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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.

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I would begin with a bet of \$1. Every time a bet is won, resets to \$1. If a bet is lost, double the bet and try again tomorrow.

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How do you define long run?

Let it ride. The payout far exceeds the risk. Betting \$1 on turn one gives an expected value of \$1.045, while say betting half gives an expected value of \$1.0225. You have found a system that is biased in your favor. Vegas will not be so kind. Maximize it by always betting the max until you are satisfied with your winnings.

Now from a practical standpoint I may do something like only betting 2/3 or something. But from a largest expected value in the long run standpoint, let it ride.

Right? :\

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If we start with \$150,and assuming that we have a very bad luck ...so we have to play as follows:

1st.day:\$1.5

2nd day:\$3

3rd day:\$6

and always bet double the amount played(thats only if we lose),other wise..i.e.if we win,we should begin from \$1.5 again.

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If we start with \$150,and assuming that we have a very bad luck ...so we have to play as follows:

1st.day:\$1.5

2nd day:\$3

3rd day:\$6

and always bet double the amount played(thats only if we lose),other wise..i.e.if we win,we should begin from \$1.5 again.

Kinda why I'm asking about long run. Are we comparing after equal # days(ie bets)?

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How do you define long run?

Let it ride. The payout far exceeds the risk. Betting \$1 on turn one gives an expected value of \$1.045, while say betting half gives an expected value of \$1.0225. You have found a system that is biased in your favor. Vegas will not be so kind. Maximize it by always betting the max until you are satisfied with your winnings.

Now from a practical standpoint I may do something like only betting 2/3 or something. But from a largest expected value in the long run standpoint, let it ride.

Right? :

I didn't do any calculations for only a small number of days. So I don't know the N for which, after N or more days, the best strategy is unique.

The let-it-ride strategy will make you broke in short order, so the best strategy would need to bet some fraction (which may or may not change)

of your money each time. Anyway, feel free to use 365 days as "the long run". I'm sure that the best strategy doesn't change from that point on.

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I agree that Let it Ride will leave you broke...likely.

But you asked re: expected value. The small chance that you win 10 straight times eg is offset by the amount you'd win. So from an expected value standpoint, I couldn't simulate a better strategy. Granted I hadn't yet gone outside the box, but...

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I'm with Maurice:You asked for expected value.

If you were to bet a fixed fraction f of your holdings, your expected value is

(1 + f*.045)^n

Clearly the largest value comes when f = 1, the Let-It-Ride strategy.

For n = 365, this amounts to 9 Million bucks

and of course, some outcomes are much larger.

Yes you could go broke.

But so what?

Suppose you use f = .01, at the end of 365 days, you've got \$1.18. That's nearly broke.

So, if you'd like to add a proviso that the player guarantee not to go broke, we'd need the definition of "go broke".

We could make f = .99, you could easily go to very tiny fractions of a dollar, but the expected value is still over 8 million.

Would you, instead, like to maximize mean time to bankruptcy?

Or optimize value while preventing your value going below some threshold T more than probability P?

So, are you sure the task is to optimize expected value? Or is it really something more complicated?

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Betting it all and letting it ride means that one event of tails loses all of your stake. A 10% bet returns \$1.49 after a year. 50% bet returns \$92. 90% returns \$3,365. 99% returns \$7,507. 99.99% returns \$8,200. The question is what is the minimum stake with which you can still bet? Although the puzzle states that money can be real numbers, it makes sense that you would lose if your stake fell below one cent. With a 90% bet, the chances of a losing streak that would take your stake below one cent is about 1%. Anything above 90% increases the chance of falling below one cent, so my guestimate is that something around the 90% ~ 95% mark gives the highest return with an acceptably low likelihood of losing the stake.

Learning to lose

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Betting it all and letting it ride means that one event of tails loses all of your stake. A 10% bet returns \$1.49 after a year. 50% bet returns \$92. 90% returns \$3,365. 99% returns \$7,507. 99.99% returns \$8,200. The question is what is the minimum stake with which you can still bet? Although the puzzle states that money can be real numbers, it makes sense that you would lose if your stake fell below one cent. With a 90% bet, the chances of a losing streak that would take your stake below one cent is about 1%. Anything above 90% increases the chance of falling below one cent, so my guestimate is that something around the 90% ~ 95% mark gives the highest return with an acceptably low likelihood of losing the stake.

