Actual interview brainteaser at a top flight shop

If it is correct that you must exit on three wins in a row and double on each win, then 14.28 seems to represent an upper limit (funny how that number came up much earlier).

Why? Suppose you happened to have 3 wins on the very first three tosses.

Capital, Bet, New Capital
100, 14.28, 114.28
114.28, 28.57, 142.85
142.85, 57.14, 200


The equation is simply 1x+2x+4x = 100
x = 14.28

If you somehow get that case of 1st three ones, and bet > 14.28, you will always exceed +100. It doesn't solve the guarantee of all three head cases = +100, but if I interpreted it correctly, it does present a limiting boundary on the 1st bet.
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I didn't get an answer to my specific questions, but maybe someone can clarify if the bet doubling on wins is a strict requirement. I gathered that from some earlier comments from OP.

The doubling requirement would also invalidate karate chops results.
 
Quote from sjfan:

Chop, you are definately on the right path. You are noticing a key requirement here. But you aren't exactly right yet.

If you want to keep going, my hint (in addition to my post previous to this) is to actually chart out all the ways the game can progress (it's not that much) and see how you might apply your insight in a more... specific manner.

I will post the full solution after xmas! (sorry, I got a lot of family stuff to attend to but I promise to post a nicely formatted one with charts and stuff but it will take some time to work it out)
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Either you're not explaining the problem right or you don't actually know the answer. Karatechop and I have the correct answer based on the rules you give.

If not then please explain EXACTLY why the solution we posted is wrong.
 
I'm with MustPlayOptions on this one. I do not see why our solution is incorrect. I'll ask you again SjFan, is there a UNIQUE solution? One that does not require us to deviate depending on the outcome of the first 2 tosses?
 
This is actually the correct answer

I didn't read that last part (when the first two diverges) when I was flipping through the back log replies.

Here's how I thought about the solution. Work backwards from all the terminal cases.

Consider HHTTH. You would end up with $200. Consider HHTTT. You would end up with zero. So, you need at least 100 on head to make this work. So at HHTT, you'll need to have $100 on hand regardless of how you got there. Now consider HHTH. You'll need to have $200 at this point. So, at HHT, you'll need at $150 to make this work.

If you work backwards from all the cases, you'll see from the point of the view of the first bet, you'll need to have $137.5 if you flip head. So you need to bet 37.5.

Since this game is fair and symmetric, you will always be at $0 if you end up in a loss situation.

Good work KarateChop. For (a) getting it right and (b) me calling it incorrect before fully reading through your post... I'll be happy to send you a finance related trivia prize (if you are comfortable sending me your addy over PM, which I totally understand if you are not, in which case, I take my hat off to you).

Quote from KarateChop:

SJFan, this is my strategy. I have not come up with a way it loses, perhaps you can.
First 2 flips:

37.50 37.50

If you won or lost BOTH of those 2 flips, you bet 25, 50, 100 UNTIL you are broke or have won game. Then you bet 0.

If you won 1 of first 2 flips, you are even ($100 bankroll),needing 2 of last 3 flips to be heads. So bet 50, 50, 100 UNTIL you have won game or game is over. You bet 0 once you have won game.
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We just need to meet the required payoff in its future value, not its present value. So , no need for the risk free rate

Quote from burritobob:

wait....dont we need to know the risk free rate
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That was an unfortunate typo on my part. The correction is on the second page or so. (Someone pointed it out to me - but by then it was over 30 minutes since the initial post so I couldn't correct it)

Quote from henry76:

doesn't question state "you will lose on second head"
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