If a gambler and the casino keep going forever, how come the casino is always the winner?
In: Mathematics
You pay me one dollar. You get to pick any 5 numbers between 1 and 11, including 1 and 11. I will use a random number generator to pick a number between 1 and 11, including 1 and 11. If the number is one of your numbers, I will pay you 2 dollars. If it isn’t, I keep your one dollar.
With 11 numbers able to be chosen, the monetary breakdown is:
6 x -$1 = -$6 (You can’t choose 6 of the 11 numbers, so you lose $1 each instance)
5 x $1 = $5 (You only win $1 with the 5 of 11 other numbers)
So every time you play this game, you will lose (-$6 + $5) / 11 = $0.09. $0.09 is a rather small number compared to the $1 bet that you placed for that one game. But if you play the game 100 times? Your expected value is -$9.
So by setting up every game to have payout odds so close to breakeven but not quite, you not only ensure that the casino will make money over the long haul, but that the player will win enough of the time to keep them coming.
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