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• 0 ## Question (Hope this hasn't been done before)

1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?

Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

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• 0 i dont think anyone considers such simple solutions so i posted this even though there was already a "winner"

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• 0 Ok here's the deal as I see it.

there are 9 coins will equal chance of heads regardless of previous outcomes

there is one coin with 100% chance of heads if picked

there is 1/10 chance of picking that one coin at random

therefore:

chance of picking head coin + chance any other coin picked falls heads = chance coin picked falls heads

1/10 + 5/10 = 60% chance any coin pics falls heads

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• 0 so i decided im going to post a different explanation of this problem every five minutes until everyone agrees with me because hey im vain and conceded and a ****. by the way im a math major so listening isnt a bad idea

as much as i understand the problems people are having with the first question. (i had some too when i first saw these kinds of problems). it is a formula in statistics and probability that the probability (from now on abbreviated P()) of something occurring lets call it A. given that B has already happened. is P(A intersect B)/P(B). (intersect means where the two probabilitys over lap meaning whats the chance that they both happened which in this case 9 heads cannot occur without 8heads first so the intersect is P(9heads). so if

P(9 heads given 8 heads)=P(9heads intersect 8 heads)/P(8heads)=P(9heads)/P(8heads)

P(8 heads)= (9/10) * (.5)^8 + 1/10 * 1^8

chosing reg coin flip 8 heads bias coin 8 heads

similarly P(9 heads)= 9/10*.5^9 + 1/10 *1^9

you can divide on your own (i know i should use a spoiler but its so late in the thread and im lazy

now another way of look at this. you have two conditions A=.4 and B=.7 and A is entirely in B meaning A cannot happing without B. so you know B happened so your in that .7 or 70%. and somone asks will A happen and you think 40% chance but wait you have more information. theres only .3 left in that .7 that you know ur already in. so you have better chance then that you have .4/.7.

a more concrete example is someone asks you "think you can make 4 free throws in a row". You say i wont take that bet I only make .75 of my shots meaning i have only a 32% chance. then you make three in a row. your friend says "i bet you cant make four in a row". Youve already made three. now you have a .75 percent chance of doing it and guess what math triumphs again that formula applied here is

P(3 row)=.75^3

P(4 row)=.75^4

P(3 row intersect 4 row)= P(4 row) (cant hit 4 in a row if you dont hit three)

P(4 row given 3 row)=.75^4/.75^3=.75

wala

now i hope i have bored you all or informed you all

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• 0 I understand the 55% math. There is a .45^8 chance that tails will be flipped 8 times in a row--or 0.0016815. There is a .1 chance that the double head was selected plus the same .0016815 chance that the other 9 legitimate coin will flip heads 8 times in a row. Therefore the chances of a 9th heads flipping is the ratio of the two likelyhoods

1-(0.0016815/(0.1 + 0.0016815) = 98.34%

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• 0 I understand the 55% math. There is a .45^8 chance that tails will be flipped 8 times in a row--or 0.0016815. There is a .1 chance that the double head was selected plus the same .0016815 chance that the other 9 legitimate coin will flip heads 8 times in a row. Therefore the chances of a 9th heads flipping is the ratio of the two likelyhoods

1-(0.0016815/(0.1 + 0.0016815) = 98.34%

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• 0 i think the answer to the first part is 1/19

Edited by monti
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• 0 (Hope this hasn't been done before)

1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?

Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

3/2

equally likely.

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• 0 yah simon thats the right answer but i have no idea y ur math worked you started in decimal and ended up in percentage which shouldnt happen plus i followed the math and i got 9.82 not 98.2 or .982 so i think you copied your math wrong, but I would like you to explain your math more clearly because somehow it worked and im curious

i think the chance of a normal coin flip 8 heads in a row is .5^8

or if your talking about chance of choosing a normal coin then it flipping 8 heads

9/10 *.5^8

or the chance of 8 heads

9/10*.5^8+1/10*1^8

the only way i can understand using the 11/21 is if you do this the long way and find the probabilty that your using the 2 head coin say p(2head) then add .5(1-p(2head)) or the probabily your using a normal coin time the chance of heads then adding p(2head). but what im trying to say is that you would have to use the 11/21 for that i think.

anyway im interested in how you reasoned the answer

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• 0 No, my math is correct. The first number in the denominator is 0.1--I suspect you entered it as 1.

