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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.

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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 $5\times 3=15$ 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 $1$.

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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 $4\cdot0.5=2$ 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 $2\cdot0.5=1$ 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.

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  • $\begingroup$ 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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