# Can somebody figure out the value of this offer? (probability q)

I’ve got this offer at a local casino. It works like this: every time I qualify I get to spin a wheel. The wheel has 11 relevant sections[ul]
[li]2 blanks[/li][li]8 unique symbols[/li][li]1 Wildcard[/li][/ul]

The aim is to collect all 8 symbols before the end of the promotion. If I really plug away at it I can get a maximum of 21 spins at the wheel during that time. Getting the wild card means I can choose any noncollected symbol instead, but this can’t be retroactively applied, I have to choose which one at the time.

Getting all 8 symbols is worth £1000, getting 7 or less is worth £0.

What is the mean value of this deal?

Bonus, I’ve actually now got 18 potential spins left and 6 symbols to collect, what’s my current value?

Also I’ve worked out the average number of spins it will take to collect 8 symbols is 20.11, (11/9+11/8+…11/2), but don’t know how to work out the total percentage of times that it will take 22 or more which it a more relevant stat to my interests.

Can you get the wildcard more than once?

It’s going to be ugly. Assuming you can use the wildcard repeatedly, my “not really thinking about it hard” guess is something like:

\$1000 * // total amount
21! / (9!(21-9)! ) * // numbers of ways of choosing 9 values
9/11 * 8/11 * … * 2/11 // Chance on each spin of getting either an unused symbol or the wildcard.

But I doubt that’s right.

Particularly since “9” should be “8” on the second line.

I wrote a little simulator and it looks like you win about 65% of the time with 21 spins. With 18 spins and 6 remaining symbols, it’s down to 63%.

Perhaps someone else can come along and verify these numbers… this was fairly hacky.

I would say your current value is 0. Unless that is casino jargon that I don’t get.

If you think that his value is 0, then would you mind offering me a chance to play that game for free, complete with the thousand-quid prize if I win?

I used the recursion
f(a, 0) = 1
f(0, b) = 0 when b non-zero
11 * f(a, b) = (b+1) * f(a-1, b-1) + (11-b-1) * f(a-1, b)
and solved in the obvious way. I was going to double-check the formula and/or simulate but …

Dr. Strangelove did that for me!

I get 65.00% and 63.06%, in very close agreement with Dr. Strangelove. The other numbers, sorted by decreasing goodness, are:

f(15, 1) = 0.950711
f(14, 1) = 0.939758
f(13, 1) = 0.926371
f(12, 1) = 0.910009
f(16, 2) = 0.891270
f(11, 1) = 0.890011
f(15, 2) = 0.868980
f(10, 1) = 0.865569
f(14, 2) = 0.842438
f(17, 3) = 0.836290
f(9, 1) = 0.835696
f(13, 2) = 0.810963
f(16, 3) = 0.804873
f(8, 1) = 0.799184
f(18, 4) = 0.790522
f(12, 2) = 0.773820
f(15, 3) = 0.768241
f(19, 5) = 0.754794
f(7, 1) = 0.754558
f(17, 4) = 0.752382
f(11, 2) = 0.730248
f(20, 6) = 0.728457
f(14, 3) = 0.725843
f(18, 5) = 0.711919
f(21, 7) = 0.710393
f(16, 4) = 0.708640
f(6, 1) = 0.700015
f(19, 6) = 0.682366
f(10, 2) = 0.679503
f(13, 3) = 0.677203
f(17, 5) = 0.663364
f(20, 7) = 0.662225
f(15, 4) = 0.658972
f(21, 8) = 0.650035
f(5, 1) = 0.633352
f(18, 6) = 0.630648
f(12, 3) = 0.621994
f(9, 2) = 0.620931
f(16, 5) = 0.609033
f(19, 7) = 0.608515
f(14, 4) = 0.603247
f(20, 8) = 0.595178
f(17, 6) = 0.573396
f(11, 3) = 0.560134
f(8, 2) = 0.554086
f(4, 1) = 0.551875
f(18, 7) = 0.549492
f(15, 5) = 0.549105
f(13, 4) = 0.541617
f(19, 8) = 0.535164
f(16, 6) = 0.511031
f(10, 3) = 0.491923
f(17, 7) = 0.485750
f(14, 5) = 0.484134
f(7, 2) = 0.478908
f(12, 4) = 0.474636
f(18, 8) = 0.470688
f(3, 1) = 0.452292
f(15, 6) = 0.444403
f(16, 7) = 0.418331
f(9, 3) = 0.418205
f(13, 5) = 0.415156
f(11, 4) = 0.403387
f(17, 8) = 0.402912
f(6, 2) = 0.395993
f(14, 6) = 0.374872
f(15, 7) = 0.348808
f(12, 5) = 0.343780
f(8, 3) = 0.340559
f(16, 8) = 0.333526
f(2, 1) = 0.330579
f(10, 4) = 0.329607
f(5, 2) = 0.306984
f(13, 6) = 0.304375
f(14, 7) = 0.279305
f(11, 5) = 0.272251
f(15, 8) = 0.264756
f(7, 3) = 0.261503
f(9, 4) = 0.255775
f(12, 6) = 0.235418
f(4, 2) = 0.215149
f(13, 7) = 0.212450
f(10, 5) = 0.203424
f(14, 8) = 0.199285
f(8, 4) = 0.185122
f(6, 3) = 0.184651
f(1, 1) = 0.181818
f(11, 6) = 0.170959
f(12, 7) = 0.151201
f(9, 5) = 0.140604
f(13, 8) = 0.140046
f(3, 2) = 0.126221
f(7, 4) = 0.121471
f(5, 3) = 0.114746
f(10, 6) = 0.114145
f(11, 7) = 0.098512
f(12, 8) = 0.089848
f(8, 5) = 0.087182
f(6, 4) = 0.068821
f(9, 6) = 0.067843
f(4, 3) = 0.057373
f(10, 7) = 0.056825
f(11, 8) = 0.050856
f(2, 2) = 0.049587
f(7, 5) = 0.046036
f(8, 6) = 0.033998
f(5, 4) = 0.030549
f(9, 7) = 0.027445
f(10, 8) = 0.023994
f(6, 5) = 0.018695
f(3, 3) = 0.018032
f(7, 6) = 0.012932
f(8, 7) = 0.009969
f(9, 8) = 0.008464
f(4, 4) = 0.008196
f(5, 5) = 0.004471
f(6, 6) = 0.002845
f(7, 7) = 0.002069
f(8, 8) = 0.001693