Learning to lose

Why is the bar set at \$.01? Because then I have no more money to bet? Well I'd argue that I'd feel the same at \$.50...or \$1.00. I'd almost rather lose my money early on than tinker with \$.25 bets here and there and hoping to get over \$5.00. To me losing the stake occurs much higher than \$.01.

edit - and again, if the question is expected value then your it doesn't matter if your odds of going broke increase or not. Think about it. You have \$1. I flip a balanced coin and say you bet what you want up to that \$1. If I flip a head I will give you 1,000,000 x your bet. If I flip a tail you lose your bet. Are you going to bet \$.99 so that you don't go broke? I'd hope not. Personally, I'd Let-it-Ride.

Edited by maurice

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Whatever you do, your expected value of money back will be 104.5% of the money you bet. There can not be any strategy to beat it. However, you can minimize the probability of losing all your money, which is a fatal event. To take the advantage of large numbers and guard the risk of losing all the money, always play with the minimum bet. Larger the number of games you play, lesser will be the chances of losing all your money. As each game is independent, basing the bet amount on the result of the previous game (like doubling etc.) will not help.

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I'd bet (.55*0.9-.45)/0.9 = 0.05 of my current amount (according to Kelly criterion)

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i agree with wiztar, bets should be based on expected value.

with a continuous bet of (.55*0.9-.45)/0.9 = 0.05

if you start off with 150 dollars, you can expect to reach 1 million dollars after about

5600 bets.

Edited by phillip1882

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i agree with wiztar, bets should be based on expected value.

with a continuous bet of (.55*0.9-.45)/0.9 = 0.05

if you start off with 150 dollars, you can expect to reach 1 million dollars after about

5600 bets.

The Let-it-Ride strategy still nets a higher expected value -> \$9.5 million

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The Let-it-Ride strategy still nets a higher expected value -> \$9.5 million

I wouldn't expect you to win 9.5 million by betting it all every time.

An unfair coin that returns 55% odds on is not unlike optimal play in blackjack (using 6-8 decks, counting cards, and with no more than 4 players + dealer). Without a team of players, the best way to place bets that I've found is start at the minimum bet (anywhere between \$5 and \$20, depending on the table and time). If you lose, you double the bet. If you win, you reset your bet to minimum. Don't forget to tip the dealer. She's hoping you win, too.

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I wouldn't expect you to win 9.5 million by betting it all every time.

An unfair coin that returns 55% odds on is not unlike optimal play in blackjack (using 6-8 decks, counting cards, and with no more than 4 players + dealer). Without a team of players, the best way to place bets that I've found is start at the minimum bet (anywhere between \$5 and \$20, depending on the table and time). If you lose, you double the bet. If you win, you reset your bet to minimum. Don't forget to tip the dealer. She's hoping you win, too.

Agreed. I certainly would not expect to win \$9.5 million ( well I do but only with .55^365 confidence and then I'd actually expect to win \$5.6 X10^101)

but that is the expected value of such a strategy. The question asked for returning the highest expected value. I don't believe it to be the optimal strategy, but then how is that to be judged?

edit - edit 2 Oh and btw what's being left out is that in blackjack the odds favor the house. In this game the odds favor the player, not because of the 55% but because of the payout (similar to pot odds in poker). You are being paid 1.9:1 with odds of 1.22:1. You mosdef should be maximizing that opportunity. Reread the OP...I may have misinterpretted something incorrectly...lemme check my numbers again.

Edited by maurice

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Agreed. I certainly would not expect to win \$9.5 million ( well I do but only with .55^365 confidence and then I'd actually expect to win \$5.6 X10^101)

but that is the expected value of such a strategy. The question asked for returning the highest expected value. I don't believe it to be the optimal strategy, but then how is that to be judged?

Perhaps the strategy that provides the highest expected average is sought?

edit - edit 2 Oh and btw what's being left out is that in blackjack the odds favor the house. In this game the odds favor the player, not because of the 55% but because of the payout (similar to pot odds in poker). You are being paid 1.9:1 with odds of 1.22:1. You mosdef should be maximizing that opportunity. Reread the OP...I may have misinterpretted something incorrectly...lemme check my numbers again.

(Overall) Blackjack odds can favour the player, depending on house rules, number of decks, and the right player.

