4
$\begingroup$

Suppose a gambler has \$100 to start with. Each time he/she has 0.4 chances of winning and 0.6 chances of losing a bet. If he/she wins he gets twice the money he put in and loses what he bet if he loses. The game stops if he/she has \$500 at hand or goes bankrupt (\$0). Which strategy will he/she choose to maximize the chance of winning(having \$500 and quit):

(1)All in each time

(2)Bet half of the current amount each time

(3)Bet \$10 each time

My first try is to use the optimal stopping time for martingales to find the probability of exit at $500, but it is a supermartingale and I am not sure how to apply OPT.

I also tried to model the problem as a Markov Chain under each strategy. The first strategy is easy. The exit distribution for the third strategy is also computable. But I am stuck on the second one.

I think there may be an easy way (maybe an intuitive way or a way that simplifies the computation) of picking the best strategy among the three.

Edited:

For strategy (1), the probability of winning is $$P(\text{winning}) = P(\text{win 3 bets in a row}) = (0.4)^3 = 0.064$$

For strategy (3), as pointed out, it is an unbalanced gambler's ruin, so $$P(\text{winning}) = P(\text{reach \$500 before \$0}) = \frac{(0.6/0.4)^{100}-1}{(0.6/0.4)^{500}-1} \approx 3.66e-71 $$

But for strategy (2), there are infinite many states $$S = \left\lbrace 100\times\left(\frac{1}{2} \right)^n\times \left(\frac{3}{2}\right)^m: \forall n,m \in \mathbf{N}^+,100\times\left(\frac{1}{2} \right)^n\times \left(\frac{3}{2}\right)^m \leq 500 \right\rbrace$$

One of my guesses is that the possibility of bankruptcy for (1) and (3) is positive. However, it takes infinite many bets for the chain to $0 in (2), so the possibility of bankruptcy is 0?

$\endgroup$
13
  • $\begingroup$ So if I start playing this game with $\$100$, and my first bet is $\$10,$ then after the first bet I will have a total of $\$110$ with probability $0.4$ (winning) and I will have $\$90$ with probability $0.6$ (losing). Is that how it works? $\endgroup$
    – David K
    May 12 at 20:42
  • $\begingroup$ Yes, that will be the third strategy. $\endgroup$
    – Gavin
    May 12 at 20:48
  • $\begingroup$ Following strategy (1) is a little excessive: if you win the first two bets, you only want to bet $\$100$ on the third bet, so you can still have a chance if you lose the third bet. So adjust it to "All in if you have less than $\$250$" and "what is still needed if you have at least $\$250$" $\endgroup$
    – Henry
    May 12 at 20:50
  • 1
    $\begingroup$ You say you have calculations for the first two. Please edit your question to include them $\endgroup$
    – Henry
    May 12 at 21:18
  • 1
    $\begingroup$ @Henry calculation added $\endgroup$
    – Gavin
    May 12 at 23:31

2 Answers 2

5
$\begingroup$

Your result of $0.4^3=0.064$ is correct for strategy 1. As expected, it turns out to be better than strategies 2 or 3 but not sensibly adapted versions of any of the three strategies.

Your calculation for strategy 3 using the Gambler's Ruin seems to use bets of $1$ each time but your question was about bets of $10$ so the calculation becomes $\frac{(0.6/0.4)^{100/10}-1}{(0.6/0.4)^{500/10}-1} \approx 8.89\times 10^{-8}$, a lot worse than strategy 1.

A better version of strategy 3 would be to bet $100$ each time, making the calculation become $\frac{(0.6/0.4)^{100/100}-1}{(0.6/0.4)^{500/100}-1} \approx 0.0758$, better than the probability for strategy 1.

