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.
(In reply to
Solution by Jer)
There's a typo in n=80*79*78*77*66. The last factor should be 76.
The value you use in the calculation is correct, giving the correct value.
But there's a consideration of what the player gets (or gets back) if he wins. The wording
the player gets back an additional $2, $25 or, $331
indicates the player also gets back the $1 he originally bet; in other words, if he wins, he doesn't offset that gain by loss of his original dollar.
(3*c3+26*c4+332*c5)/n = 0.780320944877907
still is not worth 1, but it's not as bad.
Equivalently the original figure could be used but instead of subtracting 1 to get the expected value, use the probability of a loss (times 1, the value of the loss) and subtract that.
Edited on July 12, 2022, 11:00 am
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Posted by Charlie
on 2022-07-12 10:58:04 |
re: Solution
