Cyclical Luck

[wdbaker]

Having said that, please advise me on what outcome to bet on next, in order to have an edge, in the following real-world scenarios: Having said that, please advise me on what outcome to bet on next, in order to have an edge, in the following real-world scenarios:
1. The coin is flipped and it comes out Heads 5 times in a row.

Call me crazy but if the coin isn't biased I would bet on tails and if I missed, double my bet for tails etc... until i couldn't bear the burden of the bet or I hit. Obviously you need a point to bail out at, money management, but all things considered I think it would be a good progressive bet.

And no, this is not how I play hold'em

Flame away [img]/forums/images/icons/laugh.gif[/img]

One street at a time
wdbaker Denver, Co

[Jackdaw]

As long as the coin isn't biased - there's nothing wrong with your proposed plan of action, but it won't confer any advantage. It yields the same expectation as any other series of betting decisions in this situation. The reason people feel like you do, and there are millions of them! is because the probability of 6 heads coming up in a row (before the first coin toss has been made)is completely different from the odds of a head on a single (the next) toss.

[BruceZ]

It is true that this yields no gain in expectation, but it does offer an advantage in that he will be virtually guaranteed to be ahead at some point, so long as he is allowed to double his bet enough times. For example, if he can double his bet up to 7 times, then he has over a 99% chance of being ahead by 1 bet. When that happens, he can quit a winner. He can also start over, and have a 99% chance of making another bet. The reason he does not have a positive expectation even though he is virtually guaranteed of making money is because of the small probability that he will lose 7 times in a row. Although the probability of losing 7 flips in a row is 1 in 128, if he gets greedy and plays until he does lose this many in a row, it means he will have lost 1+2+4+8+16+32+64 = 127 bets. So when this finally does happen, he will just be even on average, since the other 126 times he will have only won 1 bet each time. Of course this can happen sooner than 127 times, and it can even happen on the very first time. The more times you are allowed to double, the longer you can make money with a very small chance of losing. With an infinite bankroll, you can make money forever if you are allowed to keep doubling, even though you never have a positive expectation.

[Cyrus]

"If the coin isn't biased ..."

I never said the coin is or isn't biased!

"I would bet on tails and if I missed, double my bet for tails etc... until i couldn't bear the burden of the bet or I hit. Obviously you need a point to bail out at, but all things considered I think it would be a good progressive bet."

I am ready to offer you a game you cannot lose at! Here it is.

I will flip a coin. If it comes up Heads, you win $1. If it comes up Tails, you win nothing but you also lose nothing. If you have won the previous flip, I will flip again and with the same rules, Heads you win $1, Tails you lose nothing. And so on. I will flip you that coin until kingdom come. (Don't worry about getting paid, I'm good for it. I'm extremely wealthy!) These are all the rules!

Sounds like a good proposition to you ?

If it does, please tell me how much money you'd pay me upfront to play that game. I'm gonna auction my prop and if you the one who pays me the highest price, you gets to play.

[Cyrus]

"With an infinite bankroll, you can make money forever if you are allowed to keep doubling, even though you never have a positive expectation."

A player with an infinite BR cannot really increase his wealth.

If BR = inf, then how can BR+1u > BR ?

[img]/forums/images/icons/wink.gif[/img]

[BruceZ]

If the coin is fair, it's worth $1, so I'll give you $.99 per game.

EV = 1(1/2)^2 + 2(1/2)^3 + 3(1/2)^4 + ... = 1

That is, we win n dollars with probability (1/2)^(n+1) since we have to get n heads followed by 1 tail.

How do we know this sum is 1? Two ways. First by logic. Suppose a billion people played this game until they each got the first tail. How many heads do you expect to be thrown? Same as the number of tails since they are equal probability. Each person throws 1 tail, so there must be a billion tails and a billion heads, or an average of 1 head per person. That's the easy way.

You can also sum the series. First look at:

S = 1 + x + x^2 + x^3...
xS = x + x^2 + x^3 + ...
------------------------- subtract
S(1-x) = 1
S = 1/(1-x) this the sum of a geometric series when x<1

Now take the derivative of S:

S' = 1 + 2x + 3x^2 + 4x^3 + ... = 1/(1-x)^2

take x = 1/2
S' = 1 + 2(1/2) + 3(1/2)^2 + 4(1/2)^3 + ... = 1/(1-1/2)^2 = 4

(S'-1)/4 = 2(1/2)^3 + 3(1/2)^4 + ... = (4-1)4 = 3/4

If we add (1/2)^2 =1/4 to this, we get EV above, so this is 3/4 + 1/4 = 1. QED

[BruceZ]

I didn't say he would increase his wealth. I said he would make money forever. There is a difference.
BR(n) = BR(n-1) + money_made(n)

If you want to increase your wealth, just use OPM (other people's money). [img]/forums/images/icons/smirk.gif[/img] [img]/forums/images/icons/grin.gif[/img]

[BruceZ]

I forgot to mention the easiest way of all. If you lost a dollar for a tail, the game would be even money. Since you don't lose a dollar for the game ending tail, the EV must be 1 dollar. This is essentially the same as the logical way.

Now here's one for you. You throw 2 dice until they come up 7. If you had to bet on which roll this was to occur, which roll would you bet on?

[Cyrus]

I will accept the measly 99 cents offered by Mr Bruce Z, who is kindly asked to refrain from responding before tomorrow, to this post.

...I have an even better proposition:

I will flip the same fair coin and each time you win, you double your money! We will start with you winning $2 if the first flip is Heads. If it's Tails you win nothing (and lose nothing).

If the 2nd flip is also Heads you win $4, if 3 flips in a row come up Heads you win $8, and so on. In other words, for n consecutive wins, you win $2 to the power of n. Or ($2)^n.

Let's see what the player's EV is now:

EV = [sum of all expected payoffs of all the possible outcomes]

EV = [(probability of 1 Heads in a row)*($2)] +
[(probability of 2 Heads in a row)*($4)] +
[(probability of 3 Heads in a row)*($8)] +
...

We know that the probability of throwing 1 Heads is 50%, or (1/2) ; of throwing 2 Heads in a row is (1/2) multiplied by (1/2), which is (1/2) to the power of 2 ; of 3 Heads in a row is (1/2) to the power of 3; and so on. Therefore

EV = [(1/2)^1 * ($2)] +
[(1/2)^2 * ($4)] +
[(1/2)^3 * ($8)] +
...

EV = [(1/2) * $2] +
[(1/4) * $4] +
[(1/8) * $8] +
...

EV = $1 + $1 + $1 + ...

This goes on infinitely. That EV looks like summing to infinite $$$.

Now what?

How much should anyone pay me to play this game?

[Cyrus]

"The difference is negligible, especially for the coins and craps."

All I'm after is "a better shot at those games". This is what I wrote. In the absence of opportunities for something better, my tactic is the correct one.

For any small k above zero, however small, I can prove to you that k > 0 [img]/forums/images/icons/wink.gif[/img]

"You have a miniscule edge in these cases only when there is no penalty for assuming a bias and being wrong."

There is no penalty for being wrong. We crash-land back to the usual expectation for each game.


Frank Scobete is one writer who offers potentially beneficial advice but for all the wrong reasons. He advocates following the patterns of "rhythmic rollers" and the "5-count" in craps, but this is voodoo statistics. However, this is better than just playing randomly (if you must play craps straight), for the reason I described.