2 Coins
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How many times would i have to flip 2 coins before i've seen a heads on both coins. They don't have to come together. I just want to have seen at least 1 heads on both coins.
cheers
Trev
Quote: TrevorHow many times would i have to flip 2 coins before i've seen a heads on both coins. They don't have to come together. I just want to have seen at least 1 heads on both coins.
I'm thinking I must not even understand the question.
Issue #1: Does sentence #2 ("They don't have to come together.") mean that the two coins don't have to show heads at the same time and that you just want to see a series of results that include the head side of each coin? Example, if the two coins are a nickel and a dime, you want to flip the two until you have seen the head side of the nickel and the head side of the dime. Is that what the question is describing?
Issue #2: If the suggested interpretation of Issue #1 is correct, is sentence #1 ("How many times would I have to flip ... ?") asking about the maximum number of flips before the criteria would be met or the expected number of flips? If you're asking about the maximum, I think the answer may be infinity.
Apparently, I need more info in order to understand what is being asked.
Quote: DocI'm thinking I must not even understand the question.
Issue #1: Does sentence #2 ("They don't have to come together.") mean that the two coins don't have to show heads at the same time and that you just want to see a series of results that include the head side of each coin? Example, if the two coins are a nickel and a dime, you want to flip the two until you have seen the head side of the nickel and the head side of the dime. Is that what the question is describing?
Issue #2: If the suggested interpretation of Issue #1 is correct, is sentence #1 ("How many times would I have to flip ... ?") asking about the maximum number of flips before the criteria would be met or the expected number of flips? If you're asking about the maximum, I think the answer may be infinity.
Apparently, I need more info in order to understand what is being asked.
Fwiw, I took the question to be a simplified 2 dice question. What is your expectation for any particular number? Tt, ht, th, hh. 1/4 of the time the first roll satisfies it. 2/4 half-satisfies it. 1/4 no help. 2nd roll, depends on first roll, but 2 of 7 (th, hh) satisfies the ht roll, 2 of 7 (ht, hh) satisfies the th roll, 3 in 7 (copy ht or th, tt) doesn't help. 3rd roll, repeats that. Until you get the both heads at some point.
Not sure where 8/3 comes from.
Quote: beachbumbabsNot sure where 8/3 comes from.
It comes from the expected number of tosses needed for each coin to come up heads at least once.
The expected number of tosses before any heads at all show up is:
0 x 3/4 + 1 x 1/4 x 3/4 + 2 x (1/4)2 x 3/4 + 3 x (1/4)3 x 3/4 + ...
= 3/16 x (1 + 2 x 1/4 + 3 x (1/4)2 + 4 x (1/4)3 + ...)
= 3/16 x (1 + 1/4 + (1/4)2 + ...)2
= 3/16 x 16/9 = 1/3
The first toss that has at least one head will be two heads 1/3 of the time, so the expected number of additional tosses is 1/3.
The first toss that has at least one head will be one head and one tail 2/3 of the time; this is 1 toss plus the expected number for the other coin to come up heads, or 1 + (1/2 x 1 + (1/2)2 x 2 + (1/2)3 x 3 + ...)
= 1 + 1/2 x (1 + 1/2 x 2 + (1/2)2 x 3 + ...)
= 1 + 1/2 x (1 + 1/2 + (1/2)2 + ...)2
= 1 + 1/2 x 4 = 3
Thus, the expected number of additional tosses if the first toss with any heads has one head and one tail is 2/3 x 3 = 2, and the total number of expected tosses is 1/3 + 1/3 + 2 = 8/3.
You flip two coins at the same time until you've seen each coin land on heads. What is the expected number of flips that will take?
For example, consider the following set of flips, where the first letter is the outcome of coin 1 and the second coin 2:
HT
TT
HT
TH
This required four flips. You saw coin 1 land on heads the first flip but it took three more flips for coin 2 to land on heads.
Quote: WizardLet me rephrase the question.
