Gambler’s Fallacy - Round 2

Probability Level 4

The Las Vegas Casino Magnicifecto had some success attracting their hotel guests to play their “even value” game. Whatever bet size the player places (say $\$A$ ), there is a 50% chance that he will get $+\$A$ , and a 50% chance that he will get $- \$A$ . However, they quickly realized that they were losing money and decided to modify the rules drastically: players may not leave the game if their (total) winnings are positive.

Scrooge, who was on vacation, was eager to continue playing this game after his previous success. However, he was slightly confused about the new ruling, and decided to play it safe. He plays the second round of the game as follows:

He first makes a bet of exactly $\$10$ .
If his (total) winnings are not positive, he will leave the game.
Otherwise, he will continue to make a bet of exactly $\$10$ .

Now, what is the expected value of Scrooge’s (total) winnings from this second round?

Image credit: Wikipedia Fluteflute

- \$10

\$0

- \$ \infty

- \$5

2 solutions

A Former Brilliant Member
Apr 27, 2014

On the initial bet, there's a 50% chance he loses $10 and leaves. The other 50% of the time he wins $10. The only way he eventually leaves the game in that scenario is by dropping his net winnings down to zero. So his expectation is 0.5($10 + $0) = $5.

The most common error made is to forget to consider what happens when he loses the first game. Most spot that he would leave the game with $0 winnings once he made it past the first game and think that the answer is $0.

Calvin Lin Staff - 7 years, 1 month ago

I just dont get one thing here and that is how Scrooge's expected value of total winnings can be $-\$5$ . He places bets of $\$10$ each time and may lose or win...then the total net gain or loss should be a multiple of $10$ , ain't it ? It is because the losses and gain both are multiples of $10$ and $\$0$ should be the answer as it is the value which is not positive and he can leave after dropping the net gain to $\$0$ . So, answer $=\$0$ . @Calvin Lin Please correct me if I am wrong.

Prasun Biswas - 7 years, 1 month ago

If he loses in the first game, he will leave with -$10. This happens with probability 0.5.
If he loses in any of the subsequent games, he will leave with $0. This happens with probability 0.5.
Hence, the expected value is $0.5 \times (-\$10) + 0.5 \times (\$0) = -\$0.5$ .

Just because the values are multiples of 10 (or integers) doesn't necessarily mean that the expected value will be a multiple of 10 (or integers). You can read the note on expected value (attached to this set) if you are unfamiliar with that concept.

Calvin Lin Staff - 7 years, 1 month ago

@Calvin Lin – Oh! Thanks for the explanation. I understand it now and yes, I was unfamiliar with the concept. Thanks again !! :)

Prasun Biswas - 7 years, 1 month ago

Jerry Hill
Jan 24, 2016

By playing, Scrooge makes sure that he cannot make a profit.

Probability of winning = 50% If Scrooge wins, he bets again till he has a non-negative total winnings. Which is $0. As long as it is positive Scrooge keeps betting. It cannot go negative since the total winnings will be 0 and Scrooge will leave before it goes negative.

Probability of losing = 50% Scrooge leaves. Winnings -$10

EV = -$5

Gambler’s Fallacy - Round 2

Image credit: Wikipedia Fluteflute

2 solutions

0 pending reports