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You are offered a chance to play a game. the rules are simple. there are $100$ cards face down. Of these, $55$ say win and $45$ say lose. You begin with $10000$ dollars. You must bet $1/2$ of your money on each card turned over, and you either win or lose that amount based on what the card says. At the end of the game, all the cards have been turned over. Are you ahead or behind at the end of the card game? How much do you have at the end of your game? What arrangement of cards will earn you the most money?

My attempt:

You have a 100% chance of winning if $p>\frac{1}{2}$ and $p$ chance of winning if $p<\frac{1}{2}$. Suppose that $p>\frac{1}{2}$. Then each time you bet exactly half your money. If you have $x$ dollars, you end up with $\frac{3}{2}x$ if you win and $\frac{1}{2}x$ if you lose, hence your expected outcome is $\frac{3}{2} xp + \frac{1}{2}x(1-p)$ which equals $xp + \frac{1}{2}x$, in this case $x=10,000,$ $p=0.55,$ and $1-p=0.45$. So if $p>\frac{1}{2}$, then the total is greater than $x$.

Given that each game on average gains you money and you can play an arbitrary number of games, of course you should have 100% chance of doubling your money.

Similarly, if $p<\frac{1}{2}$, it can be shown that no matter how much we bet, we lose money on average. Then on average our money will tend to go toward zero, so we're better off just going all-in at the start, with $p$ chance of doubling our money.

I was trying to solve this problem and I am not sure if this is correct. I tried teaching this to myself, but I am not sure if I should do it this way or is there an easier way of doing this. I was working on this for a while now and I wanted to see if this concept is also correct. Can someone please help me with the solution?

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  • $\begingroup$ Is there a problem solving method that we can use? $\endgroup$ – 9999 Sep 8 '14 at 18:54
  • $\begingroup$ If anyone wants to answer feel free too $\endgroup$ – 9999 Sep 8 '14 at 19:15
  • $\begingroup$ If you bet $\dfrac{1}{10}$ of your money each turn, you would end up with about $\$16501$ after the hundred cards. But bet $\dfrac{1}{2}$ of your money each turn and you end up with about $\$1.38$ after the hundred cards. $\endgroup$ – Henry Nov 28 '14 at 19:35
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For me, you will have $10000\cdot(3/2)^{55}\cdot(1/2)^{45}$ at the end of the game whatever you do. Multiplication is commutative, so it does not matter in which order you will turn the cards.


Another way of seeing it: suppose you have $N\$$ before your turn. Your expected outcome will be $N\cdot(3/2\cdot55/100 + 1/2\cdot45/100)$. After 100 turns, you will have $10000\cdot(1.05)^{100}$, which once again does not depend on the order.

The method is completely wrong, since your expected outcome will be $N\cdot1.05$ only at the first turn. There is probably a way to prove it using a similar reasoning, but it will be tortuous.

I am letting what I've done before to highlight the fact that this cannot work (in case anyone is tempted to use this method for another problem).

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  • $\begingroup$ I suppose my answer if correct then? $\endgroup$ – 9999 Sep 8 '14 at 18:51
  • $\begingroup$ This answer is well put, and equivalent to what I would have answered had I come earlier $\endgroup$ – Asimov Sep 8 '14 at 18:57
  • $\begingroup$ @user3923424 where did you get the 1.1 again. $\endgroup$ – 9999 Sep 8 '14 at 19:07
  • $\begingroup$ It's 1.05, my mistake. $\endgroup$ – R2B2 Sep 8 '14 at 19:09
  • $\begingroup$ And I think your answer is correct. The problem is that p changes at each turn, so not very handy when you have more than one turn. $\endgroup$ – R2B2 Sep 8 '14 at 19:11
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So the problem with your method is that it doesn't take into account the fact that the probabilities change at each step because cards do not get turned back over.

So to determine your money at the last step, you need to do $10000\cdot\left(\frac{3}{2}\right)^{p}\cdot\left(\frac{1}{2}\right)^{100-p}$

This is only greater than $10000$ for approximately $p\geq 63$.

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  • $\begingroup$ so I suppose that order does not matter with the arrangement of cards? Also when solving for the expected value to find out the last step, I suppose this will tell you whether you are ahead of the game or not? $\endgroup$ – 9999 Sep 8 '14 at 19:21
  • $\begingroup$ Expected value would only work if you weren't removing cards, sorry I obviously need caffeine. $\endgroup$ – Silynn Sep 8 '14 at 19:27
  • $\begingroup$ I suppose order does not not matter then? $\endgroup$ – 9999 Sep 8 '14 at 19:33
  • $\begingroup$ Order does not matter. See "multiplication is commutative" in the answer by R2B2. $\endgroup$ – David K Sep 8 '14 at 20:09
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Let there be $n$ cards, and let exactly $m$ of the cards say "win". The others say "lose". If the cards were placed randomly, the probability that the first card is "win" is $p = \frac mn.$ Start with $N$ dollars in your possession.

In your original problem, $n = 100, m = 55, N = 10000,$ and $p = \frac{55}{100} = 0.55.$

As explained in other excellent answers, as the game progresses (and all $n$ cards are revealed) you will win exactly $m$ times and lose exactly $n - m$ times. Each win multiplies the amount of money you have by $\frac 32.$ Each loss multiplies the amount by $\frac 12.$ The result of several consecutive multiplications is independent of the order, so at the end the number of dollars you have is $$ \left(\frac 32\right)^m \left(\frac 12\right)^{n-m} N = \frac{3^m}{2^n} N = e^{m \ln 3 - n \ln 2} N. $$

So you come out ahead just when $3^m > 2^n,$ which is the same as saying $m \ln 3 > n \ln 2,$ or equivalently, $$\frac mn > \frac{\ln 2}{\ln 3} \approx 0.63093.$$

For $n = 100,$ you come out ahead if $m \ge 64,$ but you lose money if $m \le 63.$

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Define:

  • $N$ Number of cards
  • $N_1$ Number of cards of win.
  • $x_i$ result of the bet on the card $i$ where $i=1...N$ and $x_1=1$ means win and $x_1=-1$ a lose.
  • $M$ ammount of money.

First start with $x_1$, you have $N_1$ cards of win and $N-N_1$ of lose. Therefore: $$P(x_1=-1)=\frac{N-N_1}{N}$$ $$P(x_1=1)=\frac{ N_1}{N}$$

Second $x_2$ if $x_1$ was lose you will have $N-N_1-1$ of lose and $N_1$ of win, and if was win you will have $N-N_1$ of lose and $N_1-1$. Therefore you will have $N-N_1-\min\{x_1,0\}$ of cards loses and $N_1-\max\{x_1,0\}$ of cards of win therefore:

$$P(x_2=-1|x_1=k_1)=\frac{N-N_1-\min\{k_1,0\}}{N}=1-\frac{N_1-\max\{k_1,0\}}{N}$$ $$P(x_2=1|x_1=k_1)=\frac{N_1-\max\{k_1,0\}}{N-1}$$

For $x_j$ a similar reasoning shows:

$$P(x_j=-1|x_1=k_1,x_2=k_2,...,x_{j-1}=k_{j-1})=1-\frac{N_1-\min\{N_1,\sum_{l=1}^{j-1}\max\{k_l,0\}\}}{N_1-j+1}$$ $$P(x_j=1|x_1=k_1,x_2=k_2,...,x_{j-1}=k_{j-1})=\frac{N_1-\min\{N_1,\sum_{l=1}^{j-1}\max\{k_l,0\}\}}{N_1-j+1}$$

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