In the game of Keno, a player bets $1 on
precisely five of the numbers from 1 to 80 inclusively. A machine draws at random precisely 20 of the numbers from 1 to 80 inclusively.
Between the player's numbers and the machine's numbers:
• If the number of matches is less than 3, the player loses his bet.
• Otherwise, the player gets back an additional $2, $25 or, $331 if the matches are respectively 3, 4, or 5.
What is the expected value of money won by the player in a game?
*** Adapted from a problem by S.W. Golomb.
The number of matches can be any of x={0,1,2,3,4,5} There are n=80*79*78*77*76 possible drawings. So the the probability P(x)=C(x)/n where:
C(0)=60*59*58*57*56
C(1)=60*59*58*57*20*5
C(2)=60*59*58*20*19*10
C(3)=60*59*20*19*18*10
C(4)=60*20*19*18*17*5
C(5)=20*19*18*17*16
The expected value of each x is its probability multiplied by its value
The total expected value of a ticket is the sum of the x values.
[2*C(3)+25*C(4)+331*C(5)]/n = 1972190880 / 2884801920 = 19659/28756 = 0.6836
The expected money won is therefore negative: 0.6836-1=-0.3164
This game has a very low pay-off.
If matching all five numbers had a payoff $821.52 then the game would be fair. (The other payoffs remaining the same)
***Edit:
See Charlie's response for a better interpretation of the question. I hadn't considered the word 'additional' in the payouts.
Edited on July 12, 2022, 2:24 pm
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Posted by Jer
on 2022-07-12 09:30:25 |
Solution