Yes, the answer is .9835 or 98.35% chance of flipping another heads

The reason it is .45 is there are 11 heads and 9 tails. The chance of the first flip being a TAILS is 9/20 or 45%. The chances of flipping a HEADS from one of those same 9 "normal" coins is also 45% plus a 10% chance for getting HEADS from the double HEADS coin or 55%.

The chance of getting 8 heads or 8 tails in a row are very small--about 0.2%. These are overshadowed by the 10% possibility that the heads streak was caused by the double-headed coin.

yah simon thats the right answer but i have no idea y ur math worked you started in decimal and ended up in percentage which shouldnt happen plus i followed the math and i got 9.82 not 98.2 or .982 so i think you copied your math wrong, but I would like you to explain your math more clearly because somehow it worked and im curious

i think the chance of a normal coin flip 8 heads in a row is .5^8

or if your talking about chance of choosing a normal coin then it flipping 8 heads

9/10 *.5^8

or the chance of 8 heads

9/10*.5^8+1/10*1^8

the only way i can understand using the 11/21 is if you do this the long way and find the probabilty that your using the 2 head coin say p(2head) then add .5(1-p(2head)) or the probabily your using a normal coin time the chance of heads then adding p(2head). but what im trying to say is that you would have to use the 11/21 for that i think.

anyway im interested in how you reasoned the answer

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• 0 3/2

equally likely.

That first item there is a wee bit like saying I've got two kids and all three of them are girls.

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• 0 i hate to argue well thats just not true but anyway

the probability of fliping tails one time is .45 =(9/10)*1/2

but after that the coin is chosen the chance is .5 because once tails was flipped once it must be a fair coin otherwise no tails could have happened. this is where the mathematical error is coming from.

the probability of something like this is found by the chain rule or multiplication rule (i made those names up but im sure one of them is right) easiest way to do this is in steps

Probability 8 tail= first you have to chose a coin with tails on it (.9) then you have to get tails 8 time (.5^8)= (.9).5^8

probabilty 8 heads= first you have to chose a coin with heads (unidentical objects must be seperate) (.9) & (.1) then you have to get them (.5^8) & (1^8) respectively so = (9/10)(.5^8)+(.1)(1^8)

now instead of explaining the rest i refer you to previous explantions

luck is 9/10's the law

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• 0 I think I made a slight error in my calculation. The odds of flipping TAILs on any flip is 9/20 or .45. The odds of flipping HEADs on any flip is also 9/20 or .45 from one of the "normal" coins PLUS .10 from the double headed coin or .55 for HEADS

The odds of getting 8 TAILs in a row is therefore .45^8 or .001682. The odds of getting 8 HEADs in a row is 0.1 + .001682. The likelyhood of a 9th coin being another HEADS is the ratio of the of the odds of flipping 8 in a row of HEADS to the odds of Flipping all HEADS or all Tails.

In my prior examples I calculated the odds for flipping a 9th TAILS and subtracted that from 1 to get the odds for 9 HEADS--In the following, I'll calculate HEADS directly

Odds for a 9th consecutive HEADs = (0.1 + 0.45^8) / (0.1 + 2 * 0.45^8) = .9837 or 98.37%

In my prior "solution" I neglected to include the odds for flipping all TAILs in the denominator (in other words, no "2" in the prior line) which made a <0.0003 difference in the answer

Note that if there were 8 consecutive TAILs, the odds for a ninth TAIL would be 45% (9/20) since we've taken the double-headed coin out of the calculation as forcing a result as it does for HEADS

I understand the 55% math. There is a .45^8 chance that tails will be flipped 8 times in a row--or 0.0016815. There is a .1 chance that the double head was selected plus the same .0016815 chance that the other 9 legitimate coin will flip heads 8 times in a row. Therefore the chances of a 9th heads flipping is the ratio of the two likelyhoods

1-(0.0016815/(0.1 + 0.0016815) = 98.34%

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• 0 Actually I had the same initial assumption--but as explained below, the odds for ANY TAILs pick is 9/20 or 45%. While you can't get the 9th tail from the double-headed coin, it gives you two chances to break the consecutive run. The likelyhood of successive TAILs runs continues to be 0.45^n. The odds for a HEADs run at any pick continues to include the 10% chance that the double-headed coin will guarantee the result in addition to the odds it will get it's 9th HEADs in the same way the TAILS would get it's successful pick.

