We don't know.

That's an assumption too. We don't know that to be true. We assume it because its what makes the problem solvable (and it's also clearly the intent of the problem unless we're trying to make points about priors).

If I told you I had an orange tree in my backyard and you could bet a dollar at even money on whether the oranges it grows are naturally orange or bright blue would you say it's 50/50?

How can you talk about it being +EV on subsequent flips if you think no probability exists for this situation? You don't know the EV if you don't know the probability. Why are you willing to hit the all-in button after 8 heads in a row? If your philosophy is consistent, then even after 8 heads in a row you still wouldn't know if P(heads) > 50%. For all you know, maybe the prior was so heavy on Tails that those 8 heads didn't sway the probability past 50%. So maybe Heads is a -EV bet.

But the thing is, we do know. When I say it's uniform I'm saying, "I know that I don't know," because uniform is synonymous with having zero info.

Am I a human with life experience / knowledge of the world, or am I an alien from a distant galaxy? If I'm a human, then I have information and my prior distribution makes Orange a heavy favorite. If I'm an alien with no knowledge of Earth's fruits, then it's 1/(# of possible colors), so if you've already narrowed it down for me to Blue/Orange then it's 50/50.

Similarly, if you tell me two cricket teams are playing and ask me to pick the winner, my chance of guessing right is 50/50 because I have no clue about that sport. It doesn't matter if one of those teams is a 10:1 favorite. I know my chance is 50/50 even though I don't know the true chance of my team winning. Again, my claim of 50/50 amounts to, "I know that I don't know." 50/50 implies the maximum Shannon entropy, ie the least amount of information. (If Rusty is reading this, he can smack me on the head if I just shat on Shannon's grave.)

That sports example is probably something you can test via experiment if you don't believe me that it comes out to 50/50 (as opposed to, "there is no probability"). Better yet, a similar experiment can be done with a deck of cards. You can have someone remove cards from the deck and you have to guess red/black without knowing which cards they removed.

on which flip of the ten possible, would we start to really make bets and start compounding our money? Surely the first bet is zero, but we should start betting something on the second flip. How much do we bet on the second flip?

That's a good example. If you had a deck you knew to be biased red or black you wouldn't just assume the probability of a random card picked from the deck being red is uniformly distributed between 0% and 100%. That's what you're claiming is correct with the coin toss problem.

Using a uiniform distribution as the prior isn't "synonymous with having zero info". You're wrong on that.

It depends what your goal is. If it's to maximize EV (average wealth at the end), you should bet all-in any time the score isn't tied. If losing $20 means more to you than winning $20 then you should bet something less, Kelly being a reasonable candidate (which maximizes median wealth at the end). Juk's formula says the Kelly bet for the 2nd flip would be 1/3, so $6.67. If you won that bet, then on the 3rd flip you'd bet $13.33 (half of 26.67).

That's because I know a deck of cards is discrete. This coinflip scenario is continuous, per your OP.

The deck distribution would be a discrete version of uniform: 1/27, 2/28, ... 26/27 would all be equally likely (excluding 1/2). The average would be 50/50.

You're saying that wouldn't be the distribution and that it wouldn't be 50/50. I say do the experiment. The same experiment will show that the prior I just created is correct, because on the 2nd iteration, you can use Bayes to update the probability and you'll see that it gives you the correct probability.

Edit: for simplicity, let's say your friend only removes cards from one of the colors, and after each pick you make, reshuffles and then keeps removing the same number of cards. That's a good analog to this coinflip scenario because then the bias is unchanging.

Terrible assumptions.

Then do the experiment...

Edit - Hm, the experiment is flawed because it's always a human setting the deck. There could easily be a bias in how many cards a human decides to remove, which would affect the distribution of deck biases.

In this coin example, not only are you in the dark about the coin's bias, you're in the dark about the distribution of biases. The average of all the possible distributions is uniform. If we could actually perform the coin experiment, Uniform would almost always be way off, but in the long run the probabilities derived from it would be correct because the distribution shapes would average out to a flat line.

I'm not sure if there's a way to design a similar scenario IRL without a computer sim, but then the code would basically be assuming the question. But I think my previous paragraph might have cleared it up for you: the distribution doesn't have to be "definitely uniform", it only has to average out to uniform.

What would that prove? I can choose to make the prior distribution uniform or something different. (Although that alone proves that assuming a uniform distribution is incorrect.)

See edit

These are baseless assumptions

There is a lot of literature about the biased coin probability issue, so I'm curious to wait the OP's final answer.

