In a lottery game, a player picks six numbers from 1 to 23. If the player matches all six numbers, they win 20,000 dollars. Otherwise, they lose $1.
What is the expected value of this game? $
Probability help!?
2019-02-24 8:27 pm
回答 (4)
2019-02-24 8:55 pm
Probability of winning the game = C(6,6)/C(23,6) = 1/[23!/(6!17!)] = 6!17!/23!
Probability of losing the game = 1 - (6!17!/23!)
Expected value of the game = $20000 × (6!17!/23!) - $1 × [1 - (6!17!/23!)] = -$0.801866…. ≈ -$0.80
Probability of losing the game = 1 - (6!17!/23!)
Expected value of the game = $20000 × (6!17!/23!) - $1 × [1 - (6!17!/23!)] = -$0.801866…. ≈ -$0.80
2019-02-24 8:34 pm
E(x) =
potential amount lost * probability of losing + potential amount won * probability of winning =
-1 * (1 - (1/23)^6) + 20000 * (1/23)^6 =
-$0.99986489
potential amount lost * probability of losing + potential amount won * probability of winning =
-1 * (1 - (1/23)^6) + 20000 * (1/23)^6 =
-$0.99986489
2019-02-24 10:04 pm
23C6 = 100,947 possible outcomes.
One winning outcome.
Expect 19.8¢ from each $1 bet
E = lose 80.2¢
One winning outcome.
Expect 19.8¢ from each $1 bet
E = lose 80.2¢
2019-02-24 9:25 pm
$20000/C(23,6) - $1
= $20000/100947 - $1
= $0.1981 - $1 = negative 80.2 cents
= $20000/100947 - $1
= $0.1981 - $1 = negative 80.2 cents
2019-02-24 9:02 pm
23C6= 100,947
20,000/100,947 - 1≈ -$0.80
20,000/100,947 - 1≈ -$0.80
收錄日期: 2021-05-01 22:32:46
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190224122749AAbOKE6