The Game Of 3-spot Keno Is Similar To 4-spot Keno./what Is The Expected Value Of The Game?
The game of 3-spot Keno is similar to 4-spot Keno. Eighty numbers are displayed on a board, and 20 of these numbers will be chosen to be the winning numbers. To play, a player selects three numbers on his or her playing card. Suppose a player pays $1 to play and the $1 is not returned. The payoffs made by the Dunes Casino in Las Vegas for each of the possible outcomes are listed here:
Outcome Payoff
0 or 1 number $0
2 winning # $1
3 winning # $42
a) what is the expected value of the game?
b) what is the house edge for this game, expressed as a percentage of $1?
June 2nd, 2009 at 9:39 am
Two solutions which arrive at the same answer ?
The player chooses 3 numbers, leaving 77 others.
For him to have his three numbers among the 20 chosen,
the 20 numbers would have include his 3 (one way for that),
and 17 others (C(77,17) ways for that).
C(80,20) = 3.53 e+18 ways of choosing 20 numbers from 80.
C(77,17) = 4.9 e+16
So the chances of hitting 3 numbers is 4.9 e+16/3.53 e+18 = 4.9 / 353 = 0.013881
To get two winning numbers,
the player’s numbers must be 2 winners and 1 non-winner.
There are C(3,2) = 3 ways to choose two of his numbers.
Then we want the 20 numbers chosen to be those two,
plus 18 that he did not choose.
That would be C(77,18) = 1.635e+17
Multiplying that by 3, we get 4.905e+17
So his chances of getting 2 winning numbers are
4.9e+17/3.53 e+18 or 10 times the other number.
.1388
When he hits 3 numbers, he gains $41.
When he hits 2 numbers, he breaks even ($0).
When he hits 0 or 1 numbers, he loses a dollar.
So expected payoff is
0.013881 * 41 + .13881 * 0 + .8473 * -1 = -0.278179
(.8473 is 1 - .13881 - .013881).
So you lose 27.8c for every dollar you play.
Interestingly, I came up with the same answer
in my original answer, with a quite different calculation,
which was:
A player’s entry can be any of C(80,3) combinations. 82160
From the the 20 numbers there are C(20,3) winning combinations. 1140
1140/82160 = .0139
Now we have to consider the chances of getting 2 of the winning numbers.
There are C(20,2) * 60 ways to do that. 190*60 = 11400
So your expected results are:
.0139 * 41 + .139* 0 + .8471 * -1 = -.2772
The difference here is due to roundoff when
converting everything to decimal.
If you maintain and then cancel all the factorials,
it comes out exactly the same.
Most interesting!
For part b), if this is not 28 percent house advantage,
then I don’t know what it is.
=
June 2nd, 2009 at 9:39 am
Hi,
The payoffs made by the Dunes Casino in Las Vegas for each of the possible outcomes are listed here:
Outcome Payoff
0 or 1 number $0
2 winning # $1
3 winning # $42
a) What is the expected value of the game?
The probabilities of getting x correct out of 3 chosen numbers when there is a 25% likelihood of a correct answer and a 75% likelihood of choosing wrong numbers. The formula for this is:
3nCrX(.25)^X(.75)^(3-x)
#
correct probability
—————————-
0…….._.421875
1…….._.421875
2…….._.140625
3…….._.015625
Since 0 or 1 winning numbers get you nothing, 2 correct numbers get you $1 for 14.0625% of the time and you get $42 during 1.5625% of the results.
The expected value of the game is .421875(-1) + .421875(-1) + .140625(1) + .015625(42) = -.046875.
You would loose an average of about 5 cents each game.
b) What is the house edge for this game, expressed as a percentage of $1?
Therefore, in percentage form, the house edge is obtained by subtracting an actual game winning from a fair game winning and dividing the result by the fair game winning, finally multiplying by 100. Thus,
House Edge = [ (fair game winning - real game winning) / fair game winning ] * 100
House Edge = [ (100 - 79.6875) / 100 ] * 100 = 20.3125%
I hope that helps!!