50:50 cumulative odds

[ronianna]

Assuming you flick a regular coin under regular conditions then the odds of it landing as either a head or a tail is clearly 1 in 2. What would be the odds for it landing say heads twice in a row, three times in a row, four times in a row etc right the way up to 8 times in a row.

I know that each individual toss of the coin is 1 in 2, it's just that I can remember how to work out the cumulative odds.

Can anyone remember or know?

Thanks

[Hippy]

You say, "the odds of it landing as either a head or a tail is clearly 1 in 2." Wrong. The probability of it landing either head or tail is 100%, if you neglect the extremely unlikely event of it landing and remaining on its edge! The probability of it landing head up is 50% and that of tail up is 50%. So if we choose head up and then find that it lands head up twice in a row, this is 50% for the first toss then 50% for the second, so the cumulative odds are 50% of 50%, or 25% (4:1). For further cumulative odds just multiply the odds expected at each stage, so for 8 head up lands for 8 throws we have a probability of 50% * 50% * 50% * 50% * 50% * 50% * 50% * 50% = 0.39% (256:1).

For such as the lottery the odds change as each ball is drawn. The probability of matching the first ball is 49:6, (you have six balls, the lottery has 49) the second 48:5, (you now have 5 balls as one has been matched, and the lottery has 48 left) the third 47:4 and so on. The chance of matching all six main numbers is, therefore, (49 * 48 * 47 * 46 * 45 * 44) / (6 * 5 * 4 * 3 * 2 * 1) = 13,983,816 : 1

The rule is to treat each individual probability as a fraction then multiply the fractions to calculate the cumulative probability.

[Sasha13]

Just as a matter of interest, if you want to work out the odds of one thing OR another happening, i.e. betting that a 2 OR a 4 will result on rolling a die, then you add the odds of the results together, i.e. a sixth + a sixth = a third, or 33.3%

[Loosehead]

Yep Hippy is correct. Simply put the probablility of getting N heads in a row is 1 in 2**N (** means "to the power of") so 1 in 256 when N is 8 from the outset. It is a common misconception in things like this that if you where to toss 8 heads in a row then the chance of a head on the 9th attempt is something other than evens. If you've ever gone into a casino you'll always find someone marking down the numbers at roulette and anouncing that "It's been red 10 times in a row, the next spin MUST be black"! NO NO NO.

[ronianna]

Thanks everyone. Loosehead, would I be correct in assuming then in your 'roulette scenario' that if your were looking at the probability of 8 straight reds for instance, then in fact the odds go down rather than up the closer you get to the 8th spin? 1st spin - 256 to 1, 2 nd spin 128 to 1, and so on and so forth until on the last spin it would be 2 to 1? Even if I don't have the numbers right or in the right place, is the shape right? Knew I should have attended more maths lessons!!!

[Hippy]

Roulette involves a spinning wheel that is divided into 37 sections These sections are then assigned numbers (0 to 36) and colours (Red and Black). There is a Green section, for the number 0 which the house always wagers. Bets are selected in anticipation of where the ball will land on the spinning roulette wheel, with possible wagers placed on the number, colour (Red or Black), and number type (Odd or Even).

So, there are 18 Reds, and 18 Blacks, so if you bet say Red, then your probability is 37:18 or to a lower denominator 2.0555555555 :1 Not quite the two to one you might have thought because of that pesky zero, and that, of course is one of the ways in which casino owners get rich! Now the important thing to remember in all of this is that each throw of the dice, toss of the coin, or spin of the roulette wheel is a separate and unconnected event. It could be the first such occasion in the history of the universe, or the squintillionth, if it is a fair wheel then the odds will be exactly the same. None of these gambling devices has any sort of memory of what just happened.

So if you want to bet on 8 red results in a row then your probability is 2.0555555 multiplied by itself 8 times to one. To save you the pencil lead the answer is 318.738 to one, rather than the more favourable 256 to one with the coin toss example.

[Peter Pedant]

Its chapter one in absolutely every book with the words "probability and elementary statistics" in the title.

