Hi,

A gambling game is organized.

There are 3 outcomes : 1, 2 or 3.

The draw is random. Each number has the same probability to appear (1/3).

If your number is picked out of the 3 you win 2 dollars for one dollar bet.

Imagine now a large family with k members.

On each draw 3 of them bet 1 dollar each one.

Each time one of them will surely make profit of one dollar. The 2 others are going to loose.

Once one of the family members win he will stop playing and be replaced by a new member. The 2 others continue betting one dollar per draw.

They continue to bet UNTIL one of them make one dollar profit. I mean : win minus loss equals one dollar. Then another member of the family replace him.

We do not know what is the optimal number k required such as all the k members make a global profit as family.

All the members have to bet at least once.

So the question is : how many members a family must have to make a profit of at least one dollar?

Thank you for any comment.

## Puzzle : gambling game

**Moderators:** jestingrabbit, Moderators General, Prelates

### Re: Puzzle : gambling game

I might be misunderstanding the rules, so let me clarify - three players each put up $1, and two lose their dollar while the third walks away with $1 plus his original dollar? If this set up is correct, then imagine the game from the perspective of the "house". Each round, the house takes in $3 and pays out $2. Put another way, the players on aggregate put up $3 and walk away with $2, so the "profit" each round is -$1. No matter how long the players continue the game, they can never turn a profit as a whole.

If, on the other hand, the winning player gets $2 plus his bet back... Then the net gain/loss is 0 and while the family can never turn an aggregate profit, they also end up breaking even so never lose either.

If, on the other hand, the winning player gets $2 plus his bet back... Then the net gain/loss is 0 and while the family can never turn an aggregate profit, they also end up breaking even so never lose either.

### Re: Puzzle : gambling game

So every round with 3 players yields: +1 profit, -2 loss, net: -1 loss.

Every round can therefore be seen as removing the 'third' player from the table (the newest player to be seated) and adding -1 to each of the running totals for players A and B.

If K=N+2, then at the end of N rounds, the situation is: A: -N, B: -N, N:+1 gross profit being +N, gross loss being -2N, and net being -N (not yet a profit).

At this point there are only 2 players left in the family.

A: -N B: -N

Each of these players has a 1/3rds chance of choosing the right number, but there is no longer a 3/3 chance of the number being hit (only 2 players).

If, for simplicity, we allow that player A will always hit the winning number, but will only do so 2/3rds of the time, then:

After (3/2)N rounds our state will ideally be:

A: 0 B: -(5/3)N

After (3/2)N +1 rounds we will have:

A: +1 B: -(5/3)N -1 and gross profit is +N+1, gross loss being -(5/3)N-1, net being -(2/3)N

Finally we have only 1 player left, who only has a 1/3 chance of selecting the right number. The odds being against this player ever leaving the table, so I think the original question is asking: what # of family members minimizes the # of games needing to be played by the final player before the net reaches +1.

At this point, after an additional (2/3)N net wins by player B, we will have:

B: -N-1, gross profit is N+1, gross loss is N-1, the net is 0.

After one more win (or (2/3)N+1) then the net is +1, while B is at -N. This happens after: N[3 players]+(3/2)N+1[2 players]+(2/3)N net wins+1 rounds. or:

(5/3)N +2 + (2/3)N net wins rounds.

Odds of having player B nail (2/3)N net wins is directly dependent upon N. Higher N, lower chance of it happening.

I feel like the answer is K=3, but that formula up there might yield something surprising, or be completely crazy-wrong, dunno. These are my thoughts so far.

Every round can therefore be seen as removing the 'third' player from the table (the newest player to be seated) and adding -1 to each of the running totals for players A and B.

If K=N+2, then at the end of N rounds, the situation is: A: -N, B: -N, N:+1 gross profit being +N, gross loss being -2N, and net being -N (not yet a profit).

At this point there are only 2 players left in the family.

A: -N B: -N

Each of these players has a 1/3rds chance of choosing the right number, but there is no longer a 3/3 chance of the number being hit (only 2 players).

If, for simplicity, we allow that player A will always hit the winning number, but will only do so 2/3rds of the time, then:

After (3/2)N rounds our state will ideally be:

A: 0 B: -(5/3)N

After (3/2)N +1 rounds we will have:

A: +1 B: -(5/3)N -1 and gross profit is +N+1, gross loss being -(5/3)N-1, net being -(2/3)N

Finally we have only 1 player left, who only has a 1/3 chance of selecting the right number. The odds being against this player ever leaving the table, so I think the original question is asking: what # of family members minimizes the # of games needing to be played by the final player before the net reaches +1.

At this point, after an additional (2/3)N net wins by player B, we will have:

B: -N-1, gross profit is N+1, gross loss is N-1, the net is 0.

After one more win (or (2/3)N+1) then the net is +1, while B is at -N. This happens after: N[3 players]+(3/2)N+1[2 players]+(2/3)N net wins+1 rounds. or:

(5/3)N +2 + (2/3)N net wins rounds.

Odds of having player B nail (2/3)N net wins is directly dependent upon N. Higher N, lower chance of it happening.

I feel like the answer is K=3, but that formula up there might yield something surprising, or be completely crazy-wrong, dunno. These are my thoughts so far.

### Re: Puzzle : gambling game

Unless I'm misreading something this is essentially just a Martingale strategy distributed among multiple people. The answer is infinitely many; there's no way to combine a finite number of negative expectations to add up to a positive result.

### Re: Puzzle : gambling game

essentially just a Martingale strategy distributed among multiple people

I mean, maybe the original poster was asking this question, hoping to get some sort of answer that would be a sure bet for him and his friends to head down to Vegas for some fun. I dunno.

Short answer: Yes, infinite series (or finite) of negative events will always yield negative. But if you 'beat the odds' you can come out ahead on an average/players while still being down yourself.

A: +1 B: -(5/3)N -1 and gross profit is +N+1, gross loss being -(5/3)N-1, net being -(2/3)N

So let's just make K=50, N=48.

So we're at +49, -81 loss.

(5/3)N +2 + (2/3)N net wins rounds.

So we need our final player to make up 33 dollars, so that our loss becomes -48, and our net profit of 1.

Winning 33 rounds at a 1/3 chance is really slim. But if we allow him as many rounds as we'd like, there is a chance. Obviously winning 33 times in a row is 1/3^33, which is very small (1/5559060566555523) but is >0. But he could also win 32 times in a row, one loss, and then two wins. Or 31 times in a row, one loss, three wins, etc. Just need any combination of games played where the net wins over that series is (2/3)N. (+1? can't keep it straight)

This continues to work if you keep reducing the size of K, and K=1 is the best. Short story: betting is bad when the odds are <50%

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