… This all assumes the eleven sections of the wheel are equally likely. Knowing casinos, I certainly would NOT count on that.

Despite being a half-decent programmer, for some reason I never quite manage to think of applying recursion to probability problems. But it’s obvious in retrospect.

That said, it was trivial to implement this solution and verify our numbers a third time. Memoization reduced a 1-second runtime to instant (though Perl is pretty damn slow).

He’s emphasizing “current” – the OP doesn’t have anything, yet.

Thanks to Dr Stangelove and Septimus, you guys rock.

But he does: He has the opportunity to play a fairly valuable game. That opportunity is something.

To be fair I don’t have anything really yet, the possible £1000 is at the far end of a road which requires playing negative EV games a lot until I can trigger enough chances to possibly get it.

This is the key point if you really want to work out the overall EV. Clearly, if it was just something you had to go into the casino and do 18 more times, it would be worth 18 short trips out of your way to go spin that wheel, for a 63% chance at £1,000 (EV of £630, minus trivial time/travel expenses). Clearly it’s not that easy (when do casinos ever offer a +EV proposition? Never, unless they make a mistake or it’s a low-odds gamble for them, like occasional free bet vouchers to entice you to come in more often).

Do you know what the requirements are? For example, a £100 minimum Blackjack buy-in, and you get a chance to spin the wheel if you play at the Blackjack table for 30 minutes. I assume it’s something like that. Then we can run the numbers and get an actual EV for you.

It’s a “lucky card” which is put into the 4 deck Continuous Shuffle Machine when playing blackjack. I’m happy with working out my own EV whilst playing blackjack though, and was only looking at the value of the offer in a vacuum. I don’t need any further analysis.

Out of interest, do you personally have to receive the ‘lucky card’ to get a spin, or just be playing a hand at the table when it comes out? If the former, it could change the optimal strategy, especially if you were only one spin away from the prize. For example, I seem to recall whether or not to hit on a soft 18 against dealer 10 is pretty marginal in terms of the game, but if you need to get the lucky card for a spin of the wheel, it’s probably worth it!

You have to receive it personally, not the table in general. But given the hard limit on total spins it’s probably not worth altering strategy for, if you miss it you can just wait for another one. This might change in the dying hours of the promo though

Since the thread has been bumped I may as well give a progress report: Now at 15 and 4, back to the roughly 65% I started at according to septimus’s handy table.

Hitting soft 18 against dealer Ace is almost an exact tossup; so you’d definitely want to hit now. (Of course if the rule is that dealer hits soft 17, it’s right to hit soft 18 anyway.)

Checking a strategy table, I see that I’d almost always hit 12 in OP’s case. Against dealer Deuce I’d be more aggressive, splitting 2’s and 6’s and hitting 13. Making slightly wrong plays might be wise anyway — don’t they have computers now monitoring your play quality?