Once you have got the chance of HHHTH

(head head head tails heads) which as above is one in two the five - - one in 32 I think

the second part of the chapter does "what is the chance of throwing four heads and one tail in any order out of five?". and THAT part involves the expansion nCr which you were probably confused about first time round.

 

Pip pip

[Loosehead]

Well Hippy is again correct of course. To answer your question though unless you are trying to predict a series of events from the outset the the probablity of this even does not effect the next. Som ignoring the 0 for the moment 8 straight reds from the outset is the same as 8 straight heads. individually evens.

[MargeB]

If you flip a coin, apart from the chances of it landing on its edge, it is always more likely that it will land tails up, and this probability increased a few years ago.

[Quizmonster]

25% represents odds of 3/1, not 4/1. When speaking of "three to one against", there are clearly four possibilities, one of which is good...ie the bet wins...and three of which are bad...ie the bet loses. In the same way, "four to one against" involves five possibilities and represents a 20% chance.

[mews64]

Ignoring the infinitely imporobable chance of the coin landing on its edge, and also tails being more likely than heads, go back to the basic situation of  1:2 of getting heads.

I was taught to use the AND rule in dealing with probabilities which is basically the probability of getting A(a) (and) then the probability of getting B(b) is axb

with your situation of getting heads eight times it would be AxBxCxDxExFxGxH

since the probability of getting heads doesn't change after each toss then it is 1/2x1/2x1/2x1/2x1/2x1/2x1/2x1/2 which is 0.00390625

Basically if the probability A(a) of doesn't change each time the event is carried out then, it's  (a) to the power of how ever many times the event is carried out (N)        - or simply (a**N)

[Mark154]

Just to let you know mews64 didnt post the answer above. it was mark154

and if this too was posted by mews64 then we have problems

[Golem]

I'd like to contribute a literary, not mathematical, perspective.

Tom Stoppard's play Rosencrantz & Guildenstern Are Dead opens with Guildenstern spinning coins and betting on the outcome, heads or tails. The result is heads 95 times in a row! In a very clever and witty opening scene Guildernstern wonders about the possible reasons for this; "One. I'm willing it. ... Two: time has stopped dead, and the single experience of one coin being spun once has been repeated ninety times ... Three: Divine intervention ... Four: a spectacular vindication of the principle that each individual coin spun individually is as likely to come down heads as tails and therefore should cause no surprise each individual time it does."

The character note for Guildernstern is that, "he is not worried about the money, but he is worried by the implications." which seems to me an appropriate response if the laws of probability were to suddenly change for us all! The play goes on to explore this theme and the consequences it has for the two protagonists.

[JudgeJ]

Loosehead is, of course, quite right to say that gamblers who log the outcomes of roulette wheel spins are foolish to suggest that because 10 reds in a row have appeared the next spin must result in black. The wheel has no memory.

However, it is known that among the “evens” bets in roulette there does occur what are known as “long runs”. These are runs of a single outcome which veer away from the expected 50:50 distribution.

Long runs are not incredibly long but do occur quite frequently. However it is difficult to take advantage of them. If you observe 10 spins it would not be unusual to see 7 Reds and 3 Blacks. However, it would be most unusual to see 70 Reds and 30 Blacks in a hundred spins, whilst 700 Reds and 300 Blacks in a thousand spins would be almost unheard of. The higher the number of spins, the closer to 50:50 you would expect the results to be. If you can take advantage of these short-term variations from 50:50 which can occur you can overcome the operator’s 1.35% advantage (which he gains on the even money bets courtesy of the zero slot).

[JudgeJ]

There is a very good book called “Thirteen against the Bank” by Norman Leigh. It tells the tale of a team of gamblers who attempted to cash in on these “long runs”. They used a staking method known as “reverse Labouchere” and attempted to cover all the even money bets on a table over a long period. The staking method prevented them losing too much when the long runs did not occur, but they were able to cash in as soon as they did. They were said to have made so much money that they were banned from some casinos.

It is the only credible tale of a “system” that I have come across in many years of studying gambling, probability and odds.