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Roulette anyone? (Posted on 2008-07-13) Difficulty: 2 of 5
A roulette player had a system of playing one dollar 7 times, red or black, then 7 dollars 7 times, red or black, then 49 dollars 7 times, red or black, etc., each time 7 bets in increasing powers of 7.

How many times had he won if he finally won net 777,777 dollars?

For those not familiar with the roulette, in this particular bet, if you bet 1 dollar, either you lose it or gain another 1 dollar.

See The Solution Submitted by pcbouhid    
Rating: 4.0000 (2 votes)

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Solution solution | Comment 1 of 3

The powers of 7 make the total of the sets of trials act like a base-7 number, except that the possible values at each base-7 position are -7, -5, -3, -1, 1, 3, 5, 7, rather than the usual 0 thru 7.

Two things must be done to change this: calculate how much the total loss would be if the player had lost every roll of the ball and add that to the total winnings to see how much he had to make up, and then divide by two, as he gains 2 position values for each win.

First we must figure how many base-7 positions are needed. 777777 is 7^6.9706..., so we'll assume a 7-digit base-7 number. (This is an odd number, as it needs to be, as, if an even number of rounds occurs, in which each gain/loss is odd, then the winnings would be even.) If the player had lost $7 in the first set of 7 plays and $49 in the next set, etc., then he would have lost $960,799 altogether in 7 sets of 7 plays. But he didn't lose this; in fact, he gained 777,777, so he must have overcome this by 960799 + 777777 = 1738576.

As mentioned, regardless of the stakes involved in each set of plays, each win also removes a loss, so this amount needs to be halved before converting to base-7. That gives us 865288.  In base-7 that's 7250240, and as we had surmised, it's a 7-digit base-7 number.

The zero at the right-hand end represents having won zero trials in the first set of 7 spins. The 4 to its left represents four wins in the second round at $7 per roll/spin, etc.

In all, then, he had 7+2+5+0+2+4+0 = 20 wins.

Let's check if this works out:

0 wins 7 losses at      $1        -7
4 wins 3 losses at      $7        +7
2 wins 5 losses at     $49      -147
0 wins 7 losses at    $343     -2401
5 wins 2 losses at   $2401     +7203
2 wins 5 losses at  $16807    -50421
7 wins 0 losses at $117649   +823543
net winnings                  777777

  Posted by Charlie on 2008-07-13 18:31:33
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