How to figure chance of "x" consecutive losses

[Randy Gunn]

For a given hit %, what is the math to figure the maximum number of potential, or expected, consecutive losses? As an extension, I would also be interested to know the math for figuring how often I could expect a run of "x" number of losses (ie, how often can I expect to lose 8 in a row, 9 in a row, etc...).

Thank You,
Randy

[Cholly]

http://www.sbrforum.com/betting-tools/streak-calculator/

[acorn54]

For a given hit %, what is the math to figure the maximum number of potential, or expected, consecutive losses? As an extension, I would also be interested to know the math for figuring how often I could expect a run of "x" number of losses (ie, how often can I expect to lose 8 in a row, 9 in a row, etc...).

Thank You,
Randy
i think what you want is a cumulative binomial probability calculator.
that is where you set at a given win% or hit rate and find the probabilities of various runs without a winner.
i think some characters called the bernouille brothers in the 1800's came up with this if my memory serves me.

http://stattrek.com/online-calculator/binomial.aspx#single


i think this link might be of some help

[Robert Fischer]

For a given hit %, what is the math to figure the maximum number of potential, or expected, consecutive losses? As an extension, I would also be interested to know the math for figuring how often I could expect a run of "x" number of losses (ie, how often can I expect to lose 8 in a row, 9 in a row, etc...).

Thank You,
Randy

maximum consecutive losses formula
=ROUND(LN(10000)/-LN((1-0.4)),0)
Two Variables 10000 = number of total plays in series, 0.4 = Hit% of your wager.


run of "x" losses formula
=(1-0.4)^2
Two Variables 2 = number of losses in run, 0.4 = Hit% of your wager.

[Robert Fischer]

^ 10,000 plays may be a bit much... example 10 plays a day x 100 days = 1000 plays

just err a little on the high side.

[edgar]

I have no idea if this will answer your question but here goes. Consider a five horse race as an example. Each of the horses has an equal shot at winning and each horse is equally bet. The winner will pay $8 but the actual odds are (5-1) (assuming the WPS takeout is 20%). Your chances of losing should you pick one horse to win would be 4/5. Your chances of losing nine straight would be (4/5)^9 or 13.42%.

[davew]

if you know hit %, then miss % = 1- hit%

chance of losing X times in a row = (miss%)(miss %)..... X times


problem is hit% is frequently not known or inaccurate

[speculus]

This formula has served me well. It can be used in an excel sheet:

T = ROUND(EXP((N*(-Ln(P)))),0)+(N-1)


Where T is the number of trials (or in our case, number of BETS to be made),
N = longest winning streak or losing streak
P = percentage (win or loss) expressed as a probability between 0 & 1
while ROUND and EXP are MS Excel functions.

Of course, since you are looking for N, you will have to shuffle the formula to get N on the left side.

[Actor]

For a given hit %, what is the math to figure the maximum number of potential, or expected, consecutive losses?The maximum number of consecutive losses is infinity, unless the hit% = 100%, in which case the maximum number of consecutive losses is zero. If the hit% < 100% then the expectation that the maximum number of consecutive losses is finite equates to the Monte Carlo Fallacy, i.e., the belief that at some point you are "due."

As an extension, I would also be interested to know the math for figuring how often I could expect a run of "x" number of losses (ie, how often can I expect to lose 8 in a row, 9 in a row, etc...).This question is a little bit vague and depends on the interpretation of "how often." It could be interpreted as "if (1)I go to the track N days, and (2)the track always has a 9 race card, and (3)I always bet all 9 races, then how many days will I experience at least one run of X consecutive losses?" To the best of my knowledge there is a different solution for each value of X which can have values from 1 through 9. And, of course, the solution gives an expectation value, not a predictive value.

[Robert Fischer]

The maximum number of consecutive losses is infinity, unless the hit% = 100%, in which case the maximum number of consecutive losses is zero.

Technically correct. The actual title of the formula that Speculus and I posted is the maximum probable consecutive losses formula.

Using your logic a wager could have a 99% chance of winning (a 1% chance of losing), and you could POSSIBLY lose 100 in a row. The actual chances of that happening is 1 × 10-200.

It just isn't probable.

The most you could lose within the realm of probability is 2 consecutive losses.

The chances of losing two in a row with a 99% hit rate is one in ten thousand. The probability of losing 3 in a row at a 99% hit rate is one in a million.


So if you are dealing with a given hit rate, you could build strategy around the idea that "anything is possible", but in the real world there are probable limits.

Assuming you aren't cursed or doomed, it makes sense to work your finances within those limits - and by the same token if you aren't "lucky" or "due" you should respect those limits as well. :ThmbUp:

[myhorse1]

Assuming you aren't cursed or doomed

if you are cursed or doomed betting on anything is probably not the way to go.