Question for Math Experts

Nevermind, really sorry about this guys

It's hard to say without knowing your stop conditions.

HE is a single-bet value. If you keep making bets with a positive HE, "eventually" you will reach zero.

Once you reach zero in your case, you have a bankroll of 0.25 X, and must make even-money 50/50 bets until you either lose the bankroll or reach 6 (X + 0.25 X) = 7.5 X = 30 times your original bankroll. The probability of gaining 29 times your bankroll before losing it in a 50/50 game is 1/30, but the HE is effectively zero as you are making a bet at 29-1 odds that you have a 1/30 chance of winning.

Oops. brb

Quote: DingDingDing

What is the combined HE in the follwing:

You deposit Initial deposit (X) and recieve a bonus (B) of 0.25X

With orginal X only, player plays a game with a HE of .41% Until X=0.

When it is gone, player uses bonus (B) to play a Fair 1/1 game, recieving the winnings of the Bonus but not recieving a return of the bonus funds bet, until Funds are greater than (X+B)*6, or until funds are gone.

Maybe I don't understand the rules, but if you don't actually get the bonus until (and unless) you lose the initial deposit, then it doesn't matter what the house edge on the main game is. Either you lose or you don't. So then you have a free 0.25X to play with and you need to get that to 7.5X in order to be able to cash it back out, except it's not really a "fair 1/1 game" if you don't receive back your wager when you win. Something's off with the description.

erasing my embarrasment :)

I think you should describe it in English, not math, if you don't understand how to do the math. Then someone who knows math can figure it out.

That sounds like a good idea.

Quote: DingDingDing

I apologize for the confusion. I changed some things up from the first post in the second post. ....I'm ultimately trying to calculate if its a loser or winner if infinitely played.

The "expected value" of the bonus game is the amount of the bonus, as you have a 1/30 chance of winning 30x the bonus and a 29/30 chance of winning nothing, but also losing nothing as the bankroll for the bonus play was given to you.

However, in order to get to the bonus game, you have to lose the main game, which means the EV of any session where you reach the bonus game = -X + B, or -3/4 X. If your initial bankroll is 4000, and you lose it, expect to be behind 3000 when you are done. The expected value overall depends on the point at which you are willing to stop playing even before you reach the bonus.

nevermind sorry guys, im going to make a new thread, as a continuance here would casue some (more) confusion.

Quote: DingDingDing

nevermind sorry guys, im going to make a new thread, as a continuance here would casue some (more) confusion.

If you are still talking about that BJ bonus it's no good unless running 62k through to make $6 is considered good.

If you are doing 1k deposit at 6x in sports that's 7500 in action(assuming D+B)

How are you going to beat the vig only getting 25%? Now, if someone was a break even or better sports bettor it would be really good.
I'm wondering (since your balance must be zero before they will award a new bonus) Are the going to let you cash out and redeposit over and over? I'm skeptical of that unless you keep losing.

I couldn't figure out at first why they kept saying "balance must be zero" but it makes sense, they don't want people who are winning/winners to use winnings and keep getting 25% with only 6x. (Usually it's 10x minimum wagering on sports bonuses) It's more of an incentive for people who lose everything to re-up.

IIRC They used to have some really good bonuses. They probably got burnt and that's why the tightening up on the rules.

Question for ThatDonGuy,
You said:
Once you reach zero in your case, you have a bankroll of 0.25 X, and must make even-money 50/50 bets until you either lose the bankroll or reach 6 (X + 0.25 X) = 7.5 X = 30 times your original bankroll. The probability of gaining 29 times your bankroll before losing it in a 50/50 game is 1/30, but the HE is effectively zero as you are making a bet at 29-1 odds that you have a 1/30 chance of winning.

