Calculating Odds

Can someone please help me calculating odds.

I ma trying to learn the odds based on the following criteria:

"x" has a 42% chance of happening.

What are the odds that x will happen two times in a row before the 58% opposite result happens 20 times? (Not in a row but without the 2 times of "x" happening)

Another way of asking the question may be this:

If "y" has a 58% chance of happening, what are the odds of 20 "y's" happening before the opposite 42% result happens two consecutive times?

Thanks you for any and all help with this.

I had done a shoe shuffle with coding

It did count 80 hands and upto 10k shoes

And have seen none y 20 in a row without happening the
X two times


Btw: my betting strategy worked 99% towards my edge and 1% remained towards casino

Tested over 100k shoes 80hand per shoe

It is easier to figure out the probability of the 58% result happening 20 times before the 42% result happens twice in a row, then subtract that probability from 1.

Let P(a,b) be the probability of getting to the point with "a" consecutive wins and "b" total losses.
P(0,0) = 1, since that is the starting point
P(n+1, 1) = P(n, 0) x 0.42
P(n+1, 0) = (P(n, 0) x 0.58) + (P(n, 1) x 0.58) = (P(n, 0) + P(n, 1)) x 0.58

To make this easier, let p = 0.42 and q = 0.58.
P(1,0) = P(0,0) x p = 1 x p = p
P(0,1) = (P(0,0) + P(1,0)) x q = (1 x q) + (p x q) = (1 + p) x q
P(1,1) = P(0,1) x p = (1 + p) x q = (1 + p) x pq
P(0,2) = (P(0,1) + P(1,1)) x q = ((1 + p) x q) + ((1 + p) x pq) = (1 + p)² x q²
P(1,2) = P(0,2) x p = (1 + p)² x q² x p
P(0,3) = (P(0,2) + P(1,2)) x q = ((1 + p)² x q³) + ((1 + p)² x q³ x p) = (1 + p)³ x q³
Continuing in this fashion, P(0,20) = (1 + p)²⁰ x q²⁰
Since q = 1 - p, this is (1 + p)²⁰ x (1 - p)²⁰ = (1 - p²)²⁰.

The probability that the 58% event happens 20 times before the 42% event happens twice in a row = P(0,20) = (1 - p²)²⁰.
Siince p = 0.42, this is (1 - 0.1764)²⁰ = 0.8236²⁰ = about 1 / 48.5.
(For those of you looking for an exact answer, it is 2059²⁰ / 2500²⁰.)
The question asked for the reverse of this, so the solution = 1 - (1 / 48.5) = about 47.5 / 48.5, or about 95 / 97.

The exact answer is:
89073927579686057406858696660480937139686799843574718068472863197199
divided by
90949470177292823791503906250000000000000000000000000000000000000000

Quote: vegasrvp

...Another way of asking the question may be this:

If "y" has a 58% chance of happening, what are the odds of 20 "y's" happening before the opposite 42% result happens two consecutive times?

Thanks you for any and all help with this.

In blackjack the player has 48% chance of losing and 43% chance of winning, 9% to push... Thus, the odds of losing X in a row are P(Losing)^X, where X is the number of events/trails. For example, the odds of losing 5 hands in a row are .48^5 = .0255, or about 2.5%, or about 1 in 40.

The reason it's to the power is because P(Losing 3 in a row) = P(losing 1) * P(losing 1) * P(losing 1)... the P is the same, only the number of trials differs. So for problems like this, where the number of trials is the only difference, you can just use P^Trials.

I do this far too often when I sit down at a table and lose 10-13 hands in a row =/... reminding myself, hey, this shouldn't happen again for a while!

Btw: my betting strategy worked 99% towards my edge and 1% remained towards casino

Tested over 100k shoes 80hand per shoe



Do you share your system?

Btw: my betting strategy worked 99% towards my edge and 1% remained towards casino

Tested over 100k shoes 80hand per shoe

Do you share your system?

So if I am reading this correctly you are telling me I will win on the (2 x 42%) 98% of the time?

If this is a fact, Can you tell me what it would be if I made a few changes:

1. What would it be if we changed it to 48% vs 42%? (Have to hit 2 consecutive 48%'s before 20 total 51%'s)
2. Can you also tell me at the 48% what is would be if I were to change the 20 time against to each of the 20, 19, 18, 17, 16 and 15 times?

I really like where this is heading.

Thanks

So if I am reading this correctly you are telling me I will win on the (2 x 42%) 98% of the time?

If this is a fact, Can you tell me what it would be if I made a few changes:

1. What would it be if we changed it to 48% vs 42%? (Have to hit 2 consecutive 48%'s before 20 total 51%'s)
2. Can you also tell me at the 48% what is would be if I were to change the 20 time against to each of the 20, 19, 18, 17, 16 and 15 times?

I really like where this is heading.

Thanks

Quote: Romes

In blackjack the player has 48% chance of losing and 43% chance of winning, 9% to push... Thus, the odds of losing X in a row are P(Losing)^X, where X is the number of events/trails. For example, the odds of losing 5 hands in a row are .48^5 = .0255, or about 2.5%, or about 1 in 40.

The reason it's to the power is because P(Losing 3 in a row) = P(losing 1) * P(losing 1) * P(losing 1)... the P is the same, only the number of trials differs. So for problems like this, where the number of trials is the only difference, you can just use P^Trials.

I do this far too often when I sit down at a table and lose 10-13 hands in a row =/... reminding myself, hey, this shouldn't happen again for a while!

This is not for Blackjack.

Any idea how much it would change if I changed the 42% to 48% and also the 20 times to 15 times?

Quote: vegasrvp

This is not for Blackjack.

Gee, it wouldn't happen to be for the Don't Pass line in craps, would it?

That question has been answered in the other thread.

Quote: ThatDonGuy

Gee, it wouldn't happen to be for the Don't Pass line in craps, would it?

That question has been answered in the other thread.

Of Course.

Quote: vegasrvp

This is not for Blackjack.

Any idea how much it would change if I changed the 42% to 48% and also the 20 times to 15 times?

Sure... Betting systems alone are mathematically not capable of beating the game =).

Quote: Romes

Sure... Betting systems alone are mathematically not capable of beating the game =).

Some day I will find someone to help me with odds instead of telling me about betting systems.

Quote: vegasrvp

Some day I will find someone to help me with odds instead of telling me about betting systems.

Romes is about as patient and helpful as anyone you are likely to run into in these forums, IMO. If you didn't get the answer you were looking for, maybe you didn't phrase the question correctly.

I see a lot of questions asked, I see a lot of answers provided. There's some purty kewl guys that hang out round here. Don't know of any Forum rules that promised anybody would drop what they were doing to jump on a solution to whatever problem you might happen to decide to bring to the table. But there is also no rule that says you can't ask.

Good luck in your search, and welcome aboard!