Strategy 2 is more complicated to calculate, even assuming you can make arbitrarily small bets. You will never go bankrupt but the likelihood is that your bankroll will soon become very small and you will never reach your target. If you lose $n$ times then you need to win $f(n)$ times with $0.5^n1.5^{f(n)}\ge 5$, which makes $f(n)= \Big\lceil\frac{\log(5) - n \log(0.5)}{\log(1.5)}\Big\rceil$. Finding the probability of doing this is not easy because you need to find the number of patterns for each $n$to give $\text{patterns }\times 0.6^n 0.4^{f(n)}$. Summing over $n\ge 0$ seems to give the probability of winning with strategy 2 of about $0.0472$, worse than strategy 1. I used the R code, though this may contain errors

f <- function(n,bet,target){ceiling((log(target)-n*log(1-bet))/log(1+bet))}
probwin <- 0.4
target <- 5
bet <- 0.5
proboverall <- 0.4^f(0,bet,target)
ways <- rep(1,f(0,bet,target))
for (n in 1:200){
  ways <- cumsum(ways)
  ways <- c(ways, rep(ways[f(n-1,bet,target)],
                      f(n,bet,target)-f(n-1,bet,target)))
  proboverall <- proboverall + ways[f(n,bet,target)] * 
                              (1-probwin)^n * probwin^f(n,bet,target) 
  }
proboverall

We might do even better by adapting strategy 2 to bet $\sqrt[3]{5}-1 \approx 71\%$ of the current amount each time and that could make the probability of winning about $0.0773$, better than strategy 1 and better than the adapted strategy 3.

But having adapted strategies 2 and 3, perhaps we can do even better by adapting strategy 1. As I said in the comments, it is a little excessive: if you win the first two bets, you only want to bet $100$ on the third bet, so you can still have a chance if you lose the third bet; ending up with $800$ rather than $500$ is unnecessary and so unnecessarily risky. So adjust the strategy to "All in if you have less than $250$" and "what is still needed if you have at least $250$". Ross Millikan pointed out that if you have $400$ and lose $100$ you go to $300$; if you then bet $200$ and lose then you are back where you started with $100$. So if the probability of winning with this strategy is $p$ then $p=0.4^3+0.4^2 \times 0.6\times0.4 +0.4^2\times 0.6^2\times p$ which you can solve to give $p=\frac{64}{589} \approx 0.1087$.

This means that, in order of probability of winning, you have

$\begin{align} 1. & \text{ Adapted strategy }1: & 0.1087\\ 2. & \text{ Adapted strategy }2: & 0.0773\\ 3. & \text{ Adapted strategy }3: & 0.0758\\ 4. & \text{ Strategy }1: & 0.0640\\ 5. & \text{ Strategy }2: & 0.0472\\ 6. & \text{ Strategy }3: & 0.000000089\\ \end{align}$

$\endgroup$
2
$\begingroup$

The simple approach is that the game is unfair to the gambler, so s/he needs to be lucky to win. You want to bet as few times as possible to increase the variance, so the modified $1$ is the best route.

To get the probability, you have to win the first two or you are done. If you win the third as well, you win. If you lose the third you have $300$. At each point in the process one of the results is final. If you lose the fourth as well you are back at the start, so can write an equation to compute the overall probability.

$\endgroup$
5
  • $\begingroup$ Thanks. How would you go about comparing the original three strategies? Which one is the best (among the given three)? $\endgroup$
    – Gavin
    May 12 at 21:08
  • $\begingroup$ I think this approach gives an answer of $\frac{64}{589}$ $\endgroup$
    – Henry
    May 12 at 21:23
  • $\begingroup$ I would still bet $1$ is the best. You need three wins in a row, so the probability is $0.4^3=0.064$. The only competitor is $2$, but it takes so many more wins I would be shocked. $3$ is the way people play the lottery to guarantee a loss. $\endgroup$ May 12 at 22:18
  • $\begingroup$ @RossMillikan Strategy 3 can be made much better by increasing the constant bets to size $100$ $\endgroup$
    – Henry
    May 13 at 9:08
  • $\begingroup$ @Henry That is in line with what I was saying. You need to be lucky, so you want to make fewer bets. That puts it into the same scale as the other two. It is not so much how you compute how much to bet, you want to bet more each time. $\endgroup$ May 13 at 13:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.