wait, it's not quite that easy...if you bet 1/3 on 1st two flips, and split, 1 win/1 loss, you will be back to your initial $100, with only two heads remaining to be declared a "winner". At that point, you would need to change your bet to 1/2 of the $100 bankroll, or $50, in order to reach $200 in the next two flips (if you win the next two in a row). If you split the next two flips having bet $50 each, you will again be at $100, leaving you to bet it all on the final flip.
I looked at karatechop·s and it's the same as mine but he was first. the solution works and my post spells it all out.
Me three, same answer. I got it but I'm sure the interviewers would have gone home for the night by then. Sjfan, you are an ahole, I hope the sharks miss the playoffs! Just kidding but I won't be openying anymore of your posts that look like this anytime soon, I spent way too much time on this thing and I don't even want an I-bank job! I got hung up trying to do it as a system of equations like: www: 100 + B1 + B2(H) + B3(HH) = 200 and there are too many unknowns so I tried adding inequality restrictions preventing you from betting more than you had available and figuring out which ones were binding. Then I did it working backwards in a tree, figuring out how much you would need to keep playing at each node and it was a lot more intuitive.
OK, I read over most of the responses. I can see the fun of trying to figure out the math of it, but if winning is defined as having three heads somewhere in that series, and losing is defined as not getting three heads, then couldn't you wait until after three heads have already come up.....and then bet $100? For example, if it went: H,T,H,H...and at that moment, then you bet $100? You've already won at that point, and you're just playing it out. So as you define it, the answer is to be $0 until the last flip...then bet $100 unless it is a definite losing proposition (because only tails came up). Lastly...are you sure this isn't a "character defining question?" Perhaps they want to know if you'd participate in 50/50 betting. I would tell the interviewer that I wouldn't bet, and I'd look for a game where my percentage chance of winning was greater than 50/50. SM
The game wasn't well defined. The rules are as follows: 1) The game consists of up to 5 coin tosses 2) The game ends after 3 heads or 3 tails 2) You start with $100 3) You can bet on each toss and you win or lose the amount of your bet 4) You win if you have at least $200 at the end of the game
I have one question, if I win, I must double my previous bet? But if I lose, I can change the bet unit back to whatever I want? Sorry if this has been answered already, I don't want to read the previous posts in case someone has already figured it out and posted the answer.
No, you can bet whatever you want at each toss. If you win a bet you won the amount you bet or lose the amount you bet. For example: You have 100 and bet 50 If you win you now have 150 If you lose you now have 50
Just to be clear. Say I bet $10. I win, I don't have to bet $20 next turn. I can bet anything I want after I win.
Just to be clear. Say I bet $10. I win, I don't have to bet $20 next turn. I can bet anything I want after I win constrained by my purse at any given time. The reason I ask is because in the initial post, it says this
I think he was referring to the payout, which reflect the odds of a coin toss. If head, you win what you bet. If tail, you lose what your bet.