Probability of profit.
$begingroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
edited Jan 8 at 13:15
Namaste
1
asked Jan 8 at 13:06
pawelKpawelK
618
$endgroup$
add a comment |
$begingroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
edited Jan 8 at 13:15
Namaste
1
asked Jan 8 at 13:06
pawelKpawelK
618
$endgroup$
add a comment |
$begingroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
edited Jan 8 at 13:15
Namaste
1
asked Jan 8 at 13:06
pawelKpawelK
618
$endgroup$
Consider this game:
Person A has 4 coins of 1 pound. Person B has 2 coins of 2 pounds. They are once throwing their own coin. Head-the owner keeps the coin. Tails- the owner gives this coin to the player (A/B). (But this coin doesn't take part(later) in the game). At the end of the game, they count the money. Do A and B have equal chances of getting a profit?
At first, I tried to write out all the possibilities. But this is not effective.
probability
probability
edited Jan 8 at 13:15
Namaste
1
asked Jan 8 at 13:06
pawelKpawelK
618
edited Jan 8 at 13:15
Namaste
1
asked Jan 8 at 13:06
pawelKpawelK
618
edited Jan 8 at 13:15
Namaste
1
edited Jan 8 at 13:15
Namaste
1
edited Jan 8 at 13:15
Namaste
1
1
asked Jan 8 at 13:06
pawelKpawelK
618
asked Jan 8 at 13:06
pawelKpawelK
618
asked Jan 8 at 13:06
pawelKpawelK
618
618
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
43.3k25080
$endgroup$
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
$endgroup$
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
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2 Answers
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2 Answers
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$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
43.3k25080
$endgroup$
add a comment |
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
43.3k25080
$endgroup$
add a comment |
$begingroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
43.3k25080
$endgroup$
All we care about is the number of heads and tails each player throws.
For $A$ there are $5$ cases, as $A$ might throw from $0$ to $4$ tails. The probabilities are $$(.0625,.25,.375,.25,.0625)$$
For $B$ there are $3$ cases, as $B$ might throw from $0$ to $2$ tails. The probabilities are $$(.25,.5,.25)$$
Let's describe an outcome of the game as a pair $(a,b)$ where $A$ throws $a$ Tails and $B$ throws $b$ Tails. There are $5times 3=15$ possible outcomes.
$A$ turns a profit in the scenarios $$(0,1),,(0,2),, (1,1),,(1,2),, (2,2),,(3,2)$$
The probabilities of those cases sum to $boxed {.390625}$
$B$ turns a profit in the scenarios $$(1,0),,(2,0),,(3,0),,(4,0),,(3,1),,(4,1)$$ The probabilities of those cases also sum to $boxed {.390625}$
Just as a consistency test, ties occur in the scenarios $$(0,0),,(2,1),,(4,2)$$ The probabilities of those cases sum to $boxed {.21875}$ and the three total probabilities do indeed sum to $1$.
answered Jan 8 at 13:28
lulululu
43.3k25080
edited Jan 8 at 13:36
answered Jan 8 at 13:28
lulululu
43.3k25080
answered Jan 8 at 13:28
lulululu
43.3k25080
answered Jan 8 at 13:28
lulululu
43.3k25080
43.3k25080
add a comment |
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
$endgroup$
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
$endgroup$
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
$begingroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
$endgroup$
Person A throws 4 coins, and they has a $50%$ chance of keeping each one, so they expect, on average, to end the game with $4cdot0.5=2$ coins, worth 2 pounds. Similarly, Person B throws 2 coins, also with a $50%$ chance of keeping each one, and so they expect to end the game with $2cdot0.5=1$ coin, worth 2 pounds. So they both expect to end the game with 2 pounds worth of coins, meaning they each have equal chances to gain a profit from the game.
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
answered Jan 8 at 13:17
Calvin GodfreyCalvin Godfrey
633411
633411
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
$begingroup$
It's not obvious at least not to me, why saying the game has expectation $0$ for each means that they have the same chance of getting a profit. That is not true generally (though it is in this case)...imagine a game in which I have a $frac 34$ chance of getting $100$ from you and you have a $frac 14$ chance of getting $300$ from me. That has expectation $0$ for both of us but I have a much greater chance of getting a profit.
$endgroup$
– lulu
Jan 8 at 13:34
add a comment |
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