a coin tossing problem
$begingroup$
A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?
probability
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
$endgroup$
add a comment |
$begingroup$
A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?
probability
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
$endgroup$
add a comment |
$begingroup$
A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?
probability
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
$endgroup$
A and B are playing coin tossing games. If there is a 2 continuous heads, then A win. If there is a 3 continuous tails, then B win. They keep tossing the coin until one of them win. what's the probability of A wins the game?
probability
probability
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
asked Dec 7 '18 at 5:35
Mango SickMango Sick
91
91
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Hint:
For A to win, it needs to get $2H$ before B gets $3T$.
So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
$$P(HH)=frac{1}{2^2}$$
$$P(THH)=frac{1}{2^3}$$
$$P(TTHH)=frac{1}{2^4}$$
There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$
Try to generalize for $n$ tosses, and then solve for $nto infty$
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
$endgroup$
add a comment |
$begingroup$
The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)
{ inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]
and is $[frac{15}{22}, frac{7}{22}]$.
In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).
Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
$endgroup$
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
For A to win, it needs to get $2H$ before B gets $3T$.
So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
$$P(HH)=frac{1}{2^2}$$
$$P(THH)=frac{1}{2^3}$$
$$P(TTHH)=frac{1}{2^4}$$
There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$
Try to generalize for $n$ tosses, and then solve for $nto infty$
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
$endgroup$
add a comment |
$begingroup$
Hint:
For A to win, it needs to get $2H$ before B gets $3T$.
So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
$$P(HH)=frac{1}{2^2}$$
$$P(THH)=frac{1}{2^3}$$
$$P(TTHH)=frac{1}{2^4}$$
There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$
Try to generalize for $n$ tosses, and then solve for $nto infty$
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
$endgroup$
add a comment |
$begingroup$
Hint:
For A to win, it needs to get $2H$ before B gets $3T$.
So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
$$P(HH)=frac{1}{2^2}$$
$$P(THH)=frac{1}{2^3}$$
$$P(TTHH)=frac{1}{2^4}$$
There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$
Try to generalize for $n$ tosses, and then solve for $nto infty$
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
$endgroup$
Hint:
For A to win, it needs to get $2H$ before B gets $3T$.
So, possible cass may be: $$(H,H),(T,H,H), (T,T,H,H)$$
$$P(HH)=frac{1}{2^2}$$
$$P(THH)=frac{1}{2^3}$$
$$P(TTHH)=frac{1}{2^4}$$
There can be more possibilities such as $$(H,T,T,H,H),(T,T,H,T,H,H) text{ etc. }$$
Try to generalize for $n$ tosses, and then solve for $nto infty$
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
edited Dec 7 '18 at 5:59
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
answered Dec 7 '18 at 5:53
ideaidea
2,15841125
2,15841125
add a comment |
add a comment |
$begingroup$
The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)
{ inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]
and is $[frac{15}{22}, frac{7}{22}]$.
In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).
Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
$endgroup$
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
add a comment |
$begingroup$
The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)
{ inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]
and is $[frac{15}{22}, frac{7}{22}]$.
In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).
Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
$endgroup$
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
add a comment |
$begingroup$
The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)
{ inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]
and is $[frac{15}{22}, frac{7}{22}]$.
In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).
Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
$endgroup$
The probability vector $[mathbb{P}(text{A wins}), mathbb{P}(text{B wins})]$ is given by the average of the first and third row of (matrices typed like when inputted to for example to WolframAlpha)
{ inverse (1/2*[[2,-1,-1,0,0], [0,2,-1,0,0],[-1,0,2,-1,0], [-1,0,0,2,-1], [-1,0,0,0,2]] ) } [[0,0], [1/2,0], [0,0], [0,0], [0,1/2]]
and is $[frac{15}{22}, frac{7}{22}]$.
In this solution we are calculating the absorption probabilities of a Markov chain with states $H, HH, T, TT, TTT, A, B$ (with obvious meanings and transition probabilities).
Here' how the chain looks like (all transitions have probability $frac{1}{2}$ except the absorbing winning states $A$ and $B$ of course have self-transition with probability $1$) I included the starting state $S$ here but not in the calculations where this is accounted for by averaging whether you start from $H$ or $T$ (whether the first toss is heads or tails):
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
edited Dec 8 '18 at 10:32
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
answered Dec 7 '18 at 12:09
ploosu2ploosu2
4,6451024
4,6451024
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
add a comment |
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
$begingroup$
For some reason, WolframAlpha wouldn't show the result correctly if you put the last $frac{1}{2}$ inside the matrix like I have done above. Take it out and remember to divide by $2$ yourself if this also happens to you, i.e insert [[0,0], [1,0], [0,0], [0,0], [0,1]] instead in the above formula.
$endgroup$
– ploosu2
Dec 7 '18 at 12:12
add a comment |
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