Actual interview brainteaser at a top flight shop

[my7tvette]

Quote from OTCkrak:

first bet zero..

after the second head comes up, next flip bet $100

I think the first bet should be zero, but as has been pointed out, you can't wait until two heads and then go all or nothing (unless you don't achieve two heads until flip 4) because you risk "winning" the game, but not having at least $200. I haven't played out this scenario, but maybe some variant of RMarvin's strategy might work.

Bottom line, I guess I won't be getting any $300K job with a hedge fund anytime soon, because this has taken too damn long.

 

[sjfan]

I went away to do christmas shopping yesterday and look at the number of replies!

Okay, I'll respond to everyone, but right off the bat, this is wrong. You aren't accounting for all the possibilities of winning.

The trick isn't to figure out how to meet the payoff in one scenario, but in all scenarios. So the explaination of the solution MUST demonstrate by the strategy will work for all possible scenarios of winning.

Quote from KarateChop:

If I lose the first 2 flips, I lose $75. I have $25. I bet it, I win now I have 50. I bet it, I win, I have $100. I bet all that, I win, $200. Am I not understanding something here?

 

[nitro]

One solution is to bet $100 at step one. If you win, you bet zero the rest of the time and are guaranteed a win. If you lose, since you can't bet anymore, it also satisfies the conditions of the game.

Wait. The coin tosses don't stop even if I have no money to bet. So if I bet $100, and lose all my money, and the coins come HHH after coming T, I lose the game. Nevermind.

 

[Div_Arb]

i think the question is asking for a static bet amount that will keep you in the game for at least five flips, so you will have a chance to win. With this in mind, you need to remain solvent and still able to win the game if the first two flips go against you. With this in mind I come up with *******$25****** for my answer. See below:

Betting: $25
Flip Bet Result Principal
0 $25.00 ($25.00) $75.00
0 $25.00 ($25.00) $50.00
1 $25.00 $50.00 $100.00
1 $25.00 $50.00 $150.00
1 $25.00 $50.00 $200.00


You will win under all '111' scenarios. The point is, you need to remain solvent and still able to win the gasme if the first two flips go against you.

Thanks to whoever posted the nifty spreadsheet

 

[Smart Money]

Quote from sjfan:

I went away to do christmas shopping yesterday and look at the number of replies!

Okay, I'll respond to everyone, but right off the bat, this is wrong. You aren't accounting for all the possibilities of winning.

The trick isn't to figure out how to meet the payoff in one scenario, but in all scenarios. So the explaination of the solution MUST demonstrate by the strategy will work for all possible scenarios of winning.

Not possible. One of the scenarios is all tails.

I can't solve this because the rules aren't clear.

SM

 

[Smart Money]

Quote from Div_Arb:

i think the question is asking for a static bet amount that will keep you in the game for at least five flips, so you will have a chance to win. With this in mind, you need to remain solvent and still able to win the game if the first two flips go against you. With this in mind I come up with *******$25****** for my answer. See below:

Betting: $25
Flip Bet Result Principal
0 $25.00 ($25.00) $75.00
0 $25.00 ($25.00) $50.00
1 $25.00 $50.00 $100.00
1 $25.00 $50.00 $150.00
1 $25.00 $50.00 $200.00


You will win under all '111' scenarios. The point is, you need to remain solvent and still able to win the gasme if the first two flips go against you.

Thanks to whoever posted the nifty spreadsheet

What if you get all heads? It could happen. Then you have too much money. I think the rules need to be clarified better...or like I said...it is a question where you are being evaluated on how you respond...not the actual answer.

SM

 

[nitro]

My initial attempt is to represent 100 in binary, that is 1100100 base 2. 200 is 11001000 base 2, which is the whole thing shifted left by one bit.

Log2 of 100 is 6.643856.
Log2 of 200 is 7.643856.

I am just thiking out loud. Doubling is generally best represented in base 2 log since it linearizes the problem.

Form a binary number from heads and tails you see. So if it comes HT, you have 10. if TH, the number is 01. That doesn't seem to be right, since it is not position that matters, but total.

The algorithm is probably some mixture of the log of the amount + (the log of the head + tails) binary representation...

The difference between the goal and the current amount in base two binary is also possibly an idea. Initially this is clearly 100, or 1100100 base 2. XOR or something along these lines with the number of heads and tails to give the amount to wager...

Can this really be the solution though? What if the starting values were $1 and you needed to get to $2? would this schema work? Should the solution not be independent, as if it were stated, you start with $X, you need to get to $Y dollars...The algorithm should be independent of $100 and $200.

 

[Bogwaluth]

Can someone please post an answer or where to find one? The best I can come up with is winning 9 of 10 times when 3 of 5 flips is heads.

I get a starting bet of 31.76203 for those scenarios. I too will not be getting a cool job

 

[nitro]

Quote from nitro:

...
Can this really be the solution though? What if the starting values were $1 and you needed to get to $2? would this schema work? Should the solution not be independent, as if it were stated, you start with $X, you need to get to $Y dollars...The algorithm should be independent of $100 and $200.

Mapping the betting to the Fibonacci ratios can't be right either, as betting 1 / 1 has already been shown to not work...But What about starting with 1/2 2/3 etc? You have to have a choice function, otherwise you are not keeping track of the heads and tails.

 

[nitro]

Quote from nitro:

...You have to have a choice function, otherwise you are not keeping track of the heads and tails.

This is beginning to feel like a calculus problem, with the application of the chain rule. f(g(x))