Question for Don Please: Ace Prediction Math Revisited Again

[Overkill]

In a thread last year about math regarding Ace prediction that I started wherein I used the example of being 15.1% certain that my next card for my hand would be an Ace, you stated the following (I highlighted the portion of the post of interest with red font):

So, there are a couple of points to be made. Obviously, we all understand that, if an ace were certain, the edge on that round would be 51%. That's always the case: if your first card is an ace, your edge is 51%. But, as Overkill mentioned, usually, that probability is only 1/13 (7.69%), and it balances out with all the other first-card possibilities to give whatever the BS house edge is, say, 0.5%. So now, we have to rejigger the edges and the probabilities. Without any special knowledge, and to get to -0.5%, we would have 1/13(51%) + (12/13)x = -0.5%. Solving for x, we get -4.79%. So far, so good.

But now, the +51% enjoys an inflated 15.1% probability. So the new calculation for the overall edge becomes: .151(51%) + .849(-4.79%) = edge. Player edge is 3.63%.

Of course, this is the edge only when the 6 (or a ten, for that matter) appears. Naturally, it isn't your overall edge for the game, because, well, sixes appear only 1/13 of the time. So, that 3.63% edge would then be divided by 13 to give your flat-bet advantage of 0.28%. But, it goes without saying that, when the 6 (or ten) would appear, you'd bet a great deal more than your minimum bet. And, it's here that the frequency of the tens (4/13) would come in. That 0.28% would be multiplied by four (1.12%), for your flat-bet edge if you knew aces followed tens, instead of sixes, with 15.1% accuracy.

Think I've covered it all.

Don

Don, could you please explain the need for the .849(-4.79%)? Is that already accounted for when we have to reduce the 51% probability by multiplying by .151? Isn't the .151 already accounting for the possibility of getting a card other than Ace? Thanks, Overkill

[DSchles]

Don, could you please explain the need for the .849(-4.79%)? Is that already accounted for when we have to reduce the 51% probability by multiplying by .151? Isn't the .151 already accounting for the possibility of getting a card other than Ace? Thanks, Overkill

No, I don't think so. If you're 15% certain that you'll get an ace as your next card, that's one probability. And, if you do get the ace, you'll have a 51% edge on that resulting hand. Those are separate ideas. But, you have to account for the fact that you may not receive the ace. That happens with 85% probability. What, then, will be your edge? It isn't until you multiply the two probabilities (ace, no ace) by their respective edges that you get the overall edge for the game, flat betting.

Don

[aceside]

Don’s math is credible! That’s why I keep bothering him.

[Overkill]

Thanks, Don, for the thoughtful (as always) reply.

However, I respectfully disagree. (The crowd gasps! )

You say, "But you have to account for the fact that you may not receive the Ace." Again, I assert that you have already accounted for that with the 85% (the flip side of that coin is the 15%). 85 times out of 100 I will not get the Ace, but in the long run, I will 15 times out of 100. I have accounted for both scenarios.

Thanks again for replying.

Is nobody else going to join the discussion?

[DSchles]

Thanks, Don, for the thoughtful (as always) reply.

You're welcome ... I think!

However, I respectfully disagree. (The crowd gasps! )

That's surely your prerogative, ... but it doesn't mean you're right!

You say, "But you have to account for the fact that you may not receive the Ace." Again, I assert that you have already accounted for that with the 85% (the flip side of that coin is the 15%). 85 times out of 100 I will not get the Ace, but in the long run, I will 15 times out of 100. I have accounted for both scenarios.

But you haven't accounted for the player's EDGE in each of the scenarios! Suppose I say to you, tell me your e.v. for this proposition: heads you win $100, tails you lose $50. The math is simple: you multiply the probability of getting a head by $100, then the probability of getting a tail by $50, and you sum the two products. So, 0.5($100) + 0.5 (-$50) = +$25.

The 15% accounts for the probability of getting an ace. The (implied) 85% accounts for the probability of NOT getting an ace. Now, for EACH ONE of those two scenarios, you have to attribute an edge, multiply, then sum the products, as above. Do you understand?

Is nobody else going to join the discussion?

There's very little else to say!

Don