progression that utilizes the "parlay bet". A parlay bet is when you

let all your winnings ride on the very next bet.

Example: You bet \$5 on a coin toss of heads. The coin toss is heads,

which means you win \$5. Your very next bet is \$10 because you are

parlaying ("let it ride") your \$5 in winnings from the first bet. And

let's say you win that next bet and so now you win an additional \$10.

Your total winnings are now at \$15.

That's an example of winning a parlay. Here is an example of losing a

parlayed bet:

--I bet \$5 on heads and the coin toss is heads. I win \$5 and let it

ride (parlay) on the next bet. This means my second bet is \$10. And

let's say I lose that second bet because I guessed heads and the coin

toss came up tails. I lose the \$10 bet and now my total win/loss

total is minus-\$5.

Okay, so here is the scenario for which I would like a statistical

probability calculated . . .

Let's say I am betting on heads or tails and that I have nine \$5 chips

at my disposal. In other words, I have nine levels of betting, as I

plan to only bet one chip per level. One chip for each bet.

And let's say that each time I win a bet, I let my winnings ride on

the very next bet. In other words, I parlay my winnings for the

second bet after any win.

Now let's say that once I win two bets in a row, the game is over and

I am declared a winner because I won two bets in a row without going

through my nine individual bets of \$5 each.

On the other hand, if I go through all nine levels (all nine

individual \$5 bets) without winning two in a row, I am declared a

loser and the game is over.

Can you calculate the odds for me on winning or losing this series of bets?

My question is, what is the statistical probability of me being able

to guess right twice in a row before being wrong through nine levels

of betting?

Let me explain further to make this crystal clear.

Obviously, I could win my wager if I guessed right the first two

times. That would be a win on my first level (my first \$5 chip).

Likewise, I could lose my wager if I guessed wrong nine times in a

row. That would mean I lost at all nine levels.

However, the game could go on for more than just two tosses or nine

tosses --- I could be right nine different times thoughout the

progression and then be wrong on the very next guess, which means I

have not won because I was not able to guess right TWO times in a row.

I could actually win nine individual bets throughout the betting

progression but still be declared a loser if I failed to win the

ensuing bet after each one of those wins.

Make sense?

Here is an example of failing to win at least two bets in a row at the

nine different levels with some individual wins mixed in Keep in mind

that each "Level" represents a single \$5 chip:

LEVEL 1: I guess heads and the coin toss is tails. I lose my first \$5

chip. I now must bet my second $5 chip.

LEVEL 2: With my second \$5 chip, I guess heads and the coin toss is

heads. I win one in a row. Now I let it ride for my next bet.

LEVEL 2. I guess heads and the coin toss is tails. I lose my parlayed

bet and I must bet my third \$5 chip.

LEVEL 3. With my third \$5 chip, I guess tails and the coin toss is

tails. I win one in a row and parlay my next bet.

LEVEL 3. With my parlayed bet, I guess tails and the coin toss is

heads. I lose my parlayed bet and I must bet my fourth \$5 chip.

LEVEL 4. With my fourth \$5 chip, I guess heads and the coin toss is

tails. I lose my fourth \$5 chip and must bet my fifth \$5 chip.

LEVEL 5. With my fifth \$5 chip, I guess tails and the coin toss is

heads. I lose my fiffth \$5 chip and must bet my sixth \$5 chip.

LEVEL 6. With my sixth \$5 chip, I guess tails and the coin toss is

tails. I win one in a row and parlay my next bet.

LEVEL 6. With my parlayed bet, I guess tails and the coin toss is

heads. I lose my parlayed bet and must now bet my seventh \$5 chip.

LEVEL 7. With my seventh \$5 chip, I guess heads and the coin toss is

tails. I lose my seventh \$5 chip and must bet my eighth \$5 chip.

LEVEL 8. With my eighth \$5 chip, I guess heads and the coin toss is

tails. I lose my eighth \$5 chip and must bet my ninth \$5 chip.

LEVEL 9. With my ninth and final chip, I guess tails and the coin

toss is tails. I win one in a row and parlay my next bet.

LEVEL 9. With my parlayed bet, I guess tails and the coin toss is

heads. I lose my parlayed bet and since I have used all nine of my \$5

chips, the game is over and I am declared a loser.

So my question is, is there any way to calculate what my odds are of

failing to win at least one parlayed bet on a 50-50 coin toss when you

have nine individual chances to do so? Mind you, I won't always get

nine chances at a parlayed bet. As you can see from the above

example, I won some bets throughout the game but never failed to win

two in a row.

The game could be over in as few as three decisions (I win all three

bets), or it could last for as long as 18 decisions (won the first bet

but then lost all nine parlayed bets). It could also last any length

in between.

So to summarize my question --- what is the precise statistical

probability of winning or losing the game as described above? Please

note I am looking for exact statistical odds, not a guess! And please

note I am looking for the exact odds that take into account ALL

possibilities that exist in this game.

I will gladly pay anyone who can answer this correctly and precisely.

(If it's possible, and I don't know that it is.)

If calculating the odds for the described scenario is NOT

mathematically possible, then I'm not willing to pay to find that out,

but if there IS an answer, I'll be happy to pay and add a nice tip!

Thanks in advance . . .