English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Can anyone help me to solve this question? It's not my homework but i'm interested in knowing the correct ways of solving it. Which theory is most approachable?

More specifically:
Let's just say the 4D has 20 prizes. I pay $1 every time for the same 4 digit number. How many times does it take for my number to be guarenteed a place in 4D?

I know the order of number counts, so Permutation is needed here but that's all i know.

2006-12-28 18:24:23 · 3 answers · asked by SSBCSS 1 in Science & Mathematics Mathematics

3 answers

the way the question is worded is slightly disturbing. You use the word Guarenteed to win a daily Win-4 draw. In probability, as in science where the probability model is used, there are no absolutes or guarantees. Apparently, you want to know how many days, with certainty, that your 4 digit number will come up. There is no "certain solution" for this problem. However, I will give you a day number that is esstially a 50/50 proposition that that number will come up by that numbered day. In essence, you could make an over under bet with a friend, one would take before the 6,930 day and the other would take after. (That is almost 19 years for once daily drawings) Making this wager of, say 5000 dollars, would give you a better chance of coming out ahead, than playing the 4 digit number with your state lottery.
.693 is the co-log of the hyperbolic log of 2. this can be used for all such types of evaluations, like rolling double twos with dice, or a single number outcome on roulette, etc.
you take the odds 10,000 times this constant .693 =6930.
Therefore, there is a 50/50 probability that the 4 digit number will come in after 6,930 draws. There is also a 50/50 probability that it won't. In addition, this 6,930 is an expected value with variation either way. If you get to this 6,930 th day, which it is as likely as not of already coming out you would have iinvested 6,930 dollars on this number prior to this point. If it wins on that day, you will go to the lottery officials to collect your 5,000 dollars as most states pay only half of the true odds. In addition, Uncle Sam will Tax this new found fortune. So you'll get what? 3000 and change.
This is not a game that you should commit to and expect positive outcomes. For one, the house advantage is outrageous. The advantage is the equivalent of if we bet on a football game and you took side A +3 points for 100 dollars, I have side B -3 points. But if you win I will only pay you 50 dollars.
If you are going to gamble, pick out a game of skill and master it. good luck

2006-12-28 20:52:33 · answer #1 · answered by James O only logical answer D 4 · 0 0

I really don't understand what is meant by "4D" here. Is this a lottery game? Is the point to pick any 4-digit number, and try to match it? Can the number begin with zero's for digits?

If so, there are 10*10*10*10 possible numbers (one 10 for each digit), or 10,000. So the probability of yours coming up is (1/10,000) or .0001, after one play.

However, since there is a probability of losing each time you play, there is no gaurantee that you'll ever win. The only way to gaurantee winning is to play all 10,000 possible numbers at the same time.

Now if you have an infinite number of people play your strategy of playing one number a day, on average they will win after 10,000 days. But some will win after one day, and others will never win.

What you can ask is, how many days should I expect to have to play to give me a 90% chance of winning, or an X % chance of winning? For X = 50%, it's 10,000 days. For X bigger than 50%, it's longer.

If I got wrong what you meant by 4D, just substitute for 1/10,000 the appropriate value for the probability of winning one game. Then everything I said above will still apply and be correct.

2006-12-29 02:45:27 · answer #2 · answered by Hal 2 · 0 0

I assume this is a daily number game . Pick a winning combination from the same lot of numbers day after day. But you are going to play the same number daily...

The odds do not change day to day because your number did not get picked winning numbers aren't removed from the lot so the odd remain the same. If its a 1000 to 1 chance yesterday its still a 1000 to 1 today and the next etc....

You can get lucky and win tomorrow and there is no guarantee that your number will ever come up... Only that you have the stated odds of your number probably coming up as a win...

2006-12-29 02:32:57 · answer #3 · answered by MarkG 7 · 0 0

fedest.com, questions and answers