TPiR - Spelling Bee

[tvmitch]

Because most folks on here are better at math than me, and because I'm curious...

What are the odds of a win on Spelling Bee in all scenarios, at the beginning of the game, after a contestant gives answers to all three prizes - with 5 cards, 4 cards, and 3 cards? Expected-win type discussions also welcome.

Now that the cards are worth $1K each, I'd love to see what the pure odds are, since a 5-card start means that you are putting up $5K in cash at a chance at a $15-20K car. As a contestant, I would find that to be a tough call each and every time.

This may have been discussed before, and if so, I apologize for the post...but I didn't see it in a search.

[SteveR]

And could the odds of a play we saw earlier this week -- where the contestant had three R's out of five cards -- actually be smaller than winning the car?

[Fedya]

[quote name=\'mitchgroff\' post=\'201148\' date=\'Nov 6 2008, 09:02 PM\']
Because most folks on here are better at math than me, and because I'm curious...

What are the odds of a win on Spelling Bee in all scenarios, at the beginning of the game, after a contestant gives answers to all three prizes - with 5 cards, 4 cards, and 3 cards? Expected-win type discussions also welcome.

Now that the cards are worth $1K each, I'd love to see what the pure odds are, since a 5-card start means that you are putting up $5K in cash at a chance at a $15-20K car. As a contestant, I would find that to be a tough call each and every time.

This may have been discussed before, and if so, I apologize for the post...but I didn't see it in a search.
[/quote]
It's easier to figure out the probability of not winning, and then subtract that from 1 to get the probability of winning.

If memory serves, there are 11 C's, 11 A's, 6 R's, and two cards marked CAR that are automatic wins.

With five cards, you lose if you get:

C A R
-----
5 0 0
0 5 0
0 0 5
4 1 0
3 2 0
2 3 0
1 4 0
4 0 1
3 0 2
2 0 3
1 0 4
0 4 1
0 3 2
0 2 3
0 1 4

NOTE: since the number of C's and A's is equivalent, the probability of getting 4 C's and 1 R will be equivalent to the probability of getting 4 A's and one R, which will lower the number of calculations we actually have to do, since several of them can simply be multiplied by 2.

5 C's (or 5 A's):

11/30 * 10/29 * 9/28 * 8/27 * 7/26 = 11/3393 (This will be multiplied by 2.)

CCCCA (or AAAAC):

11/30 * 10/29 * 9/28 * 8/27 * 11/26 * 5 = 605/23751 (Again multiplied by 2)

Note: the multiplication by 5 is because the one A can be picked first, second, third, fourth, or fifth.

CCCAA (or AAACC)

11/30 * 10/29 * 9/28 * 11/27 * 10/26 * 10 = 3025/47502 (again multiplied by 2)

There are ten different ways to place the two A's

RRRRR:

6/30 * 5/29 * 4/28 * 3/27 * 2/26 = 1/23751

CCCCR (or AAAAR):

11/30 * 10/29 * 9/28 * 8/27 * 6/26 * 5 = 110/7917 (again multiplied by 2)

CCCRR (or AAARR):

11/30 * 10/29 * 9/28 * 6/27 * 5/26 * 10 = 275/15834 (again multiplied by 2)

CCRRR (or AARRR):

11/30 * 10/29 * 6/28 * 5/27 * 4/26 * 10 = 550/71253 (again multiplied by 2)

CRRRR (or ARRRR)

11/30 * 6/29 * 5/28 * 4/27 * 3/26 * 5 = 55/47502 (again multiplied by 2)

If you Add 'Em Up, you get a probability of losing of ~0.2651, which means a winning probability of ~0.7349, or close to 3/4.  (The actual fraction for the probability of losing is 37780/142506.)

The presence of the two CAR cards significantly increases the probability of winning, to the point that if you were running multiple simulations of the game and wanted to maximize your winnings, you'd always go for the car, even if you had only one card (assuming the car is more than $15K).

Of course, if you're a contestant, you only get one shot to play.  However, the producers need to look at an expected payout.  If raising the amount each card is worth from $500 to $1000 induces more people to take the cash, then it probably saves them in their prize budget (depending on how much they're actually paying for the car, as opposed to barter/plugs).

If any of our math teacher posters wish to correct my math, knock yourselves out.  Figuring out the odds for having only four or three cards is left as an exercise for the reader.

[Neumms]

[quote name=\'Fedya\' post=\'201161\' date=\'Nov 6 2008, 11:04 PM\']
Figuring out the odds for having only four or three cards is left as an exercise for the reader.
[/quote]

This is way cool. The fact you left an exercise for the reader is hilarious (or cruel, depending on if you still do mathematics homework).

[rjaguar3]

From an Excel spreadsheet that I created (which I will e-mail you if you want a copy), the probabilities I have are:

2 cards:  13.1%
3 cards:  37.2%
4 cards:  58.4%
5 cards:  73.5%

And, I also computed the bailout attractiveness (assuming a constant utility function).  Simply put, the contestant should not bail out unless the car is worth less than 14.5 times EDIT:  about 15.2 times the value of a single card.

[TLEberle]

[quote name=\'rjaguar3\' post=\'201186\' date=\'Nov 7 2008, 11:58 AM\']From an Excel spreadsheet that I created (which I will e-mail you if you want a copy), the probabilities I have are:
2 cards:  13.1% [/quote]I am curious about how you arrived at that number. Here's the calculatin' I did:

There are 435 ways to draw two cards. (30 * 29)/2. (It doesn't matter in which order you pick the two cards) Among those 435 ways, there are 59 ways to draw a card that has one of the two CAR cards. (And one of the two has both.) Which gives a percent of roughly 13.6%.

Where does the extra .5% come from, if indeed I did all of my maths properly?

I should be surprised that a game exists where a modicum of pricing knowledge gets you 3/4ths of the way to a brand new car, but then they also have Let 'em Roll, so who knows.

[Fedya]

Travis:

Again, it might be easier to look at the probability of NOT winning if you've only got two cards.

28/30 * 27/29 = 378/435 = .8689

You went wrong here:

Among those 435 ways, there are 59 ways to draw a card that has one of the two CAR cards. (And one of the two has both.) Which gives a percent of roughly 13.6%.

There aren't 59 ways to draw the winning card.  Assume (without loss of generality) that the CAR cards are behind #29 and #30.  There are 28 non-CAR cards (1 through 28) that go with #29, and 28 non-CAR cards that go with #30.  That's 56 winning combinations.  The 57th combination is picking #29 and #30.  57/435 = .1310.  Note that 57 + 378 = 435.

[TLEberle]

[quote name=\'Fedya\' post=\'201213\' date=\'Nov 7 2008, 07:36 PM\']Travis:

Again, it might be easier to look at the probability of NOT winning if you've only got two cards.[/quote] It certainly helps for the Punch Board, but I didn't think to back-end this problem.

28/30 * 27/29 = 378/435 = .8689

D'OH! MATHS FAIL!

You went wrong here:

Among those 435 ways, there are 59 ways to draw a card that has one of the two CAR cards. (And one of the two has both.) Which gives a percent of roughly 13.6%.

There aren't 59 ways to draw the winning card.  Assume (without loss of generality) that the CAR cards are behind #29 and #30.  There are 28 non-CAR cards (1 through 28) that go with #29, and 28 non-CAR cards that go with #30.  That's 56 winning combinations.  The 57th combination is picking #29 and #30.  57/435 = .1310.  Note that 57 + 378 = 435.

I see where I went wrong; I took the A column and the 1 row as "winning" if you set up an Excel sheet, but forgot to remove the redundant parts. Dang.