Gamblers Always Win - Round 1

Probability Level 2

The Las Vegas casino Magnicifecto was having difficulties attracting its hotel guests down to the casino floor. The empty casino prompted management to take drastic measures, and they decided to forgo the house cut. They decided to offer an “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 $- \$A$ . They felt that since the expected value of every game is 0, they should not be making or losing money in the long run.

Scrooge, who was on vacation, decided to exploit this even value game. He has an infinite bankroll (money) and decides to play the first round (entire series) in the following manner:

He first makes a bet of $\$10$ .
If he wins, he keeps his earnings and leaves.
Each time that he loses, he doubles the size of his previous bet and plays again.

Now, what is the expected value of Scrooge’s (total) winnings from this first round (which ends when he leaves)?

A "round" refers to the entire series of games played above. This question refers to the entire round. There are 4 rounds in the set of problems.

This problem is part of Go Big Or Go Home , which explores the linearity of expected value.

Image credit: Wikipedia Flipchip

- \$10

\$0

+ \$ 10

+ \$ \infty

4 solutions

Bhagirath Mehta
Apr 26, 2014

If Scrooge wins the first match, he gets $10 and goes home. If he loses, he bets $20 in a second match.

Now, if he wins his second match, he makes a net total (his earnings minus the money he lost) of $20-$10= $10.

If he loses, he bets $40 in a third match .

If he wins the third match, he will make a net total of $40-$20-$10=$10.

If he loses, he bets $80 in a fourth match, and so on.

Note that no matter which match it is, if he wins that match, he will make $10 overall.

Eventually, as the number of matches played increases, Scrooge will win one match at some point, since he has a fifty percent chance of winning and fifty percent chance of losing each time.

This means that Scrooge will eventually make a net total of $10.

Couldn't have explained it better! :)

Deepika Pandit - 7 years, 1 month ago

man! this is some way to make money. do let me know of the casino ,if you find one, with this scheme on.

yash mittal - 7 years, 1 month ago

But you have to have infinite amount of money ( or a lot) to actually gain something profitable

Jason Kang - 7 years, 1 month ago

There is a very small chance that he will play very many games, as there is only a fifty percent chance that he loses each game. (0.5)^n converges to 0 very quickly, as n increases. If we are assuming that he is playing infinite games, we should probably assume he has an infinite amount of money.

Bhagirath Mehta - 7 years, 1 month ago

@Bhagirath Mehta – „The likelihood of catastrophic loss may not even be very small. The bet size rises exponentially. This, combined with the fact that strings of consecutive losses actually occur more often than common intuition suggests, can bankrupt a gambler quickly.“

https://en.m.wikipedia.org/wiki/Martingale (betting system)

Michael B - 2 years, 10 months ago

You can't. The house most certainly would be taking a cut, and even if it doesn't you could go bankrupt for $10.

Jerry Hill - 5 years, 4 months ago

I'm just pointing out that the casino odds don't have to be 50-50 for this to work (as long as you have infinite cash)

Wen Z - 4 years, 12 months ago

The EV is $0. The question assumes the Scrooge has infinite amount of money and he is willing to bet that much. (which is not practical in any case)

Even if Scrooge has 100 billion dollars (any finite number), the EV will be 0.

I think the assumption should be stated in the question.

Jerry Hill - 5 years, 4 months ago

What is the expected no. Of matches he played?????

Jeck Sparov - 6 months, 4 weeks ago

Nitesh Kumar
Apr 27, 2014

Let the Expectation to win be E[x].

Now , the probability to win the first match only is 1/2.

As each time it loses it double its bet, and the game end after his first win.

So, E[x] = 1/2(10) + 1/2( E[x])

     E[x]   =  5   + 1/2(E[x])

    E[x] -  1/2 E[x]   =   5

     E[x]   =    10

So, he will won $10 at the first win.

Prasun Biswas
Apr 27, 2014

Let us consider that the player placed a bet $\$a$ at first (here a=10). Now, if he wins, then he takes $+\$a=\boxed{+\$10}$ home.

If he doesn't, then he plays another game with bet $\$2a$ and if he doesn't win then also, he plays with bets of $\$4a$ , then $\$8a$ and so on.

Say that he wins when he plays with bet of $\$na$ , then he gets $+\$na$ . But, the previous losses are $-\$(\frac{n}{2})a,-\$(\frac{n}{4})a,....,-\$(\frac{n}{n})a$ .

Now, observe that the total losses converges to $\$(n-1)a$ .

Then, net gain/loss $=+\$na-\$(n-1)a=(+\$na-\$na+\$a)=+\$a=\boxed{+\$10}$

NOTE : We can say that he will win the bet with $\$na$ as there is a 50% chance of winning which states that the person is bound to win on atleast one future bet of $\$na$ after which he can leave with his positive gain.

This answer makes the rather strong assumption that Scrooge has infinite money with which to fund this venture, which I feel should be stated in the question.

A Former Brilliant Member - 7 years, 1 month ago

@Anthony Hughes There is a very small chance that he will play very many games, as there is only a fifty percent chance that he loses each game. (0.5)^n converges to 0 very quickly, as n increases. However, Scrooge may have charged the bets onto a credit card or may be based on Scrooge McDuck (who practically has infinite money). Additionally, if we are assuming that he is playing infinite games, we should probably assume he has an infinite amount of money.

Bhagirath Mehta - 7 years, 1 month ago

Max B
Apr 25, 2014

if he does' not get $10 in the first round..that is 1 times what he had invested....how is he gonna play the next..quite logical it seems..wat do u say

He probably has more money in his pocket than he has placed in the first bet.

Bhagirath Mehta - 7 years, 1 month ago

People generally go into a casino with more than 10$

Matthew Stephenson - 7 years, 1 month ago

Gamblers Always Win - Round 1

Image credit: Wikipedia Flipchip

4 solutions

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