Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: July 8, 2006, 1:24 pm - IP Logged

csfb

I'm afraid such a thread would turn out like this one and a few alumni of Maven U. sould surface and explain how the first tickets has 175,000,000 + or 120,000,000 + against it, so to cut that in half play another ticket.

For MM and PB just look at the posted odds on the bvacket of the ticket and know that onley one combination of that many can hit and every ticket has the same exact odds againt that particular combination hitting.

NY United States Member #23835 October 16, 2005 3474 Posts Online

Posted: July 8, 2006, 9:09 pm - IP Logged

Quote: Originally posted by Stack47 on July 7, 2006

Coin Toss,

Its obvious true odds are meaningless when the question is "how much can I win for a buck". And you can get $5000 for your buck hitting a Pick-4 straight combination.

What happened in your other thread and in this one too, people were saying by playing 10 different straight combinations they now had 10 chances of winning $5000 so they were lowering the odds by a factor of 10. I guess they forgot by playing 10 more combinations, they changed the question to "how much can I win for 10 bucks".

We have multi-state lottery games like Powerball and Mega Millions with odds of 175 million to 1 because studies have shown more people play when the jackpots are high. I doubt that people really expect to hit the huge jackpots when they buy 10 auto picks but they can dream until the numbers are drawn. If they are buying a $10 dream, getting back $3 is a win.

The casinos are very aware of the fact people like the bonus options they have added them to many of their table games and created new games that include bonuses.

Stack

Your first two paragraphs suggest that you might be really close to getting my point, and I think you'll be there as soon as you consider that the question isn't how much can you profit, but "what are the chances of winning?" Here are some questions.

1. After thoroughly shuffling a standard deck of cards you draw a single card at random. Nobody placed a bet on the outcome and no money changed hands. Was it still possible to calculate the odds of drawing the ace of spades?

2. It seems that everybody agrees that the odds in pick 4 are 9999:1 for a single ticket. If the payout was $1000 or $9000 instead of $5000 would the odds against winning still be 9999:1 or would they change?

3. Are the odds for pick 3 999:1 for a single ticket?

4. Suppose that pick 4 tickets were 10 cents each, but it only paid $500. You buy ten tickets with the numbers 0123, 1123 2123, 3123, 4123, 5123, 6123, 7123, 8123, and 9123, thus betting $1 for a chance to win $500. Is that any different than buying a single $1 ticket for pick 3 with the number 123? If so, how?

Here's what I think you should come up with:

1. Of course. There are 51 cards that aren't the ace of spades and 1 card that is, so the odds are 51:1 against. The amount of the bet, if any, doesn't affect the odds of the event.

2. The odds remain the same. Even Coin Toss gets it right when he says "odds are odds and "math is math". Odds are the probability of the event happening. In the first question the event is drawing the ace of spades, and in this question the event is having a ticket with the winning number.

3. Yes, the odds are 999:1 999 numbers that won't match your ticket aginst 1 that will.

4. It's exactly the same probability as pick 3. Pick 4 is just pick 3 with an extra digit, from 0 through 9 in front of each of the three digit numers for a total of 10,000 possibilities instead of 1,000 possibilities. 123 is multiplied by 1 to get 0123, 123 is multiplied by 20 to get 2123, etc. That means playing the ten numbers ending in 123 in pick 4 would give you the same chance of having a winning ticket as just playing 123 in pick 3, and in this case you'd be risking the same amount of money for the same payout. That makes the odds of winning pick 4 with 10 tickets 9990:10, or 999:1, just as with pick 3.

Some posters won't get it or would just try to weasel their way out of answering by going off on a tangent about betting on next weeks game or playing baccarat, but your last post suggests you're actually thinking things through. Let's see what you get.

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10344 Posts Offline

Posted: July 9, 2006, 12:22 am - IP Logged

KY Floyd

Pick 3, Pick 4, 3 or 4 4 columns of numbers, 0 thru9 making ten numbers per column.

10 X 10 X10 = 1,000

10 X 10 X 10 X 10 = 10,000

A number is picked from each column, meaning 1 combination among 1,000 or 1 combination among 10,000 will be right, 999 or 9,999 will be wrong. If you play ten separate combinations you have reduced 999 to 989 and 9,999 to 9,989.

That's all there is to it.

Til you comprehend that you have nob suiness tell any one else how flawed their concepts of odds is or making wise guy remarks liike this:

Some posters won't get it or would just try to weasel their way out of answering by going off on a tangent about betting on next weeks game or playing baccarat,...

Sheesh

Your examples in this thread have assumed no winners among 1,000 or 10,000 comboinations, this leading you, and only you, to believe that you're up against 1,000 to 1 and reducing that by 10 brings it down to 990 to one.

That little tidbit throws out the rest of your argument, it's about equal to giving someone the wrong formula to figure out a pitcher's ERA and swearing it's correct.

Kentucky United States Member #32652 February 14, 2006 7295 Posts Offline

Posted: July 9, 2006, 3:34 am - IP Logged

Floyd: Was it still possible to calculate the odds of drawing the ace of spades?

Yep.

Floyd: There are 51 cards that aren't the ace of spades and 1 card that is, so the odds are 51:1 against.

And if you draw another card, there are now 49 cards that are not the ace of spades, so odds the are 49:1.

Floyd: would the odds against winning still be 9999:1 or would they change?

One 4 digit number is drawn and 9999 are not so the odds are always 9999:1.

Floyd: That makes the odds of winning pick 4 with 10 tickets 9990:10, or 999:1, just as with pick 3.

