Consider a game in which a coin will be flipped three times. For each heads, you will be paid \$100. Assume that the coin comes up heads with probability 2/3.

• A. Construct a table of the possibilities and probabilities in this game.
• B. Compute the expected value of the game.
• C. How much would you be willing to pay this game?
• D. Consider the effect of a change in the game so that if tails comes up two times in row, you get nothing. How would our answer to the first three parts of this question change?

We're going to call this three chances instead of coin flips, because at two-to-one odds, it's obviously not a coin.

The odds of three "heads" is 2/3 x 2/3 x 2/3 = 8/27.

The odds of three "tails" is 1/3 x 1/3 x 1/3 = 1/27.

This leaves 18/27, which split 2-to-1, 12/27 for two "heads" and 6/27 for two "tails".

Translated to money, in 27 trials you can expect: (8 x \$300 = \$2400) for three heads, (12 x \$200 = \$2400) for two heads, (6 x \$100 = \$600) for two tails.

This adds up to \$5,400, an average of \$200 per play. If you have to pay more than that, you will lose money in the long run.

If you change the game by making it a zero result if the first two chances come up "tails," that only affects one third of the two-tails chances, the ones with tails-tails-heads. That's two of six, which change from \$100 winnings to nothing.

Subtract \$200 from the \$5,400 total, and divide by 27 as before. Is the game still worth \$200 to play?

It is important to state that probability dictates there is no such thing...

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