Focusing specifically on: The probability of gaining 29 times your bankroll before losing it in a 50/50 game is 1/30

How do you calculate 1/30? (just asking so that i may learn and do the calculations myself in scenarios with different variables)

Quote: DingDingDing

Focusing specifically on: The probability of gaining 29 times your bankroll before losing it in a 50/50 game is 1/30

How do you calculate 1/30? (just asking so that i may learn and do the calculations myself in scenarios with different variables)

This is based on something called the Gambler's Ruin formula - that is, given an initial bankroll B, a target bankroll T, and bets of 1 each, what is the probability of reaching T without losing the entire bankroll first?

For a 50/50 game, the probability is B / T. In your case, B is 0.25 X and T is 7.5 X, so the probability is 0.25X / 7.5X = 1/30.

If the probability of winning a single bet is not 0.5, then, if p is the probability of winning a single bet, then the probability is:
( ((1 - p)/p)^B - 1 ) / ( ((1 - p)/p)^T - 1 )

my question originates from a prior pondering on free-play usage (and only free-play usage) i think wizard referred to it as a sort of "phantom bonus" from which the player can receive the winnings from a bet placed with the bonus, but will not receive the bonus used to make the bet after the event. Bet 1000 freeplay to win 1000, after the event the frreeplay account is 0.00 and the available balance is 1000.

i suppose then the if B were a phantom bonus, the probability is approximately reduced to half, is this correct?

Quote: DingDingDing

my question originates from a prior pondering on free-play usage (and only free-play usage) i think wizard referred to it as a sort of "phantom bonus" from which the player can receive the winnings from a bet placed with the bonus, but will not receive the bonus used to make the bet after the event. Bet 1000 freeplay to win 1000, after the event the frreeplay account is 0.00 and the available balance is 1000.

i suppose then the if B were a phantom bonus, the probability is approximately reduced to half, is this correct?

No. The probability is the same. The expected value is what is reduced.

The probability of a fair coin toss coming up heads is 1/2, whether it pays even money, 1-2, or anything else.

New question. Baccarat.

Just played this game long enogh to learn the payouts, can you help me with the math on the player and abker edges?

i bet 5 on player and won 5
i bet 5 on banker and won 4.75

The following i copy pasted from the "rules"

Baccarat

Baccarat House Rules

If an extra card comes out of the shoe after the game is completed and that card is visible to the players, that card will be burned.
Whenever possible, a dealer or supervisor will shuffle the cards in the shoe and leave it ready for the dealer to be used whenever the yellow card (shuffle card) appears. This is done to maximize the hands dealt and in order for the player to enjoy a continuous gaming experience. Once the yellow card (shuffle card) is shown, the regular procedure will be followed with the dealer announcing when will the shuffle or change of cards be completed. After the shuffle procedure, a card will be burned as usual.


Basic Rules

Win by predicting which of the two hands (Player or Banker) will have a point value closest to 9.

Ace is worth 1, cards from 2 to 9 are counted at face value, face cards and 10s have a value of zero. The values are summed and once it reaches double digits, it drops to the left digit (12 becomes 2, 21 becomes 1). Suits do not make any difference and only single values count.

The dealer gives 4 cards in order: player, banker, then player, then banker. A third card may be dealt according to the rules in Third card section.

Pairs

In Baccarat, SuperSix and Punto Banco, the following side bets can be placed in addition to, or instead of Player/Banker wagers: Player Pair, Banker Pair, Any pair. These side bets pay according to payout table if the first 2 cards dealt to the Player or the Banker constitute a pair (8-8, 10-10 or J-J) and loses on all other outcomes.

Punto Banco

Punto = Player.
Banco = Bank.

Third Card (instead of links Player and Banker)

Player 1 to 1

Player gets a third card when the first two cards total 0, 1, 2, 3, 4, or 5, and the Banker hand has a point value of 7 or less. Otherwise, the Player does not get a third card.


Banker 1 to 1 (due to commission)

Banker gets a third card depending on the value of the Player's third card (if any).

If Player does not get a third card, its hand value is 7 or less, and the Banker's first two cards total 0, 1, 2, 3, 4, or 5, then the Banker gets a third card.

