Need math help
July 3rd, 2020 at 9:53:49 AM
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There are 2 ways I wonder and would like explained in simple terms about the following. The first is based on past history of American roulette spins (because the past 20 or 100 spins are recorded for you) and the other is based on random numbers used in the same manner.
Using the past history to determine the bet, if I placed bets on the last 5 numbers shown, I’d have a .1315 chance of matching any of those numbers. However using the birthday paradox, it states there is a .33 chance of selecting those past 5 numbers, along with a sixth, the outcome of the spin to win or lose. 6 numbers of a pool of 38, would match .3297 percent of the time.
Knowing if that happened, ie. the winning spin was 7, and the past 5 numbers were 31,4,7,16,32, I now assume the next spin will match because the recent 5 change to 7,31,4,7,16. This would add to the past history of w/l, because that’s 2 out of 2 or way over what the paradox states or .33.
The question would be, is there a number of losses of matching 6 numbers that would justify betting that would be below the .33 for a series of spins making a 7/1 payout worth it. Say if it were 20 losses in a row and the next spin was a win in the first position, (spin 8, previous 5 are 8,3,14,17,30), and no other matches other than the 8 came up for the next 5 spins, it would still be 6/26 or .23; below the.33 of birthday paradox chances.
Is there a simple equation I can use to calculate this, or is it the same old my chance is 1/38 paying 35/1...
Using the past history to determine the bet, if I placed bets on the last 5 numbers shown, I’d have a .1315 chance of matching any of those numbers. However using the birthday paradox, it states there is a .33 chance of selecting those past 5 numbers, along with a sixth, the outcome of the spin to win or lose. 6 numbers of a pool of 38, would match .3297 percent of the time.
Knowing if that happened, ie. the winning spin was 7, and the past 5 numbers were 31,4,7,16,32, I now assume the next spin will match because the recent 5 change to 7,31,4,7,16. This would add to the past history of w/l, because that’s 2 out of 2 or way over what the paradox states or .33.
The question would be, is there a number of losses of matching 6 numbers that would justify betting that would be below the .33 for a series of spins making a 7/1 payout worth it. Say if it were 20 losses in a row and the next spin was a win in the first position, (spin 8, previous 5 are 8,3,14,17,30), and no other matches other than the 8 came up for the next 5 spins, it would still be 6/26 or .23; below the.33 of birthday paradox chances.
Is there a simple equation I can use to calculate this, or is it the same old my chance is 1/38 paying 35/1...
July 3rd, 2020 at 10:00:57 AM
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This.Quote: Sirjoeyk...or is it the same old my chance is 1/38 paying 35/1...
Psalm 25:16
Turn to me and be gracious to me, for I am lonely and afflicted.
Proverbs 18:2
A fool finds no satisfaction in trying to understand, for he would rather express his own opinion.
July 3rd, 2020 at 12:04:41 PM
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Quote: SirjoeykUsing the past history to determine the bet, if I placed bets on the last 5 numbers shown, I’d have a .1315 chance of matching any of those numbers. However using the birthday paradox, it states there is a .33 chance of selecting those past 5 numbers, along with a sixth, the outcome of the spin to win or lose. 6 numbers of a pool of 38, would match .3297 percent of the time.
The birthday paradox applies only if you apply it to the entire group at once - not, for example, if there are already five people in the room, you know all of their birthdays are different, and a sixth walks in.
As I like to put it: the probability of N red spins in a row is identical to the probability of (N - 1) red spins in a row followed by a black spin.
July 3rd, 2020 at 12:11:26 PM
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You can use any method you like to pick your number(s).
It doesn't matter if you use
Every time you put $1 on a number your chances are 1 in 38 (or 37 single zero) and the payout if you win will be 35 to 1. Similar logic, sometimes a different house Edge, applies to other bets.
Just make sure you have fun.
Waiting for a blue moon or pig to fly by might be cheaper in the long run, especially if the drinks are free while you wait!
It doesn't matter if you use
- numbers that have just come up
- numbers that haven't appeared for ages
- wait for a blue moon and bet the day of the month
- count how many times a flying pig goes by
- use friends' birthdays
Every time you put $1 on a number your chances are 1 in 38 (or 37 single zero) and the payout if you win will be 35 to 1. Similar logic, sometimes a different house Edge, applies to other bets.
Just make sure you have fun.
Waiting for a blue moon or pig to fly by might be cheaper in the long run, especially if the drinks are free while you wait!
