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Expected profits in Blackjack - tricky probability problem

  • Thread starter Nikitin
  • Start date
  • #1
726
27

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?
 
Last edited:

Answers and Replies

  • #2
34,366
10,438
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
726
27
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
726
27
Pls help i got my exam tomorrow
 
  • #5
34,366
10,438
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
726
27
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
34,366
10,438
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
726
27
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
 

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