If the Player gets a third card, then the Banker gets a third card when the Banker's first two cards total:

0, 1, or 2; or
3 and the Player's third card is not an 8; or
4 and the Player's third card is 2, 3, 4, 5, 6, or 7; or
5 and the Player's third card is 4, 5, 6, or 7; or
6 and the Player's third card is 6 or 7.

Payouts Payouts Baccarat
Player 1 to 1 PAYOUT
Player 1 to 1 1 to 1
Banker (Win with 6) .5 to 1 .95 to 1 (due to commission)
Tie 8 to 1 8 to 1
Player Pair 11 to 1 11 to 1
Banker Pair 5 to 1 11 to 1
Any Pair 5 to 1

The "quick version" is this: assuming you have an "infinite deck" - that is, a deck where the number of aces, 2s, 3s, ..., queens, kings is always the same, it is calculated like this:
There are 13 possible values for the player's first card
For each of these, there are 13 possible values for the player's second card
For each of these 13 x 13 combinations, there are 13 possible values for the banker's first card
For each of these 13 x 13 x 13 combinations, there are 13 possible values for the banker's second card
Of these 13 x 13 x 13 x 13 combinations, there are 13 possible values for the player's third card, should the player need it; if the player does not need it, then it is, in effect, treated as a zero card
Of these 13 x 13 x 13 x 13 x 13 combinations, there are 13 possible values for the banker's third card, should the banker need it; if the banker does not need it, then it is, in effect, treated as a zero card
Thus, there are 13 x 13 x 13 x 13 x 13 x 13 = 4,826,809 sets of six cards in a particular order
Go through the rules, and, for each one:
If you are calculating the HE for a player bet, count +1 for each one where the player wins, -1 for each one where the banker wins, and 0 for each one that is a tie
If you are calculating the HE for a banker bet, count +0.95 for each one where the banker wins, -1 for each one where the player wins, and 0 for each one that is a tie

I am not going to go through all that, but, fortunately for all of us, The Wizard has already done this, and in fact has done it for 8-deck and 6-deck shoes (which are not "infinite"; if the player's first card in an 8-deck shoe is an ace, there are now only 31 aces as opposed to 32 of each other card, which does affect the HE, although only slightly).

Another 1st grade math question....

if the probability of six consecutive heads is 1/64

is that

1 time in 64 flips?
or
1time in 64 x 6 = 384 flips?

or neither?

Quote: DingDingDing

Another 1st grade math question....

if the probability of six consecutive heads is 1/64

is that

1 time in 64 flips?
or
1time in 64 x 6 = 384 flips?

or neither?

I am a little confused as to how you expressed it, but I think what you meant by the second one is correct.
In 64 separate sets of 6 tosses, where no toss is considered to be in more than one set (i.e. tosses 1-6 are a set, tosses 7-12 are another set, tosses 13-18 are a third set, ..., tosses 379-384 are the 64th set), you are expected to have one set that is all heads.

As for 64 consecutive tosses, this doesn't sound right, but, in fact, you will get at least one set of six consecutive heads within the 64 about 39% of the time. I don't think there's an "easy" way to determine this other than something called a Markov chain, where you start at a particular condition and work out the probabilities of each subsequent condition until you reach the desired result.
In this case, let P(t,h) be the probability of having h consecutive heads after t tosses
You start with 0 tosses and 0 heads, so P(0,0) = 1
At each toss, you have probability 1/2 of adding 1 to the consecutive heads and 1/2 of going back to zero, so for each t > 0:
P(t,0) = 1/2 P(t-1, 0) + 1/2 P(t-1, 1) + 1/2 P(t-1, 2) + 1/2 P(t-1, 3) + 1/2 P(t-1, 4) + 1/2 P(t-1, 5)
P(t,1) = 1/2 P(t-1,0)
P(t,2) = 1/2 P(t-1,1)
P(t,3) = 1/2 P(t-1,2)
P(t,4) = 1/2 P(t-1,3)
P(t,5) = 1/2 P(t-1,4)
Since P(t-1,6) is a state where we already have 6 in a row, include this in the calculation of P(t,6):
P(t,6) = P(t-1,6) + 1/2 P(t,5)
The solution is P(64,6) = about 0.38929, or 38.929%,

i was just shocked today to find out that the average expected number of flips required for:

