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143 votes
Accepted

±1-random walk from 5 until 20 or broke

You can use symmetry here - Starting at $5$, it is equally likely to get to $0$ first or to $10$ first. Now, if you get to $10$ first, then it is equally likely to get to $0$ first or to $20$ first. ...
Michael Biro's user avatar
  • 13.8k
142 votes

±1-random walk from 5 until 20 or broke

It is a fair game, so your expected value at the end has to be $5$ like you started. You must have $\frac 34$ chance to go broke and $\frac 14$ chance to end with $20$.
Ross Millikan's user avatar
50 votes

When to stop rolling a die in a game where 6 loses everything

Before deciding whether to stop or roll, suppose you have a non-negative integer number of points $n$. How many more rolls should you make to maximise the expected gain over stopping (zero)? ...
Parcly Taxel's user avatar
30 votes
Accepted

When to stop rolling a die in a game where 6 loses everything

In the last round you can get $\frac{1+2+3+4+5}{6}$ or lose $p\frac 1 6$, whenever the second is more than the first you should stop. So once you have scored more than 15 you should stop. If you score ...
Djura Marinkov's user avatar
26 votes

±1-random walk from 5 until 20 or broke

Hint: for $0 \le n \le 20$, let $p_n$ be the probability that you go broke if you start with $n$ points. You have $p_0=1$ and $p_{20}=0$. For $0 < n < 20$ you have $$p_n = \frac{1}{2} p_{n-1} + \...
angryavian's user avatar
  • 90.8k
25 votes

When to stop rolling a die in a game where 6 loses everything

The question is missing the concept of utility, a function that specifies how much you value each possible outcome. In the game, the utility of ending the game with a certain score would be the value ...
Timothy Shields's user avatar
12 votes
Accepted

The gambler makes 100 bets and wins 10. How much money does he have at the end?

Odds of 8 to 1 means that for every \$1 you wager, you could either win \$8 in addition to receive your \$1 wager back lose your \$1 wager So from his 10 wins, he gets \$80, and from his 90 losses, ...
msitt's user avatar
  • 470
9 votes
Accepted

Expected value of repeatedly betting on a coin flip?

The deception Here's an explanation for the deception: on average, we know that we will win in each round. That's what happens on a round-to-round basis. However, in the long run, that's not what ...
Sarvesh Ravichandran Iyer's user avatar
9 votes

Price of Option in Betting Game

The error is here: If this is the case, I choose to regenerate my number to try and win, which gives me another 50% chance of winning. No, it doesn't. It would give you an extra $50\%$ chance of ...
Especially Lime's user avatar
8 votes

Asymmetric ruin probability

As indicated, the following recursion holds $$p(n+2)-2p(n)+p(n-1)=0.$$ It only has a solutions on the form $r^n$, where $r$ is a root of the characteristic polynomial $$r^3-2r^1+r^0=0.$$ Finding the ...
mbe's user avatar
  • 947
8 votes

Price of Option in Betting Game

Let $U_1$, $U_2$ and $U_3$ be the three numbers the probability of you to win $$P(U_1> U_2) + P(U_3 > U_2 > U_1) = \frac12 + \frac16 = \frac23$$
Kroki's user avatar
  • 13.2k
7 votes
Accepted

3 person bet based on the perceived likelihoods of an outcome

This is an interesting problem. I thought about it for some time and I believe I have a complete solution. First, let's see how we would answer the question if we had only 2 people. Say, Alice ...
Thanassis's user avatar
  • 3,175
7 votes

±1-random walk from 5 until 20 or broke

All answers so far are great but some readers seem to feel they lack an intuitive explanation - and perhaps the maths to back it up. Consider that you have an equal chance of moving up or down along ...
FreeElk's user avatar
  • 231
7 votes

Kelly Criterion for simultaneous independent bets

Any elaborate answer appears to be helplessly convoluted. Nonetheless, Withrow (2007) concluded: When the number of bets is small, the optimal sizes of bet seem to be almost exactly proportional to ...
useranonis's user avatar
6 votes

When to stop rolling a die in a game where 6 loses everything

There is unfortunately many mistakes in the calculation of the expectation proposed: 1) The expectation does not take into account the optionality of the player deciding to play or not. 2) Also ...
Litteul's user avatar
  • 86
6 votes

The gambler makes 100 bets and wins 10. How much money does he have at the end?

