Probability of profit.












0












$begingroup$


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







share|cite|improve this question











$endgroup$

















    0












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 13:15









      Namaste

      1




      1










      asked Jan 8 at 13:06









      pawelKpawelK

      618




      618






















          2 Answers
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          0












          $begingroup$

          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 $1$.






          share|cite|improve this answer











          $endgroup$





















            0












            $begingroup$

            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.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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














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            2 Answers
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            2 Answers
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            0












            $begingroup$

            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 $1$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              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 $1$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                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 $1$.






                share|cite|improve this answer











                $endgroup$



                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 $1$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 8 at 13:36

























                answered Jan 8 at 13:28









                lulululu

                43.3k25080




                43.3k25080























                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      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


















                    0












                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$













                    • $begingroup$
                      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
















                    0












                    0








                    0





                    $begingroup$

                    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.






                    share|cite|improve this answer









                    $endgroup$



                    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.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    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




















                    • $begingroup$
                      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




















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