Expected profits in Blackjack - tricky probability problem

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In summary, Lars is vacationing in Las Vegas and plays blackjack until he wins, doubling his bet for each game. The probability distribution for the number of times he plays before quitting is X~f(x) = 0.7^{x-1} 0.3. If he hasn't won after 5 games, he stops playing and his winnings, W, have an expected value of 6.59 dollars. His expected loss, E(L), is calculated by multiplying the probability of each outcome by the amount of money lost, starting from 0 dollars and increasing by 2 dollars for each loss. The expected value of Y, the winnings Lars is left with after his bets are deducted, is equal to the expected
  • #1
Nikitin
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Homework Statement


Lars is vacationing in Las Vegas and goes into the largest casino he comes across.
He sits down at the blackjack table and decides to play until he wins and to double
the bet for each game. He bets one dollar in the first game, two dollars in the
second game, and so on until he wins.

Assume (for simplicity) that he always gets back twice what he bet if he wins, that
his probability of winning is 0.3 in every game, and that Lars stops playing after
winning once.

a) Let X be the number of times Lars player before he quits. What is the
probability distribution of X?

b) Assume in the rest of this problem that Lars runs out of money and thus
stops playing if he has not won after playing five games. Let W be the
number of dollars Lars wins at the casino (regardless of how much he has
bet). What is the expected value of W?

c) Let Y be the winnings Lars is left with after his bets are deducted. What is
the expected value of Y ?

The Attempt at a Solution



I've done a) ##X##~## f(x) = 0.7^{x-1} 0.3 ##, and b) ##W = 2^{X-1} \cdot 2 \Rightarrow E(W) = \Sigma_{x=1}^{5} f(x) 2^x = 6.59$##.

I'm completely stuck on c though. I keep getting the wrong answer, as I try to calculate the expected loss. For instance, why is the following wrong?

##E(L) =##
##+ 0.3\cdot (0)##
##+ 0.7 \cdot 0.3 \cdot (1)##
##+ 0.7 \cdot 0.7 \cdot 0.3 \cdot (1+2)##
##+ 0.7 \cdot 0.7 \cdot 0.7 \cdot 0.3 \cdot (1+2+4)##
##+ 0.7 \cdot 0.7 \cdot 0.7 \cdot 0.7 \cdot 0.3 (1+2+4+8)##
##+ 0.7\cdot 0.7 \cdot 0.7 \cdot 0.7 \cdot 0.7 \cdot (1+2+4+8+16)##
##= 7.5$##

I mean, these are all the possibilities if Lars is only playing a maximum of 5 times?
 
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  • #2
Are E(L) the expected bets? Then the first line (Lard wins the first round) should not be zero as he still has to put his bet.
Is E(L) the expected total result? Then the first line (Lard wins the first round) should not be zero as he gains money.

The other lines look wrong as well for the same reason.
 
  • #3
thx for reply.

E(L) = the expected loss, so that

Expected Profit = Expected gains - Expected losses = E(W)-E(L).

Is that a wrong line of thinking?
 
  • #4
Pls help i got my exam tomorrow
 
  • #5
See my previous post...
Then the first line (Lard wins the first round) should not be zero as he still has to put his bet.
 
  • #6
It's not the expected bet, but the expected loss. Each line is the probability for a certain outcome multiplied by the amount of money that would be lost in said outcome.

So in the first line, Lars wins on the first try and loses nothing. In the second line, Lars loses on the first try but wins on the second, losing one dollar. and so on until Lars loses 5 times in a row and gives up.

However, I get the wrong expected loss of money with this approach.
 
  • #7
So in the first line, Lars wins on the first try and loses nothing.
Yeah, but he still has to place his bet of 1 dollar to play the first round.
In the second line, Lars loses on the first try but wins on the second, losing one dollar.
He uses 1+2 dollars (and wins 4).
 
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  • #8
Ah yes, of course. He doesn't get his starting money back even if he wins (he just gets twice the original amount). That's where I was stuck. OK, i get the correct answer now. thanks
 

Related to Expected profits in Blackjack - tricky probability problem

1. What is the expected profit in Blackjack?

The expected profit in Blackjack is dependent on various factors such as the number of decks used, the specific rules of the game, and the player's strategy. In general, the expected profit for the casino is around 0.5% of the total amount bet, while the expected profit for a skilled player can be slightly higher.

2. How is the expected profit calculated in Blackjack?

The expected profit in Blackjack is calculated by considering the probability of winning or losing each hand, along with the corresponding payouts. This can be a complex calculation, as it involves taking into account the dealer's up card, the player's hand, and the remaining cards in the deck.

3. Is it possible to have a positive expected profit in Blackjack?

Yes, it is possible to have a positive expected profit in Blackjack if the player uses a perfect strategy and plays at a table with favorable rules. However, the expected profit is still relatively low, and there is always a risk of losing in any gambling game.

4. Can the expected profit change during a Blackjack game?

Yes, the expected profit can change during a Blackjack game, as the probability of certain outcomes can change with each dealt card. This is why card counting strategies can be effective, as they allow the player to adjust their bets based on the changing probability of winning.

5. How can understanding expected profit in Blackjack improve my chances of winning?

Understanding expected profit in Blackjack can help a player make more informed decisions when it comes to betting and strategy. By knowing the probability of winning or losing in each situation, players can avoid making risky bets and increase their chances of making a profit in the long run.

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