Edited by Molly Mae

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Agreed. I certainly would not expect to win \$9.5 million ( well I do but only with .55^365 confidence and then I'd actually expect to win \$5.6 X10^101)

but that is the expected value of such a strategy. The question asked for returning the highest expected value. I don't believe it to be the optimal strategy, but then how is that to be judged?

edit - edit 2 Oh and btw what's being left out is that in blackjack the odds favor the house. In this game the odds favor the player, not because of the 55% but because of the payout (similar to pot odds in poker). You are being paid 1.9:1 with odds of 1.22:1. You mosdef should be maximizing that opportunity. Reread the OP...I may have misinterpretted something incorrectly...lemme check my numbers again.

yeah scratch that - you are being paid out 45:50 so you want odds of at least 50:45...your odds are 55:45. Still a rare scenario where the House is not favored.

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Perhaps the strategy that provides the highest expected average is sought?

How do you mean? How does that differ from highest expected value?

Edited by maurice

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not sure i agree maurice

even at only 1 dollar, if you bet it all each time, one loss and your broke.

if you bet 5% each time, the chances of you making a profit are much higher.

admitedly, you may only make 5 dollars. but that's better than losing!

unfortunately this is a bad idea. a simple computer sim will demonstrate why.

total = 1.0

bet = total*0.05

for i in range(0,365):

chance = random.random()

if chance <0.55:

total += bet*0.9

bet = total*0.05

else:

total -= bet

bet = bet*2

if total < 0.01:

break

print(total)

here i assume the min bet to be .05% of your total so far. i busted before i got to 365 running it. where as...

total = 1.0

for i in range(0,365):

bet = total*0.05

chance = random.random()

if chance < 0.55:

total += bet*0.9

else:

total -= bet

print(total)

gives me 3.65.

Edited by phillip1882

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I'm with Maurice:You asked for expected value.

If you were to bet a fixed fraction f of your holdings, your expected value is

(1 + f*.045)^n

Clearly the largest value comes when f = 1, the Let-It-Ride strategy.

For n = 365, this amounts to 9 Million bucks

and of course, some outcomes are much larger.

Yes you could go broke.

But so what?

Suppose you use f = .01, at the end of 365 days, you've got \$1.18. That's nearly broke.

So, if you'd like to add a proviso that the player guarantee not to go broke, we'd need the definition of "go broke".

We could make f = .99, you could easily go to very tiny fractions of a dollar, but the expected value is still over 8 million.

Would you, instead, like to maximize mean time to bankruptcy?

Or optimize value while preventing your value going below some threshold T more than probability P?

So, are you sure the task is to optimize expected value? Or is it really something more complicated?

You can not do this way.

If you bet all, Pr( any bet is Tail) will be close 1. and you will lose all money.

But, Pr(all bet is head) is close to 0. You can not win forever.

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To reach the largest expected payoff, we need to make sure us never broke up.

If the first time, we bet \$1.0, and we lost it.

To have possible to recover to initial balance, bet 1.0/0.9. if win, recove to 150. if lost again. We are losing 1 + 1/0.9.

If first and second both lost, we bet (1 + 1/0.9) / 0.9 in the third time. This will still guarente us back to 150.

This strategy will let us stay longer in the bet.

Considering H is 55%m T is 45% prob, to make money.

If first 1.0 bet is lost, we need to not only recover back but also make profit. that requires 1.0/0.9 + x. if the second lost. we totally lose 1 + 1.0/0.9 + x. and next time bet ( 1 + 1.0/0.9 + x) /0.9 and possibly recover us to 150 + x. Because the possibility of consecutive Tail is 0.45^n. it because very little if we choose x as mall as possible. that is 0.01. Therefore, take enough time, we will make 0.01. and repeate again. finally we will go to 150 + 0.01 * infinty.

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in some way i understand morice's point, if your only allowed very few bets, betting it all makes sense.

but the question asks: what's the best strategy in the long run. ie. you're allowed as many bets as you want.

betting 5% each time may not make you a millionare overnight, but its the most garenteed way of eventually reaching that goal in a reasonable amount of time.

Edited by phillip1882

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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?

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Maurice: In your example, are you betting 100% of it right out?

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not sure i agree maurice

even at only 1 dollar, if you bet it all each time, one loss and your broke.

if you bet 5% each time, the chances of you making a profit are much higher.

admitedly, you may only make 5 dollars. but that's better than losing!

I have acknowledged I will likely go broke. I have acknowledged how unlikely it would be to win 365 successive times. The fact remains that the expected value of my \$1 utilizing this strategy (@yuli yes betting %100 each time) is \$9.5million.

The first time it would be 1.9 x .55 + 0 x .45 = 1.045. Since I only have a positive outcome when I win each time the expected value after n bets is 1.9^n x .55^n. When n =365 E~9,500,500

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