General Betting
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23 Nov 09
14:47
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22 Jun 06
One for those with greater maths knowledge than me.
What are the chances/percentage odds of a consecutive sequence of 4 heads (or tails) occuring in 100 coin tosses-
a) Once
b) Twice
c) Three times
d) Four times
TIA
What are the chances/percentage odds of a consecutive sequence of 4 heads (or tails) occuring in 100 coin tosses-
a) Once
b) Twice
c) Three times
d) Four times
TIA
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1) 1-(1/16)^100
2) 1-(2/16)^100 3) 1-(3/16)^100 4) 1-(4/16)^100 no idea what the odds of that happening 5 times in a row would be though. |
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not a or b
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100 go's @ a 15-1 chance = 6.66 it should occur
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Don't think I worded it very well. In other words - on average, how many times should I expect the sequence of 4 to occur in every 100 tosses.
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Each toss is 99/1.......4 5 6 in a row still 99/1
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4% give or take . on averge ...............
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zipper, so you'll give me 99/1 each of 2. Can I open an account please.
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Well its 1 in 4 for it to happen at any time, so 4/1
Then 100 flips means 100/4 = 25 So you should get four in a row 25 times every 100 flips Hope this helps |
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They all have a 50/50 chance the coin does not know what it landed on 1st or 2nd time
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jack, if you had 4 in row 25 times in a 100 spins, that in itself would be 100 spins. I would say about 10.
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Yes you can mugs are always welcome ....... 100 balls in a tub .. all with numbers .. 1 to a 100 the true odds of anyone picking the named number is 99/1 .. you ok with that ...... so lets say the guy said number 56 and it was 56 .. so he has another pick ..... its still 99/1 ....... if he picked 10 on the bounce ( lucky mug ) its still 99/1 on the 11th pick ...
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Seeeebooo!!
Are you saying that you should expect the consecutive sequence of 4 of the same to occur on 6.66 separate occasions during the 100 tosses? |
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Anyone backing 99/1 shots needs to find another job .. but in this case the odds are true .... toss of a coin heads or tails its evens ......... it dont matter if you tossed 50 heads on the bounce .. the next toss is evens.......... heads or tails .
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What are the chances/percentage odds of a consecutive sequence of 4 heads (or tails) occuring in 100 coin tosses-
About 99.9999999999999%. Give or take a couple of niners at the end. |
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Ok, first one was a guess.
Pocket calculator says: a) 99.8% if you choose heads or tails before the experiment b) 99.9998% if you want want a sequence of 4 in a row, regardless if heads or tails. |
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A most pleasing thread :-)
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sharp minds
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If you want the number of such sequences, it is about six.
You have 97 sequeces of 4 a piece with 100 flips. Start 1 to 4. Ending 97 to 100. Chances for any sequence being 4 heads are 1/16. Dividing 97 by 16 does the rest. If you want any sequence, heads or tails, double the number. So 12 times within 100 flips you will have 4 in a row. (Remark: Wenn flipping 5 heads in a row, you actually have 4 in row twice. Flip 1 to 4 and flip 2 to 5.) |
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1) 1-(1/16)^100
2) 1-(2/16)^100 3) 1-(3/16)^100 4) 1-(4/16)^100 no idea what the odds of that happening 5 times in a row would be though. Surely it's 1-(1/16)^97. 100 tosses gives you 97 chances to get 4. No? |
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(above gives chances of NOT getting 4 in a row- so it's 1- to get the probability that you will)
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Full explaination:
any 4 tosses gives a 1/16 probability of 4 heads, therefore 1-(1/16) of no heads. In a sequence of 100 tosses you get 97 sequeces of 4 results, so the chances of NONE of them being 4 consecutive heads are 1-(1/16)^97. That works out as .0019 (ish). There is therefore a 99.8% (ish) chance that you will get at least one sequence of 4 heads. |
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therefore 1-(1/16) of no heads
Sorry- I meant that there's a 1-(1/16) probability that the sequence wont be 4 consecuticve heads- not NO heads). Hope that clears it up. |
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And consider this:
Although there is a 99.9% chance that you'll get at least one sequecnce of 4 consecutive heads there is only a 51% chance that you'll get 7 in a row. Such is the power of 2. |
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Not convinced. The 97 groups are not independent.