How do you come up with 10 winning tickets? Because only one 4 digit number is drawn, buying 10 different numbers only reduces the number of losing outcomes by 10. 9990 ways to lose and only 1 way to win; 9990:1.

How many different ways can it be explained to you; when you buy 10 different numbers on straight pick-4 wager, 9 of them are guaranteed to be losers?

NY United States Member #23835 October 16, 2005 3474 Posts Online

Posted: July 9, 2006, 9:35 pm - IP Logged

Quote: Originally posted by Stack47 on July 9, 2006

Floyd: Was it still possible to calculate the odds of drawing the ace of spades?

Yep.

Floyd: There are 51 cards that aren't the ace of spades and 1 card that is, so the odds are 51:1 against.

And if you draw another card, there are now 49 cards that are not the ace of spades, so odds the are 49:1.

Floyd: would the odds against winning still be 9999:1 or would they change?

One 4 digit number is drawn and 9999 are not so the odds are always 9999:1.

Floyd: That makes the odds of winning pick 4 with 10 tickets 9990:10, or 999:1, just as with pick 3.

How do you come up with 10 winning tickets? Because only one 4 digit number is drawn, buying 10 different numbers only reduces the number of losing outcomes by 10. 9990 ways to lose and only 1 way to win; 9990:1.

How many different ways can it be explained to you; when you buy 10 different numbers on straight pick-4 wager, 9 of them are guaranteed to be losers?

Stack

How many different ways can it be explained to you that when you buy 10 different numbers on straight pick-4 that there are 10 different chances to match the winning number? I think I've tried most of them, but I'll try once more. Let's start with the deck of cards again.

Me: There are 51 cards that aren't the ace of spades and 1 card that is, so the odds are 51:1 against.

you: And if you draw another card, there are now 49 cards that are not the ace of spades, so odds the are 49:1.

You apparently see that the second card gives you better odds than the first (but only after the first card is shown to lose), but you got those odds wrong. When I draw 1 card from a normal deck there are 51 cards left. If the first card isn't the ace of spades and I draw a second card I'm leaving 50, so my odds are 50:1. You seem to be using Coin Toss' math, where you subtract the number of choices played from the original odds, but that's not how it works. Now let's make the cards analagous to pick 4. You draw two cards but don't check them. There's a stack of 50 cards and a stack of 2 cards, and the ace of spades is in one of them. It could be in the 50 in which case you have 50 chances to lose, or it could be in the stack of 2. If you check the first card and if it isn't the ace of spades the second card could still be the ace of spades. If you can't see that that means you have two chances with two cards you can save yourself the trouble of reading on.

you: when you buy 10 different numbers on straight pick-4 wager, 9 of them are guaranteed to be losers?

If the thread were called "completely obvious 101" you'd be right on track. If you have 1 pick 4 ticket there's a 99.99% chance that you have a losing ticket and if you have two pick 4 tickets you have a 100% chance of having a losing ticket. If you have more than 2 tickets the odds of having a losing ticket are still 100%, but that's not what we're interested in.People buy tickets to try and get a winning ticket, so what we're interested in are the odds of having a winning ticket, and each ticket has a chance of matching the winning number.

you: How do you come up with 10 winning tickets?

I don't, and I've never said that. That' something that apparently came out of your own grasp of things. Only a moron would suggest that 10 tickets is a chance to win 10 times. I've said that 10 tickets is 10 chances to win. With ticket #1 you have 1 chance to win and with ticket # 2 you have another chance to win, so that's 2 chances. With 10 tickets you have 10 chances. Let's separate the numbers that result in losing from the ones that result in winning.

Number 0000 (there's no 1 at the front, but that's the 10,000th possible number) Number 9999 Number 9998 Number 9997 . . . Number 0011 Above the line are the 9990 numbers you don't have tickets for If any of those numbers is drawn you won't have a ticket with the winning numbers.

_______________

Below the line are the numbers that you do have tickets for. Notice that there are 10 of them.

Number 0010 If the winning number is 0010 you have a winning ticket. Your first chance to win Number 0009 If the winning number is 0009 you have a winning ticket. 2nd chance to win. Number 0008 If the winning number is 0008 you have a winning ticket. 3rd chance to win. Number 0007 If the winning number is 0007 you have a winning ticket. 4th chance to win. Number 0006 If the winning number is 0006 you have a winning ticket. 5th chance to win. Number 0005 If the winning number is 0005 you have a winning ticket. 6th chance to win. Number 0004 If the winning number is 0004 you have a winning ticket. 7th chance to win. Number 0003 If the winning number is 0003 you have a winning ticket. 8th chance to win. Number 0002 If the winning number is 0002 you have a winning ticket. 9th chance to win. Number 0001 If the winning number is 0001 you have a winning ticket. 10th chance to win.

Now let's do the 3rd grade math I've talked about:

10,000 The possible numbers that could be drawn. - 10 The numbers you have tickets for. 9990 The numbers you don't have tickets for.

The winning number could be among the 9990 numbers for which you don't have a ticket. That means there are 9990 numbers with which you will lose and you'll have 10 losing tickets. The winning number could also be among the 10 numbers for which you do have a ticket. That means there are 10 numbers which could give you a winning ticket. You'll also have 9 losing tickets, but the 1 winning ticket means you won.

10,000 possible numbers

9990 numbers that lose.

10 numbers that could win

9990 to 10.

That's a ratio and if we perform the same function on both sides we won't change the ratio, so dividing by 10 we get 999:1 It's that simple.