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gambling strategy


jasen
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In a gambling game, the rules are :

you bring an amount of money,
you can bet any amount of money you like,

the game stop if :
 1. your money have doubled (exactly 2x your first amount of money)
 2. or you lose all of your money. 

you can't stop in the middle of the game.

example : if you bring 10 dollars:

1. If you bet all your money in the first bet, so if you win you get 20 dollars and game over, or if you lose, you lose all your money and game over.
2  If you have already win 19 dollars, so in the next game you maximum bet is only 1 dollar. 

what is your strategy to win the game ?

Edited by jasen
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I don't know of a strategy to win coin tosses. I would expect to go broke or double my money with equal likelihood. Even so, here's what I would probably do:

Run a Martingale scheme. Bet $1 and keep betting $1 as long as I win. When I lose I double up. So long as I don't lose 3 times in a row, I will double my $10. Even so, I will expect to double or go bust an equal number of times.

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Now I am confused.

I think I followed the strategy you described yesterday, when you postulated that it would deliver 100% chance of winning.

But, now your simulation it gives you a 50% chance of winning.

 

Did I miss something??

I was quoting bonanova solution/answer, not my proposed strategy.

I was working on the strategy that you describe:

  • Bet half my money [rounded to an integer]; or
  • bet the "ceiling" [twice my original money - Current money].

Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.

Your strategy is not applicable, you can't bet more money than you have,  if your current money is 50$, your original money is 100$, so 2*100-50 = 150$, which you don't have.

Now I explain, my proposed strategy.

If I have 100$, then I bet 50$, if I lose, my money remains 50$
Now I have 50$, then I bet 25$, if I lose, my money remains 25$
Now I have 25$, then I bet 12.5$, if I lose, my money remains 12.5$

this keep happens, the game will never over, to infinitely small amount of money.

but :

if I have 100$, then I bet 50$, if I win, my money is 150$
Now I have 150$, then I bet 50$, if I win, my money is 200$

game over, and I win.

so. it seem that I will win 100%.

But, My proposed strategy although theoretically true, it is not applicable.
If I'm lucky I will win in a few step, but if not, it will take an infinite time to win.
then I will give up, an bet rest of my money.

so I conclude, there is no strategy to win, we only have 50% chance to win.

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I was working on the strategy that you describe:

  • Bet half my money [rounded to an integer]; or
  • bet the "ceiling" [twice my original money - Current money].

Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.

Curious about your result. If you average over many trials are your expected winnings positive?

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I was working on the strategy that you describe:

  • Bet half my money [rounded to an integer]; or
  • bet the "ceiling" [twice my original money - Current money].

Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.

Curious about your result. If you average over many trials are your expected winnings positive?

As I mentioned, I only did a small number of runs [about 100].

BTW - I like your title "Retired Expert"!

 

I think that if you do a large number of runs, you eventually get to about 50%.

Now I am confused.

I think I followed the strategy you described yesterday, when you postulated that it would deliver 100% chance of winning.

But, now your simulation it gives you a 50% chance of winning.

 

Did I miss something??

I was quoting bonanova solution/answer, not my proposed strategy.

I was working on the strategy that you describe:

  • Bet half my money [rounded to an integer]; or
  • bet the "ceiling" [twice my original money - Current money].

Testing this in a spread sheet [with only about 100 trials] I am winning about 62% of the time.

Your strategy is not applicable, you can't bet more money than you have,  if your current money is 50$, your original money is 100$, so 2*100-50 = 150$, which you don't have.

Now I explain, my proposed strategy.

Hidden Content

yes, if you [and the house] are willing to bet micro, nano or pico dollars, then the game can go on forever. But your money is gone, so it is theoretically correct, but not very practical.

 

 

Edited by dgreening
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