Question 1194145: In a 6/35 lottery game, a player pays $1 and selects six numbers from 1 to 35. Any player who has chosen the six winning numbers wins $1,000,000. Assuming that this is the only way to win, what is the expected value of this game? (round to the nearest cent)
Answer by math_tutor2020(3817)   (Show Source):
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We have n = 35 numbers to pick from and r = 6 slots to fill.
The order doesn't matter.
Use the nCr combination formula.
n C r = (n!)/(r!(n-r)!)
35 C 6 = (35!)/(6!*(35-6)!)
35 C 6 = (35!)/(6!*29!)
35 C 6 = (35*34*33*32*31*30*29!)/(6!*29!)
35 C 6 = (35*34*33*32*31*30)/(6!)
35 C 6 = (35*34*33*32*31*30)/(6*5*4*3*2*1)
35 C 6 = (1,168,675,200)/(720)
35 C 6 = 1,623,160
There are 1,623,160 different combinations possible.
This is the number of possible ways to play the lotto game.
Of this, there's only one way to win the grand prize since you must get all 6 matching numbers. Again the order doesn't matter.
The odds of winning are 1,623,160 to 1 or about 1.6 million to 1.
We write that probability as the fraction 1/(1,623,160) or 1/1623160
Type that fraction into your calculator to get the approximate decimal of 0.00000061608221 which is very small. There are six "0"s between the decimal point and the digit "6".
The probability of winning is roughly 0.00000061608221
The probability of losing the lotto game is roughly
1-0.00000061608221 = 0.9999993839178
Make a table showing the two outcomes and their probabilities
X = net pay out = amount the person walks away with
X is either -1 or 999,999
The person either loses $1, or wins $1 million to walk away with 1,000,000-1 = $999,999
| X | P(X) | | -1 | 0.9999993839178 | | 999,999 | 0.00000061608221 |
We'll form a new column called X*P(X)
This is where we multiply each X and P(X) across the rows
| X | P(X) | X*P(X) | | -1 | 0.9999993839178 | -0.9999993839178 | | 999,999 | 0.00000061608221 | 0.6160815939178 |
Now add the values in that third column to get the expected value
-0.9999993839178 + 0.6160815939178 = -0.38391779
that rounds to -0.38
The player is expected to lose on average 38 cents per game.
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