2 consecutive heads is 6 flips
3 consecutive heads is 14 flips
4 consecutive heads is 30 flips
5 consecutive heads is 62 flips

I always thought it was:

2 consecutive heads require 4 flips
3 consecutive heads require 8 flips
4 consecutive heads require 16 flips
5 consecutive heads require 32 flips

Why is the probabitlty and expected number of flips not the same?

im guessing if one set of 6 can be a part of another set 6, the expected number of flips for 6 consecutice heads is 126 flips?

Quote: DingDingDing

im guessing if one set of 6 can be a part of another set 6, the expected number of flips for 6 consecutice heads is 126 flips?

Yes.

Click here for a proof that the expected number of tosses to get N consecutive heads is 2^N+1 - 2.

that a really easy equation for my limited math mind., thanks,

is there a way to tweak it for probabilitys other than .50 ?


if so, for example, how would one find the expected number of flips to get 5 consecutive wins if the probability of winning once is 42%

Yes, plug in your new p to equation d2.

Yes - the same link shows that, if the probability of heads coming up in one toss is p, the expected number of flips needed to net N heads in a row is:

(1 - p^N) / (p^N x (1 - p))

For p = 0.42 and N = 5, this is (1 - 0.42⁵) / (0.42⁵ x 0.58) = about 130.2.

what is the defined difference in "probabilty" and "Expected Number of Events Nessesary for a given outcome"

and why do the two differ numerically?


I had always imagined/assumed that probability meant: "1444 times is the expected number of roulette spins to hit a single number twice in a row on an American wheel"

Sometimes when you flip a coin, it takes more than two tosses to get a head.

That is the difference.

Quote: DingDingDing

what is the defined difference in "probabilty" and "Expected Number of Events Nessesary for a given outcome"

and why do the two differ numerically?


I had always imagined/assumed that probability meant: "1444 times is the expected number of roulette spins to hit a single number twice in a row on an American wheel"

Probability is the likelihood of a particular event happening. You can use it to determine the expected number of times an event will happen in a given number of trials, but the expected number of trials needed for an event to take place can be something entirely different.

The difference usually appears when trying to determine the number of trials needed for an event to appear a certain number of times in a row, since you have to allow for runs of fewer than the target number of successes. For example, in the case of a particular number coming up twice in a row, you have to include all of the cases where the number comes up just once.

take 4 consecutive coin heads, .5 x .5 x .5 x .5 = 6.25 Percent probability

Percent in the dictionary is defined as: the rate, number, or amount in each hundred ____________.

Per---Cent (century) etc.


But one hundred of what, 100 single flips, 100 groups of 6 flips, 100 different 6 flip combinations.

it always seemed to me, before today, that 6.25 percent meant that in 100 events a 6.25 probability event should occur 6,25 times. or 12.5 in 200 times ect....

but if 1 occurance of 4 consecutive heads = 30 expected trails, then, that must mean?

6.25 occurances of 4 consecutive heads require 187.5 flips: which seems not to be 6.25 per(cent(100)) but rather 6.25 per(187.5)

im not yet able to wrap my head around why 6.25% is One "Perthirty"

fascinating :)

Try to keep it simpler, two coin flips.

There are 4 possibilities that happen with equal probability

TT
HT
TH
HH

But there is no guarantee that in 4 trials of two coin tosses, that each of these possibilities happens exactly once.

If you are flipping a coin and waiting for a head, you might flip it right away, you might flip a whole lot of tails first.

Since there is no guarantee that when you do a number of trials equal to the number of possibilities that you will see all of the possibilities exactly once, it is logical that the average number of trials required to see all of the possibilities is higher than just the average of the probibilities.

That isn't explained particularly well.

In the case of a single coin tossed, you might wait for a head for a while. That pulls the average number of tosses needed before you see one higher than just the probability of the possible outcomes.