As shown below the calculation is

(0.1+.45^8)/(0.1 + 2 * .45^8) = .9837 or 98.37% chance of drawing a 9th HEADs

i hate to argue well thats just not true but anyway

the probability of fliping tails one time is .45 =(9/10)*1/2

but after that the coin is chosen the chance is .5 because once tails was flipped once it must be a fair coin otherwise no tails could have happened. this is where the mathematical error is coming from.

the probability of something like this is found by the chain rule or multiplication rule (i made those names up but im sure one of them is right) easiest way to do this is in steps

Probability 8 tail= first you have to chose a coin with tails on it (.9) then you have to get tails 8 time (.5^8)= (.9).5^8

probabilty 8 heads= first you have to chose a coin with heads (unidentical objects must be seperate) (.9) & (.1) then you have to get them (.5^8) & (1^8) respectively so = (9/10)(.5^8)+(.1)(1^8)

now instead of explaining the rest i refer you to previous explantions

luck is 9/10's the law

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• 0 Keep in mind it is ONE coin which is being flipped for the 9th time. Many of the proposed solutions don't take that into account. The 10% liklihood that THAT coin is the double-headed one plays very heavily vs. the very slim odds that one would roll 8 heads in a row from a true coin. Thus the approximately 98% chance that the selected coin was the double headed one to produce that result. ##### Share on other sites
• 0 you answer is pretty right but all im saying is this

this isnt entirely whats wrong with you method but if you concede this it will change how your doing the problem

in this problem you pick a coin at the beginning and stay with it. the chance of flipping a ninth tails is .5 not .45. Once youve fliped 8 tails all youve proven is that you dont have the double headed coin so one flip would have done that. so you know you have one of the fair coins so 50-50.

i dont know if i misunderstood this part but if not

1-P(8 heads) does not equal P(8tails) or the other way around theres mixes in there (heads and tails).

also i think u say you answer is just P(8heads)/(P(8heads)+P(8tails))

so what if you tried this formulas on 2heads (having already flipped 1) so P(2heads)/(P2(heads)+P(2 tails))

so with you math i think would be (.1+.45^2)/((.1+.45^2)+.45^2))=.599

by my math (but your formula) would be (.1+.9*(.5^2))/((.1+.9(.5^2)+.9(.5^2))=.5909

now .5909 is right but y. to find this you are suppose to use a formula of P(2heads given 1 heads)=P(2 heads)/P(1heads)

and it just so happens that .9*.5 (1 heads)=.9*.5*2(2 heads *2) so luck

and your math only works on 9 consectutive because you .05^9 is so little that it doesnt matter.

think of this spatially even draw it and it will help alot

u can think of probabilities as a square space the whole square is 1. or 100%. you have the area of chance of 8 heads a small portion. We tell you that 8 heads already happened so you know your in that small area and thats all that you need to consider. And then we ask whats the chance of 9 heads. And you know your already in that .10351... of the square and within that is a small portion of .101757... that is the chance of 9 heads. so you say there is this .101757 entirely within the .10351 that i know im in. so then i have a .101.../.10351 chance that it will happen. its conditional probability

think of this way .10351 u know ur in so make that your 100% or multiply by its reciprical 1/.10351=9.660377

so 9 head is in there so that needs to be enlarged .101757*9.660377=.9830...

this might not be that clear its hards to explain this stuff over type, but hey i try

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• 0 Ok, I agree that the P(2tails) is .45 * .5. I was forgetting my own reminder that the same coin is used throughout--so there are 9 tails out of 18 (not 20) possibilities.

