hod do you work out odds?

[rjms1977]

Everyone knows when you flip a coin their is a 50% chance it will be heads and a %50 chance it will be tails, but what i want to know is how do you work out the chances of there being 10 heads in a row or 15 heads in a row etc, as the chances will be a lot slimmer than 50%! I guess you have to multiply the odds for each toss of the coin but by how much? This has been doing my head in for days and when i look for answers on the net i always end up at a pitch page offering me some sort of gambling scam, so any help would be much appreciated!

[Hymie]

Without going in to the detailed maths, for a simple toss of a coin, the odds are 1 in 2 to the power n, where n is the number of tosses.

Therefore the odds of 10 heads in a row is 1 in 2 to the power 10 = 1,024.

15 heads in a row is 1 in 32,768 (according to my calculator).

[answer001]

If you want the odds of something AND something else happening, you times the odds together (eg the odds of getting a head, and a head, and a head ie 3 heads in a row, is 1in2 x 1in2 x 1in2 = 1in8, or 12.5% )
The chance of rolling a six on a dice AND then getting tails on a coin toss is 1/6 x 1/2 = 1/12 (1in12 or 8.3%)

[sp1214]

This is where I believe probability theory is all wrong. The standard reply is that you ignore everything that has gone on before and each toss is a new event. But if you have had 5 heads on the trot you would expect sooner or later that there would be some equalization and therefore a tail should be imminent.

In my opinion there are two facts to consider.

1. past events
2. the outcome of a single toss ignoring everything that has gone before.

The text books won't tell you this of course and we are brainwashed into accepting the current theory.

[R1Geezer]

**** eyed b0llox SP. Coin has no memory, same for a roulette wheel that's why the casinos love those plonkers who go 1, 2, 4, 8, 16, 32 etc on black or red. 10 blacks in a row, the next one must be red, Right? WRONG. the wheel has no memory. Now calculating from the outset as the 2 guys above have shown that's different.

[sp1214]

Thank goodness Britain is a country where you can speak your mind even if it differs from others. When the bigots run the country like you Geezer all hope is lost.

[Gef]

The first 2 answers are correct.

Consider tossing a coin 3 times. the possible outcomes are:-
HHH
HHT
HTH
HTT
TTT
TTH
THT
THH

There is a one in 8 chance (2x2x2) of 3 heads in a row.

Hope this helps.

[ll_billym]

In fact 15 heads has exactly the same chance of happening as any other sequence such as:

THHHTTTTHHTHTHTT

or

HTTTTHHTHHHTHTHT

[sp1214]

Yes but the longer the sequence you will probably find it nearer to 50/50%. In other words a run on heads or tails will still adjust to the mean of 50%.

Have you watched that game on Ch 4 in the afternoons. Near to the end of the game it will either show a surplus of reds or blues. But on most occasions the contestant is finally left with just 1 red and 1 blue. In other words if the choice he is left with is 5 blues and 1 red then the odds of picking that red are 1 in 6 not 50% red or blue.

[R1Geezer]

yes SP but this is not one of those things where you give an opinion, It's a solid mathmatical fact, not a debate at all you either know or you don't, you are right or wrong, no grey areas.

[beso]

Television games are notorious for being rigged.

Without a doubt sp1214 is wrong. The previous outcomes have absolutely no bearing on subsequent outcomes. Each toss is completely independent.

Yes the more tosses the more likely the totals will be closer to equal but this has nothing to do with the previous outcomes. It is to do with the statistics of probability where the chance of a significant deviation from equal outcomes falls with the number of trials.

[sp1214]

Of course I'm wrong, the text books say so. But just apply a bit of savvy to the real world. My earlier point about 'deal or no deal' happened on Ch 4 this afternoon. The contestant was left with 1 blue and 6 reds. Even one of the box holders said the odds of picking one of the 6 red boxes was almost a certainty. This was after a run of blues being selected without a red. Compare the question with this.

[Hymie]

SP1214 - of course, the odds of picking a red was 6 in 7, using the same logic as above.
You are now looking at a slightly different scenario, where there is a limited �population� from which the selection is being made.

I would be willing to bet if Joe Public were asked which of the following 6 numbers is more likely to win the national lottery (4, 11, 23, 35, 39, 45 or 22, 23, 24, 25, 26, 27), most would select the first six � whereas both have an equal chance.

[ll_billym]

SP, at the point in the game you talk about there was a 1 in 7 chance that a Blue box would get picked, or 14%.

Blue = 1 in 7 or 14%
Red - 6 in 7 or 86%

What does the colour of the last box to be picked, be it Red or Blue, have to do with those numbers?

There will still be 1 Blue box and 6 Red boxes left to pick. The chance of picking a Red box is much higher full stop.

[beso]

In the case of the TV show It is not a random selection and the outcomes are not independent of the previous choices.

Statistics do not apply in this case because the contestents are making a conscious decision and are influenced by what they can see and their impression of previous selections.

The outcome depends on the superstitions of the contestent. It is not the same as flipping coins.

[beso]

Maybe I misunderstand. I haven't seen the UK version of "Deal of No Deal". In Australia the contestent chooses among a set of breifcases.

What I said in my last post only applies if the contestent is selecting a box and has the choice of colour.

[beso]

Even assuming the contestent has no choice over colour the case of Deal or No Deal is not the same as tossing coins. The outcomes of the box selection are not independent.

The more blue boxes taken the more likely a red one will be the next. It is more like th case of having 50 pennies set up, half with heads and half with tails. Indeed once all the selections have been made the outcome will always be half each because that is how it was set up.

In the case of tossing a coin 50 times the outcome follows a random statistical distribution because the previous tosses have not used up one of a fixed number of outcomes available.