Question

A gambler who has P1,000 plays the single-die game with the following system: At the first toss of the die, he bets P300 on even numbers and quits if he wins. If he loses, he bets P400 on even numbers on the second toss and quits if he wins. If he loses again, he bets his final P300 on even numbers on the third toss. Is the game fair?

Answer #1

For it to be a fair game, the expected winnings for the person should be 0 from the whole process.

Let us calculate the probabilities and expected winnings for each of the 4 cases: (We know that the probability of getting an even number on any dice throw is 1/2 = 0.5)

- Win on the first toss: Profit = 300, Prob = 0.5
- Loss on second toss but win on second Toss, Profit = -300 + 400 = 100, Prob. = 0.5*0.5 = 0.25
- Loss on first 2 tosses but win on the third toss, Profit = -300 - 400 + 300 = -400, Prob. = 0.5*0.5*0.5 = 0.125
- Loss on all 3 tosses, Profit = -300 - 400 - 300 = -1000, Prob. = 0.5*0.5*0.5 = 0.125

The expected winning amount is computed here as:

= 0.5*300 + 0.25*100 + 0.125*(-400) + 0.125*(-1000)

= 150 + 25 - 50 - 125 = 0

As the expected winnings from the whole process is 0,
**therefore yes it is a fair game.**

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