Odds of 8 to 1 mean you make a bet of \$1, then you win \$8. You also receive your original \$1 back, so if you win an 8 to 1 bet, you pay \$1 and receive back \$9. In regards to your problem, the ...
John Doe's user avatar
  • 14.6k
6 votes
Accepted

When should I stop playing this dice game?

Your calculation is incorrect because $2^{n}\cdot \left(\frac 56 \right)^n$ is the expected profit of rolling $n$ times and quitting. You should not subtract the $\frac 162^{n}$ because the loss of ...
Ross Millikan's user avatar
6 votes

Best strategy to reach $500 for a gambling situation in a casino

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 ...
Henry's user avatar
  • 158k
5 votes

Gambler's fallacy and the Law of large numbers

Any sequence has the same probability as any other, but there are more sequences that are "balanced" than any other given proportion. For example, if I flip a coin 4 times then there are 6 ways to get ...
Deusovi's user avatar
  • 2,857
5 votes

When to stop rolling a die in a game where 6 loses everything

It's not hard to see that the optimal strategy in this game is "stop on $k$ or higher," for some positive integer $k$. This is because what is optimal is only determined by your current score, not ...
Mike Earnest's user avatar
  • 77.5k
5 votes
Accepted

Why don't billionaires (or multi-millionaires for that matter) use the Martingale betting system?

Two main reasons: It is actually unprofitable in the very long run, no matter how large one's finite wealth is. In your example, the player is one loss away from going bust as early as round 17 (edit:...
kviiri's user avatar
  • 1,305
5 votes

The perfect gambler - would it work?

This is a gambling system called the "Martingale System" and it has a reasonable chance of working in the short term, but the issue is that in the long term you will always lose. if you just bet one ...
Alex Robinson's user avatar
5 votes
Accepted

The perfect gambler - would it work?

The flaw is that if you lose seven in a row you don't have enough coins to double your bet again. It is a losing bet, so the more you play the more certain you are to lose. No betting strategy can ...
Ross Millikan's user avatar
5 votes
Accepted

How Much is "Large" in the Law of Large Numbers?

The Law of Large Numbers is not an empirical result. You cannot demonstrate this with your own finite-trial experiment. It is a mathematical statement which states that - if you're familiar with ...
Clarinetist's user avatar
  • 19.6k
5 votes
Accepted

Infinite Coin toss: Borel-Cantelli and Kolmogorov

It suffices to show that, with probability 1, B gives 1 dollar to A $n$ times in a row (or less if the game will end). We can split up $A_1,A_2,\dots$ into blocks of $n$, i.e., let $E_1$ be the event ...
mathworker21's user avatar
  • 34.6k
5 votes

How many tickets should Paul buy?

I've found a proof for $n=\textbf{196}$. In fact, Paul can guarantee a third with the following strategy. Observe that if you consider the set $G=\{1, 2, ..., 49 \}$ as the union of three sets $A, B$ ...
Dr. Mathva's user avatar
  • 8,641
5 votes
Accepted

A winning wager that loses over time

Here are all the possible outcomes if you stop after ten tosses of the coin. I have rounded the numbers for display, but the calculations were at the full precision of the software on which they were ...
David K's user avatar
  • 99.5k
5 votes

Expected value and Gambler's fallacy

You’re mixing up various things here. The fact that the expected value is zero doesn’t imply that we end up at any particular point; it’s just an expected value. To disentangle the ...
joriki's user avatar
  • 239k
5 votes

Optimal strategy in betting game

I would argue that the optimal initial betting is $\frac{2 - \sqrt{2}}{2} \sim 0.29$. To do so, we need a few arguments: Whatever the initial bet is that you use to go from $0$ to $+1$, the second ...
E-A's user avatar
  • 5,987

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