It is fairly straightforward to calculate what the winnings might be in a simple betting game.
Let us postulate an urne containing 4 balls, three white and one black. Balls are put back into the jar after each draw. Tickets are $10 each, and the payout is as follows :
Ball 1, white a loss, -10$
Ball 2, white a loss, -10$
Ball 3, white a loss, -10$
Ball 4, black a win, +10$
If a white ball is drawn, I loose. If the black ball is drawn, I am given 20$, a 10$ win after the recovery of my ticket price. On average, I expect to loose 5$ per draw. I never actually loose 5$; this is an average.
If I bet for 20 consecutive draws, I would then expect to loose 100$.
I am a betting kind of person. Would I expect to loose precisely 100$ on 20 draws; could I bet on this. Not really. 5$ is the money average loss of the system. But the outcome choices are three white and one black. To take this into account, I need to calculate the error this introduces to my money expectations.
Wednesday, January 19, 2011
Betting Game(2)

Thus on a twenty draw splurge, I would expect to loose 5$+- 1,94 per drawing or
100$+-38,75 for the lot.
Note that this error margin decreases with the number of draws. For a 100 draw situation
I would expect to loose 5$+-.87 per, or 500$+- 87, a less bumpy ride.
Of mathematical interest, the sine value of a 60 degree angle is .866… Squared, this is ¾.
Example from : Jacques Allard, Concepts fondamentaux de statistique, Addison-Wesley, 1992.
100$+-38,75 for the lot.
Note that this error margin decreases with the number of draws. For a 100 draw situation
I would expect to loose 5$+-.87 per, or 500$+- 87, a less bumpy ride.
Of mathematical interest, the sine value of a 60 degree angle is .866… Squared, this is ¾.
Example from : Jacques Allard, Concepts fondamentaux de statistique, Addison-Wesley, 1992.
Tuesday, January 18, 2011
Monday, January 17, 2011
Saturday, January 15, 2011
Thursday, January 6, 2011
Subscribe to:
Posts (Atom)







