| (4 intermediate revisions by 3 users not shown) | |||
| Line 2: | Line 2: | ||
Is this another way to interpret it? | Is this another way to interpret it? | ||
*What is the probability that a person chooses the six correct lotto numbers from numbers 1 through 50? | *What is the probability that a person chooses the six correct lotto numbers from numbers 1 through 50? | ||
| + | --[[User:Hschonho|Mike Schonhoff]] 13:21, 15 October 2008 (UTC) | ||
| + | |||
| + | I believe the expected value is basically like asking, "What would your average winnings be if you play this lottery many times?" For example, the expected value of rolling a fair die is 3.5 because you will roll each value in 1-6 an equal number of times, making the average roll 3.5. The formula for expected value is <math>E(X)=\sum_{s \in S} p(s)X(s)</math>, where ''p(s)'' and ''X(s)'' are the probability and outcome, respectively, of the ''s''th trial. So for this problem, the outcome of nearly all trials is $-1, except for the winning ticket, which is $9,999,999 (if you're picky about the $1 ticket cost). Hope this helps. | ||
| + | --[[User:Mkorb|Mkorb]] 14:35, 15 October 2008 (UTC) | ||
| + | |||
| + | |||
| + | ---- | ||
| + | |||
| + | The way I would suggest doing it is saying "expected loss" = "expected winnings from a ticket - "expected cost of a ticket" | ||
| + | |||
| + | Expected cost of a ticket is just 1, since tickets cost always cost 1 dollar. | ||
| + | |||
| + | If the number of tickets possible is T, then the probability of choosing the winning ticket is 1/T. Expected winnings from a ticket = ten million * 1/T + 0/T + 0/T+... (T-1 losing tickets account for the T-1 0/T terms). To add fractions with the same denominator, just add the numerators. So the expected winnings from a ticket = (ten million + 0 + 0...) / T = ten million / T. | ||
| + | --[[User:Norlow|Norlow]] 16:17, 15 October 2008 (UTC) | ||
Latest revision as of 11:17, 15 October 2008
So for this first one... I'm confused on what the question is asking. Is this another way to interpret it?
- What is the probability that a person chooses the six correct lotto numbers from numbers 1 through 50?
--Mike Schonhoff 13:21, 15 October 2008 (UTC)
I believe the expected value is basically like asking, "What would your average winnings be if you play this lottery many times?" For example, the expected value of rolling a fair die is 3.5 because you will roll each value in 1-6 an equal number of times, making the average roll 3.5. The formula for expected value is $ E(X)=\sum_{s \in S} p(s)X(s) $, where p(s) and X(s) are the probability and outcome, respectively, of the sth trial. So for this problem, the outcome of nearly all trials is $-1, except for the winning ticket, which is $9,999,999 (if you're picky about the $1 ticket cost). Hope this helps. --Mkorb 14:35, 15 October 2008 (UTC)
The way I would suggest doing it is saying "expected loss" = "expected winnings from a ticket - "expected cost of a ticket"
Expected cost of a ticket is just 1, since tickets cost always cost 1 dollar.
If the number of tickets possible is T, then the probability of choosing the winning ticket is 1/T. Expected winnings from a ticket = ten million * 1/T + 0/T + 0/T+... (T-1 losing tickets account for the T-1 0/T terms). To add fractions with the same denominator, just add the numerators. So the expected winnings from a ticket = (ten million + 0 + 0...) / T = ten million / T. --Norlow 16:17, 15 October 2008 (UTC)
