Alembert Joined: Jan 14, 2013
• Posts: 57
August 5th, 2014 at 6:53:48 AM permalink
Early on it was asked where you can find a 50/50 bet. In New Mexico, no Vig on 4/10 becomes a 0% wager. Bet the 4 and 10 in tandem for a true 50/50 wager. 3 ways to win 4, 3 ways to win 10, 6 ways to lose both.
HughJass Joined: Jul 20, 2014
• Posts: 40
August 5th, 2014 at 8:03:16 AM permalink
Quote: JyBrd0403

Yeah, maybe I didn't make myself clear. I was looking for the lengthy mathematical formula or proof. The mathematicians I talked to said it was too complicated for them, so if you've got a formula, lay it on me.

Let

b_k = bet value at the kth level
p_k = probability that series terminates with a win at the kth level, having been preceded by k-1 losses in a row
n - 1 = greatest number of losses in a row that can be sustained
e = player's expectation

e = + p_1b_1 + p_2(b_2 - b_1) + p_3(b_3 - b_2 - b_1) +...
+ p_n(b_n - b_(n-1) - ... - b_1)
+ (1 - p_1 - p_2-...
- p_n)(-b_n - b_(n-1) - ... - b_1)

The terms on the first line represent products of the probability that the series will terminate with a win at each successive level times the net profit at that level. The term on the third line gives the product of the probability that the series ends in failure at the nth level times the net loss.

regroup the terms:

e = 2p_1b_1 + (2p_2 + p_1)b_2 + (2p_3 + p_2 + p_1)b_3 + ...
+ (2p_n + p_(n-1) + ... + p_2 + p_1)b_n
- (b_1 + b_2 + ... + b_(n-1) + b_n)

p_k = (1 - p)^(k-1)p, where p is the probability of a win on any individual play and 1 - p is the probability of a loss.

substituting:

e = [2p]b_1 + [2p(1 - p) + p]b_2 + [2p(1 - p)^2 + p(1 - p) + p]b_3 + ...
+ [2p(1 - p)^(n-1) + p(1 - p)^(n-2) + ... + p(1 - p)^2 + p(1 - p)^1
+ p(1 - p)^0]b_n - (b_1 + b_2 + ... b_(n-1) + b_n)

factor out p so that the kth term is rewritten as:

p[(1 - p)^(k-1) + (1 - p)^(k-1) + (1 - p)^(k-2) + ... + (1 - p)^2
+ (1 - p)^1 + (1 - p)^0]b_k

use the formula for the sum of a geometric series and rewrite the kth term:

p[(1 - p)^(k-1) + (((1 - p)^k - 1)/((1 - p) - 1))]b_k
= [(2p - 1)(1 - p)^(k-1) + 1]b_k

e = sum_{k = 1}^{n} [2p - 1)(1 - p)^(k-1) + 1]b_k - sum_{k = 1}^{n} b_k
e = sum_{k = 1}^{n} [2p - 1)(1 - p)^(k-1)b_k + sum_{k = 1}^{n} b_k - sum_{k = 1}^{n} b_k

cancel the last two summations and factor out (2p - 1):

e = (2p - 1) sum_{k = 1}^{n} (1 - p)^(k-1)b_k

Since (1 - p) is positive and b_k is positive, the summation is positive. Therefore, the sign of e depends on the sign of (2p - 1). In an even-payoff unfair game, p < .5 and (2p - 1) is negative. In a fair game, p = .5 and (2p - 1) = 0.
bobsims Joined: Apr 8, 2014
• Posts: 316
August 5th, 2014 at 7:28:46 PM permalink
Quote: 24Bingo

No.

We have calculated the EV on the D'Alembert.

On a 50/50 game, with a finite time limit or finite bankroll limit, the EV is 0.

.

LOL. Yeah I got that part by 22 years old. What MY question is-how does D'alembert work, not on a 50/50 game but a 50.68-49.32 game-the bank bet at baccarat where over the "long term" the gambler, unlike craps, roulette, BJ, etc. CAN expect his winning bets to overwhelm his losing bets. Yes there is the little matter of the commission...

JyBrd0403 Joined: Jan 25, 2010
• Posts: 548
August 7th, 2014 at 3:19:28 PM permalink
Quote: HughJass

Let

b_k = bet value at the kth level
p_k = probability that series terminates with a win at the kth level, having been preceded by k-1 losses in a row
n - 1 = greatest number of losses in a row that can be sustained
e = player's expectation

e = + p_1b_1 + p_2(b_2 - b_1) + p_3(b_3 - b_2 - b_1) +...
+ p_n(b_n - b_(n-1) - ... - b_1)
+ (1 - p_1 - p_2-...
- p_n)(-b_n - b_(n-1) - ... - b_1)

The terms on the first line represent products of the probability that the series will terminate with a win at each successive level times the net profit at that level. The term on the third line gives the product of the probability that the series ends in failure at the nth level times the net loss.

regroup the terms:

e = 2p_1b_1 + (2p_2 + p_1)b_2 + (2p_3 + p_2 + p_1)b_3 + ...
+ (2p_n + p_(n-1) + ... + p_2 + p_1)b_n
- (b_1 + b_2 + ... + b_(n-1) + b_n)

p_k = (1 - p)^(k-1)p, where p is the probability of a win on any individual play and 1 - p is the probability of a loss.

