**Solutions:**

**1.** If the advisors' vote is two against one, then bet x on the majority. If you lose, then the advisor in the minority must be the truth-teller so you can bet everything on the next two bets by following the advice of the truth-teller. This strategy gives you winnings of 4 * (100 - x).

If you win by following the majority the first time, then disregard the advisor in the minority and you'll have (100 + x) after the first bet. In the next bet if the remaining two agree, then you should bet everything because one of them must always tell the truth. But if the remaining two disagree, you'll find out who is telling the truth by betting nothing. On the last bet, you follow the truth teller and end up with at least 2 * (100 + x).

Setting 4 * (100 - x) = 2 * (100 + x), we conclude that 6x = 200 or x = 33 1/3.

Plugging that into either equation corresponds to handling both cases when the vote is two against one. Thus, in either case, we end up with 2 2/3 of what we start with -- a total of $233.33.

**2.** In this case, unfortunately, you cannot guarantee anything. The reason is that your "advisors" can arrange it so that you never have a reason to choose one outcome over the other. Call the advisors A, B, C and D, and imagine this set of results.

*Bet 1:* AB advise 0; CD advise 1.

*Outcome:* 1, so A and B have each lied once.

*Bet 2:* AC advise 0: BD advise 1.

*Outcome:* 1, which means A has lied twice and so is a definite liar.

B and C have each lied once.

D has not yet lied.

*Bet 3:* BD advise 0; AC advise 1.

*Outcome:* 1. So, B is a definite liar. C and D have each lied once.

*Bet 4:* DC advise 0 and BD advise 1.

Notice that, in bet 2, if you had bet on 1 but the outcome had been 0 (because the advisors had changed the written result), then B would be out, A and C would have lied once and C would have never lied. This result is symmetric to what happened with just a rearrangement of labels. So, you would have learned no more if the outcome had been 0. Similarly, in the third bet, if the outcome were 0, then D and B would both be possible partial truth tellers in bet 4. Thus, no series of outcomes ever teaches you whom to trust. The advisors can arrange it so that every bet you make is a loser.

**3.** Even with five bets, the advisors can keep you from winning much. Their strategy starts similarly to their strategy when you had only four bets.

*Bet 1:* AB advise 0; CD advise 1.

*Outcome:* 1, so A and B have each lied once.

*Bet 2:* AC advise 0: BD advise 1.

*Outcome:* 1, so A has lied twice so is a definite liar.

B and C have each lied once.

D has not yet lied.

*Bet 3:* BC advise 0; AD advise 1.

Now, we must do a case analysis.

Case 1: Suppose you bet nothing on bet 3. Suppose the outcome is 1. Then we know that D is the partial truth teller because everyone else has lied twice. However, D may still lie once. If D advises 1 in bet 4 and you bet $x on 1, then if x < 50 and D tells the truth in bet 4, you cannot bet anything on bet 5 because D could lie then. If x >= 50, then D could lie in bet 4 and you will never do better than regain your original capital. So, if you bet nothing, you cannot attain $150.

Case 2: Suppose you bet something on 0 in bet 3. Then your situation is as in case 1 only worse.

Case 3: Suppose you bet $x on 1 in bet 3. If x < 12.50, then if the outcome is in fact 1 and you bet $y in bet 4 on the outcome offered by D and x + y < 50, then if you win in bet 4, you cannot bet in bet 5 (because D may lie). If x + y > 50, then y > 37.50, so if you lose bet 4, you will have less than $75, so cannot attain $150 in bet 5. Therefore x >= 12.50.

Now if the outcome is 0. Then you have (100 - x) and you know that B, C and D have all lied once. In bet 4, two of them must advise one outcome and one must advise the other. Let's say you bet $z on the outcome suggested by the majority. Then x + z <= 25, otherwise losing in bets 3 and 4 would leave you less than $75, making it impossible to attain $150 in bet 5. But this reasoning implies z <= $12.50, and therefore if you win in bet 4, you will have at most $100 and D can still lie in bet 5. If you bet with the minority in bet 4, the majority could be telling the truth and then there would remain two advisors who have lied only once.

Therefore, you cannot guarantee to attain $150. In fact, a refinement of this argument shows you cannot even attain $134.