the sake of interest that we toss another 9000 times and by some miracle we get 4500 heads and 4500 tails.
That's a total of 4500+497 Heads in 10,000 tosses. That's 4997/10000 = 49.97% Heads: Really getting close to 50% now. damn those pesky 3 extra tails.

Let's go crazy: Let's toss another 990,000 times where by some absolute miracle we had 50% heads exactly.
That's a total of 495,000 + 4500 + 497 Heads in 1,000,000 tosses. That's 499,997/1,000,000 = 49.9997% Heads.

49.9997% is very much closer to the the 50% mean than the 49.97% result after 10,000 tosses or the 49.7% after 500 tosses and very very much closer than the crappy 47% after 100 tosses.

Clearly we are reverting quickly to a mean of 50%. But we were always 3 heads behind. The numerical disparity stayed there even as the percentage disparity faded away to next to nothing.

And all this ignores the fact that our numerical disparity was likely to increase in one direction or another as we tossed more and more often. We could easily have reached 499900 heads and 500100 tails after our million tosses. That's a numerical disparity of 100 extra tails but the percentage heads is still 49.99% which is closer to 50% than we had at 100 or 1000 tosses with only 3 extra tails.

Real coins don't have memories. 50:50 chances will revert towards 50:50 and get closer and closer to 50.00000:50.00000 as each coin toss happens. There is no reason to suspect that exactly 50:50 will be the end point of any predetermined sequence of tosses.

Exactly 50:50 may occur after some arbitrary count of coin toss events, but it's equally likely that it never will (I figure a 50% chance of ever passing exactly through 50%, over the life of the coin, but I leave that as a challenge to prove or disprove)

What kind of coin is this? A magical one which counts how often it has landed each way and frantically plays catchup as it approaches 1000?

Yes, we are ignoring the HA and the gambler's fallacy and the obvious fact that 1000 coin flips won't yield perfect 50-50 results for simplicity sake.

There are applications to the question. A rough example is a NFL Vegas League where you might choose certain middle of the pack teams to go 8-8 versus the spread throughout. We don't know ahead of time how the distribution will end up, but the question remains how to devise a strategy to maximize ROR if the (theoretical) endpoint is a 50-50 distribution.

Quote: hawkcatcher

.. we are ignoring...the obvious fact that 1000 coin flips won't yield perfect 50-50 results for simplicity sake.

That was a central assumption in your original question. Now you ask a different question entirely.

That is still the central assumption - a 50-50 distribution.

If you get paid even money, then always bet on whichever is more likely to come up (whichever has been rolled a lesser amount of times). Essentially, you have a bag full of marbles, 500 labelled "heads" and the other 500 labelled "tails".

Say there have been 50 tails and 60 heads with 890 remaining flips, 450 ways to win (bet tails) and 440 ways to lose (440). Your advantage is 10/890, or ~1.1% advantage if you bet tails. Bet 1% of your BR on tails.

Quote: RS

...Essentially, you have a bag full of marbles, 500 labelled "heads" and the other 500 labelled "tails".
Say there have been 50 tails and 60 heads with 890 remaining flips, 450 ways to win (bet tails) and 440 ways to lose (440). Your advantage is 10/890, or ~1.1% advantage if you bet tails. Bet 1% of your BR on tails.

Comparing back to the original coin tossing analogy, the marble gets returned to the bag after each toss (or draw). So there's no advantage. Though RS's analogy works where we know that the next toss (or draw) is one step closer to exact 50:50 parity. On the assertion of the original model where 50:50 was the defined outcome, I still say sit back and wait till 500 heads or 500 tails have been seen. Then bet the house :)

The question needs to be defined before the answer can be worked out. It's either exactly defined 50:50 at 1000 events or it isn't. The events either do, or do not depend on prior events. And why would any external agency or bookie give even money after a number of events have shown the outcome to have a player advantage?

It seems to me that the optimal strategy would be to wait until there have been either 500 heads or 500 tails, then bet everything you can on the other for the remaining flips. Is there a betting limit involved?

Exactly Don. Or as I put it

Quote: OnceDear

Sit out and watch the first 999 tosses and keep very careful note of the result, meanwhile arrange to sell or mortgage all your possessions. Then stake everything you own in the world on that last toss. You are absolutely guaranteed to double your wealth :)
You may not have to wait for the last coin toss in the situation where maybe it has landed 490 times heads and 500 times tails, where the prudent strategy is to bet everything you own in the world on those last 10 tosses, doubling up each time!!!!

The OP came up with an analogy where we had 1000 coin tosses ending in exactly 500 heads and 500 tails. That was a given and when asked on that point reconfirmed it twice. Later he came back with a totally different analogy betting on sports events. He needs to define the question again or completely accept our correct answers to the question originally asked.

Quote: OnceDear

Exactly Don. Or as I put it

The OP came up with an analogy where we had 1000 coin tosses ending in exactly 500 heads and 500 tails. That was a given and when asked on that point reconfirmed it twice. Later he came back with a totally different analogy betting on sports events. He needs to define the question again or completely accept our correct answers to the question originally asked.

I don't think so. I think you misread his statement that "we are ignoring the obvious fact that 1000 coin flips won't yield perfect 50-50 results"; he wants us to assume that, "for simplicity's sake," the 1000 coin flips will yield 500 heads and 500 tails.

In that case, our solution - bet nothing until 500 heads or 500 tails come up, and then bet as much as possible on the other result for the remaining flips - is correct.

In fact, if the limitation of 500 heads and 500 tails applies to any 1000 consecutive flips, then once 500 of either are observed, we can determine the result of every flip from that point on in advance.
Proof:
Let there be N flips before we start looking.
Flips N+1 through N+1000 have 500 heads and 500 tails.
Flips N+2 through N+1001 also have 500 heads and 500 tails.
If flip N+1 was heads, then flips N+2 through N+1000 had 499 heads and 500 tails; in order for N+2 through N+1001 to have 500 heads, flip N+1001 must be heads as well.
Similarly, if flip N+1 was tails, then flips N+2 through N+1000 had 500 heads and 499 tails; in order for N+2 through N+1001 to have 500 tails, flip N+1001 must be tails as well.
Therefore, flips N+1 and N+1001 must be the same. Since N can be any non-negative integer, then if flip N is heads, then flips N+1000, N+2000, N+3000, and so on must be heads as well, and if flip N is tails, then N+1000, N+2000, N+3000, and so on must be tails as well.
Therefore, the strategy is:
(a) Note the results of each flip, in order, until 500 are heads or 500 are tails
(b) Bet the maximum on whichever did not reach 500 until flip 1000
(c) At this point, for each upcoming 1000 flips, bet the same pattern as how the first 1000 flips resulted. Every bet must be a winner, as otherwise the most recent 1000 flips would not have 500 heads and 500 tails.

I did not ask a different question. I merely attempted to put it in a different context for OnceDear. The real life results of 1000 coin flips have nothing to do with my question.

I do accept the answers given. Thank you.

If you have an unlimited bank roll, start with one unit on the side that is behind, increase by one unit for each bet, win or lose until you have a profit, then start over. If you know there is going to be a reversion to the mean and there is no table limit this should work well for you.