Actual interview brainteaser at a top flight shop

at 0 W 2 L you must have at least $25

at 1 W 2 L you must have at least $50

at 2 w 2 L you must have at least $100

at 2 W 1 L you must have at least $150

at 2 W 0 L you must have at least $175


I think anyway....don't know if that helps at all
 
Okay, I'm not doing any more guesses without an explanation of why it would work for all cases. Because the actual solution is slightly more complicated in that it needs to describe what to do in all cases (though, it's not tedious, because there's a very simple logic and pattern in all cases)

Quote from KarateChop:

Ok my last guess. 37.50 37.50 If you win first 2 flips or lose first 2 flips continue with 25 50 100. If you win ONLY 1 of first 2 flips, continue with 50 50 100.
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The numbers aren't entirely right and the cases aren't entirely right, but I think you might be on the most correct path of thinking so far.

Quote from 4tenTrader:

...

I think anyway....don't know if that helps at all
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Well, you need to have 25 dollars minimum to bet on flip 3. So the most you can lose in first 2 flips is 75. Assuming you lose first 2, you need to win last 3, so you can just bet it all on every flip hence the 25, 50, 100 strategy. If you need the last flip to be heads, you need to have $100 to bet and win the game. So we need min $25 for flip 3, $50 for flip 4 and $100 for flip 5(or have already won the game). With the 37.50 37.50 you are either break even, down to 25 or up to 175. If you are up to 175, you need to come out +25 on the next 3 flips should a head appear. As soon as a head appears, you stop betting. You either win or end up broke. Now, the other scenarios. If you are at 25 at the end of 2 flips, you need to win last 3 and so you just bet everything evey flip. If you have split the first 2 flips, you need 2 of the final 3 flips to be heads. You have $100, and you are going to bet 50, 50, 100.
If you lose the third flip, you need the last 2 flips to be heads and are in the same position as the guy who lost the first 2 flips. If you win the third flip, you need 1 of the last 2 to come up heads. So you bet 50 to either win on next flip or be in the position to have it all($100) riding on the last flip.
 
Really? So if you have $25 dollar, you've seen two heads so far, you got it all. Heads come up. you won the game. but you only have $50 instead of $200.

Otherwise, I think you are beginning to think about the problem in the right way

Quote from KarateChop:

Well, you need to have 25 dollars minimum to bet on flip 3....
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How do I end up with $25 having seen two heads already? I would have $175 on flip 3 with my betting pattern? I am saying you must leave yourself with at least $25 bankroll regardless of outcome of first 2 flips. I'm betting 0 on any remaining flips once I've won the game.
 
This is what I don't understand... so when do you end up in a situation where you only have $25 left? If there's still a possibility of winning on the next flip or two flips, then how would $25 ever get you to the $200 you need?

Quote from KarateChop:

How do I end up with $25 having seen two heads already? I would have $175 on flip 3 with my betting pattern? I am saying you must leave yourself with at least $25 bankroll regardless of outcome of first 2 flips. I'm betting 0 on any remaining flips once I've won the game.
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Hint: it might prove very very useful to draw a tree of all the way the game can progress it's actually not a very large at all
 
Here is another try:

Do not bet anything for the first 2 flips.

00-111(W), 00-anything else is (L)
After the first 2 flips resulted in 2 tails, you bet everything, because even if the 4th and 5th flips are head, you still lose.

10-011(W), 10-101(W), 10-110(W), 00-anything else is (L)
After the first 2 flips resulted in 1 tail and 1 head, bet 50% on the 3rd flip, this is because it need to be equal to the bet in the 4th flip. If 3rd and 4th flips worked out, you got your $100 return and won the game. If either one of 3rd flip or 4th flip didn't work out, they would offset each other +50-50, if not which mean both 3rd and 4th flip didn't work out, than there won't be enough head flip to win the game anyway. If the 3rd and 4th flip offset each other, bet everything in the 5th flip (you will either win $100 or lose the game).

11-000(L), 11-anything else (W)
If the first 2 flips resulted in 2 heads, you take your money and run! :p
 
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?
 
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