Gambling

benworth

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Apr 26, 2021
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Hello. I seem to be creating a storm on a gambling website. But I don't dispute math. How can those two statements both be true? ?

Because you now have some idea about where I'm coming from, I have some questions that may not be exhausted by this post:

1. What qualification do respondents of mathematical questions have, on this website?

The storm I'm kinda engulfed in, is to do with math as applied to gambling. Specifically, the game of blackjack (BJ). This game, according to stats, has a house edge of about 0.5%. If we remove ties (where the player and dealer draw the same card total, resulting in no win or loss to either), then the house edge is about 8% (54% in favour of the dealer, 46% chance of winning for the player).

If a player uses a Fibonacci betting strategy (a negative progression strategy), then generally speaking, he would only need to win approx. 50% of hands dealt, in order to return him to his starting pot. Agreed?

The way I've described the game of BJ is somewhat narrow, though, because although the above may be true, one needs to accept the natural variation inherent in the game. Two of the main variations being a number of streaks of losses, favouring the house or player. And if a large number of losses were to be experienced by the player, who is using a negative progression strategy, then this may mean the player runs out of money.

However, what if a player is insanely wealthy and has a line of credit at a casino, worth $100 billion. And what if this same player were to start with a small bet of just $5. My next question is:

2. Because the player starting bet is so small ($5) compared to his bankroll ($100 billion), would it not be feasible that this player should always come away from a casino a winner, in regard to his starting and finishing pot (assuming time is never an object, and he applies the Fibonacci betting strategy)?

You may want to note that from what I've read, the only reason that a good BJ player loses using a Fibonacci betting strategy, is because he experiences a catastrophic losing streak, and he runs out of money. But for the above example, the player would need to experience a seriously abnormal losing streak to run out of money. A losing streak that would exceed the natural variation of the game of BJ.

Best regards,

David
 
What you said is flawed because the BJ tables have a max bet, so having a $100 billion will not help.
Just card count and you'll get a small (very small) percent advantage. Read the book, Beat The Dealer by Ed Thorpe.
 
No, I don't want to buy and read a book. Imagine there were no table limits. Is the question too hard?
 
Let’s assume you are making a fair bet. You stake a dollar; you lose it with probability 50%, you get it back plus an additional dollar with probability 50%.

I am going to ignore table limits and limited financial resources. But I am going to assume a finite number of bets.

Let’s say a maximum of one. What is your expected gain. 0. Probability of gain of 1 = 50%. Probability of loss of 1 = 50%.

Maximum of two. What are the possible outcomes?
Win first bet and quit. Probability of gain of 1 = 50%.
Lose first bet, win second bet. Probability of breakeven = 50% * 50% = 25%.
Lose first and second bet. Probability of loss of 2 = 25%.
Your expected gain is zero 50(1) + 25(0) + 25(-2) = 50 - 50 = 0.

Maximum of three. What are the possible outcomes?
Win first bet and quit. Probability of gain of 1 = 50%.
Lose first bet, win second bet, and quit. Probability of breakeven = 50% * 50% = 25%.
Lose first and second bet and bet 2.
Win third bet. Breakeven. Probability 12.5%.
Lose third bet. Loss of 4. Probability 12.5%.
Your expected gain 50(1) + 37.5(0) + 12.5(-4) = 50 - 50 = 0.

Maximum of four. What are the possible outcomes?
Win 1st bet and quit. Gain 1. Probability = 50%.
Lose 1st bet, win 2nd, and quit. Gain 0. Probability 25%.
Lose first and second bet. Bet 2. Win 3rd bet, bet 1, and win. - 1 - 1 + 2 + 1 = 1 in gains. Probability 6.25%.
Lose 1st and second. Bet 2. Win third bet, bet 1, and lose. - 1 - 1 + 2 -1 = - 1 in gains. and fourth. Probability 6.25%.
Lose first three bets. Bet 3 and win. - 1 - 1 - 2 + 3 = - 1. Probability 6.25%.
Lose all four bets. - 1 - 1 - 2 - 3 = - 7. Probability 6.25%.
Your expected gain 50(1) + 25(0) + 6.25(1 - 1 - 1 - 7) = 50 -8 * 6.25 = 50 - 50 = 0.

You should see the pattern by now. The expected value is always zero no matter how great the number of bets is, provided the number is finite.You can prove that by mathematical induction if you know that technique. So you are correct that at a fair game the fibonacci approach leads to an expected value of zero, but that is not an expected gain. And of course, you are not talking about a fair bet in a casino.
 
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Upon re-reading my previous post, I realized that the OP probably does not recognize the hidden assumption in his reasoning. That last big but winning bet still leaves you in the hole. I suspect that the OP assumed that the final winning bet resulted in a net gain. A very large winning bet can only occur after a long string of losing bets, which have dug a very deep hole.

It is also possible that the OP does not realize that strict adherence to the strategy means that the probability of making fewer than three bets is 75%. This long string of bets that the OP is contemplating is very unlikely to occur.

It amazes me that sensible people can believe that a series of fair bets can have a probability of gain. I think it comes from failure to actually do numerical experiments. They make qualitative arguments about quantitative processes and talk themselves into knots.
 
While taking Linear Algebra I solved the following problem which I made up. (I think that a horse race has 9 horses)Can you come up with different bets on each of the 9 horses in a race, where the amount bet depends on the odds of each horse, such that you either break even or win. I was disappointed with my results!
 
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