Gambler’s Continued Fallacy - Round 3

Probability Level 2

The Las Vegas Casino Magnicifecto offers an “even value” game, with the catch that players may not leave the game if their (total) winnings are positive. (See previous question for more details.)

Scrooge has the gambling itch and decided that he could still afford to play, since he won in the first round. He recalled his brilliant strategy of doubling up each time, and decided to play the third round as follows:

He first makes a bet of $ 10 \$10 .
If he wins, he will double the size of his bet.
If his (total) winnings is ever non-positive, he will leave the game.

Now, what is the expected dollar amount of Scrooge's (total) winnings from this third round?

− $ ∞ - \$\infty $ 0 \$0 − $ 20 - \$20 − $ 10 -\$10

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Calvin Lin by Calvin Lin

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2 solutions

Siam Habib
Apr 26, 2014

Lets simplify the problem first by assuming that Scrooge won the first n n games and therefore his total winnings is 10 $ × ∑ i = 1 n 2 n = 10 ( 2 n + 1 − 1 ) $ 10\$ \times\displaystyle \sum_{i=1}^n 2^n = 10(2^{n+1}-1)\$ . If he loses the next turn, his total winning will be 10 ( 2 n + 1 − 1 − 2 n + 1 ) $ = − 10 $ 10(2^{n+1}-1 - 2^{n+1})\$=-10\$ and therefore he would stop playing.

So, the total expected winnings in any turn n n is − 10 / 2 n $ -10/2^{n}\$ . So, the expected value is ∑ i = 1 ∞ − 10 / 2 n $ = − 10 $ \displaystyle \sum_{i=1}^\infty -10/2^n\$ = -10\$ .

So, the answer is − 10 $ -10\$ .

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In the first game he either continues to play or gets out of the round with -10$. In the second game, again (now the amount of bet placed is 20, since he won in the last round), either he continues to play or gets out of the round with 10(previous round win)-20(this rounds loss) = -10$, and so on..his only option is to get out of the game with -10$ score or keep playing.

more mathematically, in nth game:

past profit from n-1 games = (10+102+102^2+...102^(n-2)) = 10(2^(n-1) - 1)

bet in nth game = 10*2^(n-1)

if he does get out of the round by loosing in this game, his net score = loss in nth game + past profits

= -10*2^(n-1) + 10( 2^(n-1) -1) = -10$

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