You flip two coins at the same time until you've seen each coin land on heads. What is the expected number of flips that will take?
For example, consider the following set of flips, where the first letter is the outcome of coin 1 and the second coin 2:
HT
TT
HT
TH
This required four flips. You saw coin 1 land on heads the first flip but it took three more flips for coin 2 to land on heads.
Hi Guys,
Yes. What he said.
How did you come up with 8/3.
Trev
The expected number of tosses before any heads at all show up is:
..... =1/3
I thought it would be 3/4. There are 4 possible permutations with 3 giving at least 1 head.
Quote: TrevorThatDonGuy said:
The expected number of tosses before any heads at all show up is:
..... =1/3
I thought it would be 3/4. There are 4 possible permutations with 3 giving at least 1 head.
3/4 of the time, the first toss will have at least one head; this is 0 tosses "before any heads show up"
1/4 x 3/4 of the time, the first toss will be two tails, and the second toss will be at least one head; this is 1 toss.
1/4 x 1/4 x 3/4 of the time, the first two tosses will be two tails, and the third toss will be at least one head; this is 2 tosses.
And so on.
The expected number of tosses at the start that are both tails is:
0 x 3/4 + 1 x 1/4 x 3/4 + 2 x (1/4)2 x 3/4 + 3 x (1/4)3 x 3/4 + ... = 1/3.
1 + 2 x + 3 x2 + 4 x3 + ... = (1 + x + x2 + x3 + ...)2
Also, for -1 < x < 1, 1 + 1/x + (1/x)2 + (1/x)2 + ... = 1 / (1-x)
Quote: ThatDonGuy3/4 of the time, the first toss will have at least one head; this is 0 tosses "before any heads show up"
1/4 x 3/4 of the time, the first toss will be two tails, and the second toss will be at least one head; this is 1 toss.
1/4 x 1/4 x 3/4 of the time, the first two tosses will be two tails, and the third toss will be at least one head; this is 2 tosses.
And so on.
The expected number of tosses at the start that are both tails is:
0 x 3/4 + 1 x 1/4 x 3/4 + 2 x (1/4)2 x 3/4 + 3 x (1/4)3 x 3/4 + ... = 1/3.
1 + 2 x + 3 x2 + 4 x3 + ... = (1 + x + x2 + x3 + ...)2
Also, for -1 < x < 1, 1 + 1/x + (1/x)2 + (1/x)2 + ... = 1 / (1-x)
Oh i see. Thanks ThatDonGuy
at 1st read I did not get what the event was as you described.Quote: TrevorHow many times would i have to flip 2 coins before i've seen a heads on both coins. They don't have to come together. I just want to have seen at least 1 heads on both coins.
now I got it.
sounds like a 3x3 matrix card game I play (maybe later)
I also get 2.67 rolls on average
but from the distribution (i used a transition matrix in Excel)...
(that is interesting to me)
| flip | prob | cumulative prob | flip*prob |
|---|---|---|---|
| 1 | 0.25 | 0.25 | 0.25 |
| 2 | 0.3125 | 0.5625 | 0.625 |
| 3 | 0.203125 | 0.765625 | 0.609375 |
| 4 | 0.11328125 | 0.87890625 | 0.453125 |
| 5 | 0.059570313 | 0.938476563 | 0.297851563 |
| 6 | 0.030517578 | 0.968994141 | 0.183105469 |
| 7 | 0.015441895 | 0.984436035 | 0.108093262 |
| 8 | 0.007766724 | 0.992202759 | 0.062133789 |
| 9 | 0.003894806 | 0.996097565 | 0.035053253 |
| 10 | 0.001950264 | 0.998047829 | 0.01950264 |
| 11 | 0.000975847 | 0.999023676 | 0.01073432 |
| 12 | 0.000488102 | 0.999511778 | 0.005857229 |
| 13 | 0.000244096 | 0.999755874 | 0.003173247 |
| 14 | 0.000122059 | 0.999877933 | 0.001708828 |
| 15 | 6.10324E-05 | 0.999938966 | 0.000915485 |
| 16 | 3.05169E-05 | 0.999969483 | 0.00048827 |
| 17 | 1.52586E-05 | 0.999984741 | 0.000259396 |
| 18 | 7.62935E-06 | 0.999992371 | 0.000137328 |
| 19 | 3.81469E-06 | 0.999996185 | 7.2479E-05 |
| 20 | 1.90735E-06 | 0.999998093 | 3.81469E-05 |
| . | sum | avg # flips | 2.666624705 |
why the question?