You see where this is wrong, right? This is for a Marty, not a D'Alembert.
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5368
August 7th, 2014 at 4:46:37 PM permalink
Quote: JyBrd0403

You see where this is wrong, right? This is for a Marty, not a D'Alembert.

It appears to apply for any method. Note that it is the expected value for playing until you either get a win or reach a loss limit (which is expressed in a number of losses, but can be converted from a dollar limit).
For Martingale, p_1 = 1, p_2 = 2, p_3 = 4, p_4 = 8, p_5 = 16, p_6 = 32, and so on.
For D'Alembert, p_1 is variable, p_2 = p_1 + 1, p_3 = p_2 + 1, p_4 = p_3 + 1, and so on.
However, it does show that EV in this case is negative (exactly how negative has to be calculated...) if the probability of winning is less than 1/2, and zero if it equals 1/2.
JyBrd0403 Joined: Jan 25, 2010
• Posts: 548
August 9th, 2014 at 3:03:03 PM permalink
Quote: ThatDonGuy

It appears to apply for any method. Note that it is the expected value for playing until you either get a win or reach a loss limit (which is expressed in a number of losses, but can be converted from a dollar limit).
For Martingale, p_1 = 1, p_2 = 2, p_3 = 4, p_4 = 8, p_5 = 16, p_6 = 32, and so on.
For D'Alembert, p_1 is variable, p_2 = p_1 + 1, p_3 = p_2 + 1, p_4 = p_3 + 1, and so on.

D'Alembert doesn't start over after a win.
Buzzard Joined: Oct 28, 2012
• Posts: 6814
August 9th, 2014 at 3:35:34 PM permalink
Looks like a classic ROOSKIE. Nothing new here.
Shed not for her the bitter tear Nor give the heart to vain regret Tis but the casket that lies here, The gem that filled it Sparkles yet
ThatDonGuy Joined: Jun 22, 2011
• Posts: 5368
August 9th, 2014 at 6:09:53 PM permalink
Quote: JyBrd0403

D'Alembert doesn't start over after a win.

That's why p_1 is variable.

However, when I did some Monte Carlo analysis on D'Alembert under the stated stop condition (the first bet wins, or the number of wins = the number of losses), I did notice something:

When there is no other stop condition, EV is positive even for a 50% game - but some of the "runs" required are quite large; for example, one took three billion bets (and the bets reached almost 100,000x the original bet - have fun finding a table that allows that) before it finally evened out. (The mean number of bets needed to even out is about 200,000 - and this does include the fact that it takes only one or two bets to win 75% of the time.)

When there is a stop condition - either bankroll-based or time-based - then EV is 0.

If you remove the "total loss" condition from HughJass's calculations:

e = p1 b1
+ p2 (b2 - b1)
+ p3 (b3 - b2 - b1)
+ ...
+ pn (bn - bn-1 -...- b1)

e = b1 * (p1 - p2 - p3 -...- pn)
+ b2 * (p2 - p3 - p4 -...- pn)
+ b3 * (p3 - p4 - p5 -...- pn)
+ ...
+ bn-1 * (pn-1 - pn)
+ bn * pn

e = b1 * (p - p (1-p) - p (1-p)2 -...- p (1-p)n-1)
+ b2 * (p (1-p) - p (1-p)2 -...- p (1-p)n-1)
+ b3 * (p (1-p)2 -...- p (1-p)n-1)
+ ...
+ bn-1 * (p (1-p)n-2 - p (1-p)n-1)
+ bn * p (1-p)n-1

e = b1 * p * (2 - 1 -(1-p) - (1-p)2 -...- (1-p)n-1)
+ b2 * p * (2(1-p) - (1-p) - (1-p)2 -...- (1-p)n-1)
+ b3 * p * (2(1-p)2 - (1-p)2 -...- p (1-p)n-1)
+ ...
+ bn-1 * (2 (1-p)n-2 - (1-p)n-2 - (1-p)n-1)
+ bn * p * (2 (1-p)n-1 - (1-p)n-1)

e = b1 * p * (2 - 1 - (1-p) - (1-p)2 -...- (1-p)n-1)
+ b2 * p (1-p) * (2 - 1 - (1-p) - (1-p)2 -...- (1-p)n-2)
+ b3 * p (1-p)2 * (2 - 1 - (1-p) - (1-p)2 -...- (1-p)n-3)
+ ...
+ bn-1 * p (1-p)n-2 * (2 - 1 - (1-p))
+ bn * p (1-p)n-1 * (2 - 1)

