BrainDen.com - Brain Teasers
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## Question

You have a biased coin which falls heads 55% of the time and

tails 45% of the time. You start with \$1 and, once a day, you

may bet any fraction of your current amount that the coin will

land heads. If it lands tails, you lose the amount that you

bet; if it lands heads you get 190% of your bet back. So, for

example, say on some day you have \$150 and decide to bet \$10 of

it that day. If the coin lands tails, you lose your bet and then

have \$140 available for the next day. If it lands heads, you

win \$9 plus your \$10 bet back (190% of \$10) and then have \$159

available for the next day. What is the optimal betting strategy?

That is, what strategy will give you the largest expected value

in the long run?

Please Note: For simplicity, assume that money values are real

numbers. So, for example, one may have \$124.60012898735.

## Recommended Posts

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to use morice's example, if i have 1 dollar, and the payout is 100,000,000 x my bet amount if i win, and simply lose if i lose, with a 50/50 chance of winning, it may seem reasonable to bet it all. however, if after i go broke, i'm not allowed to bring any more money to the bet, it seems to me i would want to keep a certain percentage around in case the first coin toss doesn't go my way. and even if the first coin toss does go my way, i most certianly wouldn't bet it all again and keep doing so, would you?

I agree...my strategy is not practical, it just satisfies super's quest for highest expected value...and again I've not simulated a strategy that nets a higher one but neither has anyone else...I am open to the thought that I am overlooking an out of the box idea.

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Like witzar, I had thought that the Kelly Criterion gives the largest expected value of the final bankroll.

It actually maximizes the expected growth of the bankroll. So, in the case of this topic, if p is the

proportion of the bankroll to be bet each time, the expected bankroll increase for each play is:

(1+.9×p).55×(1-p).45 which attains its maximum when p=.05. So, I was wrong in my answer (#6) to maurice.

In short I was looking for the answer to a different problem than the one I posed. Sorry about that,

especially to maurice! Maurice has the correct answer to the problem I posed -- the let-it-ride strategy

does indeed maximize the expected value of the final bankroll.

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No need to apologize sp...especially as I figured all along that you were meaning something else than how I "interpreted" the question. And especially as I agree that Let-it-Ride is not an optimal strategy over 365 bets. But mostly especially because I learned something...

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From the discussion, I decided to test my intuition and put it into a spreadsheet.

to test scenarios, I made formulas for two methods. First method, always betting a set percentage (P) of current holdings. Second, If win, using set percentage of current holdings as above, if lose, bet a factor (F) of the previous bet. In the first case, the probability of going bust seemed to be related to the percentage of losses overall while the second case seemed to be more related to a single long losing streak. I did each to produce result for 1000 bets, looked at value of holdings on day 365 and day 1000. Plugging in different values for P and F and recalculation several times, comparing numbers I was surprised to find that, using the same P value I was able to find a value for F that would tend to give better results with no greater chance of going bust over the first method. I experimented for .01<=P<.=2 and for 1.1<=F<=2. In general, higher P & F values increased the probability of going bust Using P = .02 and F = 1.4 appeared to be about optimum for the second method, in general requiring about 11 losses in a row to go bust, which is 0.00015% probability. However, the resulting winnings over repeated runs vary too much to give a firm answer. It would be interesting if someone could come up with a formula to calculate the probability of winning a set amount (W) after 365 days for different values of P and F.

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