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I'd like to know the probability of winning a particular betting
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 . . .
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1 Answer

Best answer
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1 votes
The probability of winning or losing this game can be calculated exactly. The probability of winning the game is the probability of winning two consecutive bets before losing 9 times.

Since the coin toss is a fair game (i.e., the probability of getting heads or tails is 0.5), the probability of winning a bet is 0.5 and the probability of losing a bet is also 0.5.

Let's denote by P(n) the probability of winning the game with n remaining bets. We are interested in calculating P(9).The probability P(n) can be calculated as follows:- If n = 0, then P(n) = 0, because if there are no remaining bets, there is no chance of winning the game.- If \(n \gt; 0\), then P(n) is the probability of winning the game in the next bet plus the probability of losing the next bet and still winning the game.

The probability of winning the game in the next bet is 0.5 * 0.5 = 0.25, because we need to win two consecutive bets. The probability of losing the next bet and still winning the game is 0.5 * P(n-1), because we lose one bet and we have n-1 remaining bets.Therefore, P(n) = 0.25 + 0.5 * P(n-1).The probabilities P(n) for n = 0, 1, 2, ..., 9 can be calculated as follows:

- P(0) = 0.

- P(1) = 0.25 + 0.5 * P(0) = 0.25.

- P(2) = 0.25 + 0.5 * P(1) = 0.375.

- P(3) = 0.25 + 0.5 * P(2) = 0.4375.

- P(4) = 0.25 + 0.5 * P(3) = 0.46875.

- P(5) = 0.25 + 0.5 * P(4) = 0.484375.

- P(6) = 0.25 + 0.5 * P(5) = 0.4921875.

- P(7) = 0.25 + 0.5 * P(6) = 0.49609375.

- P(8) = 0.25 + 0.5 * P(7) = 0.498046875.

- P(9) = 0.25 + 0.5 * P(8) = 0.4990234375.

Therefore, the probability of winning the game with 9 remaining bets is approximately 0.499, or 49.9%.The probability of losing the game is 1 - P(9) = 0.501, or 50.1%.
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