e = b1 * p * (2 - 1 - (1-p) - (1-p)2 -...- (1-p)n-1)
+ b2 * p (1-p) * (2 - 1 - (1-p) - (1-p)2 -...- (1-p)n-2)
+ b3 * p (1-p)2 * (2 - 1 - (1-p) - (1-p)2 -...- (1-p)n-3)
+ ...
+ bn-1 * p (1-p)n-2 * (2 - 1 - (1-p))
+ bn * p (1-p)n-1 * (2 - 1)

e = b1 * p * (2 - [(1-p)n - 1] / [(1-p) - 1])
+ b2 * p (1-p) * (2 - [(1-p)n-1 - 1] / [(1-p) - 1])
+ b3 * p (1-p)2 * (2 - [(1-p)n-2 - 1] / [(1-p) - 1])
+ ...
+ bn-1 * p (1-p)n-2 * (2[(1-p)2 - 1] / [(1-p) - 1])
+ bn * p (1-p)n-1 * (2 - [(1-p)^1 - 1] / [(1-p) - 1])

e = b1 * p * (2 + [(1-p)n - 1] / p)
+ b2 * p (1-p) * (2 + [(1-p)n-1 - 1] / p)
+ b3 * p (1-p)2 * (2 + [(1-p)n-2 - 1] / p)
+ ...
+ bn-1 * p (1-p)n-2 * (2 + [(1-p)2 - 1] / p)
+ bn * p (1-p)n-1 * (2 + [(1-p)^1 - 1] / p])

e = b1 * (2p + [(1-p)n - 1])
+ b2 * (2p (1-p) + (1-p) [(1-p)n-1 - 1])
+ b3 * (2p (1-p)2 + (1-p)2 [(1-p)n-2 - 1])
+ ...
+ bn-1 * (2p (1-p)n-2 + (1-p)n-2 [(1-p)2 - 1])
+ bn * (2p (1-p)n-1 + (1-p)n-1 [(1-p) - 1])

e = b1 * (2p + (1-p)n - 1)
+ b2 * (2p (1-p) + (1-p)n - (1-p))
+ b3 * (2p (1-p)2 + (1-p)n - (1-p)2)
+ ...
+ bn-1 * (2p (1-p)n-2 + (1-p)n - (1-p)n-2)
+ bn * (2p (1-p)n-1 + (1-p)n - (1-p)n-1)

e = b1 * ((2p-1) + (1-p)n)
+ b2 * ((2p-1) (1-p) + (1-p)n)
+ b3 * ((2p-1) (1-p)2 + (1-p)n)
+ ...
+ bn-1 * ((2p-1)(1-p)n-2 + (1-p)n)
+ bn * ((2p-1) (1-p)n-1 + (1-p)n)

If p = 0.5, then all of the 2p-1 terms = 0, and e = 0.5^n * (b1 + b2 + ... + bn)
When you start, bk = (k-1) + b1, which is not necessarily 1 (e.g. if your first five bets are LLLLW, the first bet of the next "sequence" is 4, not 1)

e = 0.5n * (b1 n + (n-1) n / 2) = (b1 n + (n-1) n / 2) / 2n

Then again, as n approaches positive infinity, this approaches (b1 + n) / (ln 2 * 2n) by L'Hopital's Rule; in turn, this approaches 1 / (ln 4 * 2n), which approaches zero...assuming, of course, that b1 is finite. That's what happens when you deal with "infinity."
24Bingo Joined: Jul 4, 2012
• Posts: 1348
August 11th, 2014 at 12:56:03 AM permalink

Let's take a 1:1 game where p is your chance of winning.

Your credit card company is having a special where they'll lend out all the money you need, no fees, no questions asked... but only for one [T]. After that, you've got nothing but a bill.

That is to say, infinite bankroll, but you only have T hands.

So let's go back to the reasoning from before. First the trivial case: W and you're out, 1 unit up. So we add p to the expected value. (We won't be considering pairs of hands this time, so there's no point singling out LW.)

Now, again we'll get recursive, but unlike before, we're taking it one hand at a time, and again unlike before, we have a counter; when it reaches 0, game over.

So we've got the function f(d, t), and it runs as follows:

if (t == 0 || d == 1) return 0;
else return (p*(f(d-1,t-1) + d) + (1-p)*(f(d+1,t-1) - d));

And our ultimate answer will be p + (1-p)*f(2,T-1).

Now this... technically... gives you a way to calculate the expected value, but since it'll take you a billion iterations for a time limit of thirty hands, I wouldn't recommend it.

I'll work on a closed-form representation tomorrow night.

EDIT: Frankly... screw it.
The trick to poker is learning not to beat yourself up for your mistakes too much, and certainly not too little, but just the right amount.
JyBrd0403 Joined: Jan 25, 2010