I think that changes my formula to

Probability of a 9th consecutive Heads = (0.1+.45^8)/(.1+0.45^8+0.45*0.5^7) = 0.96658 or 96.66% instead of 98.37%

It's late and I'll have to mull it over tomorrow.

you answer is pretty right but all im saying is this

this isnt entirely whats wrong with you method but if you concede this it will change how your doing the problem

in this problem you pick a coin at the beginning and stay with it. the chance of flipping a ninth tails is .5 not .45. Once youve fliped 8 tails all youve proven is that you dont have the double headed coin so one flip would have done that. so you know you have one of the fair coins so 50-50.

i dont know if i misunderstood this part but if not

1-P(8 heads) does not equal P(8tails) or the other way around theres mixes in there (heads and tails).

also i think u say you answer is just P(8heads)/(P(8heads)+P(8tails))

so what if you tried this formulas on 2heads (having already flipped 1) so P(2heads)/(P2(heads)+P(2 tails))

so with you math i think would be (.1+.45^2)/((.1+.45^2)+.45^2))=.599

by my math (but your formula) would be (.1+.9*(.5^2))/((.1+.9(.5^2)+.9(.5^2))=.5909

now .5909 is right but y. to find this you are suppose to use a formula of P(2heads given 1 heads)=P(2 heads)/P(1heads)

and it just so happens that .9*.5 (1 heads)=.9*.5*2(2 heads *2) so luck

and your math only works on 9 consectutive because you .05^9 is so little that it doesnt matter.

think of this spatially even draw it and it will help alot

u can think of probabilities as a square space the whole square is 1. or 100%. you have the area of chance of 8 heads a small portion. We tell you that 8 heads already happened so you know your in that small area and thats all that you need to consider. And then we ask whats the chance of 9 heads. And you know your already in that .10351... of the square and within that is a small portion of .101757... that is the chance of 9 heads. so you say there is this .101757 entirely within the .10351 that i know im in. so then i have a .101.../.10351 chance that it will happen. its conditional probability

think of this way .10351 u know ur in so make that your 100% or multiply by its reciprical 1/.10351=9.660377

so 9 head is in there so that needs to be enlarged .101757*9.660377=.9830...

this might not be that clear its hards to explain this stuff over type, but hey i try

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• 0 ok i thought of a little better ways to explain this and i know im not that good at explaning (theres a reason im not a teacher)

i might have already said this but just skip it if i did 9heads can not happen without 8heads right so if were thinking of this spatially the 9heads area would be entirely in the 8 heads area

if not spatially

then the P(9heads)must be part of the P(8heads) in fact P(9heads)+P(8headsthen tails)=p(8heads) i mean those are the only choices just like 1tails plus 1 heads = 1 its whole area

if 8 heads has already happened whats the chance of 9 heads

so then the question is what is the ratio of 9heads to 8heads

next try at being a half decent explantion

i posted this earlier

a more concrete example is someone asks you "think you can make 4 free throws in a row". You say i wont take that bet I only make .75 of my shots meaning i have only a 32% chance. then you make three in a row. your friend says "i bet you cant make four in a row". Youve already made three. now you have a .75 percent chance of doing it and guess what math triumphs again that formula applied here is

P(3 row)=.75^3

P(4 row)=.75^4

P(3 row intersect 4 row)= P(4 row) (cant hit 4 in a row if you dont hit three)

P(4 row given 3 row)=.75^4/.75^3=.75

wala

so by your formula if i understand it it would be .75^4/(.75^4+.25^4)=.9878

i think that would be how you would apply your formula anyway for another day

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• 0 There are two processes that are happening:

1. selection of the coin

2. flipping of the selected coin

The p(8 heads in a row) = (9/10)*(.50^8)+ (1/10)*1^8 = 521/5120

Given this the 9th flip will result in a heads = (9/10)*(.50^9)+ (1/10)*1^9 = 1033/10240

Hence the probability that the 9th flip will yield heads given the first 8 are all heads = p(9)/p(8) = 1033/1042 = 0.9914

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