Logic/Probability Question Alice and Bob play the following game: There are three fair coins available, a 5p coin, a 10p coin and a 20p coin. Alice owns one of these coins, and Bob owns the other two coins. All three toins are tossed simultaneously. It is agreed that any coin falling tails counts as zero for its owner. Any coin falling heads counts its value in pence for the owner. The person that gets the highest score wins all three coins. If all coins come up tails no one wins and the toss is repeated. If at least one coin comes up heads, the game ends. Does it matter which coin Alice owns?
 A: No, it doesn't matter, the game is fair (expected outcome 0p) no matter which coin Alice owns. 
We have 7 equiprobable outcomes (all 3 tails is a fair reroll, so the remaining 7 outcomes are still equiprobable).
Note that due to the denominations used, the largest coin denomination coming up as heads will always win.
So what happens when Alice owns the 5p coin? There is only one outcome that will make her win: HTT (5p coin, 10p coin and 20p coin in that order, used from now on). That's a $\frac17$ probability. That means her expcetd winnings are
$$w_5=\frac17(5+10+20) + \frac67\times 0 - 5 = 5 - 5 =0$$
The "-5" at the end comes from the fact that Alice owned the 5p coin before; if she agrees to the game, it is her "bet".
If Alice owns the 10p coin, then the 10p coin must show heads for Alice to win, and the 20p must show tails. What the 5p coin does is irrelevant. So now there are now 2 outcomes that make Alice win: HHT and THT. So Alice now wins with a probability of $\frac27$. Her expected winnings are
$$w_{10}=\frac27(5+10+20) + \frac57\times 0 - 10 = 10 - 10 =0,$$
which is again $0$.
If Alice owns the 20p coin, now all that matters is that it comes up heads, so there are now 4 outcomes where Alice wins: XXH, where each X is independently replaced by H or T. That means her win probability increases to $\frac47$. However, her expectation is still
$$w_{20}=\frac47(5+10+20) + \frac37\times 0 - 20 = 20 - 20 =0.$$
So the expected winnings of Alice are $0p$ in each case, which means it does not matter which coin Alice owns. 
This counter intuitive result can be explained by the fact that the higher the coin denomination that she previously owns, the higher her loss if she actually looses. That is offset (and this case, exactly) by the higher chance to win at all.