as 2.67 is not 2 or 3 flips
Sally
the math for those so interested
| A | B | C | D | prob | prob |
|---|---|---|---|---|---|
| no H 1 coin | 1 coin * 2(A*2) | 1coin^2(A*A) | B-C (no HH) | both coins H cumulative | both coins H on X flip |
| 0.5 | 1 | 0.25 | 0.75 | 0.25 | 0.25 |
| 0.25 | 0.5 | 0.0625 | 0.4375 | 0.5625 | 0.3125 |
| 0.125 | 0.25 | 0.015625 | 0.234375 | 0.765625 | 0.203125 |
| 0.0625 | 0.125 | 0.00390625 | 0.12109375 | 0.87890625 | 0.11328125 |
| 0.03125 | 0.0625 | 0.000976563 | 0.061523438 | 0.938476563 | 0.059570313 |
| 0.015625 | 0.03125 | 0.000244141 | 0.031005859 | 0.968994141 | 0.030517578 |
| 0.0078125 | 0.015625 | 6.10352E-05 | 0.015563965 | 0.984436035 | 0.015441895 |
| 0.00390625 | 0.0078125 | 1.52588E-05 | 0.007797241 | 0.992202759 | 0.007766724 |
| 0.001953125 | 0.00390625 | 3.8147E-06 | 0.003902435 | 0.996097565 | 0.003894806 |
| 0.000976563 | 0.001953125 | 9.53674E-07 | 0.001952171 | 0.998047829 | 0.001950264 |
| 0.000488281 | 0.000976563 | 2.38419E-07 | 0.000976324 | 0.999023676 | 0.000975847 |
| 0.000244141 | 0.000488281 | 5.96046E-08 | 0.000488222 | 0.999511778 | 0.000488102 |
| 0.00012207 | 0.000244141 | 1.49012E-08 | 0.000244126 | 0.999755874 | 0.000244096 |
| 6.10352E-05 | 0.00012207 | 3.72529E-09 | 0.000122067 | 0.999877933 | 0.000122059 |
| 3.05176E-05 | 6.10352E-05 | 9.31323E-10 | 6.10342E-05 | 0.999938966 | 6.10324E-05 |
| 1.52588E-05 | 3.05176E-05 | 2.32831E-10 | 3.05173E-05 | 0.999969483 | 3.05169E-05 |
| 7.62939E-06 | 1.52588E-05 | 5.82077E-11 | 1.52587E-05 | 0.999984741 | 1.52586E-05 |
| 3.8147E-06 | 7.62939E-06 | 1.45519E-11 | 7.62938E-06 | 0.999992371 | 7.62935E-06 |
| 1.90735E-06 | 3.8147E-06 | 3.63798E-12 | 3.81469E-06 | 0.999996185 | 3.81469E-06 |
| 9.53674E-07 | 1.90735E-06 | 9.09495E-13 | 1.90735E-06 | 0.999998093 | 1.90735E-06 |
The expect number of flips for coin A to get a head is obviously 2.
Using a simple recursive equation of x = 1/4 x + 1/4, you get x= 1/3 chance that coin B comes up heads after coin A does.
So the expectation is 2 flips for A plus 1/3 chance of another 2 flips for B.
2 + 1/3 * 2 = 8/3
