Probability Help
Quote: nenedutyWhat an excellent website you have, very informative and full of knowledge. I am not new to roulette but am new to trying to get the correct probability on certain bets in roulette. I am trying to figure out the probability of one of the dozens coming up several times in a row. So say I want to figure out the probability of the 2nd dozen coming up 4 times in a row. Would that be 1/3x1/3x1/3x1/3= would this be 1 in 81 that a specific dozen will come up 4 times in row? How about a double street. 1/6x1/6x1/6x1/6=1 in 1296. And finally how about a street(3 numbers) 1/12x1/12x1/12x1/12=1 in 20736. Is my math correct? Just trying to figure out the odds of something specific happening 4 times in a row. Thanks Alex ps I know it doesn't include the zero(s) Thanks again for your time Mr. and God Bless
Yes..... but to get a correct answer you must include the zeros!
For one zero wheel its 12/37 x 12/37 x 12/37 x 12/37 = 20,736/1,874,161, which is around 1 in 90.
For a two zero wheel its 12/38 x 12/38 x 12/38 x 12/38 = 20,736/2,085,136 which is around 1 in 101.
P(column 1) = NumberOfWins / TotalNumberOfSpots = 12/38 = 6/19 = ~32% (.3157).
In statistics, whenever you speak the problem, but you use the word "and" it's multipliciative. What's the odds of This AND That must happen... This * That...
Whenever you speak the problem, but you use the word "or", it's additive... What's the odds of This OR That? This + That...
So... Let's speak your problem: What are the odds column 1 lands AND column 1 lands AND column 1 lands AND column 1 lands?
From above we know P(column 1) = (12/38)... So P(column 1 4 times) = P(column 1) * P(column 1) * P(column 1) * P(column 1)... Which is the same as P(column 1)^4.
The short cut, when asking about "repeating" events that are the same, is to take the number you're curious about and that's the exponential power the probability gets raised to (since we're just multiplying them by themselves over and over). So to put it short...
P(column 1 4 times in a row) = P(column 1) ^ 4 = .3157^4 = .00995... or about 1 in 100.5.... I think SOOPOO rounded =).
So P(column 1 10 times in a row) = P(column 1) ^ 10... and so on.
you seem to be a smart person on probability .
Can you please help me .
What is the probability ( % ) of 2 heads come out in a row in every 9 times you flip a coin ?
What is the probability ( % ) of 2 heads or 2 tails ( % ) come out in a row in every 9 times you flip a coin ?
Thank you for your help .
Quote: kimma2366Hi Romes ,
you seem to be a smart person on probability .
Can you please help me .
What is the probability ( % ) of 2 heads come out in a row in every 9 times you flip a coin ?
What is the probability ( % ) of 2 heads or 2 tails ( % ) come out in a row in every 9 times you flip a coin ?
Thank you for your help .
kimma,
You asked this identical question in 3 different threads, though addressed to different people. We consider this flooding, a type of spam. Please stop it and wait for an answer. Thank you.
Quote: kimma2366What is the probability ( % ) of 2 heads come out in a row in every 9 times you flip a coin ?
What is the probability ( % ) of 2 heads or 2 tails ( % ) come out in a row in every 9 times you flip a coin ?
I am interpreting the questions as:
(a) what is the probability that, when you toss a coin 9 times, two (or more) heads will come up in a row at least once?
(b) what is the probability that, when you toss a coin 9 times, two (or more) heads or two (or more) tails (or both) will come up in a row at least once?
For (a), it is easier to count the number of ways you can toss 9 coins without 2 heads in a row.
These can be divided into groups:
All nine tails (TTTTTTTTT); there is only one way to do this
One head, eight tails; any of the nine tosses can be the head, so there are 9
Two heads, seven tails; the two heads cannot be consecutive, so you have, for the two heads:
1-3, 1-4, 1-5, 1-6, 1-7, 1-8, 1-9
2-4, 2-5, 2-6, 2-7, 2-8, 2-9
3-5, 3-6, 3-7, 3-8, 3-9
4-6, 4-7, 4-8, 4-9
5-7, 5-8, 5-9
6-8, 6-9
7-9
That is 28 possible ways
For 3 heads and 6 tails, the heads can be:
1-3-5 through 1-3-9 (5 ways)
1-4-6 through 1-4-9 (4)
1-5-7 through 1-5-9 (3)
1-6-8, 1-6-9 (2)
1-7-9 (1)
2-4-6 through 2-4-9 (4)
2-5-7 through 2-5-9 (3)
2-6-8, 2-6-9 (2)
2-7-9 (1)
3-5-7 through 3-5-9 (3)
3-6-8, 3-6-9 (2)
3-7-9 (1)
4-6-8, 4-6-9 (2)
4-7-9 (1)
5-7-9 (1)
That is 35 ways
For 4 heads and 5 tails:
1-3-5-7, 1-3-5-8, 1-3-5-9
1-3-6-8, 1-3,6-9
1-3-7-9
1-4-6-8, 1-4-6-9
1-4-7-9
1-5-7-9
2-4-6-8, 2-4-6-9
2-4-7-9
2-5-7-9
3-5-7-9
That is 15 ways
The only way with 5 heads is 1-3-5-7-9
This is a total of 1 + 9 + 28 + 35 + 15 + 1 = 89 ways
Thus, of the 512 ways that a coin can be tossed 9 times, there are 512 - 89 = 423 ways to get at least two heads in a row
The probability is 423/512 = about 82.62%
(b) is easier; the only way you don't get either two heads or two tails is if the tosses alternate heads, tails, heads, tails,...
There are two ways this can happen: HTHTHTHTH and THTHTHTHT
There are 512 - 2 = 510 ways that you get two heads or two tails in a row.
The probability that it does happen = 510/512 = 255/256 = about 99.61%
Your answers are very clear .
Thank you again for your help .
Quote: beachbumbabskimma,
You asked this identical question in 3 different threads, though addressed to different people. We consider this flooding, a type of spam. Please stop it and wait for an answer. Thank you.
Got it .
My first time here .
Sorry .