Frankly I keep thinking that my best strategy is to bet my entire bankroll on the 10th roll or to bet 1/3 of the bankroll on the 9th roll and 2/3 on the final roll.

Maybe adopting a progressive "no limit" betting should be another option, providing an infinite series of 10 rolls.

I'm betting all in on rolls 2-10. It's the highest EV strategy and I'm happy with the risk/reward on all of the bets. That's my final answer. I don't think any other answer is correct. I was just curious how other gamblers approach this. There was some decent discussion.

The followup question would be how do we determine what initial bankroll would be required before we want to start scaling down bets. It's interesting because it's mathematically intractable without assumptions about priors but perhaps too abstract to really have any practical implications.

Based on what you've been saying ITT, how do you know it's the highest EV strategy or that it's even +EV? I know it's +EV because I can actually tell you the EV. You can't even put a number on the EV because you have some weird philosophical objection.

If you don't know what to use for the prior, then you don't know if one coinflip is enough to sway the probability past 50%, nor two coinflips, nor 9. So to turn around and say betting is +EV is a contradiction.

Also how do you know the 1st flip is 50%? If you don't know the prior and you don't know the average of the possible priors, you shouldn't be able to put a number on the 1st flip's probability. According to you, you don't know the median of the distribution, nor anything about its shape or whether it's symmetric. To say the 1st flip is 50% implies you know the median.

No, you can't tell me the EV. You can only tell me the EV under certain assumptions which are incorrect.

There's no contradiction.

If I treat heads and tails the same none of this matters.

ie. if my strategy is to take the result of the first flip and bet all in on that on flips 2-10, without caring if it's heads or tails, I know the strategy is +EV. It doesn't matter what the prior is.

If that's not apparent, try to come up with a prior that makes my strategy -EV.

You don't know that it's 50%. You know that it's not 50%.

Easy, there are countless examples. For instance, suppose the prior says there's a 99.999% chance that P(Heads)=60% and a .001% chance that P(Tails)=51%.

If the first flip is Tails, then on the 2nd flip, P(bias is 51% tails) = .51*.00001 / (.99999*.4 + .00001*.51) = .00001275

So there's still a 99.998725% chance that Heads is a 60% favorite, and yet you're about to shove all-in on Tails.

No, there are zero examples.


In this example, my strategy gets paid $20*2^9 = $10240 about .6^10 * 0.99999 = 0.6046% of the time which is an EV of $61.92, far more than the $20 we start with.

(That's just for heads, ignoring what we get paid on tails, which is like a buck in EV)

I think you misread my post somewhere. Tails is winning 1-0, so you're going all-in on Tails, despite the fact that there's a >99.99% chance that you're getting it in as a 40% underdog. That's -EV

I think you misread the part where ~$60 > $20 tho

The actual EV of my strategy is $62.99 which is not -EV by my calculations. Looking forward to seeing your higher EV strategy.

And no, of course it's not surprising that most of the EV comes when we bet heads on a heads-biased coin.

Another interesting thing about this problem is that a strategy that randomly chooses between "all in on tails every flip" or "all in on heads every flip" with 50% frequency for each, is +EV.

So you can bet profitably just by knowing that the coin is biased, without knowing what the bias is and without ever seeing a single flip.

This is the part I overlooked. Even if I change the numbers so that the chance of Heads is <50% (causing our first nonzero bet to most likely be on the underdog), the net EV will still be positive because the larger EV gains will compensate for the more frequent smaller EV losses.

Kudos to you if you predicted that dynamic. I tested it by writing out the EV of the 2nd flip (given the proposed strategy) as a function of 3 variables: heads bias, tails bias, and P(bias = heads bias). There is no solution to make the EV ≤ 0. That's for a prior with only two possible biases, but if there were a prior that made the game -EV, I see no reason it would require more than two possible biases.

So at this point I don't think I can philosophically convert you. And the fact that I can't think of a good experiment means it might really be a philosophical matter.

The problem with my argument is, when I said the possible distributions average out to uniform, you were right that that was an assumption. Namely, it assumed the distribution of distributions is equally weighted. All I did was shift the problem up a level, and we could play that game forever because it's turtles all the way up.

Maybe a real mathematician / Bayesian wizard can make a better case for a uniform prior than I've mustered up.

I've enjoyed this discussion and gained from it.

What if we run the analysis on a uniform range of (.5, 1) simultaneously running side by side with an analysis of (0,.5). Then, we use Bayes to tell us in which universe we reside. Then we make our last bets according to our most likely universe.