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no, they're not independent trials, and this assumption of independence is implicit in aye robot's approach
as soon as you have a non head you reset the problem back to the start, but you waste a flip in the process, so you don't get 97 attempts at all it's a tricky problem imo, not easy to just write the solution down - i don't agree with aye robot's answer |
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something not stated clearly in the question, does "HHHHH" count as two overlapping sequences of "HHHH" or are we just counting strings of at least four H in a row?
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a very dirty, approximate solution along similar lines, expected number of tosses until first tail is 2 - so an average trial (going for HHHH) costs 2 tosses, so on average you get 97/2 attempts at it
then prob 0 sequence of HHHH is (1-1/16)^48.5 = 4.37% |
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The probability is affected by the energy of he who tosses the coin, as all energy is constantly changing and he projects onto it a vibrational alignment based on his own perceptions and belief and expectations. Quantum physics. That which is like unto itself is drawn. he sees a reflection of his own inner core beliefs shown to him.
A negative or positive energy within he who puts his energy to the coin will affect his choice making in how to toss the coin and when. That determines his alignment with his expectation and belief, which in turn manifests into his life experience. In other words, there is no exact answer and there never can be as every individual is unique and energy constantly changes and expands. |
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i dont like my dirty solution now either, it allows for four flip sequences which are not HHHH but which still end with H for each mini trial
you could set up a markov chain to describe this, then diagonalise the matrix describing the jump probs, that would work |
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dont you have to do monte carlo or something for this? i have never seen a good formula for working it out
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you don't have to monte carlo it, but an analytic solution is not easy, the markov chain diagonalisation approach is degree level maths
some further maths boards would do simpler cases at a level 10+ years ago, they probably don't go further than simultaneous equations for special papers these days there might be a smarter way to find an analytic solution, but this is a tough problem to find a neat solution imo |
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Eldrick is spot on. The neatest solution involves using a diagonalised Markov chain.
I've just coded it, and the answer (probability of getting at least 4 heads in a row is: 0.9727. |
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I'm quite surprised at the volume of replies on this with no one asking a question about the coin.
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If you want to know how to work this out in Excel, say, it's quite easy. Just set up a grid of 100 rows (one for each trial) and 5 columns. The first row is where you've already tossed the coin 100 times, and columns 1 - 5 are where you've scored (so far) 0, 1, 2, 3 and 4 heads in a row, and the cells are filled with the probabilities of achieving the result (4 in a row) from that point in the game.
The first row is easy to fill in. Obviously, if you've already flipped 100 times and you've scored 0 so far, the prob of success is 0 (cell 1). And similarly for 1, 2, and 3. In cell 5, though, the prob is obviously 1 (because you've already succeeded. Then do the next row. This corresponds to the state of the game after 99 tosses. Obviously the prob is zero for the 1st 2nd and 3rd cells, and 1 for the last. For the 4th, you can succeed if you flip a head, so the prob is obviously 0.5. In general, the game is such that if you throw a head (prob 0.5), you have one less throw to go, but you have one less head to get (score increases by one), and if you throw a tail (prob 0.5), you have one less throw to go, but your score goes back to zero. You can use this information to get a general recursive principle to fill in the whole of the Excel sheet: The prob of winning from score S with n flips left = 0.5 * (prob of winning from score S + 1 with n-1 flips left) + 0.5 * (prob of winning from score 0 with n-1 flips left). |
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"I've just coded it, and the answer (probability of getting at least 4 heads in a row is:
0.9727." Snap. Unfortunately I think we have drifted a little from the initial question. |
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Thanks for the considered replies chaps. Eldrick, I was meaning separate chains of 4 so HHHHH would count as only 1 (assuming a T at each end) not 2. Before asking this my basic maths assumed the answer would be something around 4 which is what i wanted to hear. But I will try and read up on some of the theories explained here.
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I'm sorry not to be involved in this- I've been out for the day. Anyhow, in what way is my approach flawed, each coin toss is independent and there are 97 sequences of 4 coins- each of which starts with a possible head.
A genuine question- why is this wrong? |
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"so you don't get 97 attempts at all"
Eldrick - why not? In this approach you effectively start again at each toss whether you have thrown a head or not. |
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each toss is independent, each set of 4 is not as they all overlap
you don't get 97 fresh attempts, far less in practice |